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AP Chemistry · ⁨AP化学⁩

Tips · ⁨ヒント⁩

AP化学では、原子構造と周期律から始まり、結合と分子形状、分子間力、化学量論と溶液、反応速度論、熱力学、平衡、酸塩基、電気化学に至ります。一貫したテーマは粒子レベルでの説明です。ほとんどの設問は「なぜか説明せよ」で終わり、答えは粒子、力、エネルギーについて述べる必要があります。

2つの概念が最も多くの得点を占めます。 分子間力は沸点、溶解度、クロマトグラフィーを説明します。ル・シャトリエの原理はすべての平衡移動を説明します。これらを適切に学べば、試験の大部分が解けるようになります。

単位と有効数字のない計算は不十分です。

本ノートは9つのCED単元に沿っており、各単元には worked examples と試験の式シートに記載されている関係性が含まれています。ライブラリには過去のFRQと採点基準があり、化学の長文設問では粒子レベルでの振る舞いの説明に重点が置かれるため、モデル回答はその文体で書かれています。

  • 1

    Atomic Structure and Properties · ⁨原子構造と性質⁩

    Watch lesson · ⁨レッスンを視聴⁩
    1.1

    Moles and Molar Mass · ⁨モルとモル質量⁩

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 1.1.A: Calculate quantities of a substance or its relative number of particles using dimensional analysis and the mole concept.

    • 1.1.A.1 One cannot count particles directly while performing laboratory work. Thus, there must be a connection between the masses of substances reacting and the actual number of particles undergoing chemical changes.
    • 1.1.A.2 Avogadro's number ($N_{\mathrm{A}} = 6.022 \times 10^{23}\ \mathrm{mol}^{-1}$) provides the connection between the number of moles in a pure sample of a substance and the number of constituent particles (or formula units) of that substance.
    • 1.1.A.3 Expressing the mass of an individual atom or molecule in atomic mass units (amu) is useful because the average mass in amu of one particle (atom or molecule) or formula unit of a substance will always be numerically equal to the molar mass of that substance in grams. Thus, there is a quantitative connection between the mass of a substance and the number of particles that the substance contains.
      • Equation: $n = m/M$
    日本語

    学習目標 1.1.A: 次元解析およびモルの概念を用いて、物質の量やその相対的な粒子数を計算する。

    • 1.1.A.1 実験作業中に粒子を直接数えることはできない。したがって、反応する物質の質量と化学変化を起こす実際の粒子数の間には、何らかの関連性が必要である。
    • 1.1.A.2 アボガドロの数($N_{\mathrm{A}} = 6.022 \times 10^{23}\ \mathrm{mol}^{-1}$)は、純粋な試料中のモル数と、その物質の構成粒子(または式単位)の数との間に関連性を示している。
    • 1.1.A.3 個々の原子や分子の質量を原子質量単位(amu)で表すことは有用である。なぜなら、1つの粒子(原子または分子)または式単位の平均質量(amu)は、数値的に常にその物質のモル質量(g/mol)に等しいからである。したがって、物質の質量とそれが含む粒子数の間には量的な関係がある。
      • 式: $n = m/M$

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Because atoms are far too small to count, chemists count in moles 摩尔. One mole is Avogadro's number 阿伏伽德罗常数 of particles, $N_A=6.022\times10^{23}$. The molar mass 摩尔质量 (grams per mole, read off the periodic table) converts between mass and moles:

    $$n=\frac{m}{M}.$$
    Moles are the bridge between the lab (grams you weigh) and the equation (particles that react).

    Worked example. How many moles, and how many molecules, are in $36.0\ \text{g}$ of water ($M=18.0\ \text{g/mol}$)?

    $$n=\frac{m}{M}=\frac{36.0}{18.0}=2.0\ \text{mol},\qquad N=n\,N_A=2.0\times6.022\times10^{23}=1.2\times10^{24}\ \text{molecules}.$$

    日本語

    原子は非常に小さすぎて数えることができないため、化学者は モル で単位数量を表す。1モルは アボガドロの定数 $N_A=6.022\times10^{23}$ 個の粒子にあたる。モル質量(周期表から読み取れるg/mol)は、質量とモル数を相互に変換するために用いる:

    $$n=\frac{m}{M}.$$
    モルは、実験室で秤量する質量(g)と、反応する粒子数を結びつける橋渡しの役割を果たしている。

    モールはハブとなる: 質量、粒子数、気体体積、濃度の変換
    モールはハブとなる: 質量、粒子数、気体体積、濃度の変換

    ** worked example.** $36.0\ \text{g}$ の水($M=18.0\ \text{g/mol}$)には、モル数がいくつ、分子数はいくつあるか?

    $$n=\frac{m}{M}=\frac{36.0}{18.0}=2.0\ \text{mol},\qquad N=n\,N_A=2.0\times6.022\times10^{23}=1.2\times10^{24}\ \text{molecules}.$$

    Explore · ⁨探索⁩

    Link mass, moles and molar mass

    Molar mass $M$ is the bridge between the mass you weigh and the number of moles: $m = M \times n$. Because $M$ is fixed for a substance, mass is proportional to moles — double the moles, double the mass.

    1.2

    Mass Spectra of Elements · ⁨元素の質量スペクトル⁩

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 1.2.A: Explain the quantitative relationship between the mass spectrum of an element and the masses of the element's isotopes.

    • 1.2.A.1 The mass spectrum of a sample containing a single element can be used to determine the identity of the isotopes of that element and the relative abundance of each isotope in nature.
    • 1.2.A.2 The average atomic mass of an element can be estimated from the weighted average of the isotopic masses using the mass of each isotope and its relative abundance.
      • Exclusion Statement: Interpreting mass spectra of samples containing multiple elements or peaks arising from species other than singly charged monatomic ions will not be assessed on the AP Exam.
    日本語

    学習目標 1.2.A: 元素の質量スペクトルと、その元素の同位体の質量との間の量的な関係を説明する。

    • 1.2.A.1 単一元素を含む試料の質量スペクトルは、その元素の同位体の同定および各同位体の自然界における相対的存在率の決定に用いられる。
    • 1.2.A.2 各同位体の質量とその相対的存在率を用いて、同位体質量の加重平均を計算することで、元素の平均原子質量を推定できる。
      • 除外事項: 複数の元素を含む試料の質量スペクトルの解釈や、単一電荷の単原子イオン以外の種によるピークの解釈については、AP試験では評価されない。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    Mass spectrometry

    A mass spectrometer 质谱仪 separates atoms by mass, giving a mass spectrum 质谱: peaks at each isotope 同位素 (same element, different neutron count) with heights showing their relative abundance 相对丰度. The element's average atomic mass is the abundance-weighted average of its isotope masses.

    Worked example. Chlorine is $75.8\%$ chlorine-35 and $24.2\%$ chlorine-37. Its average atomic mass is

    $$A_r=(35)(0.758)+(37)(0.242)=26.5+8.95=35.5.$$
    The average lies closer to $35$ because that isotope is more abundant – which is why the periodic-table value is $35.5$, not a whole number.

    日本語
    質量分析

    質量分析計は質量に基づいて原子を分離し、質量スペクトルを提供する。各同位体(同一元素だが中性子数が異なる)におけるピークの高さはその相対存在度を示しており、元素の平均原子質量は同位体の質量の存在度重み平均である。

    相対原子質量は、同位体の存在比で重み付けした平均値である
    相対原子質量は、同位体の存在比で重み付けした平均値である
    塩素の質量スペクトル: 存在比の異なる2つの同位体
    塩素の質量スペクトル: 存在比の異なる2つの同位体

    ** worked example.** 塩素は $75.8\%$ が塩素-35、$24.2\%$ が塩素-37である。その平均原子質量は

    $$A_r=(35)(0.758)+(37)(0.242)=26.5+8.95=35.5.$$
    平均値は $35$ に近い。なぜなら、この同位体の方が存在比が高いからである。そのため周期表の値は整数ではなく $35.5$ となる。

    Explore · ⁨探索⁩

    Explore isotopes — same element, different mass

    Keep the protons fixed but change the neutrons to build the two chlorine isotopes ($^{35}$Cl and $^{37}$Cl). Same element, different mass — exactly the peaks a mass spectrum shows.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    moles/məʊlz/ 物質量
    Avogadro's number/ˌævəˈɡædrəʊz ˈnʌmbə/ アボガドロ定数
    molar mass/ˈməʊlə mæs/ モル質量
    mass spectrometer/mæs spekˈtrɒmɪtə/ 質量分析計
    mass spectrum/mæs ˈspektrəm/ 質量スペクトル
    isotope/ˈaɪsətəʊp/ 同位体 (isotope)
    relative abundance/ˈrelətɪv əˈbʌndəns/ 相対的な豊度
    percent composition/pəˈsent ˌkɒmpəˈzɪʃn/ 組成百分率
    1.3

    Elemental Composition of Pure Substances · ⁨純物質の元素組成⁩

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 1.3.A: Explain the quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance.

    • 1.3.A.1 Some pure substances are composed of individual molecules, while others consist of atoms or ions held together in fixed proportions as described by a formula unit.
    • 1.3.A.2 According to the law of definite proportions, the ratio of the masses of the constituent elements in any pure sample of that compound is always the same.
    • 1.3.A.3 The chemical formula that lists the lowest whole number ratio of atoms of the elements in a compound is the empirical formula.
    日本語

    学習目標 1.3.A: 純物質の質量による元素組成と実験式との間の量的な関係を説明する。

    • 1.3.A.1 純物質の中には、個別の分子から構成されるものもあれば、式単位によって記述される一定の比率で結合した原子やイオンから構成されるものもある。
    • 1.3.A.2 定比例の法則によれば、その化合物の純粋な試料に含まれる構成元素の質量比は常に一定である。
    • 1.3.A.3 化合物内の元素の原子の最も簡単な整数比を列挙した化学式を、実験式という。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    The percent composition 百分组成 of a compound is each element's mass fraction. From it you find the empirical formula 实验式 (simplest whole-number ratio of atoms) by converting each element's mass to moles and dividing by the smallest. The molecular formula 分子式 is a whole-number multiple of the empirical formula, found from the molar mass.

    Worked example. A compound is $40.0\%$ C, $6.7\%$ H, and $53.3\%$ O by mass. Assuming $100\ \text{g}$, convert each mass to moles and divide by the smallest:

    $$\text{C}:\frac{40.0}{12}=3.33,\quad \text{H}:\frac{6.7}{1}=6.7,\quad \text{O}:\frac{53.3}{16}=3.33 \;\xrightarrow{\div 3.33}\; 1:2:1,$$
    so the empirical formula is $\text{CH}_2\text{O}$. If the molar mass were $180\ \text{g/mol}$ ($=6\times30$), the molecular formula would be $\text{C}_6\text{H}_{12}\text{O}_6$ – glucose.

    日本語

    化合物の百分組成とは、各元素の質量分率のことです。これを用いて、各元素の質量を物質量に変換し、最も小さい値で割ることで、原子の最も単純な整数比である実験式を求められます。分子式は、実験式の整数倍であり、モル質量から導かれます。

    簡易化学式の導出: 質量→モル数、最小値で割り、比率を読み取る
    簡易化学式の導出: 質量→モル数、最小値で割り、比率を読み取る

    ** worked example.** ある化合物は質量基準で $40.0\%$ C、$6.7\%$ H、$53.3\%$ O を含み、総質量は $100\ \text{g}$ と仮定する。各質量をモル数に変換し、最小値で割る:

    $$\text{C}:\frac{40.0}{12}=3.33,\quad \text{H}:\frac{6.7}{1}=6.7,\quad \text{O}:\frac{53.3}{16}=3.33 \;\xrightarrow{\div 3.33}\; 1:2:1,$$
    したがって簡易化学式は $\text{CH}_2\text{O}$ である。もしモル質量が $180\ \text{g/mol}$ ($=6\times30$)であれば、分子式は $\text{C}_6\text{H}_{12}\text{O}_6$ となり、これはグルコースである。

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    empirical formula/emˈpɪrɪkl ˈfɔːmjʊlə/ 経験式
    molecular formula/məˈlekjʊlə ˈfɔːmjʊlə/ 分子式
    mixture/ˈmɪkstʃə/ 混合物
    1.4

    Composition of Mixtures · ⁨混合物の組成⁩

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 1.4.A: Explain the quantitative relationship between the elemental composition by mass and the composition of substances in a mixture.

    • 1.4.A.1 Pure substances contain atoms, molecules, or formula units of a single type. Mixtures contain atoms, molecules, or formula units of two or more types, whose relative proportions can vary.
    • 1.4.A.2 Elemental analysis can be used to determine the relative numbers of atoms in a substance and to determine its purity.
    日本語

    学習目標 1.4.A: 質量による元素組成と混合物中の物質の組成との間の量的な関係を説明する。

    • 1.4.A.1 純物質は単一の種類の原子、分子、または式単位を含む。混合物は2つ以上の種類の原子、分子、または式単位を含み、それらの相対的な比率は変化する可能性がある。
    • 1.4.A.2 元素分析を用いて、物質に含まれる原子の相対的な数を特定し、純度を判定することができる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Unlike a pure compound, a mixture 混合物 has variable composition – its parts keep their own identities. Describe a mixture by the mass or mole fraction of each component; these do not follow a fixed formula. Spectroscopy (like PES or absorption) can measure how much of each component is present.

    日本語

    純粋な化合物と異なり、混合物は組成が一定ではなく、構成成分それぞれが独自の性質を保っています。混合物は各成分の質量分率や物質量分率で記述されますが、固定された化学式に従いません。分光法(PESや吸収スペクトロscopyなど)を用いることで、各成分の含有量を測定できます。

    1.5

    Atomic Structure and Electron Configuration · ⁨原子構造と電子配置⁩

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 1.5.A: Represent the ground-state electron configuration of an atom of an element or its ions using the Aufbau principle.

    • 1.5.A.1 The atom is composed of negatively charged electrons and a positively charged nucleus that is made of protons and neutrons.
    • 1.5.A.2 Coulomb's law is used to calculate the force between two charged particles.
      • Equation: $F_{coulombic} \propto \dfrac{q_1 q_2}{r^2}$
    • 1.5.A.3 In atoms and ions, the electrons can be thought of as being in "shells (energy levels)" and "subshells (sublevels)," as described by the ground-state electron configuration. Inner electrons are called core electrons, and outer electrons are called valence electrons. The electron configuration is explained by quantum mechanics, as delineated in the Aufbau principle and exemplified in the periodic table of the elements.
      • Exclusion Statement: The assignment of quantum numbers to electrons in subshells of an atom will not be assessed on the AP Exam.
    • 1.5.A.4 The relative energy required to remove an electron from different subshells of an atom or ion or from the same subshell in different atoms or ions (ionization energy) can be estimated through a qualitative application of Coulomb's law. This energy is related to the distance from the nucleus and the effective (shield) charge of the nucleus.
    日本語

    学習目標 1.5.A: オフーバウ原理を用いて、元素の原子またはイオンの基底状態電子配置を表す。

    • 1.5.A.1 原子は、負の電荷を持つ電子と、陽子と中性子からなる正の電荷を持つ原子核で構成されている。
    • 1.5.A.2 クーロンの法則は、2つの帯電粒子間の力を計算するために用いられる。
      • 式: $F_{coulombic} \propto \dfrac{q_1 q_2}{r^2}$
    • 1.5.A.3 原子やイオンにおいて、電子は「殻(エネルギー準位)」および「副殻(サブレベル)」にあると考えられるが、これは基底状態電子配置によって説明される。内側の電子はコア電子と呼ばれ、外側の電子は価電子と呼ばれる。電子配置は量子力学によって説明され、オフーバウ原理で規定され、元素周期表で例示されている。
      • 除外事項: 原子の副殻内の電子に対する量子数の割り当てについては、AP試験では評価されない。
    • 1.5.A.4 異なる副殻からの電子の除去、あるいは同じ副殻でも異なる原子やイオンからの電子の除去に必要な相対的なエネルギー(第一イオン化エネルギーなど)は、クーロンの法則の定性的な適用によって推定できる。このエネルギーは、原子核からの距離および原子核の有効(遮蔽)電荷に関係している。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    An atom is a tiny nucleus 原子核 (protons and neutrons) surrounded by electrons in shells and subshells (s, p, d, f). The electron configuration 电子排布 lists how electrons fill these, lowest energy first (e.g. $1s^2\,2s^2\,2p^6$). The outermost, highest-energy electrons – the valence electrons 价电子 – control chemistry. Coulomb's law explains their energies: electrons closer to, and less shielded from, the nucleus are held more tightly.

    Worked example. Write the electron configuration of sulfur ($Z=16$). Fill subshells in order until $16$ electrons are placed: $1s^2\,2s^2\,2p^6\,3s^2\,3p^4$. The $3s$ and $3p$ electrons (six in total) are the valence electrons, so sulfur tends to gain two electrons to complete its octet.

    日本語

    原子は、電子が殻および亜殻(s, p, d, f)に配置されている微小な原子核(陽子と中性子)から成ります。電子配置は、これらの軌道に電子が充填される様子を示し、エネルギーが低い順に並べます(例:$1s^2\,2s^2\,2p^6$)。最外殻の最高エネルギー電子——価電子——が化学的性質を決定します。クーロンの法則により、そのエネルギーが説明されます。原子核に近い、あるいは遮蔽効果が少ない電子ほど強く引きつけられます。

    An atom: a tiny nucleus of protons and neutrons, with electrons in shells
    An atom: a tiny nucleus of protons and neutrons, with electrons in shells

    9つの炎を焚いた容器が並び、それぞれ異なる鮮やかな色(深紅、オレンジ、黄色、緑、青、紫)を放つ *炎色反応: 各金属イオンは独自の色を示す。熱によって電子が励起され、基底状態に戻る際に特定の波長の光を放出するためである。

    ** worked example.** スズ($Z=16$)の電子配置を書け。サブシェルをエネルギー順に埋めていき、$16$ 個の電子を配置するまで続ける: $1s^2\,2s^2\,2p^6\,3s^2\,3p^4$。$3s$ と $3p$ の電子(合計6個)は価電子であるため、スズはオクテットを満たすために電子を2個受け取りやすい。

    周期表は原子構造に基づき元素を整理しており、電子配置がそのパターンを説明する
    周期表は原子構造に基づき元素を整理しており、電子配置がそのパターンを説明する
    Explore · ⁨探索⁩

    Explore how electrons fill the shells

    Change the atomic number $Z$ and watch the electrons fill the shells lowest-energy-first (the Aufbau principle 构造原理). The outermost electrons are the valence electrons that drive bonding.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    nucleus/ˈnjuːklɪəs/ 核
    electron configuration/ɪˈlektrɒn kənˌfɪɡjəˈreɪʃn/ 電子配置
    valence electrons/ˈveɪləns ɪˈlektrɒnz/ 最外殻電子
    Photoelectron spectroscopy/ˌfəʊtəʊɪˈlektrɒn spekˈtrɒskəpi/ 光電子分光法
    Atomic radius/əˈtɒmɪk ˈreɪdɪəs/ 原子半径
    Ionization energy/ˌaɪənaɪˈzeɪʃn ˈenədʒi/ イオン化エネルギー
    1.6

    Photoelectron Spectroscopy · ⁨光電子分光法⁩

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 1.6.A: Explain the relationship between the photoelectron spectrum of an atom or ion and: i. The ground-state electron configuration of the species. ii. The interactions between the electrons and the nucleus.

    • 1.6.A.1 The energies of the electrons in a given shell can be measured experimentally with photoelectron spectroscopy (PES). The position of each peak in the PES spectrum is related to the energy required to remove an electron from the corresponding subshell, and the relative height of each peak is (ideally) proportional to the number of electrons in that subshell.
    日本語

    学習目標 1.6.A: 原子またはイオンの光電子スペクトルと以下の関係について説明する: i. その種の基底状態電子配置。 ii. 電子と原子核との相互作用。

    • 1.6.A.1 特定の殻内の電子のエネルギーは、光電子分光法(PES)によって実験的に測定できる。PESスペクトルにおける各ピークの位置は、対応する副殻から電子を取り除くために必要なエネルギーに関連しており、各ピークの相対的な高さは、理想的にはその副殻内の電子の数に比例する。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Photoelectron spectroscopy 光电子能谱 (PES) measures the energy needed to remove electrons from each subshell. Each peak is a subshell: its position gives the binding energy (how tightly held) and its height gives the number of electrons in it. PES data let you read an element's electron configuration and confirm shell structure directly.

    日本語

    光電子分光法 (PES) は、各サブシェルから電子を取り除くために必要なエネルギーを測定します。各ピークはサブシェルを表し、その位置は結合エネルギー(どれだけ強く保持されているか)を、その高さはその中の電子数を示します。PESデータを用いることで、元素の電子配置を読み取り、殻構造を直接確認できます。

    連続電離エネルギーの大きな飛躍が殻構造を示す
    連続電離エネルギーの大きな飛躍が殻構造を示す
    ネオンの光電子スペクトル: サブシェルごとに1つのピーク、高さは電子数に対応
    ネオンの光電子スペクトル: サブシェルごとに1つのピーク、高さは電子数に対応
    1.7

    Periodic Trends · ⁨周期傾向⁩

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 1.7.A: Explain the relationship between trends in atomic properties of elements and electronic structure and periodicity.

    • 1.7.A.1 The organization of the periodic table is based on patterns of recurring properties of the elements, which are explained by patterns of ground-state electron configurations and the presence of completely or partially filled shells (and subshells) of electrons in atoms.
      • Exclusion Statement: Writing the electron configuration of elements that are exceptions to the aufbau principle will not be assessed on the AP Exam.
    • 1.7.A.2 Trends in atomic properties within the periodic table (periodicity) can be predicted by the position of the element on the periodic table and qualitatively understood using Coulomb's law, the shell model, and the concepts of shielding and effective nuclear charge. These properties include:
      • i. Ionization energy
      • ii. Atomic and ionic radii
      • iii. Electron affinity
      • iv. Electronegativity.
    • 1.7.A.3 The periodicity (in 1.7.A.2) is useful to predict/estimate values of properties in the absence of data.
    日本語

    学習目標 1.7.A: 元素の原子性質の傾向と、電子構造および周期性との関係について説明する。

    • 1.7.A.1 元素周期表の編成は、元素の反復する性質のパターンに基づいているが、これらは基底状態電子配置のパターンおよび原子内の完全にまたは部分的に満たされた殻(および副殻)の存在によって説明される。
      • 除外事項: オフブル原理の例外となる元素の電子配置を記述することは、AP試験の評価対象外です。
    • 1.7.A.2 周期表における原子特性の傾向(周期性)は、その元素が周期表に位置する場所によって予測でき、クーロンの法則、殻モデル、遮蔽効果および有効核電荷の概念を用いて定性的に理解できます。これらの特性には以下が含まれます:
      • i. 電離エネルギー
      • ii. 原子半径およびイオン半径
      • iii. 電子親和力
      • iv. 電気陰性度。
    • 1.7.A.3 周期性(1.7.A.2参照)は、データが存在しない場合に特性の値を予測・推定するために有用です。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Trends across the periodic table follow from nuclear charge and shielding:

    • Atomic radius 原子半径 decreases across a period (stronger pull) and increases down a group (more shells).
    • Ionization energy 电离能 (energy to remove an electron) increases across, decreases down – opposite to radius.
    • Electronegativity 电负性 (pull on shared electrons) increases across and up, toward fluorine.
    日本語

    周期表における傾向は、核電荷と遮蔽効果に基づいています。

    電気陰性度は周期方向に増加し、族方向に減少する
    電気陰性度は周期方向に増加し、族方向に減少する
    • 原子半径は周期内では減少する(強い引力)、族内では増加する(電子層が増える)。
    • イオン化エネルギー(電子を除去するエネルギー)は周期内で増加し、族内で減少する – 半径とは逆の傾向。
    • 電気陰性度(共有電子に対する引力)は周期内で増加し、族内では上昇する(フッ素に近づく)。
    周期法則:半径、イオン化エネルギー、電気陰性度が周期内および族内でどのように変化するか
    周期法則:半径、イオン化エネルギー、電気陰性度が周期内および族内でどのように変化するか
    Explore · ⁨探索⁩

    Explore atomic radius across a period

    Step across Period 3 and watch the atomic radius shrink — each added proton raises the effective nuclear charge and pulls the same shell in tighter.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Electronegativity/ɪˌlektrəʊŋɡəˈtɪvɪti/ 電気陰性度
    cations/ˈkætaɪənz/ カチオン
    anions/ˈænaɪənz/ アニオン
    ionic compound/aɪˈɒnɪk ˈkɒmpaʊnd/ イオン化合物
    1.8

    Valence Electrons and Ionic Compounds · ⁨価電子とイオン化合物⁩

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 1.8.A: Explain the relationship between trends in the reactivity of elements and periodicity.

    • 1.8.A.1 The likelihood that two elements will form a chemical bond is determined by the interactions between the valence electrons and nuclei of elements.
    • 1.8.A.2 Elements in the same column of the periodic table tend to form analogous compounds.
    • 1.8.A.3 Typical charges of atoms in ionic compounds are governed by the number of valence electrons and predicted by their location on the periodic table.
    日本語

    学習目標 1.8.A: 元素の反応性の傾向と周期性との関係について説明する。

    • 1.8.A.1 2つの元素が化学結合を形成する可能性は、価電子と元素の原子核との相互作用によって決定される。
    • 1.8.A.2 周期表の同じ列にある元素は、類似した化合物を形成する傾向がある。
    • 1.8.A.3 イオン化合物中の原子の一般的な電荷は、価電子の数によって支配され、周期表上の位置によって予測される。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Atoms gain, lose, or share valence electrons to reach stable configurations. Metals (low ionization energy) lose electrons to form cations 阳离子; nonmetals gain electrons to form anions 阴离子. Oppositely charged ions attract into an ionic compound 离子化合物, whose formula balances the charges to make the whole neutral. For example, aluminium ($3+$) and oxide ($2-$) combine as $\text{Al}_2\text{O}_3$ so the $+6$ and $-6$ cancel.

    日本語

    原子は安定した電子配置になるために価電子を得る、失う、または共有する。金属(低いイオン化エネルギー)は電子を失ってカチオンを形成し、非金属は電子を得てアニオンを形成する。異なる電荷を持つイオンが引き合ってイオン化合物となり、その化学式は全体の電荷を中和させるようにバランスしている。例えば、アルミニウム($3+$)と酸化物イオン($2-$)は $\text{Al}_2\text{O}_3$ を形成し、$+6$ と $-6$ が相殺される。

    Explore · ⁨探索⁩

    Watch an ionic bond form by electron transfer

    A metal gives up its valence electron(s) and a non-metal takes them, so both reach a full shell. The atoms become oppositely charged ions that attract — that electrostatic pull is the ionic bond.

    1.8

    Exam tips · ⁨試験対策⁩

    English
    • Use the mole as the hub: convert grams↔moles with $n=m/M$ and moles↔particles with Avogadro's number.
    • Relative atomic mass is the abundance-weighted average of the isotopes — it lies closer to the more abundant one, not halfway.
    • For an empirical formula, turn each element's mass into moles and divide by the smallest; scale up to whole numbers.
    • Read periodic trends from nuclear charge and shielding: atomic radius decreases across a period, ionisation energy and electronegativity increase across (and up).
    • Fill electron configurations in energy order; the outer valence electrons control the chemistry.
    日本語
    • モールを中核として使用:$n=m/M$ でグラム↔モール変換を行い、アボガドロ定数でモール↔粒子変換を行う。
    • 相対原子質量は同位体の存在率加重平均であり、より存在率の高い方に近い位置にある(真ん中ではない)。
    • 実験式の場合、各元素の質量をモール数に変換し、最も小さい値で割る;整数にするために倍数する。
    • 核電荷と遮蔽効果から周期法則を読み取る:原子半径は周期内で減少し、イオン化エネルギーと電気陰性度は周期内(および族内の上昇方向)で増加する。
    • 電子配置をエネルギー順に埋めていく;最外殻の価電子が化学的性質を支配する。
  • 2

    Compound Structure and Properties · ⁨化合物の構造と性質⁩

    Watch lesson · ⁨レッスンを視聴⁩
    2.1

    Types of Chemical Bonds

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 2.1.A: Explain the relationship between the type of bonding and the properties of the elements participating in the bond.

    • 2.1.A.1 Electronegativity values for the representative elements increase going from left to right across a period and decrease going down a group. These trends can be understood qualitatively through the electronic structure of the atoms, the shell model, and Coulomb's law.
    • 2.1.A.2 Valence electrons shared between atoms of similar electronegativity constitute a nonpolar covalent bond. For example, bonds between carbon and hydrogen are effectively nonpolar even though carbon is slightly more electronegative than hydrogen.
    • 2.1.A.3 Valence electrons shared between atoms of unequal electronegativity constitute a polar covalent bond.
      • i. The atom with a higher electronegativity will develop a partial negative charge relative to the other atom in the bond.
      • ii. In single bonds, greater differences in electronegativity lead to greater bond dipoles.
      • iii. All polar bonds have some ionic character, and the difference between ionic and covalent bonding is not distinct but rather a continuum.
    • 2.1.A.4 The difference in electronegativity is not the only factor in determining if a bond should be designated as ionic or covalent. Generally, bonds between a metal and nonmetal are ionic, and bonds between two nonmetals are covalent. Examination of the properties of a compound is the best way to characterize the type of bonding.
    • 2.1.A.5 In a metallic solid, the valence electrons from the metal atoms are considered to be delocalized and not associated with any individual atom.
    日本語

    学習目標 2.1.A: 結合の種類と結合に関与する元素の性質との関係について説明する。

    • 2.1.A.1 主族元素の電気陰性度は、周期内では左から右へ増加し、族内では上から下へ減少する。これらの傾向は、原子の電子構造、殻モデル、およびクーロンの法則によって定性的に理解できる。
    • 2.1.A.2 電気陰性度が近い原子間で共有される価電子は、非極性共有結合を形成する。例えば、炭素は水素よりわずかに電気陰性度が高いにもかかわらず、炭素と水素の間の結合は実質的に非極性である。
    • 2.1.A.3 電気陰性度が異なる原子間で共有される価電子は、極性共有結合を形成する。
      • i. 電気陰性度の高い原子は、結合内の他の原子に対して部分負電荷を帯びる。
      • ii. 単結合において、電気陰性度の差が大きいほど、大きな結合双極子が生じる。
      • iii. 全ての極性結合にはある程度のイオン性があり、イオン結合と共有結合の違いは明確ではなく、連続体として捉えられる。
    • 2.1.A.4 電気陰性度の違いが、結合をイオン結合または共有結合と分類する唯一の要因ではない。一般的に、金属と非金属の間の結合はイオン結合であり、2つの非金属の間の結合は共有結合である。化合物の性質を検討することが、結合の種類を同定する最良の方法である。
    • 2.1.A.5 金属固体では、金属原子からの価電子は局在化せず、特定の原子に属さない自由電子として扱われる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    A chemical bond 化学键 is an attraction that holds atoms together. Which type forms depends on the atoms' electronegativities:

    • Ionic bond 离子键: electrons transfer from a metal to a nonmetal (large electronegativity difference).
    • Covalent bond 共价键: nonmetals share electrons (small difference). A big-but-not-huge difference gives a polar covalent 极性共价 bond.
    • Metallic bond 金属键: metal atoms share a "sea" of mobile electrons.
    日本語
    Sodium chloride crystals: ionic solids form a lattice of oppositely charged ions
    Sodium chloride crystals: ionic solids form a lattice of oppositely charged ions

    A chemical bond 化学键 is an attraction that holds atoms together. Which type forms depends on the atoms' electronegativities:

    Ionic bonding: a metal transfers its outer electrons to a non-metal
    Ionic bonding: a metal transfers its outer electrons to a non-metal
    • Ionic bond 离子键: electrons transfer from a metal to a nonmetal (large electronegativity difference).
    • Covalent bond 共价键: nonmetals share electrons (small difference). A big-but-not-huge difference gives a polar covalent 极性共价 bond.
    • Metallic bond 金属键: metal atoms share a "sea" of mobile electrons.
    A diamond crystal: giant covalent networks explain extreme hardness and high melting points
    A diamond crystal: giant covalent networks explain extreme hardness and high melting points
    Explore · ⁨探索⁩

    Form an ionic bond by electron transfer · ⁨電子移動によるイオン結合の形成⁩

    An ionic bond forms when a metal gives electrons to a non-metal, making oppositely charged ions that attract; a covalent bond shares electrons instead. · ⁨イオン結合 は、金属が非金属に電子を与えて異符号のイオンを作り、それらが引き合うことで形成されます。一方、共有結合 は電子を共有することで形成されます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    chemical bond/ˈkemɪkl bɒnd/ 化学結合
    Ionic bond/aɪˈɒnɪk bɒnd/ イオン結合
    Covalent bond/ˈkəʊvələnt bɒnd/ 共有結合
    polar covalent/ˈpəʊlə ˈkəʊvələnt/ 極性共有結合
    Metallic bond/məˈtælɪk bɒnd/ 金属結合
    2.2

    Intramolecular Force and Potential Energy

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 2.2.A: Represent the relationship between potential energy and distance between atoms, based on factors that influence the interaction strength.

    • 2.2.A.1 A graph of potential energy versus the distance between atoms (internuclear distance) is a useful representation for describing the interactions between atoms. Such graphs illustrate both the equilibrium bond length (the separation between atoms at which the potential energy is lowest) and the bond energy (the energy required to separate the atoms).
    • 2.2.A.2 In a covalent bond, the bond length is influenced by both the size of the atom's core and the bond order (i.e., single, double, triple). Bonds with a higher order are shorter and have larger bond energies.
    • 2.2.A.3 Coulomb's law can be used to understand the strength of interactions between cations and anions.
      • i. Because the interaction strength is proportional to the charge on each ion, larger charges lead to stronger interactions.
      • ii. Because the interaction strength increases as the distance between the centers of the ions (nuclei) decreases, smaller ions lead to stronger interactions.
    日本語

    学習目標 2.2.A: 相互作用の強さに影響を与える要因に基づき、ポテンシャルエネルギーと原子間の距離との関係を表現する。

    • 2.2.A.1 ポテンシャルエネルギーと原子間の距離(核間距離)のプロットは、原子間の相互作用を説明するための有用な表現である。このようなグラフは、平衡結合長(ポテンシャルエネルギーが最小となる原子間の距離)および結合エネルギー(原子を分離するために必要なエネルギー)を示す。
    • 2.2.A.2 共有結合において、結合長は原子のコアサイズと結合次数(単結合、二重結合、三重結合など)の両方によって影響を受ける。結合次数が高いほど結合長は短く、結合エネルギーも大きくなる。
    • 2.2.A.3 クーロンの法則を用いて、陽イオンと陰イオンの間の相互作用の強さを理解することができる。
      • i. 相互作用の強さは各イオンの電荷に比例するため、電荷が大きいほど相互作用が強くなる。
      • ii. 相互作用の強さはイオン中心(原子核)間の距離が小さくなると増大するため、イオンが小さいほど相互作用が強くなる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    As two atoms approach, a potential energy 势能 curve captures the balance of attraction and repulsion. It dips to a minimum at the bond length 键长 (the stable separation) whose depth is the bond energy 键能. Shorter, stronger bonds sit in deeper, tighter wells; more shared pairs (double, triple bonds) give shorter, stronger bonds.

    日本語

    As two atoms approach, a potential energy 势能 curve captures the balance of attraction and repulsion. It dips to a minimum at the bond length 键长 (the stable separation) whose depth is the bond energy 键能. Shorter, stronger bonds sit in deeper, tighter wells; more shared pairs (double, triple bonds) give shorter, stronger bonds.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    potential energy/pəˈtenʃl ˈenədʒi/ 位置エネルギー
    bond length/bɒnd leŋθ/ 結合長
    bond energy/bɒnd ˈenədʒi/ 結合エネルギー
    2.3

    Structure of Ionic Solids

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 2.3.A: Represent an ionic solid with a particulate model that is consistent with Coulomb's law and the properties of the constituent ions.

    • 2.3.A.1 The cations and anions in an ionic crystal are arranged in a systematic, periodic 3-D array that maximizes the attractive forces among cations and anions while minimizing the repulsive forces.
      • Exclusion statement: Knowledge of specific crystal structures is not essential to an understanding of the learning objective and will not be assessed on the AP Exam.
    日本語

    学習目標 2.3.A: クーロンの法則および構成イオンの性質と一致する粒子モデルを用いてイオン固体を表現する。

    • 2.3.A.1 イオン結晶中の陽イオンと陰イオンは、陽イオンと陰イオンの間の引力を最大化し、反発力を最小限に抑えるように、体系的かつ周期的な3次元配列で配置されている。
      • 除外事項: 具体的な結晶構造に関する知識は、学習目標を理解するために必須ではなく、AP試験の評価対象外です。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    An ionic solid 离子固体 is a repeating 3-D lattice 晶格 of alternating cations and anions, held by strong electrostatic attraction. This explains their high melting points, brittleness, and why they conduct only when molten or dissolved (ions freed to move). The lattice energy rises with larger ion charges and smaller ions, so MgO (both $2+/2-$) melts far higher than NaCl (both $1+/1-$).

    日本語

    An ionic solid 离子固体 is a repeating 3-D lattice 晶格 of alternating cations and anions, held by strong electrostatic attraction. This explains their high melting points, brittleness, and why they conduct only when molten or dissolved (ions freed to move). The lattice energy rises with larger ion charges and smaller ions, so MgO (both $2+/2-$) melts far higher than NaCl (both $1+/1-$).

    Ions pack into a giant lattice of alternating positive and negative ions
    Ions pack into a giant lattice of alternating positive and negative ions
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    ionic solid/aɪˈɒnɪk ˈsɒlɪd/ イオン性固体
    lattice/ˈlætɪs/ 格子
    2.4

    Structure of Metals and Alloys

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 2.4.A: Represent a metallic solid and/or alloy using a model to show essential characteristics of the structure and interactions present in the substance.

    • 2.4.A.1 Metallic bonding can be represented as an array of positive metal ions surrounded by delocalized valence electrons (i.e., a "sea of electrons").
    • 2.4.A.2 Interstitial alloys form between atoms of significantly different radii, where the smaller atoms fill the interstitial spaces between the larger atoms (e.g., with steel in which carbon occupies the interstices in iron).
    • 2.4.A.3 Substitutional alloys form between atoms of comparable radius, where one atom substitutes for the other in the lattice. (e.g., in certain brass alloys, other elements, usually zinc, substitute for copper.)
    日本語

    学習目標 2.4.A: 物質に含まれる構造および相互作用の本質的な特徴を示すために、モデルを用いて金属固体および/または合金を表現する。

    • 2.4.A.1 金属結合は、正の金属イオンの配列が自由電子(「電子の海」)に囲まれているとして表現できる。
    • 2.4.A.2 间隙型合金は、半径が大きく異なる原子間に形成され、小さな原子が大きな原子の間の空隙を埋める(例:炭素が鉄の空隙占据する鋼)。
    • 2.4.A.3 置換型合金は、半径が同等の原子間に形成され、一方の原子が格子内で他方を置き換える(例:特定の真鍮合金では、通常亜鉛などの他の元素が銅を置き換える)。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    In a metal, cations sit in a lattice bathed in delocalized 离域 electrons, which explains conductivity, malleability, and luster. An alloy 合金 mixes metals: a substitutional alloy swaps in similar-sized atoms; an interstitial alloy (like steel) fits small atoms into the gaps, making it harder.

    日本語

    In a metal, cations sit in a lattice bathed in delocalized 离域 electrons, which explains conductivity, malleability, and luster. An alloy 合金 mixes metals: a substitutional alloy swaps in similar-sized atoms; an interstitial alloy (like steel) fits small atoms into the gaps, making it harder.

    Different-sized atoms in an alloy stop the layers sliding, so it is harder
    Different-sized atoms in an alloy stop the layers sliding, so it is harder
    Metallic bonding: positive ions in a sea of delocalised electrons
    Metallic bonding: positive ions in a sea of delocalised electrons
    Native copper: metallic bonding gives a shiny, malleable solid with delocalised electrons
    Native copper: metallic bonding gives a shiny, malleable solid with delocalised electrons
    Explore · ⁨探索⁩

    Slide layers in a metallic lattice · ⁨金属格子内の層をずらす⁩

    A metal is positive ions in a sea of delocalised electrons. The layers can slide without breaking the bond, so metals are malleable and conduct. · ⁨金属 は、自由電子の海の中に正イオンが配置されています。層同士が滑っても結合が切断されないため、金属は展性や延性があり、電気伝導性を示します。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    delocalized/dɪˈlɒkəlaɪzd/ 非局在化している
    alloy/ˈælɔɪ/ 合金 (alloy)
    2.5

    Lewis Diagrams

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 2.5.A: Represent a molecule with a Lewis diagram.

    • 2.5.A.1 Lewis diagrams can be constructed according to an established set of principles.
    日本語

    学習目標 2.5.A: レイス図で分子を表す。

    • 2.5.A.1 レイス図は確立された一連の原則に従って描くことができる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    A Lewis diagram 路易斯结构 shows valence electrons as bonding pairs and lone pairs 孤对电子, giving most atoms an octet 八隅体 (8 valence electrons; H wants 2). Steps: count total valence electrons, connect atoms with single bonds, complete octets on outer atoms, then form multiple bonds if the central atom is short.

    Worked example. Draw carbon dioxide, $\text{CO}_2$. Total valence electrons $=4+2(6)=16$. Put C in the centre; single bonds to each O use $4$ electrons and leave the outer O atoms short. Completing octets forces two double bonds, $\text{O}=\text{C}=\text{O}$: each O then has two lone pairs, C has none, and all $16$ electrons are placed with every atom at an octet.

    日本語

    A Lewis diagram 路易斯结构 shows valence electrons as bonding pairs and lone pairs 孤对电子, giving most atoms an octet 八隅体 (8 valence electrons; H wants 2). Steps: count total valence electrons, connect atoms with single bonds, complete octets on outer atoms, then form multiple bonds if the central atom is short.

    Dot-and-cross diagrams show the bonding pairs and lone pairs in a molecule
    Dot-and-cross diagrams show the bonding pairs and lone pairs in a molecule

    Worked example. Draw carbon dioxide, $\text{CO}_2$. Total valence electrons $=4+2(6)=16$. Put C in the centre; single bonds to each O use $4$ electrons and leave the outer O atoms short. Completing octets forces two double bonds, $\text{O}=\text{C}=\text{O}$: each O then has two lone pairs, C has none, and all $16$ electrons are placed with every atom at an octet.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Lewis diagram/ˈluːɪs ˈdaɪəɡræm/ ルイス図
    lone pairs/ləʊn peəz/ 非結合電子対
    octet/ɒkˈtet/ オクテット
    2.6

    Resonance and Formal Charge

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 2.6.A: Represent a molecule with a Lewis diagram that accounts for resonance between equivalent structures or that uses formal charge to select between nonequivalent structures.

    • 2.6.A.1 In cases where more than one equivalent Lewis structure can be constructed, resonance must be included as a refinement to the Lewis structure. In many such cases, this refinement is needed to provide qualitatively accurate predictions of molecular structure and properties.
    • 2.6.A.2 The octet rule and formal charge can be used as criteria for determining which of several possible valid Lewis diagrams provides the best model for predicting molecular structure and properties.
    • 2.6.A.3 As with any model, there are limitations to the use of the Lewis structure model, particularly in cases with an odd number of valence electrons.
    日本語

    学習目標 2.6.A: 等価な構造間の共鳴を考慮したレース図、あるいは非等価な構造の選択に形式電荷を用いたレース図で分子を表す。

    • 2.6.A.1 複数の等価なレース構造を描ける場合、共鳴をレース構造の補足として含める必要がある。多くの此类の場合、この補足は分子の構造および性質について定性的に正確な予測を行うために必要である。
    • 2.6.A.2 オクテット則と形式電荷は、複数の有効なレース図のうち、分子の構造と性質の予測に最も適したモデルを提供するものを選ぶための基準として用いられる。
    • 2.6.A.3 いかなるモデルと同様に、レース構造モデルにも限界があり、特に価電子が奇数個であるケースではその適用に限界が生じる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    When two or more valid Lewis diagrams differ only in electron placement, the true structure is an average – resonance 共振. Formal charge 形式电荷 (valence electrons minus lone-pair electrons minus half the bonding electrons) picks the best structure: the one with formal charges closest to zero, and any negative charge on the most electronegative atom.

    Worked example. Assign formal charges in the nitrate ion, $\text{NO}_3^{-}$ (one double bond, two single bonds). For N (4 bonds, no lone pairs): $5-0-4=+1$. For the double-bonded O (2 lone pairs): $6-4-2=0$. For each single-bonded O (3 lone pairs): $6-6-1=-1$. The total is $+1+0+(-1)+(-1)=-1$, matching the ion's overall charge – a good check that the structure is drawn correctly. Because the three O atoms are equivalent by resonance, the real ion has three identical bonds.

    日本語

    When two or more valid Lewis diagrams differ only in electron placement, the true structure is an average – resonance 共振. Formal charge 形式电荷 (valence electrons minus lone-pair electrons minus half the bonding electrons) picks the best structure: the one with formal charges closest to zero, and any negative charge on the most electronegative atom.

    Worked example. Assign formal charges in the nitrate ion, $\text{NO}_3^{-}$ (one double bond, two single bonds). For N (4 bonds, no lone pairs): $5-0-4=+1$. For the double-bonded O (2 lone pairs): $6-4-2=0$. For each single-bonded O (3 lone pairs): $6-6-1=-1$. The total is $+1+0+(-1)+(-1)=-1$, matching the ion's overall charge – a good check that the structure is drawn correctly. Because the three O atoms are equivalent by resonance, the real ion has three identical bonds.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    resonance/ˈrezənəns/ 共振
    Formal charge/ˈfɔːml tʃɑːdʒ/ 形式電荷
    2.7

    VSEPR and Bond Hybridization

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 2.7.A: Based on the relationship between Lewis diagrams, VSEPR theory, bond orders, and bond polarities:

    • i. Explain structural properties of molecules.

    • ii. Explain electron properties of molecules.

    • 2.7.A.1 VSEPR theory uses the Coulombic repulsion between electrons as a basis for predicting the arrangement of electron pairs around a central atom.

    • 2.7.A.2 Both Lewis diagrams and VSEPR theory must be used for predicting electronic and structural properties of many covalently bonded molecules and polyatomic ions, including the following:

      • i. Molecular geometry (linear, trigonal planar, tetrahedral, trigonal pyramidal, bent, trigonal bipyramidal, seesaw, T-shaped, octahedral, square pyramidal, square planar)
      • ii. Bond angles
      • iii. Relative bond energies based on bond order
      • iv. Relative bond lengths (multiple bonds, effects of atomic radius)
      • v. Presence of a dipole moment
      • vi. Hybridization of valence orbitals for atoms within a molecule or polyatomic ion
    • 2.7.A.3 The terms "hybridization" and "hybrid atomic orbital" are used to describe the arrangement of electrons around a central atom. When the central atom is $sp$ hybridized, its ideal bond angles are $180°$; for $sp^2$ hybridized atoms the bond angles are $120°$; and for $sp^3$ hybridized atoms the bond angles are $109.5°$.

      • Exclusion statement: An understanding of the derivation and depiction of hybrid orbitals will not be assessed on the AP Exam. The course includes the distinction between sigma and pi bonding, the use of VSEPR to explain the shapes of molecules, and the $sp$, $sp^2$, and $sp^3$ nomenclature.
      • Exclusion statement: Hybridization involving d orbitals will not be assessed on the AP Exam. When an atom has more than four pairs of electrons surrounding the central atom, students are only responsible for the shape of the resulting molecule.
    • 2.7.A.4 Bond formation is associated with overlap between atomic orbitals. In multiple bonds, such overlap leads to the formation of both sigma and pi bonds. The overlap is stronger in sigma than pi bonds, which is reflected in sigma bonds having greater bond energy than pi bonds. The presence of a pi bond also prevents the rotation of the bond and leads to geometric isomers.

      • Exclusion statement: Molecular orbital theory is recommended as a way to provide deeper insight into bonding. However, the AP Exam will neither explicitly assess molecular orbital diagrams, filling of molecular orbitals, nor the distinction between bonding, nonbonding, and antibonding orbitals.
    日本語

    学習目標 2.7.A: レイス図、VSEPR理論、結合次数、結合極性の関係に基づいて:

    • i. 分子の構造的性質を説明する。

    • ii. 分子の電子的性质を説明する。

    • 2.7.A.1 VSEPR理論は、電子間のクーロン反発を基盤として、中心原子周囲の電子対の配置を予測するために用いられる。

    • 2.7.A.2 以下の共有結合分子や多原子イオンを含む、多くの共有結合分子および多原子イオンの電子および構造的性質を予測するには、レース図とVSEPR理論の両方を用いる必要がある:

      • i. 分子の幾何学的形状(直線形、平面三角形、正四面体形、三角錐形、湾曲形、三角双錐形、シーソー形、T字形、正八面体形、四角錐形、正方形平面形)
      • ii. 結合角
      • iii. 結合次数に基づく相対的な結合エネルギー
      • iv. 相対的な結合長(多重結合、原子半径の影響)
      • v. 双極子モーメントの有無
      • vi. 分子または多原子イオン内の原子における価電子軌道の混成
    • 2.7.A.3 「混成」と「混成原子軌道」という用語は、中心原子周囲の電子の配置を記述するために用いられる。中心原子が $sp$ 混成である場合、理想的な結合角は $180°$ であり、$sp^2$ 混成の原子では結合角は $120°$ であり、$sp^3$ 混成の原子では結合角は $109.5°$ である。

      • 除外事項: 混成軌道の導出および描写に関する理解はAP試験では評価されない。本課程では、シグマ結合とピ結合の違い、VSEPRを用いた分子形状の説明、そして $sp$, $sp^2$, $sp^3$ の命名法が含まれる。
      • 除外事項: d軌道を含む混成はAP試験では評価されない。中心原子を取り巻く電子対が4対を超える場合、学生は生成される分子の形状のみについて責任を負う。
    • 2.7.A.4 結合形成は原子軌道間の重なりに伴う。多重結合では、このような重なりによりシグマ結合とピ結合の両方が形成される。シグマ結合の方がピ結合よりも重なりが強いため、シグマ結合の方が結合エネルギーが大きいことが反映されている。ピ結合の存在は結合の回転を妨げ、幾何異性体の形成をもたらす。

      • 除外事項: 分子軌道理論は、結合に対するより深い洞察を得る方法として推奨される。しかし、AP試験では分子軌道図、分子軌道の電子充填、および結合性・非結合性・反結合性軌道との区別については明記して評価しない。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    VSEPR: molecular shapes

    VSEPR 价层电子对互斥 theory predicts shape: electron pairs (bonds and lone pairs) around a central atom spread out as far apart as possible. Counting electron domains gives the geometry (linear, trigonal planar, tetrahedral, …); lone pairs push bonds closer, bending the shape. Hybridization 杂化 ($sp$, $sp^2$, $sp^3$) describes the mixed orbitals matching that geometry. Molecular shape and bond polarity together decide whether the whole molecule is polar.

    Worked example. Predict the shape of ammonia, $\text{NH}_3$. Nitrogen has $3$ bonding pairs and $1$ lone pair – four electron domains, so the electron geometry is tetrahedral and the hybridization is $sp^3$. The lone pair is invisible in the shape but still pushes the bonds together, so the molecular shape is trigonal pyramidal with a bond angle of about $107^{\circ}$ (a little less than the ideal $109.5^{\circ}$). The three N–H dipoles do not cancel, so the molecule is polar.

    Bonds form when atomic orbitals overlap. Every single bond is one sigma bond σ键 – orbitals overlapping head-on along the line joining the two atoms. A multiple bond adds a pi bond π键, made by $p$ orbitals overlapping sideways above and below that line: a double bond is one sigma + one pi, a triple bond one sigma + two pi. Head-on overlap is more effective, so a sigma bond is stronger (higher bond energy) than a pi bond – which is why a double bond is stronger than a single bond but not twice as strong. A pi bond also locks the two atoms so they cannot rotate about the bond; a C=C double bond therefore has fixed cis and trans forms – geometric isomers 几何异构体 that a freely-rotating single bond could never show.

    Worked example. Count the bonds in ethene, $\text{H}_2\text{C}=\text{CH}_2$. The four C–H bonds are single bonds (one sigma each); the C=C is one sigma plus one pi. So ethene has 5 sigma and 1 pi bond, and that single pi bond is exactly what stops the two $\text{CH}_2$ ends from twisting relative to each other.

    日本語
    VSEPR: molecular shapes

    VSEPR 价层电子对互斥 theory predicts shape: electron pairs (bonds and lone pairs) around a central atom spread out as far apart as possible. Counting electron domains gives the geometry (linear, trigonal planar, tetrahedral, …); lone pairs push bonds closer, bending the shape. Hybridization 杂化 ($sp$, $sp^2$, $sp^3$) describes the mixed orbitals matching that geometry. Molecular shape and bond polarity together decide whether the whole molecule is polar.

    Hybridisation and shape: sp3 tetrahedral, sp2 planar, sp linear
    Hybridisation and shape: sp3 tetrahedral, sp2 planar, sp linear
    The common VSEPR shapes and their bond angles
    The common VSEPR shapes and their bond angles

    Worked example. Predict the shape of ammonia, $\text{NH}_3$. Nitrogen has $3$ bonding pairs and $1$ lone pair – four electron domains, so the electron geometry is tetrahedral and the hybridization is $sp^3$. The lone pair is invisible in the shape but still pushes the bonds together, so the molecular shape is trigonal pyramidal with a bond angle of about $107^{\circ}$ (a little less than the ideal $109.5^{\circ}$). The three N–H dipoles do not cancel, so the molecule is polar.

    Bonds form when atomic orbitals overlap. Every single bond is one sigma bond σ键 – orbitals overlapping head-on along the line joining the two atoms. A multiple bond adds a pi bond π键, made by $p$ orbitals overlapping sideways above and below that line: a double bond is one sigma + one pi, a triple bond one sigma + two pi. Head-on overlap is more effective, so a sigma bond is stronger (higher bond energy) than a pi bond – which is why a double bond is stronger than a single bond but not twice as strong. A pi bond also locks the two atoms so they cannot rotate about the bond; a C=C double bond therefore has fixed cis and trans forms – geometric isomers 几何异构体 that a freely-rotating single bond could never show.

    Worked example. Count the bonds in ethene, $\text{H}_2\text{C}=\text{CH}_2$. The four C–H bonds are single bonds (one sigma each); the C=C is one sigma plus one pi. So ethene has 5 sigma and 1 pi bond, and that single pi bond is exactly what stops the two $\text{CH}_2$ ends from twisting relative to each other.

    Explore · ⁨探索⁩

    Predict molecular shape with VSEPR · ⁨VSEPR理論を用いて分子形状を予測する⁩

    VSEPR: electron pairs repel and spread as far apart as possible, setting the molecule's shape. Add bonding and lone pairs and watch the geometry change. · ⁨VSEPR: 電子対同士は反発し合い、可能な限り離れて配置することで分子の形状が決まります。結合電子対と非共有電子対を追加すると、幾何学的配置が変化するのが見えます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    VSEPR/ˈvespə/ VSEPR
    Hybridization/ˌhaɪbrɪdaɪˈzeɪʃn/ 混成軌道
    sigma bond/ˈsɪɡmə bɒnd/ シグマ結合
    pi bond/paɪ bɒnd/ π結合
    geometric isomers/ˌdʒiːəʊˈmetrɪk ˈaɪsəməz/ 幾何異性体
    2.7

    Exam tips

    • Decide the bond type from the atoms: ionic (metal + non-metal, electron transfer), covalent (non-metals, sharing), metallic (sea of delocalised electrons).
    • An ionic solid conducts only when molten or dissolved (ions free to move), never as a solid.
    • Draw Lewis structures to satisfy the octet (H wants 2), then use VSEPR — electron pairs spread as far apart as possible — to predict the shape.
    • Lone pairs take up space and push bonds closer, bending the shape (water is bent, ammonia pyramidal).
    • A molecule can have polar bonds yet be non-polar overall if its symmetry makes the dipoles cancel ($\text{CO}_2$).
    • A single bond is 1 sigma bond; a double bond is 1 sigma + 1 pi, a triple bond 1 sigma + 2 pi. Sigma is stronger than pi, and a pi bond blocks rotation, giving cis/trans geometric isomers.
  • 3

    Properties of Substances and Mixtures · ⁨物質と混合物の性質⁩

    Watch lesson · ⁨レッスンを視聴⁩
    3.1

    Intermolecular and Interparticle Forces

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.1.A: Explain the relationship between the chemical structures of molecules and the relative strength of their intermolecular forces when: i. The molecules are of the same chemical species. ii. The molecules are of two different chemical species.

    • 3.1.A.1 London dispersion forces are a result of the Coulombic interactions between temporary, fluctuating dipoles. London dispersion forces are often the strongest net intermolecular force between large molecules.
      • i. Dispersion forces increase with increasing contact area between molecules and with increasing polarizability of the molecules.
      • ii. The polarizability of a molecule increases with an increasing number of electrons in the molecule and the size of the electron cloud. It is enhanced by the presence of pi bonding.
      • iii. The term "London dispersion forces" should not be used synonymously with the term "van der Waals forces."
    • 3.1.A.2 The dipole moment of a polar molecule leads to additional interactions with other chemical species.
      • i. Dipole-induced dipole interactions are present between a polar and nonpolar molecule. These forces are always attractive. The strength of these forces increases with the magnitude of the dipole of the polar molecule and with the polarizability of the nonpolar molecule.
      • ii. Dipole-dipole interactions are present between polar molecules. The interaction strength depends on the magnitudes of the dipoles and their relative orientation. Interactions between polar molecules are typically greater than those between nonpolar molecules of comparable size because these interactions act in addition to London dispersion forces.
      • iii. Ion-dipole forces of attraction are present between ions and polar molecules. These tend to be stronger than dipole-dipole forces.
    • 3.1.A.3 The relative strength and orientation dependence of dipole-dipole and ion-dipole forces can be understood qualitatively by considering the sign of the partial charges responsible for the molecular dipole moment, and how these partial charges interact with an ion or with an adjacent dipole.
    • 3.1.A.4 Hydrogen bonding is a strong type of intermolecular interaction that exists when hydrogen atoms covalently bonded to the highly electronegative atoms (N, O, and F) are attracted to the negative end of a dipole formed by the electronegative atom (N, O, and F) in a different molecule, or a different part of the same molecule.
    • 3.1.A.5 In large biomolecules, noncovalent interactions may occur between different molecules or between different regions of the same large biomolecule.
    日本語

    学習目標 3.1.A: 次の場合に、分子の化学構造とそれらの分子間力の相対的な強さとの関係を説明する: i. 分子が同じ化学種である場合。 ii. 分子が2種類の異なる化学種である場合。

    • 3.1.A.1 ロンドン分散力は、一時的に変動する双極子間のクーロン相互作用の結果である。ロンドン分散力は、大きな分子間の主要な分子間力となることが多い。
      • i. 分散力は、分子間の接触面積の増加と分子の分極率の増加とともに増大する。
      • ii. 分子の分極率は、分子内の電子数の増加と電子雲のサイズの増加とともに増大する。π結合の存在によって促進される。
      • iii. 「ロンドン分散力」という用語を「ファンデルワールス力」と同義語として使用すべきではない。
    • 3.1.A.2 極性分子の双極子モーメントは、他の化学種との追加の相互作用をもたらす。
      • i. 双極子-誘起双極子相互作用は、極性分子と非極性分子の間に存在する。これらの力は常に引力である。これらの力の強さは、極性分子の双極子の大きさの増大と非極性分子の分極率の増大とともに増加する。
      • ii. 双極子-双極子相互作用は、極性分子の間に存在する。相互作用の強さは双極子の大きさとそれらの相対的な配向に依存する。極性分子間の相互作用は、同程度の大きさの非極性分子間の相互作用よりも通常大きい。なぜなら、これらの相互作用はロンドン分散力に加わって働くからである。
      • iii. イオン-双極子引力は、イオンと極性分子の間に存在する。これらの力は双極子-双極子力よりも強い傾向がある。
    • 3.1.A.3 双極子-双極子力およびイオン-双極子力の相対的な強さと方向依存性は、分子の双極子モーメントを引き起こす部分電荷の符号、およびそれらがイオンや隣接する双極子とどのように相互作用するかを考慮することで定性的に理解できる。
    • 3.1.A.4 水素結合は、強い種類の分子間相互作用であり、高電気陰性度原子(N, O, F)に共有結合している水素原子が、別の分子、あるいは同じ分子内の別の部分における同様の電気陰性度原子(N, O, F)によって形成される双極子の負の端に引き寄せられるときに存在する。
    • 3.1.A.5 大きな生体分子において、非共有結合性相互作用は、異なる分子間、または同一の大きな生体分子内の異なる領域間で発生することがある。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Intermolecular forces 分子间作用力 (IMFs) are attractions between molecules – much weaker than bonds, but they set melting/boiling points. From weakest to strongest:

    • London dispersion forces 伦敦色散力: present in all molecules; they arise from Coulombic attraction between temporary, fluctuating dipoles, and are stronger for larger, more polarizable electron clouds - often the strongest net force between large molecules.
    • Dipole–dipole 偶极-偶极: between polar molecules. Their strength depends on the size of the dipoles and their relative orientation - a $\delta+$ end lining up with a neighbour's $\delta-$ end attracts, so these act in addition to dispersion and make polar molecules stickier than nonpolar ones of similar size.
    • Hydrogen bonding 氢键: a strong dipole force when H is bonded to N, O, or F.
    • Ion–dipole 离子-偶极: between an ion and a polar molecule (the ion pulls on the oppositely-charged end of the dipole). These are the strongest of the four here (stronger even than hydrogen bonds), and are exactly what lets water dissolve an ionic solid - each $\text{Na}^+$ is surrounded by the $\delta-$ oxygen ends of water molecules.

    The whole ladder is understood qualitatively by looking at the sign of the partial charges: a larger or better-aligned partial charge, or a full ionic charge, gives a stronger attraction. Stronger IMFs mean higher boiling points and lower vapor pressure. This is why water ($18\ \text{g/mol}$, hydrogen-bonded) boils at $100\,{}^{\circ}\text{C}$ while methane ($16\ \text{g/mol}$, dispersion only) boils at $-162\,{}^{\circ}\text{C}$.

    日本語

    Intermolecular forces 分子间作用力 (IMFs) are attractions between molecules – much weaker than bonds, but they set melting/boiling points. From weakest to strongest:

    London dispersion: an instantaneous dipole induces a dipole in a neighbour
    London dispersion: an instantaneous dipole induces a dipole in a neighbour
    Hydrogen bonding: an H on N/O/F is attracted to a lone pair on another molecule
    Hydrogen bonding: an H on N/O/F is attracted to a lone pair on another molecule
    • London dispersion forces 伦敦色散力: present in all molecules; they arise from Coulombic attraction between temporary, fluctuating dipoles, and are stronger for larger, more polarizable electron clouds - often the strongest net force between large molecules.
    • Dipole–dipole 偶极-偶极: between polar molecules. Their strength depends on the size of the dipoles and their relative orientation - a $\delta+$ end lining up with a neighbour's $\delta-$ end attracts, so these act in addition to dispersion and make polar molecules stickier than nonpolar ones of similar size.
    • Hydrogen bonding 氢键: a strong dipole force when H is bonded to N, O, or F.
    • Ion–dipole 离子-偶极: between an ion and a polar molecule (the ion pulls on the oppositely-charged end of the dipole). These are the strongest of the four here (stronger even than hydrogen bonds), and are exactly what lets water dissolve an ionic solid - each $\text{Na}^+$ is surrounded by the $\delta-$ oxygen ends of water molecules.

    The whole ladder is understood qualitatively by looking at the sign of the partial charges: a larger or better-aligned partial charge, or a full ionic charge, gives a stronger attraction. Stronger IMFs mean higher boiling points and lower vapor pressure. This is why water ($18\ \text{g/mol}$, hydrogen-bonded) boils at $100\,{}^{\circ}\text{C}$ while methane ($16\ \text{g/mol}$, dispersion only) boils at $-162\,{}^{\circ}\text{C}$.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Intermolecular forces/ˌɪntəməˈlekjʊlə ˈfɔːsɪz/ 分子間力
    London dispersion forces/ˈlʌndn dɪˈspɜːʃn ˈfɔːsɪz/ ロンダン分散力
    Dipole–dipole/ˈdaɪpəʊl ˈdaɪpəʊl/ 双極子-双極子
    Ion–dipole/ˈaɪɒn ˈdaɪpəʊl/ イオン-双極子
    Hydrogen bonding/ˈhaɪdrədʒn ˈbɒndɪŋ/ 水素結合
    3.2

    Properties of Solids

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.2.A: Explain the relationship among the macroscopic properties of a substance, the particulate-level structure of the substance, and the interactions between these particles.

    • 3.2.A.1 Many properties of liquids and solids are determined by the strengths and types of intermolecular forces present. Because intermolecular interactions are overcome completely when a substance vaporizes, the vapor pressure and boiling point are directly related to the strength of those interactions. Melting points also tend to correlate with interaction strength, but because the interactions are only rearranged, in melting, the relations can be more subtle.
    • 3.2.A.2 Particulate-level representations, showing multiple interacting chemical species, are a useful means to communicate or understand how intermolecular interactions help to establish macroscopic properties.
    • 3.2.A.3 Due to strong interactions between ions, ionic solids tend to have low vapor pressures, high melting points, and high boiling points. They tend to be brittle due to the repulsion of like charges caused when one layer slides across another layer. They conduct electricity only when the ions are mobile, as when the ionic solid is melted (i.e., in a molten state) or dissolved in water or another solvent.
    • 3.2.A.4 In covalent network solids, the atoms are covalently bonded together into a three-dimensional network (e.g., diamond) or layers of two-dimensional networks (e.g., graphite). These are only formed from nonmetals and metalloids: elemental (e.g., diamond, graphite) or binary compounds (e.g., silicon dioxide and silicon carbide). Due to the strong covalent interactions, covalent solids have high melting points. Three-dimensional network solids are also rigid and hard, because the covalent bond angles are fixed. However, graphite is soft because adjacent layers can slide past each other relatively easily.
    • 3.2.A.5 Molecular solids are composed of distinct, individual units of covalently-bonded molecules attracted to each other through relatively weak intermolecular forces. Molecular solids generally have a low melting point because of the relatively weak intermolecular forces present between the molecules. They do not conduct electricity because their valence electrons are tightly held within the covalent bonds and the lone pairs of each constituent molecule. Molecular solids are sometimes composed of very large molecules or polymers.
    • 3.2.A.6 Metallic solids are good conductors of electricity and heat, due to the presence of free valence electrons. They also tend to be malleable and ductile, due to the ease with which the metal cores can rearrange their structure. In an interstitial alloy, interstitial atoms tend to make the lattice more rigid, decreasing malleability and ductility. Alloys typically retain a sea of mobile electrons and so remain conducting.
    • 3.2.A.7 In large biomolecules or polymers, noncovalent interactions may occur between different molecules or between different regions of the same large biomolecule. The functionality and properties of such molecules depend strongly on the shape of the molecule, which is largely dictated by noncovalent interactions.
    日本語

    学習目標 3.2.A: 物質の巨視的性質、物質の微視的構造、およびこれらの粒子間の相互作用の関係について説明する。

    • 3.2.A.1 液体および固体の多くの性質は、存在する分子間力の強さと種類によって決定される。物質が気化する際、分子間相互作用は完全に打ち破られるため、蒸気圧および沸点はこれらの相互作用の強さに直接関係している。融点も相互作用の強さと相関する傾向があるが、相互作用が単に再配置されるに過ぎないため、融解においては関係性がより複雑であることがある。
    • 3.2.A.2 複数の相互作用する化学種を示す微視的表現は、分子間相互作用が巨視的性質の確立にどのように寄与するかを伝達したり理解したりするための有用な手段である。
    • 3.2.A.3 イオン間の強い相互作用のため、イオン結晶は蒸気圧が低く、融点および沸点が高い傾向がある。一つの層が他の層の上を滑る際に生じる同種電荷の反発により、脆い傾向がある。イオンが移動可能である場合、つまりイオン結晶が溶融(溶融状態)しているとき、または水或其他溶媒に溶解している場合にのみ、電気伝導を行う。
    • 3.2.A.4 共価網目結晶では、原子は三次元網目(例:ダイヤモンド)または二次元網目の層(例:グラファイト)として共有結合されている。これらは非金属および半金属のみから形成される:元素(例:ダイヤモンド、グラファイト)または二元化合物(例:二酸化ケイ素および炭化ケイ素)。強い共有結合のため、共価結晶は高い融点を持つ。共価結合角が固定されているため、三次元網目結晶は剛直で硬い。しかし、グラファイトは隣接する層が互いに比較的容易に滑り合うため柔らかい。
    • 3.2.A.5 分子固体は、共有結合によって結ばれた個々の分子からなる明確な単体であり、それらは比較的弱い分子間力によって互いに引き合っています。分子固体は、分子間に存在する比較的弱い分子間力のために、一般的に低い融点を持ちます。価電子が共有結合および各構成分子の非共有電子対に強く保持されているため、電気伝導性を示しません。分子固体は、sometimes非常に大きな分子やポリマーで構成されることもあります。
    • 3.2.A.6 金属固体は、自由な価電子が存在するため、電気および熱の良好な伝導体です。また、金属核が構造を容易に再配置できるため、展性および延性に富む傾向があります。间隙合金において、间隙原子は格子をより剛性化させ、展性と延性を低下させる傾向があります。合金は通常、移動可能な電子の海を維持するため、導電性を保ち続けます。
    • 3.2.A.7 巨大生体分子やポリマーでは、異なる分子間、あるいは同一の巨大生体分子内の異なる領域間で、非共有結合相互作用が生じることがあります。このような分子の機能と性質は、分子の形状に強く依存しており、その形状は主に非共有結合相互作用によって決定されます。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    A solid's properties reflect the particles and forces holding it: ionic, covalent-network (a 3D network like diamond, very hard and high-melting; graphite is a layered exception — high-melting but soft, because its 2D layers slide over one another), metallic, and molecular solids (held by weak IMFs, soft, low-melting). Matching a solid's properties to its structure is a common exam task.

    Metallic solids conduct electricity and heat and are malleable 可锻的 and ductile 可延展的, all because their free valence electrons 自由价电子 move easily and let the metal ions slide past one another without breaking the bonding. Solids are also either crystalline 晶体 (particles in a regular, repeating 3-D arrangement) or amorphous 非晶体 (no long-range order, like glass).

    日本語

    A solid's properties reflect the particles and forces holding it: ionic, covalent-network (a 3D network like diamond, very hard and high-melting; graphite is a layered exception — high-melting but soft, because its 2D layers slide over one another), metallic, and molecular solids (held by weak IMFs, soft, low-melting). Matching a solid's properties to its structure is a common exam task.

    Metallic solids conduct electricity and heat and are malleable 可锻的 and ductile 可延展的, all because their free valence electrons 自由价电子 move easily and let the metal ions slide past one another without breaking the bonding. Solids are also either crystalline 晶体 (particles in a regular, repeating 3-D arrangement) or amorphous 非晶体 (no long-range order, like glass).

    The four solid structures: giant ionic, simple molecular, giant covalent, and metallic
    The four solid structures: giant ionic, simple molecular, giant covalent, and metallic
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    malleable/ˈmæləbl/ 展性がある
    ductile/ˈdʌktaɪl/ 延性がある
    free valence electrons/friː ˈveɪləns ɪˈlektrɒnz/ 自由電子
    crystalline/ˈkrɪstəlaɪn/ 結晶性
    amorphous/əˈmɔːfəs/ 非晶質
    3.3

    Solids, Liquids, and Gases

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.3.A: Represent the differences between solid, liquid, and gas phases using a particulate-level model.

    • 3.3.A.1 Solids can be crystalline, where the particles are arranged in a regular three-dimensional structure, or they can be amorphous, where the particles do not have a regular, orderly arrangement. In both cases, the motion of the individual particles is limited, and the particles do not undergo overall translation with respect to each other. The structure of the solid is influenced by interparticle interactions and the ability of the particles to pack together.
    • 3.3.A.2 The constituent particles in liquids are in close contact with each other, and they are continually moving and colliding. The arrangement and movement of particles are influenced by the nature and strength of the forces (e.g., polarity, hydrogen bonding, and temperature) between the particles.
    • 3.3.A.3 The solid and liquid phases for a particular substance typically have similar molar volume because, in both phases, the constituent particles are in close contact at all times.
    • 3.3.A.4 In the gas phase, the particles are in constant motion. Their frequencies of collision and the average spacing between them are dependent on temperature, pressure, and volume. Because of this constant motion, and minimal effects of forces between particles, a gas has neither a definite volume nor a definite shape.
      • Exclusion Statement: Understanding/interpreting phase diagrams will not be assessed on the AP Exam.
    日本語

    学習目標 3.3.A: 粒子レベルモデルを用いて、固体、液体、気体の相間の違いを表すこと。

    • 3.3.A.1 固体は、粒子が規則的な三次元構造で配列している結晶状であるか、または粒子が規則的で整然とした配列を持たない無定形状であるかのいずれかです。どちらの場合でも、個々の粒子の運動は制限されており、粒子同士に対して全体的な並進運動を行うことはありません。固体の構造は、粒子間の相互作用および粒子が詰まり合う能力に影響を受けます。
    • 3.3.A.2 液体を構成する粒子は互いに密着しており、絶えず運動し衝突しています。粒子の配列および運動は、粒子間の力の性質と強さ(例:極性、水素結合、温度)によって影響を受けます。
    • 3.3.A.3 特定の物質の固体および液体相は、通常、類似したモル体積を持ちます。なぜなら、両方の相において構成粒子が常に密着しているからです。
    • 3.3.A.4 気体相では、粒子は絶えず運動しています。衝突頻度および平均的な粒子間の距離は、温度、圧力、体積に依存します。この絶えずの運動および粒子間の力の影響が極めて小さいため、気体は一定の体積も一定の形状も持ちません。
      • 除外事項: AP試験では、相図の理解・解釈は評価対象外です。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    The three states differ in how tightly particles are held. Rising temperature raises average kinetic energy; when it overcomes the attractions, the substance melts or boils. Gases are mostly empty space, so they are compressible and fill their container.

    Particles are packed in a solid, close but mobile in a liquid, and far apart in a gas
    Particles are packed in a solid, close but mobile in a liquid, and far apart in a gas
    Steam rising from boiling water: gases expand to fill space and obey the ideal gas law at high T and low P
    Steam rising from boiling water: gases expand to fill space and obey the ideal gas law at high T and low P
    Explore · ⁨探索⁩

    Melt and boil by adding heat · ⁨熱を加えて融解・沸騰させる⁩

    Temperature sets the average kinetic energy of the particles. Warm a solid and the particles break their fixed pattern (melt), then spread right out (boil). · ⁨温度は粒子の平均 運動エネルギー を決めます。固体を温めると粒子は固定された配列を崩して(融解)、さらに広がりながら(沸騰)動きます。⁩

    3.4

    The Ideal Gas Law

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.4.A: Explain the relationship between the macroscopic properties of a sample of gas or mixture of gases using the ideal gas law.

    • 3.4.A.1 The macroscopic properties of ideal gases are related through the ideal gas law:
      • EQN: $PV = nRT$.
    • 3.4.A.2 In a sample containing a mixture of ideal gases, the pressure exerted by each component (the partial pressure) is independent of the other components. Therefore, the partial pressure of a gas within the mixture is proportional to its mole fraction ($X$), and the total pressure of the sample is the sum of the partial pressures.
      • EQN: $P_{A} = P_{total} \times X_{A}$, where $X_{A} =$ moles A/total moles;
      • EQN: $P_{total} = P_{A} + P_{B} + P_{C} + \ldots$
    • 3.4.A.3 Graphical representations of the relationships between $P$, $V$, $T$, and $n$ are useful to describe gas behavior.
    日本語

    学習目標 3.4.A: 理想気体法則を用いて、気体または混合気体の試料の巨視的性質間の関係について説明すること。

    • 3.4.A.1 理想気体の巨視的性質は、理想気体法則によって関連付けられます:
      • EQN: $PV = nRT$.
    • 3.4.A.2 理想気体の混合を含む試料において、各成分が及ぼす圧力(分圧)は、他の成分とは独立しています。したがって、混合物中の気体の分圧はそのモル分率($X$)に比例し、試料の総圧は分圧の和です。
      • 式: $P_{A} = P_{total} \times X_{A}$, ここで $X_{A} =$ は A のモル数/総モル数;
      • 式: $P_{total} = P_{A} + P_{B} + P_{C} + \ldots$
    • 3.4.A.3 $P$、$V$、$T$、および $n$ の間の関係のグラフ表現は、気体の挙動を説明する際に有用です。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    An ideal gas obeys

    $$PV=nRT,$$
    linking pressure, volume, moles, and absolute temperature. Use it to find any one quantity from the others, or (holding some constant) to predict how a gas responds to a change.

    An ideal gas is a model of point particles with no forces between them
    An ideal gas is a model of point particles with no forces between them

    Worked example. How many moles of gas fill a $2.0\ \text{L}$ container at $300\ \text{K}$ and $1.5\ \text{atm}$? Using $R=0.0821\ \text{L atm/(mol K)}$,

    $$n=\frac{PV}{RT}=\frac{1.5\times2.0}{0.0821\times300}=0.12\ \text{mol}.$$
    Always use kelvin for $T$ and match the units of $R$ to your pressure and volume.

    Explore · ⁨探索⁩

    Compress a gas and watch the pressure · ⁨ガスを圧縮して圧力の変化を見る⁩

    $PV = nRT$. At fixed temperature, squeezing the gas into a smaller volume packs the molecules closer, so they hit the walls more often and the pressure rises. · ⁨$PV = nRT$。温度が一定の場合、ガスをより小さな 体積 に圧縮すると分子間隔が狭まり、容器壁への衝突頻度が高まるため、圧力 が上昇します。⁩

    3.5

    Kinetic Molecular Theory

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.5.A: Explain the relationship between the motion of particles and the macroscopic properties of gases with: i. The kinetic molecular theory (KMT). ii. A particulate model. iii. A graphical representation.

    • 3.5.A.1 The kinetic molecular theory (KMT) relates the macroscopic properties of gases to motions of the particles in the gas. The Maxwell-Boltzmann distribution describes the distribution of the kinetic energies of particles at a given temperature.
    • 3.5.A.2 All the particles in a sample of matter are in continuous, random motion. The average kinetic energy of a particle is related to its average velocity by the equation:
      • EQN: $KE = \frac{1}{2}\,mv^{2}$.
    • 3.5.A.3 The Kelvin temperature of a sample of matter is proportional to the average kinetic energy of the particles in the sample.
    • 3.5.A.4 The Maxwell-Boltzmann distribution provides a graphical representation of the energies/velocities of particles at a given temperature.
    日本語

    学習目標 3.5.A: 粒子の運動と気体の巨視的性質との関係を、次のものを用いて説明すること: i. 運動論的分子理論 (KMT)。 ii. 粒子モデル。 iii. グラフによる表現。

    • 3.5.A.1 運動論的分子理論 (KMT) は、気体の巨視的性質と気体中の粒子の運動を結びつけます。マクスウェル・ボルツマン分布は、ある温度における粒子の運動エネルギーの分布を記述します。
    • 3.5.A.2 物質の試料にあるすべての粒子は、絶えずランダムな運動をしています。粒子の平均運動エネルギーは、以下の式により平均速度と関連付けられます:
      • EQN: $KE = \frac{1}{2}\,mv^{2}$.
    • 3.5.A.3 物質の試料のケルビン温度は、試料中の粒子の平均運動エネルギーに比例します。
    • 3.5.A.4 マクスウェル・ボルツマン分布は、ある温度における粒子のエネルギー/速度のグラフによる表現を提供します。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    Kinetic theory: gas pressure

    Kinetic molecular theory 分子运动论 explains gas behavior: particles are tiny, in constant random motion, with negligible volume and no attractions, and collisions are elastic. Temperature is proportional to average kinetic energy, so at a given temperature lighter molecules move faster (Graham's law of effusion).

    日本語
    Kinetic theory: gas pressure

    Kinetic molecular theory 分子运动论 explains gas behavior: particles are tiny, in constant random motion, with negligible volume and no attractions, and collisions are elastic. Temperature is proportional to average kinetic energy, so at a given temperature lighter molecules move faster (Graham's law of effusion).

    The Maxwell-Boltzmann distribution of molecular speeds shifts right when heated
    The Maxwell-Boltzmann distribution of molecular speeds shifts right when heated
    Explore · ⁨探索⁩

    Heat a gas and watch the speed spread · ⁨ガスを加熱して速度分布の変化を見る⁩

    Gas molecules have a range of speeds. Raising the temperature shifts the whole distribution to higher speeds and flattens it, so more molecules move fast. · ⁨ガス分子には 速度の範囲 があります。温度 を上げると、速度分布全体が高速側へシフトし平坦化するため、速く動く分子の割合が増加します。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Kinetic molecular theory/kɪˈnetɪk məˈlekjʊlə ˈθɪəri/ 運動論的分子説
    3.6

    Deviation from the Ideal Gas Law

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.6.A: Explain the relationship among non-ideal behaviors of gases, interparticle forces, and/or volumes.

    • 3.6.A.1 The ideal gas law does not explain the actual behavior of real gases. Deviations from the ideal gas law may result from interparticle attractions among gas molecules, particularly at conditions that are close to those resulting in condensation. Deviations may also arise from particle volumes, particularly at extremely high pressures.
    日本語

    学習目標 3.6.A: 気体の非理想的な振る舞い、粒子間力、および/または体積との関係について説明すること。

    • 3.6.A.1 理想気体法則は、実在気体の実際の振る舞いを説明できません。理想気体法則からの逸脱は、気体分子間の粒子間引力によって生じる可能性があります。特に凝縮に至る条件に近い場合です。また、粒子の体積によって逸脱が生じる可能性もあります。特に極めて高い圧力の場合です。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    Real gases deviate from ideal behavior at high pressure and low temperature, where molecules are close enough that their real volume and their attractions matter. Attractions lower the pressure below ideal; molecular volume raises it.

    3.7

    Solutions and Mixtures

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.7.A: Calculate the number of solute particles, volume, or molarity of solutions.

    • 3.7.A.1 Solutions, also sometimes called homogeneous mixtures, can be solids, liquids, or gases. In a solution, the macroscopic properties do not vary throughout the sample. In a heterogeneous mixture, the macroscopic properties depend on location in the mixture.
    • 3.7.A.2 Solution composition can be expressed in a variety of ways; molarity is the most common method used in the laboratory.
      • EQN: $M = n_{solute}/L_{solution}$
    日本語

    学習目標 3.7.A: 溶質粒子の数、体積、またはモル濃度を計算すること。

    • 3.7.A.1 溶液は、均一混合物とも呼ばれることがあります。固体、液体、気体のいずれかです。溶液中では、試料全体を通じて巨視的性質は変化しません。不均一混合物では、巨視的性質は混合物内での位置に依存します。
    • 3.7.A.2 溶液の組成は多様な方法で表すことができ、モル濃度は実験室で最も一般的に用いられる方法である。
      • 式: $M = n_{solute}/L_{solution}$

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    A solution 溶液 is a homogeneous mixture of a solute 溶质 dissolved in a solvent 溶剂. Concentration is usually molarity 摩尔浓度:

    $$M=\frac{\text{moles of solute}}{\text{liters of solution}}.$$
    Dilution conserves moles: $M_1V_1=M_2V_2$.

    Worked example. What volume of water must you add to $50\ \text{mL}$ of $6.0\ \text{M}$ HCl to make it $2.0\ \text{M}$? The moles of HCl are unchanged, so $M_1V_1=M_2V_2$ gives the final volume $V_2=\dfrac{M_1V_1}{M_2}=\dfrac{6.0\times50}{2.0}=150\ \text{mL}$. You therefore add $150-50=100\ \text{mL}$ of water.

    日本語

    A solution 溶液 is a homogeneous mixture of a solute 溶质 dissolved in a solvent 溶剂. Concentration is usually molarity 摩尔浓度:

    $$M=\frac{\text{moles of solute}}{\text{liters of solution}}.$$
    Dilution conserves moles: $M_1V_1=M_2V_2$.

    Worked example. What volume of water must you add to $50\ \text{mL}$ of $6.0\ \text{M}$ HCl to make it $2.0\ \text{M}$? The moles of HCl are unchanged, so $M_1V_1=M_2V_2$ gives the final volume $V_2=\dfrac{M_1V_1}{M_2}=\dfrac{6.0\times50}{2.0}=150\ \text{mL}$. You therefore add $150-50=100\ \text{mL}$ of water.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    solution/səˈluːʃn/ 溶液
    solute/ˈsɒljuːt/ 溶質
    solvent/ˈsɒlvənt/ 溶媒
    molarity/məʊˈlærɪti/ モル浓度
    3.8

    Representations of Solutions

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.8.A: Using particulate models for mixtures: i. Represent interactions between components. ii. Represent concentrations of components.

    • 3.8.A.1 Particulate representations of solutions communicate the structure and properties of solutions, by illustration of the relative concentrations of the components in the solution and/or drawings that show interactions among the components.
      • Exclusion Statement: Colligative properties will not be assessed on the AP Exam.
      • Exclusion Statement: Calculations of molality, percent by mass, and percent by volume for solutions will not be assessed on the AP Exam.
    日本語

    学習目標 3.8.A: 混合体の粒子モデルを用いて: i. 成分間の相互作用を表す。 ii. 成分の濃度を表す。

    • 3.8.A.1 溶液の粒子による表現は、溶液内の成分の相対的な濃度を示す図や、成分間の相互作用を示す描画を通じて、溶液の構造と性質を伝える。
      • 除外事項: AP試験では、集束性物性(コリゲイティブ・プロパティ)は評価されない。
      • 除外事項: 溶液のモル濃度、質量百分率、体積百分率の計算はAP試験では評価されない。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    A particulate diagram shows the solute and solvent particles. For an ionic solute, show it fully dissociated into separate ions surrounded by solvent; count particles to reason about concentration and conductivity.

    3.9

    Separation of Solutions and Mixtures

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.9.A: Explain the results of a separation experiment based on intermolecular interactions.

    • 3.9.A.1 The components of a liquid solution cannot be separated by filtration. They can, however, be separated using processes that take advantage of differences in the intermolecular interactions of the components.
      • i. Chromatography (paper, thin-layer, and column) separates chemical species by taking advantage of the differential strength of intermolecular interactions between and among the components of the solution (the mobile phase) and with the surface components of the stationary phase. The resulting chromatogram can be used to infer the relative polarities of components in a mixture.
      • ii. Distillation separates chemical species by taking advantage of the differential strength of intermolecular interactions between and among the components and the effects these interactions have on the vapor pressures of the components in the mixture.
    日本語

    学習目標 3.9.A: 分子間相互作用に基づき、分離実験の結果を説明する。

    • 3.9.A.1 液体溶液の成分はろ過では分離できない。しかし、成分の分子間相互作用の違いを利用したプロセスを用いることで分離できる。
      • i. クロマトグラフィー(紙、薄層、カラム)は、溶液(移動相)の成分間および成分と固定相の表面成分との間の分子間相互作用の強さの違いを利用して化学種を分離する。得られたクロマトグラムは、混合物中の成分の相対的な極性を推定するために使用できる。
      • ii. 蒸留は、成分間の分子間相互作用の強さの違いと、これらの相互作用が混合物中の成分の蒸気圧に与える影響を利用して化学種を分離する。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Because a mixture's components keep their properties, physical methods separate them: filtration (by particle size), distillation (by boiling point), and chromatography 色谱法 (by how strongly each component sticks to a stationary phase versus moving with a solvent).

    日本語

    Because a mixture's components keep their properties, physical methods separate them: filtration (by particle size), distillation (by boiling point), and chromatography 色谱法 (by how strongly each component sticks to a stationary phase versus moving with a solvent).

    Paper chromatography separates a mixture as the solvent rises up the paper
    Paper chromatography separates a mixture as the solvent rises up the paper
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    chromatography/krəʊməˈtɒɡrəfi/ クロマトグラフィー
    3.10

    Solubility

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.10.A: Explain the relationship between the solubility of ionic and molecular compounds in aqueous and nonaqueous solvents, and the intermolecular interactions between particles.

    • 3.10.A.1 Substances with similar intermolecular interactions tend to be miscible or soluble in one another.
    日本語

    学習目標 3.10.A: 水溶性および非水溶性溶媒中でのイオン化合物および分子化合物の溶解性と、粒子間の分子間相互作用との関係を説明する。

    • 3.10.A.1 類似した分子間相互作用を持つ物質は、互いに混和したり溶解したりしやすい傾向がある。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Solubility 溶解度 is how much solute dissolves. "Like dissolves like": polar (and ionic) solutes dissolve in polar solvents; nonpolar in nonpolar. Dissolving happens when solute–solvent attractions are comparable to the attractions being broken.

    日本語

    Solubility 溶解度 is how much solute dissolves. "Like dissolves like": polar (and ionic) solutes dissolve in polar solvents; nonpolar in nonpolar. Dissolving happens when solute–solvent attractions are comparable to the attractions being broken.

    A cluster of large, glassy, bright blue crystals of copper sulfate
    Blue copper(II) sulfate crystals grown from solution: a saturated solution left to evaporate deposits its dissolved solid back out as regular crystals
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Solubility/ˌsɒljuːˈbɪlɪti/ 溶解度
    3.11

    Spectroscopy and the Electromagnetic Spectrum

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.11.A: Explain the relationship between a region of the electromagnetic spectrum and the types of molecular or electronic transitions associated with that region.

    • 3.11.A.1 Differences in absorption or emission of photons in different spectral regions are related to the different types of molecular motion or electronic transition:
      • i. Microwave radiation is associated with transitions in molecular rotational levels.
      • ii. Infrared radiation is associated with transitions in molecular vibrational levels.
      • iii. Ultraviolet/visible radiation is associated with transitions in electronic energy levels.
    日本語

    学習目標 3.11.A: 電磁スペクトルの各領域と、その領域に関連する分子遷移や電子遷移の種類との関係を説明する。

    • 3.11.A.1 異なるスペクトル領域における光子の吸収や放出の違いは、異なる種類の分子運動や電子遷移に関係している:
      • i. マイクロ波放射は、分子回転準位の遷移に関連している。
      • ii. 赤外線放射は、分子振動準位の遷移に関連している。
      • iii. 紫外線/可視光放射は、電子エネルギー準位の遷移に関連している。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Spectroscopy 光谱学 studies how matter absorbs or emits light. Different regions of the electromagnetic spectrum probe different changes: microwaves (rotation), infrared (bond vibrations), ultraviolet–visible (electron transitions). The light absorbed reveals structure.

    日本語

    Spectroscopy 光谱学 studies how matter absorbs or emits light. Different regions of the electromagnetic spectrum probe different changes: microwaves (rotation), infrared (bond vibrations), ultraviolet–visible (electron transitions). The light absorbed reveals structure.

    Explore · ⁨探索⁩

    Scan across the electromagnetic spectrum · ⁨電磁波スペクトルをスキャンする⁩

    Light is a wave with a range of wavelengths. Shorter wavelength means higher frequency and more energy per photon, from radio waves up to gamma rays. · ⁨光は 波長 の異なる波です。波長が短いほど 周波数 が高くなり、光子あたりのエネルギーが大きくなります。ラジオ波からガンマ線まで続きます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Spectroscopy/spekˈtrɒskəpi/ 分光法
    3.12

    Properties of Photons

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.12.A: Explain the properties of an absorbed or emitted photon in relationship to an electronic transition in an atom or molecule.

    • 3.12.A.1 When a photon is absorbed (or emitted) by an atom or molecule, the energy of the species is increased (or decreased) by an amount equal to the energy of the photon.

    • 3.12.A.2 The wavelength of the electromagnetic wave is related to its frequency and the speed of light by the equation:

      • EQN: $c = \lambda\nu$.

      The energy of a photon is related to the frequency of the electromagnetic wave through Planck's equation:

      • EQN: $E = h\nu$.
    日本語

    学習目標 3.12.A: 原子や分子の電子遷移に対する、吸収または放出された光子の性質を説明する。

    • 3.12.A.1 光子が原子や分子によって吸収(または放出)されると、その種のエネルギーは光子のエネルギーに等しい量だけ増加(または減少)する。

    • 3.12.A.2 電磁波の波長は、周波数および光速と以下の式で関係している:

      • EQN: $c = \lambda\nu$.

      光子のエネルギーは、プランクの式を通じて電磁波の周波数と関係している:

      • EQN: $E = h\nu$.

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Light is carried by photons 光子, each with energy $E=h\nu=\dfrac{hc}{\lambda}$. Higher frequency (shorter wavelength) means higher energy. A molecule absorbs a photon only when its energy matches an allowed energy gap.

    日本語

    Light is carried by photons 光子, each with energy $E=h\nu=\dfrac{hc}{\lambda}$. Higher frequency (shorter wavelength) means higher energy. A molecule absorbs a photon only when its energy matches an allowed energy gap.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    photons/ˈfəʊtɒnz/ 光子
    3.13

    The Beer-Lambert Law

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 3.13.A: Explain the amount of light absorbed by a solution of molecules or ions in relationship to the concentration, path length, and molar absorptivity.

    • 3.13.A.1 The Beer-Lambert law relates the absorption of light by a solution to three variables according to the equation:

      • EQN: $A = \varepsilon bc$.

      The molar absorptivity, $\varepsilon$, describes how intensely a chemical species absorbs light of a specific wavelength. The path length, $b$, and concentration, $c$, are proportional to the number of light-absorbing particles in the light path.

    • 3.13.A.2 In most experiments the path length and wavelength of light are held constant. In such cases, the absorbance is proportional only to the concentration of absorbing molecules or ions. The spectrophotometer is typically set to the wavelength of maximum absorbance (optimum wavelength) for the species being analyzed to ensure the maximum sensitivity of measurement.

    日本語

    学習目標 3.13.A: 分子やイオンの溶液が吸収する光の量が、濃度、光路長さ、およびモル吸光度とどのように関係しているかを説明する。

    • 3.13.A.1 ベール・ランベルト則は、以下の式により、溶液による光の吸収と3つの変数を関連づける:

      • EQN: $A = \varepsilon bc$.

      モル吸光度 $\varepsilon$ は、化学種が特定の波長の光をどれだけ強く吸収するかを示す。光路長さ $b$ と濃度 $c$ は、光路中の光吸収粒子の数に比例する。

    • 3.13.A.2 ほとんどの実験では、光路長と光の波長は一定に保たれる。このような場合、吸光度は吸収する分子やイオンの濃度にのみ比例する。分光光度計は通常、分析対象の種に対して最大吸光度(最適波長)になるように設定され、測定感度を最大化する。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    The Beer–Lambert law 比尔-朗伯定律 relates how much light a solution absorbs to its concentration:

    $$A=\varepsilon\,b\,c,$$
    where $A$ is absorbance, $\varepsilon$ the molar absorptivity, $b$ the path length, and $c$ the concentration. Since $A$ is proportional to $c$, measuring absorbance is a fast way to find an unknown concentration.

    Worked example. A dye has molar absorptivity $\varepsilon=2000\ \text{L/(mol cm)}$; in a $1.0\ \text{cm}$ cell a sample reads absorbance $A=0.40$. Its concentration is $c=\dfrac{A}{\varepsilon b}=\dfrac{0.40}{2000\times1.0}=2.0\times10^{-4}\ \text{M}$. Because $A\propto c$, a solution twice as concentrated would read $A=0.80$ – the basis of a calibration curve.

    日本語

    The Beer–Lambert law 比尔-朗伯定律 relates how much light a solution absorbs to its concentration:

    $$A=\varepsilon\,b\,c,$$
    where $A$ is absorbance, $\varepsilon$ the molar absorptivity, $b$ the path length, and $c$ the concentration. Since $A$ is proportional to $c$, measuring absorbance is a fast way to find an unknown concentration.

    Worked example. A dye has molar absorptivity $\varepsilon=2000\ \text{L/(mol cm)}$; in a $1.0\ \text{cm}$ cell a sample reads absorbance $A=0.40$. Its concentration is $c=\dfrac{A}{\varepsilon b}=\dfrac{0.40}{2000\times1.0}=2.0\times10^{-4}\ \text{M}$. Because $A\propto c$, a solution twice as concentrated would read $A=0.80$ – the basis of a calibration curve.

    Explore · ⁨探索⁩

    Link absorbance to concentration · ⁨吸光度と濃度の関係を示す⁩

    The Beer-Lambert law says absorbance $A = \varepsilon b c$: absorbance is proportional to concentration, so a calibration line lets you read an unknown concentration. · ⁨ビール-ランベルトの法則 によると、吸光度 $A = \varepsilon b c$ :吸光度は濃度に 比例 するため、検量線を用いて未知の濃度を決定できます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Beer–Lambert law/bɪə ˈlæmbət lɔː/ Beer-Lambertの法則
    3.13

    Exam tips

    • Intermolecular forces (dispersion < dipole–dipole < hydrogen bonding) set boiling points — they are much weaker than the bonds inside a molecule.
    • Boiling breaks the forces between molecules, not the covalent bonds within them.
    • Use $PV=nRT$ with temperature in kelvin and $R$ matched to your pressure/volume units.
    • For solutions use molarity $M=\text{mol}/\text{L}$; dilution conserves moles, so $M_1V_1=M_2V_2$.
    • "Like dissolves like" — polar/ionic solutes dissolve in polar solvents, non-polar in non-polar.
  • 4

    Chemical Reactions · ⁨化学反応⁩

    Watch lesson · ⁨レッスンを視聴⁩
    4.1

    Recognizing a Chemical Reaction

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 4.1.A: Identify evidence of chemical and physical changes in matter.

    • 4.1.A.1 A physical change occurs when a substance undergoes a change in properties but not a change in composition. Changes in the phase of a substance (solid, liquid, gas) or formation/separation of mixtures of substances are common physical changes.
    • 4.1.A.2 A chemical change occurs when substances are transformed into new substances, typically with different compositions. Production of heat or light, formation of a gas, formation of a precipitate, and/or color change provide possible evidence that a chemical change has occurred.
    日本語

    学習目標 4.1.A: 物質における化学変化および物理変化の証拠を特定する。

    • 4.1.A.1 物理変化とは、物質が性質の変化を経るが組成の変化は伴わない現象である。物質の相(固体、液体、気体)の変化や、物質の混合体の形成・分離は一般的な物理変化である。
    • 4.1.A.2 化学変化とは、物質が新しい物質に変換される現象であり、通常は異なる組成を伴う。熱または光の放出、気体の生成、沈殿の生成、および/または色の変化は、化学変化が起こった可能性のある証拠となる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    A chemical reaction 化学反应 rearranges atoms into new substances. Signs one has happened: a color change, a gas or precipitate 沉淀 forming, or a temperature change. Atoms are conserved, so an equation must be balanced – the same count of each element on both sides.

    日本語
    Magnesium burning: a chemical reaction rearranges atoms and releases energy as light and heat
    Magnesium burning: a chemical reaction rearranges atoms and releases energy as light and heat

    A chemical reaction 化学反应 rearranges atoms into new substances. Signs one has happened: a color change, a gas or precipitate 沉淀 forming, or a temperature change. Atoms are conserved, so an equation must be balanced – the same count of each element on both sides.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    chemical reaction/ˈkemɪkl rɪˈækʃn/ 化学反応
    precipitate/prɪˈsɪpɪteɪt/ 沈殿物
    4.2

    Net Ionic Equations

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 4.2.A: Represent changes in matter with a balanced chemical or net ionic equation: i. For physical changes. ii. For given information about the identity of the reactants and/or product. iii. For ions in a given chemical reaction.

    • 4.2.A.1 All physical and chemical processes can be represented symbolically by balanced equations.
    • 4.2.A.2 Chemical equations represent chemical changes. These changes are the result of a rearrangement of atoms into new combinations; thus, any representation of a chemical change must contain equal numbers of atoms of every element before and after the change occurred. Equations thus demonstrate that mass and charge are conserved in chemical reactions.
    • 4.2.A.3 Balanced molecular, complete ionic, and net ionic equations are differing symbolic forms used to represent a chemical reaction. The form used to represent the reaction depends on the context in which it is to be used.
    日本語

    学習目標 4.2.A: 平衡化学方程式または正味イオン方程式を用いて物質の変化を表す: i. 物理変化の場合。 ii. 反応物および/または生成物の正体に関する情報がある場合。 iii. 特定の化学反応におけるイオンの場合。

    • 4.2.A.1 すべての物理変化および化学過程は、平衡方程式によって記号的に表すことができる。
    • 4.2.A.2 化学方程式は化学変化を表す。これらの変化は、原子が新しい組み合わせに再配置される結果であるため、化学変化前後においてすべての元素の原子数が等しくなければならない。したがって、方程式は質量と電荷が化学反応において保存されることを示している。
    • 4.2.A.3 平衡分子方程式、完全イオン方程式、正味イオン方程式は、化学反応を表すために用いられる異なる記号的形式である。使用する形式は、その文脈に応じて決まる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    For reactions in water, ionic compounds split into ions. A net ionic equation 净离子方程式 shows only the species that actually change, leaving out the spectator ions 旁观离子 that appear unchanged on both sides. It captures the real chemistry (e.g. $\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}(s)$).

    日本語

    For reactions in water, ionic compounds split into ions. A net ionic equation 净离子方程式 shows only the species that actually change, leaving out the spectator ions 旁观离子 that appear unchanged on both sides. It captures the real chemistry (e.g. $\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}(s)$).

    Mixing two solutions can form an insoluble precipitate
    Mixing two solutions can form an insoluble precipitate
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    net ionic equation/net aɪˈɒnɪk ɪˈkweɪʒn/ 正イオン方程式
    spectator ions/spekˈteɪtə ˈaɪɒnz/ 観測イオン
    4.3

    Three Ways to Represent a Reaction

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 4.3.A: Represent a given chemical reaction or physical process with a consistent particulate model.

    • 4.3.A.1 Balanced chemical equations in their various forms can be translated into symbolic particulate representations.
    日本語

    学習目標 4.3.A: 特定の化学反応または物理過程を、一貫性のある粒子モデルで表す。

    • 4.3.A.1 さまざまな形式の平衡化学方程式は、記号的な粒子表現に翻訳できる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    The same reaction can be shown as a symbolic equation, a particulate drawing (atoms and molecules), and a macroscopic observation (what you see). Moving between these levels – connecting the equation to the particles to the beaker – is a core skill.

    A balanced equation has the same number of each atom on both sides
    A balanced equation has the same number of each atom on both sides
    4.4

    Physical Changes versus Chemical Changes

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 4.4.A: Explain the relationship between macroscopic characteristics and bond interactions for: i. Chemical processes. ii. Physical processes.

    • 4.4.A.1 Processes that involve the breaking and/or formation of chemical bonds are typically classified as chemical processes. Processes that involve only changes in intermolecular interactions, such as phase changes, are typically classified as physical processes.
    • 4.4.A.2 Sometimes physical processes involve the breaking of chemical bonds. For example, plausible arguments could be made for the dissolution of a salt in water, as either a physical or chemical process, involves breaking of ionic bonds, and the formation of ion-dipole interactions between ions and solvent.
    日本語

    学習目標 4.4.A: 以下のmacroscopic characteristics(宏視的性質)と結合相互作用の関係について説明する: i. 化学過程。 ii. 物理過程。

    • 4.4.A.1 化学結合の切断および/または形成を伴う過程は、通常、化学過程として分類される。相変化のように分子間相互作用の変化のみを伴う過程は、通常、物理過程として分類される。
    • 4.4.A.2 時折、物理過程には化学結合の切断が含まれる。例えば、塩の水中での溶解は、物理過程でも化学過程でもあり得るが、いずれの場合もイオン結合の切断と、イオンと溶媒との間のイオン-双極子相互作用の形成を伴うと合理的に議論できる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    A physical change 物理变化 alters form but not identity (melting, dissolving); a chemical change 化学变化 makes new substances by breaking and forming bonds. Dissolving salt is physical; the salt is unchanged and recoverable.

    日本語

    A physical change 物理变化 alters form but not identity (melting, dissolving); a chemical change 化学变化 makes new substances by breaking and forming bonds. Dissolving salt is physical; the salt is unchanged and recoverable.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    physical change/ˈfɪzɪkl tʃeɪndʒ/ 物理変化
    chemical change/ˈkemɪkl tʃeɪndʒ/ 化学変化
    4.5

    Stoichiometry

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 4.5.A: Explain changes in the amounts of reactants and products based on the balanced reaction equation for a chemical process.

    • 4.5.A.1 Because atoms must be conserved during a chemical process, it is possible to calculate product amounts by using known reactant amounts, or to calculate reactant amounts given known product amounts.
    • 4.5.A.2 Coefficients of balanced chemical equations contain information regarding the proportionality of the amounts of substances involved in the reaction. These values can be used in chemical calculations involving the mole concept.
    • 4.5.A.3 Stoichiometric calculations can be combined with the ideal gas law and calculations involving molarity to quantitatively study gases and solutions.
    日本語

    学習目標 4.5.A: 化学過程の平衡反応式に基づき、反応物および生成物の量の変化を説明する。

    • 4.5.A.1 化学過程中に原子は保存されなければならないため、既知の反応物の量を用いて生成物の量を計算したり、既知の生成物の量から反応物の量を計算したりすることができる。
    • 4.5.A.2 平衡化学方程式の係数は、反応に関与する物質の量の比例関係に関する情報を含んでいる。これらの値は、モル概念に関連する化学計算に使用できる。
    • 4.5.A.3 化学量論計算は理想気体法則およびモル濃度に関連する計算と組み合わせて用いることができ、気体および溶液を定量的に研究することができる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Stoichiometry 化学计量 uses the balanced equation's mole ratios to relate amounts of reactants and products. The path is always grams → moles → (mole ratio) → moles → grams. The limiting reactant 限量反应物 runs out first and sets the maximum product (the theoretical yield 理论产量); the percent yield compares actual to theoretical.

    Worked example. How much water forms when $4.0\ \text{g}$ of hydrogen burns in excess oxygen? $2\text{H}_2+\text{O}_2\rightarrow2\text{H}_2\text{O}$. Convert to moles, cross the mole ratio ($2:2=1:1$ here), convert back:

    $$n(\text{H}_2)=\frac{4.0}{2.0}=2.0\ \text{mol}\;\Rightarrow\;n(\text{H}_2\text{O})=2.0\ \text{mol}\;\Rightarrow\;m=2.0\times18=36\ \text{g}.$$

    Worked example (limiting reactant). $10.0\ \text{g}$ of N$_2$ reacts with $5.0\ \text{g}$ of H$_2$ ($\text{N}_2+3\text{H}_2\rightarrow2\text{NH}_3$). Which runs out? Moles: $n(\text{N}_2)=10.0/28=0.36$, $n(\text{H}_2)=5.0/2=2.5$. The reaction needs $3$ H$_2$ per N$_2$; you have $2.5/0.36=7.0$, far more than $3$, so N$_2$ is limiting. It makes $2\times0.36=0.72\ \text{mol}$ of ammonia, with hydrogen left over.

    日本語

    Stoichiometry 化学计量 uses the balanced equation's mole ratios to relate amounts of reactants and products. The path is always grams → moles → (mole ratio) → moles → grams. The limiting reactant 限量反应物 runs out first and sets the maximum product (the theoretical yield 理论产量); the percent yield compares actual to theoretical.

    The limiting reactant runs out first and decides how much product forms
    The limiting reactant runs out first and decides how much product forms

    Worked example. How much water forms when $4.0\ \text{g}$ of hydrogen burns in excess oxygen? $2\text{H}_2+\text{O}_2\rightarrow2\text{H}_2\text{O}$. Convert to moles, cross the mole ratio ($2:2=1:1$ here), convert back:

    $$n(\text{H}_2)=\frac{4.0}{2.0}=2.0\ \text{mol}\;\Rightarrow\;n(\text{H}_2\text{O})=2.0\ \text{mol}\;\Rightarrow\;m=2.0\times18=36\ \text{g}.$$

    Worked example (limiting reactant). $10.0\ \text{g}$ of N$_2$ reacts with $5.0\ \text{g}$ of H$_2$ ($\text{N}_2+3\text{H}_2\rightarrow2\text{NH}_3$). Which runs out? Moles: $n(\text{N}_2)=10.0/28=0.36$, $n(\text{H}_2)=5.0/2=2.5$. The reaction needs $3$ H$_2$ per N$_2$; you have $2.5/0.36=7.0$, far more than $3$, so N$_2$ is limiting. It makes $2\times0.36=0.72\ \text{mol}$ of ammonia, with hydrogen left over.

    Explore · ⁨探索⁩

    Scale reactants and products by the mole ratio

    A balanced equation fixes the mole ratio between species. Product amount is proportional to the limiting reactant, scaled by that ratio.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Stoichiometry/ˌstəʊɪkɪˈɒmətri/ 化学量論
    limiting reactant/ˈlɪmɪtɪŋ rɪˈæktənt/ limiting reactant(限定試薬)
    theoretical yield/θɪəˈretɪkl jiːld/ 理論収量
    4.6

    Introduction to Titration

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 4.6.A: Identify the equivalence point in a titration based on the amounts of the titrant and analyte, assuming the titration reaction goes to completion.

    • 4.6.A.1 Titrations may be used to determine the amount of an analyte in solution. The titrant has a known concentration of a species that reacts specifically and quantitatively with the analyte. The equivalence point of the titration occurs when the analyte is totally consumed by the reacting species in the titrant. The equivalence point is often indicated by a change in a property (such as color) that occurs when the equivalence point is reached. This observable event is called the endpoint of the titration.
    日本語

    学習目標 4.6.A: 滴定反応が完了すると仮定して、滴定剤および被検物の量に基づき、滴定における等量点を特定する。

    • 4.6.A.1 滴定法は、溶液中の分析対象物の量を決定するために使用される。滴定剤には、分析対象物と特異的かつ定量的に反応する物質が既知の濃度で含まれている。滴定の等量点とは、滴定剤中の反応種によって分析対象物が完全に消費された点を指す。等量点は、その到達時に生じる性質の変化(例えば色の変化など)によって示されることが多い。この観測可能な現象を滴定の終了点という。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    An acid–base titration curve

    A titration 滴定 finds an unknown concentration by reacting it with a solution of known concentration until they reach the equivalence point 等当点 (stoichiometrically equal). From the known volume and concentration, use the mole ratio to find the unknown. An indicator or a pH curve signals the endpoint.

    Worked example. $25.0\ \text{mL}$ of hydrochloric acid is exactly neutralized by $30.0\ \text{mL}$ of $0.100\ \text{M}$ NaOH. Find the acid's concentration. The reaction is $1:1$, so the moles match:

    $$n(\text{NaOH})=0.0300\times0.100=3.00\times10^{-3}\ \text{mol}=n(\text{HCl}),\qquad [\text{HCl}]=\frac{3.00\times10^{-3}}{0.0250}=0.120\ \text{M}.$$

    日本語
    An acid–base titration curve

    A titration 滴定 finds an unknown concentration by reacting it with a solution of known concentration until they reach the equivalence point 等当点 (stoichiometrically equal). From the known volume and concentration, use the mole ratio to find the unknown. An indicator or a pH curve signals the endpoint.

    Titration apparatus: a burette delivers a known solution into a conical flask
    Titration apparatus: a burette delivers a known solution into a conical flask

    Worked example. $25.0\ \text{mL}$ of hydrochloric acid is exactly neutralized by $30.0\ \text{mL}$ of $0.100\ \text{M}$ NaOH. Find the acid's concentration. The reaction is $1:1$, so the moles match:

    $$n(\text{NaOH})=0.0300\times0.100=3.00\times10^{-3}\ \text{mol}=n(\text{HCl}),\qquad [\text{HCl}]=\frac{3.00\times10^{-3}}{0.0250}=0.120\ \text{M}.$$

    A titration setup: reacting volumes from a burette find the concentration of an unknown solution
    A titration setup: reacting volumes from a burette find the concentration of an unknown solution
    Explore · ⁨探索⁩

    Titrate an acid and find the equivalence point

    Adding base to acid raises the pH slowly, then sharply at the equivalence point where moles of acid and base are equal. The steep jump locates that volume.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    titration/taɪˈtreɪʃn/ 滴定
    equivalence point/ɪˈkwɪvələns pɔɪnt/ 当量点
    4.7

    Types of Chemical Reactions

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 4.7.A: Identify a reaction as acid-base, oxidation-reduction, or precipitation.

    • 4.7.A.1 Acid-base reactions involve transfer of one or more protons ($\text{H}^+$ ions) between chemical species.
    • 4.7.A.2 Oxidation-reduction (redox) reactions involve transfer of one or more electrons between chemical species, as indicated by changes in oxidation numbers of the involved species. Combustion is an important subclass of oxidation-reduction reactions, in which a species reacts with oxygen gas. In the case of hydrocarbons, carbon dioxide and water are products of complete combustion.
    • 4.7.A.3 In a redox reaction, electrons are transferred from the species that is oxidized to the species that is reduced.
      • Exclusion statement: The meaning of the terms "reducing agent" and "oxidizing agent" will not be assessed on the AP Exam.
    • 4.7.A.4 Oxidation numbers may be assigned to each of the atoms in the reactants and products; this is often an effective way to identify the oxidized and reduced species in a redox reaction.
    • 4.7.A.5 Precipitation reactions frequently involve mixing ions in aqueous solution to produce an insoluble or sparingly soluble ionic compound. All sodium, potassium, ammonium, and nitrate salts are soluble in water.
      • Exclusion statement: Rote memorization of "solubility rules" other than those implied in 4.7.A.5 will not be assessed on the AP Exam.
    日本語

    学習目標 4.7.A: 反応を酸塩基反応、酸化還元反応、または沈殿反応として識別する。

    • 4.7.A.1 酸塩基反応では、化学種間でのプロトン($\text{H}^+$イオン)の移動が関与する。
    • 4.7.A.2 酸化還元(レドックス)反応は、関与する種間の酸化数の変化によって示されるように、化学種間で電子が移動することを伴う。燃焼は酸化還元反応の重要な下位分類であり、ある種が酸素ガスと反応する。炭化水素の場合、完全燃焼の生成物は二酸化炭素と水である。
    • 4.7.A.3 レドックス反応において、電子は酸化された種から還元された種へ移動する。
      • 除外事項: 「還元剤」と「酸化剤」の用語の意味については、AP試験の評価対象外である。
    • 4.7.A.4 酸化数は、反応物および生成物の各原子に割り当てることができる。これは、レドックス反応における酸化・還元された種を特定する有効な方法となることが多い。
    • 4.7.A.5 沈殿反応は、水溶液中のイオンを混合して難溶性または微溶性のイオン化合物を生成することが多い。すべてのナトリウム、カリウム、アンモニウム、および硝酸塩は水に可溶である。
      • 除外事項: 4.7.A.5で言及されているもの以外の「溶解度の法則」の暗記については、AP試験の評価対象外である。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    Common patterns include synthesis (combining), decomposition (breaking apart), combustion (with oxygen, releasing energy), precipitation (forming an insoluble solid), acid–base (proton transfer), and redox (electron transfer). Recognizing the type helps predict the products.

    4.8

    Acid-Base Reactions

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 4.8.A: Identify species as Brønsted-Lowry acids, bases, and/or conjugate acid-base pairs, based on proton-transfer involving those species.

    • 4.8.A.1 By definition, a Brønsted-Lowry acid is a proton donor and a Brønsted-Lowry base is a proton acceptor.
    • 4.8.A.2 Only in aqueous solutions, water plays an important role in many acid-base reactions, as its molecular structure allows it to accept protons from and donate protons to dissolved species.
    • 4.8.A.3 When an acid or base ionizes in water, the conjugate acid-base pairs can be identified and their relative strengths compared.
      • Exclusion statement: Lewis acid-base concepts will not be assessed on the AP Exam. The emphasis in AP Chemistry is on reactions in aqueous solution.
    日本語

    学習目標 4.8.A: プロトン移動に関与するこれらの種に基づき、Brønsted-Lowry酸、塩基、および/または共役酸塩基対として種を識別する。

    • 4.8.A.1 定義により、Brønsted-Lowry酸はプロトンを供与する物質であり、Brønsted-Lowry塩基はプロトンを受け取る物質である。
    • 4.8.A.2 水溶液中のみ、水の分子構造が溶解した種からプロトンを受け取り、プロトンを供与できるため、多くの酸塩基反応において重要な役割を果たす。
    • 4.8.A.3 酸または塩基が水中で電離すると、共役酸塩基対を同定し、それらの相対的な強さを比較することができる。
      • 除外事項: ルイス酸塩基概念については、AP試験の評価対象外である。AP化学では水溶液内の反応に重点が置かれている。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    An acid 酸 donates a proton ($\text{H}^+$); a base 碱 accepts one (Brønsted–Lowry). They react to form water and a salt: $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$. Strong acids and bases dissociate completely; weak ones only partly.

    日本語

    An acid 酸 donates a proton ($\text{H}^+$); a base 碱 accepts one (Brønsted–Lowry). They react to form water and a salt: $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$. Strong acids and bases dissociate completely; weak ones only partly.

    Bronsted-Lowry: an acid donates a proton to a base, forming two conjugate pairs
    Bronsted-Lowry: an acid donates a proton to a base, forming two conjugate pairs
    Explore · ⁨探索⁩

    Move along the pH scale

    pH measures how acidic or basic a solution is. Each step of 1 pH is a tenfold change in hydrogen-ion concentration; 7 is neutral, below is acidic, above is basic.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    acid/ˈæsɪd/ 酸
    base/beɪs/ 底
    4.9

    Oxidation-Reduction (Redox) Reactions

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 4.9.A: Represent a balanced redox reaction equation using half-reactions.

    • 4.9.A.1 Balanced chemical equations for redox reactions can be constructed from half-reactions.
    日本語

    学習目標 4.9.A: 半反応式を用いて、バランス取れたレドックス反応方程式を表す。

    • 4.9.A.1 レドックス反応のバランス取れた化学方程式は、半反応式から作成できる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    In a redox 氧化还原 reaction, electrons transfer between species. Oxidation 氧化 is loss of electrons (oxidation number rises); reduction 还原 is gain (oxidation number falls). Track changes with oxidation numbers, and remember every oxidation is paired with a reduction – the electrons lost equal the electrons gained.

    Worked example. Find the oxidation number of manganese in permanganate, $\text{KMnO}_4$. Potassium is $+1$ and each oxygen is $-2$ (four of them, $-8$). The whole formula is neutral, so $(+1)+\text{Mn}+(-8)=0$, giving $\text{Mn}=+7$ – its maximum, which is why permanganate is a powerful oxidizing agent (it can only gain electrons).

    日本語

    In a redox 氧化还原 reaction, electrons transfer between species. Oxidation 氧化 is loss of electrons (oxidation number rises); reduction 还原 is gain (oxidation number falls). Track changes with oxidation numbers, and remember every oxidation is paired with a reduction – the electrons lost equal the electrons gained.

    Redox is electron transfer: the reducing agent is oxidised, the oxidising agent reduced
    Redox is electron transfer: the reducing agent is oxidised, the oxidising agent reduced

    Worked example. Find the oxidation number of manganese in permanganate, $\text{KMnO}_4$. Potassium is $+1$ and each oxygen is $-2$ (four of them, $-8$). The whole formula is neutral, so $(+1)+\text{Mn}+(-8)=0$, giving $\text{Mn}=+7$ – its maximum, which is why permanganate is a powerful oxidizing agent (it can only gain electrons).

    Explore · ⁨探索⁩

    Watch electrons transfer in a redox reaction

    In a redox reaction one species is oxidised (loses electrons) and another is reduced (gains them). Follow the electrons move from the metal to the non-metal.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    redox/rɪˈdɒks/ 酸化還元
    Oxidation/ˌɒksɪˈdeɪʃn/ 酸化
    reduction/rɪˈdʌkʃn/ 還元
    4.9

    Exam tips

    • Balance every equation and work in moles — convert grams→moles, cross the mole ratio, then convert back.
    • Find the limiting reactant by comparing mole ratios; it sets the maximum product (theoretical yield).
    • In redox, remember OIL RIG: oxidation is loss of electrons, reduction is gain; every oxidation is paired with a reduction.
    • For titrations use the balanced ratio to link the known and unknown at the equivalence point.
    • A net ionic equation shows only the species that change; leave out spectator ions.
  • 5

    Kinetics · ⁨反応速度論⁩

    Watch lesson · ⁨レッスンを視聴⁩
    5.1

    How Fast a Reaction Goes

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 5.1.A: Explain the relationship between the rate of a chemical reaction and experimental parameters.

    • 5.1.A.1 The kinetics of a chemical reaction is defined as the rate at which an amount of reactants is converted to products per unit of time.
    • 5.1.A.2 The rates of change of reactant and product concentrations are determined by the stoichiometry in the balanced chemical equation.
    • 5.1.A.3 The rate of a reaction is influenced by reactant concentrations, temperature, surface area, catalysts, and other environmental factors.
    日本語

    学習目標 5.1.A: 化学反応の速度と実験パラメータとの関係について説明する。

    • 5.1.A.1 化学反応の動力学とは、単位時間あたりに反応物が生成物に変換される量として定義される。
    • 5.1.A.2 反応物および生成物の濃度の変化率は、バランス取れた化学方程式の化学量論によって決定される。
    • 5.1.A.3 反応の速度は、反応物濃度、温度、表面積、触媒、その他の環境要因に影響を受ける。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    The Maxwell–Boltzmann distribution

    Kinetics 动力学 studies reaction rate 反应速率 – how fast reactants become products. Rate is the change in concentration per unit time, and it typically decreases as reactants are used up. Rate rises with higher concentration, higher temperature, greater surface area, and a catalyst.

    日本語
    A catalytic converter: catalysts speed rates by lowering activation energy without being consumed
    A catalytic converter: catalysts speed rates by lowering activation energy without being consumed
    The Maxwell–Boltzmann distribution

    Kinetics 动力学 studies reaction rate 反应速率 – how fast reactants become products. Rate is the change in concentration per unit time, and it typically decreases as reactants are used up. Rate rises with higher concentration, higher temperature, greater surface area, and a catalyst.

    More particles in the same volume collide more often, so the rate rises
    More particles in the same volume collide more often, so the rate rises
    A gas syringe: measuring gas volume over time gives reaction rate from the slope
    A gas syringe: measuring gas volume over time gives reaction rate from the slope
    Explore · ⁨探索⁩

    Change conditions and watch the rate · ⁨条件を変化させて反応速度を見る⁩

    Reaction rate rises with temperature, concentration, and a catalyst — each gives more frequent or more successful collisions. Change each and watch the reaction speed up. · ⁨反応 速度 は 温度 、濃度 、触媒 の増加とともに速くなります。これらは衝突回数を増やすか、成功する衝突の割合を増やすためです。各条件を変化させると反応が加速する様子がわかります。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Kinetics/kɪˈnetɪks/ 反応速度論
    reaction rate/rɪˈækʃn reɪt/ 反応速度
    5.2

    Writing the Rate Law

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 5.2.A: Represent experimental data with a consistent rate law expression.

    • 5.2.A.1 Experimental methods can be used to monitor the amounts of reactants and/or products of a reaction over time and to determine the rate of the reaction.
    • 5.2.A.2 The rate law expresses the rate of a reaction as proportional to the concentration of each reactant raised to a power.
    • 5.2.A.3 The power of each reactant in the rate law is the order of the reaction with respect to that reactant. The sum of the powers of the reactant concentrations in the rate law is the overall order of the reaction.
    • 5.2.A.4 The proportionality constant in the rate law is called the rate constant. The value of this constant is temperature dependent and the units reflect the overall reaction order.
    • 5.2.A.5 Comparing initial rates of a reaction is a method to determine the order with respect to each reactant.
    日本語

    学習目標 5.2.A: 実験データを一致する速度則の式で表す。

    • 5.2.A.1 実験手法を用いて、反応の時間経過に伴う試薬および/または生成物の量をモニタリングし、反応速度を決定することができる。
    • 5.2.A.2 速度式は、反応速度が各試薬の濃度をあるべき乗に raised した値に比例することを表す。
    • 5.2.A.3 速度式における各試薬のべき乗は、その試薬に対する反応次数である。速度式における試薬濃度のべき乗の総和は、全反応次数である。
    • 5.2.A.4 速度式における比例定数は速度定数と呼ばれる。この定数の値は温度に依存し、その単位は全反応次数を反映している。
    • 5.2.A.5 反応の初期速度を比較することは、各試薬に対する反応次数を決定するための方法である。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    The rate law 速率方程 relates rate to reactant concentrations:

    $$\text{rate}=k[\text{A}]^m[\text{B}]^n,$$
    where $k$ is the rate constant 速率常数 and $m,n$ are the orders 级数. Orders are found experimentally (not from the balanced coefficients) by seeing how the rate changes when you change one concentration at a time.

    Worked example. In experiments, doubling $[\text{A}]$ makes the rate four times larger, while doubling $[\text{B}]$ leaves the rate unchanged. So the reaction is second order in A ($2^2=4$) and zero order in B, giving $\text{rate}=k[\text{A}]^2$. The overall order is $2+0=2$. Never read the orders off the balanced coefficients – only experiment gives them.

    日本語

    The rate law 速率方程 relates rate to reactant concentrations:

    $$\text{rate}=k[\text{A}]^m[\text{B}]^n,$$
    where $k$ is the rate constant 速率常数 and $m,n$ are the orders 级数. Orders are found experimentally (not from the balanced coefficients) by seeing how the rate changes when you change one concentration at a time.

    Rate against concentration for zero-, first-, and second-order reactions
    Rate against concentration for zero-, first-, and second-order reactions

    Worked example. In experiments, doubling $[\text{A}]$ makes the rate four times larger, while doubling $[\text{B}]$ leaves the rate unchanged. So the reaction is second order in A ($2^2=4$) and zero order in B, giving $\text{rate}=k[\text{A}]^2$. The overall order is $2+0=2$. Never read the orders off the balanced coefficients – only experiment gives them.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    rate law/reɪt lɔː/ rate law(速度式)
    rate constant/reɪt ˈkɒnstənt/ 速度定数
    orders/ˈɔːdəz/ 秩序
    5.3

    Concentration Over Time

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 5.3.A: Identify the rate law expression of a chemical reaction using data that show how the concentrations of reaction species change over time.

    • 5.3.A.1 The order of a reaction can be inferred from a graph of concentration of reactant versus time.
    • 5.3.A.2 If a reaction is first order with respect to a reactant being monitored, a plot of the natural log (ln) of the reactant concentration as a function of time will be linear.
    • 5.3.A.3 If a reaction is second order with respect to a reactant being monitored, a plot of the reciprocal of the concentration of that reactant versus time will be linear.
    • 5.3.A.4 The slopes of the concentration versus time data for zeroth, first, and second order reactions can be used to determine the rate constant for the reaction.
      • Zeroth order:
      • Equation: $[\mathrm{A}]_t - [\mathrm{A}]_0 = -kt$
      • First order:
      • Equation: $\ln[\mathrm{A}]_t - \ln[\mathrm{A}]_0 = -kt$
      • Second order:
      • Equation: $1/[\mathrm{A}]_t - 1/[\mathrm{A}]_0 = kt$
    • 5.3.A.5 Half-life is a critical parameter for first order reactions because the half-life is constant and related to the rate constant for the reaction by the equation:
      • Equation: $t_{1/2} = 0.693/k.$
    • 5.3.A.6 Radioactive decay processes provide an important illustration of first order kinetics.
    日本語

    学習目標 5.3.A: 反応種の濃度が時間とともにどのように変化するかを示すデータを用いて、化学反応の速度式の表現を特定する。

    • 5.3.A.1 反応の次数は、試薬の濃度を横軸(または縦軸)に時間をプロットしたグラフから推定できる。
    • 5.3.A.2 モニタリング中の試薬に対して反応が一次である場合、試薬濃度の自然対数 (ln) を時間の関数としてプロットすると直線となる。
    • 5.3.A.3 モニタリング中の試薬に対して反応が二次である場合、その試薬の濃度の逆数を時間の関数としてプロットすると直線となる。
    • 5.3.A.4 零次反応、一次反応、および二次反応の濃度対時間プロットの傾きは、反応の速度定数を求めるために使用できます。
      • 零次反応:
      • 式: $[\mathrm{A}]_t - [\mathrm{A}]_0 = -kt$
      • 一次反応:
      • 式: $\ln[\mathrm{A}]_t - \ln[\mathrm{A}]_0 = -kt$
      • 二次反応:
      • 式: $1/[\mathrm{A}]_t - 1/[\mathrm{A}]_0 = kt$
    • 5.3.A.5 半減期は一次反応にとって重要なパラメータです。なぜなら、半減期は一定であり、次の式により反応の速度定数と関連しているからです:
      • 式: $t_{1/2} = 0.693/k.$
    • 5.3.A.6 放射性崩壊過程は、一次反応動力学の重要な例示を提供します。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    The integrated rate laws describe how concentration falls with time and give straight-line tests:

    • Zero order: $[\text{A}]$ vs $t$ is linear.
    • First order: $\ln[\text{A}]$ vs $t$ is linear; constant half-life 半衰期.
    • Second order: $\dfrac{1}{[\text{A}]}$ vs $t$ is linear.

    Whichever plot is straight tells you the order and gives $k$ from its slope.

    Worked example. A first-order reaction has rate constant $k=0.030\ \text{s}^{-1}$. Its half-life is

    $$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{0.030}=23\ \text{s},$$
    and, being first order, that half-life stays the same no matter how much reactant remains – so after $46\ \text{s}$ ($2$ half-lives) one quarter is left.

    日本語

    The integrated rate laws describe how concentration falls with time and give straight-line tests:

    A first-order reaction has a constant half-life
    A first-order reaction has a constant half-life
    • Zero order: $[\text{A}]$ vs $t$ is linear.
    • First order: $\ln[\text{A}]$ vs $t$ is linear; constant half-life 半衰期.
    • Second order: $\dfrac{1}{[\text{A}]}$ vs $t$ is linear.

    Whichever plot is straight tells you the order and gives $k$ from its slope.

    Worked example. A first-order reaction has rate constant $k=0.030\ \text{s}^{-1}$. Its half-life is

    $$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{0.030}=23\ \text{s},$$
    and, being first order, that half-life stays the same no matter how much reactant remains – so after $46\ \text{s}$ ($2$ half-lives) one quarter is left.

    Explore · ⁨探索⁩

    Track concentration as a reaction runs · ⁨反応進行に伴う濃度の変化を追跡する⁩

    As reactants are used up the rate slows, so a concentration-time curve is steep at first and flattens out. Raising temperature or adding a catalyst steepens it. · ⁨反応物が消費されるにつれ 速度は遅くなり 、濃度-時間曲線は当初急峻で後に水平化します。温度を上げたり触媒を加えたりすると曲線は急峻になります。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    half-life/hɑːf laɪf/ 半減期
    5.4

    The Steps a Reaction Really Takes

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 5.4.A: Represent an elementary reaction as a rate law expression using stoichiometry.

    • 5.4.A.1 The rate law of an elementary reaction can be inferred from the stoichiometry of the particles participating in a collision.
    • 5.4.A.2 Elementary reactions involving the simultaneous collision of three or more particles are rare.
    日本語

    学習目標 5.4.A: 化学量論を用いて、素反応を速度式の形で表現する。

    • 5.4.A.1 衝突に関与する粒子の化学量論から、素反応の速度式を推定できる。
    • 5.4.A.2 3つ以上の粒子が同時に衝突する素反応は稀である。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    A reaction mechanism 反应机理 is the sequence of elementary steps 基元反应 that actually occur. Their molecularity (how many particles collide in a step) sets that step's rate law directly. Species made in one step and used up in a later one are intermediates 中间体.

    日本語

    A reaction mechanism 反应机理 is the sequence of elementary steps 基元反应 that actually occur. Their molecularity (how many particles collide in a step) sets that step's rate law directly. Species made in one step and used up in a later one are intermediates 中间体.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    reaction mechanism/rɪˈækʃn ˈmekənɪzəm/ 反応機構
    elementary steps/ˌelɪˈmentəri steps/ 素反応
    intermediates/ˌɪntəˈmiːdɪəts/ 中間体
    5.5

    Why Collisions Do or Do Not React

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 5.5.A: Explain the relationship between the rate of an elementary reaction and the frequency, energy, and orientation of particle collisions.

    • 5.5.A.1 For an elementary reaction to successfully produce products, reactants must successfully collide to initiate bond-breaking and bond-making events.
    • 5.5.A.2 In most reactions, only a small fraction of the collisions leads to a reaction. Successful collisions have both sufficient energy to overcome the activation energy requirements and orientations that allow the bonds to rearrange in the required manner.
    • 5.5.A.3 The Maxwell-Boltzmann distribution curve describes the distribution of particle energies; this distribution can be used to gain a qualitative estimate of the fraction of collisions with sufficient energy to lead to a reaction, and also how that fraction depends on temperature.
    日本語

    学習目標 5.5.A: 素反応の速度と、粒子間の衝突の頻度、エネルギー、および配向との関係を説明する。

    • 5.5.A.1 素反応が正しく生成物を生成するためには、試薬が結合の切断および形成のイベントを開始するために正しく衝突する必要がある。
    • 5.5.A.2 ほとんどの反応において、衝突のわずかな割合しか反応に至らない。成功した衝突は、活性化エネルギーの要件を克服するのに十分なエネルギーを持ち、かつ、必要な方法で結合が再配置されることを可能にする配向を持っている。
    • 5.5.A.3 マクスウェル・ボルツマン分布曲線は粒子のエネルギーの分布を表しており、この分布を用いることで、反応につながるのに十分なエネルギーを持つ衝突の割合を定性的に推定でき、またその割合が温度にどのように依存するかを理解できる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    Collision theory

    Collision theory 碰撞理论: molecules must collide with enough energy (the activation energy 活化能, $E_a$) and the correct orientation to react. Higher temperature means more molecules exceed $E_a$, so the reaction speeds up sharply.

    日本語
    Collision theory

    Collision theory 碰撞理论: molecules must collide with enough energy (the activation energy 活化能, $E_a$) and the correct orientation to react. Higher temperature means more molecules exceed $E_a$, so the reaction speeds up sharply.

    A collision only reacts with the right orientation and enough energy
    A collision only reacts with the right orientation and enough energy
    Explore · ⁨探索⁩

    Which molecules clear the activation energy · ⁨活性化エネルギーを超える分子を選ぶ⁩

    Only collisions with energy above the activation energy react. Heating shifts the speed distribution right, so a much larger fraction of molecules can react. · ⁨活性化エネルギー より高いエネルギーを持つ衝突のみが反応します。加熱すると速度分布が右へシフトするため、反応できる分子の割合が大幅に増加します。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Collision theory/kəˈlɪʒn ˈθɪəri/ 衝突理論
    activation energy/ˌæktɪˈveɪʃn ˈenədʒi/ 活性化エネルギー
    5.6

    Reading an Energy Profile

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 5.6.A: Represent the activation energy and overall energy change in an elementary reaction using a reaction energy profile.

    • 5.6.A.1 Elementary reactions typically involve the breaking of some bonds and the forming of new ones.
    • 5.6.A.2 The reaction coordinate is the axis along which the complex set of motions involved in rearranging reactants to form products can be plotted.
    • 5.6.A.3 The energy profile gives the energy along the reaction coordinate, which typically proceeds from reactants, through a transition state, to products. The energy difference between the reactants and the transition state is the activation energy for the forward reaction.
    • 5.6.A.4 The rate of an elementary reaction is temperature dependent because the proportion of particle collisions that are energetic enough to reach the transition state varies with temperature. The Arrhenius equation relates the temperature dependence of the rate of an elementary reaction to the activation energy needed by molecular collisions to reach the transition state.
      • Exclusion statement: Calculations involving the Arrhenius equation will not be assessed on the AP Exam.
    日本語

    学習目標 5.6.A: 反応エネルギー図を用いて、素反応における活性化エネルギーと全エネルギー変化を表現する。

    • 5.6.A.1 素反応は通常、いくつかの結合の切断と新しい結合の形成を伴う。
    • 5.6.A.2 反応座標とは、試薬を生成物へ再配置するために必要な複雑な運動のセットをプロットする軸である。
    • 5.6.A.3 エネルギー図は、反応座標に沿ったエネルギーを示す。これは通常、試薬から始まり、遷移状態を経て生成物へと進む。試薬と遷移状態との間のエネルギー差が、正方向反応の活性化エネルギーである。
    • 5.6.A.4 素反応の速度は温度に依存する。なぜなら、遷移状態に到達するのに十分なエネルギーを持つ粒子間の衝突の割合は温度によって異なるからである。アレニウスの式は、素反応の速度の温度依存性と、遷移状態に到達するために分子間衝突が必要とする活性化エネルギーとの関係を示す。
      • 除外事項: アレニウスの式に関連する計算はAP試験では評価されない。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    Reaction energy profile

    An energy profile 能量图 plots energy along the reaction path. The peak is the transition state 过渡态; the climb from reactants to the peak is $E_a$; the difference between reactant and product energies is the enthalpy change $\Delta H$ (down for exothermic).

    日本語
    Reaction energy profile

    An energy profile 能量图 plots energy along the reaction path. The peak is the transition state 过渡态; the climb from reactants to the peak is $E_a$; the difference between reactant and product energies is the enthalpy change $\Delta H$ (down for exothermic).

    Exothermic reactions end lower than the reactants; endothermic end higher
    Exothermic reactions end lower than the reactants; endothermic end higher
    Explore · ⁨探索⁩

    Read activation energy off the profile · ⁨エネルギープロファイルから活性化エネルギーを読む⁩

    An energy profile plots energy along the reaction. The hump is the activation energy; the drop from reactants to products is $\Delta H$. A catalyst lowers the hump. · ⁨エネルギープロファイル は反応経路に沿ったエネルギーを描いたものです。山頂が 活性化エネルギー であり、反応物から生成物へのエネルギー低下が $\Delta H$ です。触媒は山頂を下げます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    energy profile/ˈenədʒi ˈprəʊfaɪl/ エネルギー図
    transition state/trænˈsɪʃn steɪt/ 制限因子
    5.7

    The Sequence of Steps

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 5.7.A: Identify the components of a reaction mechanism.

    • 5.7.A.1 A reaction mechanism consists of a series of elementary reactions, or steps, that occur in sequence. The components may include reactants, intermediates, products, and catalysts.
    • 5.7.A.2 The elementary steps when combined should align with the overall balanced equation of a chemical reaction.
    • 5.7.A.3 A reaction intermediate is produced by some elementary steps and consumed by others, such that it is present only while a reaction is occurring.
    • 5.7.A.4 Experimental detection of a reaction intermediate is a common way to build evidence in support of one reaction mechanism over an alternative mechanism.
      • Exclusion statement: Collection of data pertaining to detection of a reaction intermediate will not be assessed on the AP Exam.
    日本語

    学習目標 5.7.A: 反応メカニズムの構成要素を特定する。

    • 5.7.A.1 反応メカニズムは、順に起こる一連の素反応、すなわちステップからなる。構成要素には、試薬、中間体、生成物、触媒が含まれる可能性がある。
    • 5.7.A.2 素反応ステップを合わせると、化学反応の全体的な係数付き平衡方程式と整合するはずである。
    • 5.7.A.3 反応中間体は、ある素反応ステップによって生成され、他の素反応ステップによって消費されるため、反応が行われている間にのみ存在する。
    • 5.7.A.4 反応中間体の実験的検出は、ある反応メカニズムを別の代替メカニズムよりも支持する証拠を構築するための一般的な方法である。
      • 除外事項: 反応中間体の検出に関するデータの収集はAP試験では評価されない。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    In a multi-step mechanism, the steps must add up to the overall balanced equation (intermediates cancel). Each step has its own energy hill; the tallest hill is the slowest step.

    The slow step with the higher barrier is rate-determining
    The slow step with the higher barrier is rate-determining
    5.8

    Finding the Rate Law From a Mechanism

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 5.8.A: Identify the rate law for a reaction from a mechanism in which the first step is rate limiting.

    • 5.8.A.1 For reaction mechanisms in which each elementary step is irreversible, or in which the first step is rate limiting, the rate law of the reaction is set by the molecularity of the slowest elementary step (i.e., the rate-limiting step).
      • Exclusion statement: Collection of data pertaining to detection of a reaction intermediate will not be assessed on the AP Exam.
    日本語

    学習目標 5.8.A: 最初のステップが速度決定ステップであるメカニズムから、反応の速度式を特定する。

    • 5.8.A.1 各素反応が不可逆である、あるいは第一素反応が律速素反応である反応機構において、反応の速度式は最速素反応(すなわち律速素反応)の分子数によって決定される。
      • 除外事項: 反応中間体の検出に関するデータの収集はAP試験では評価されない。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    The rate-determining step 决速步骤 is the slowest step; its molecularity gives the overall rate law. A valid mechanism must (1) add to the overall reaction and (2) predict the experimentally observed rate law.

    Worked example. Suppose the slow (rate-determining) step is the bimolecular collision $\text{NO}_2+\text{NO}_2\rightarrow\text{NO}_3+\text{NO}$. Its molecularity gives the rate law directly: $\text{rate}=k[\text{NO}_2]^2$. If experiment shows exactly this, the proposed mechanism is consistent; if experiment gave $\text{rate}=k[\text{NO}_2]$, the mechanism would be wrong.

    日本語

    The rate-determining step 决速步骤 is the slowest step; its molecularity gives the overall rate law. A valid mechanism must (1) add to the overall reaction and (2) predict the experimentally observed rate law.

    Worked example. Suppose the slow (rate-determining) step is the bimolecular collision $\text{NO}_2+\text{NO}_2\rightarrow\text{NO}_3+\text{NO}$. Its molecularity gives the rate law directly: $\text{rate}=k[\text{NO}_2]^2$. If experiment shows exactly this, the proposed mechanism is consistent; if experiment gave $\text{rate}=k[\text{NO}_2]$, the mechanism would be wrong.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    rate-determining step/reɪt dɪˈtɜːmɪnɪŋ step/ 律速段階
    5.9

    When the First Step Is Fast

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 5.9.A: Identify the rate law for a reaction from a mechanism in which the first step is not rate limiting.

    • 5.9.A.1 If the first elementary reaction is not rate limiting, approximations (such as pre-equilibrium) must be made to determine a rate law expression.
    日本語

    学習目標 5.9.A: 第一素反応が律速素反応ではない反応機構から、反応の速度式を特定する。

    • 5.9.A.1 第一素反応が律速素反応でない場合、速度式を導くために、予平衡などの近似を用いる必要がある。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    If a fast step precedes the slow one, an intermediate appears in the slow step's rate law. Use the fast pre-equilibrium to rewrite that intermediate in terms of reactants, so the final rate law contains only measurable species.

    5.10

    Energy Profiles for Many Steps

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 5.10.A: Represent the activation energy and overall energy change in a multistep reaction with a reaction energy profile.

    • 5.10.A.1 Knowledge of the energetics of each elementary reaction in a mechanism allows for the construction of an energy profile for a multistep reaction.
    日本語

    学習目標 5.10.A: 多段階反応において、反応エネルギープロファイルを用いて活性化エネルギーおよび全体的なエネルギー変化を表す。

    • 5.10.A.1 メカニズム内の各素反応の熱力学に関する知識は、多段階反応のためのエネルギープロファイルを作成する上で有用である。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    A multi-step reaction's profile shows several peaks (one per step) with valleys (intermediates) between them. The highest peak is the rate-determining transition state – it controls the overall rate.

    5.11

    How Catalysts Speed Things Up

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 5.11.A: Explain the relationship between the effect of a catalyst on a reaction and changes in the reaction mechanism.

    • 5.11.A.1 In order for a catalyst to increase the rate of a reaction, the addition of the catalyst must increase the number of effective collisions and/or provide a reaction path with a lower activation energy relative to the original reaction coordinate.
    • 5.11.A.2 In a reaction mechanism containing a catalyst, the net concentration of the catalyst is constant. However, the catalyst will frequently be consumed in the rate-determining step of the reaction, only to be regenerated in a subsequent step in the mechanism.
    • 5.11.A.3 Some catalysts accelerate a reaction by binding to the reactant(s). The reactants are either oriented more favorably or react with lower activation energy. There is often a new reaction intermediate in which the catalyst is bound to the reactant(s). Many enzymes function in this manner.
    • 5.11.A.4 Some catalysts involve covalent bonding between the catalyst and the reactant(s). An example is acid-base catalysis, in which a reactant or intermediate either gains or loses a proton. This introduces a new reaction intermediate and new elementary reactions involving that intermediate.
    • 5.11.A.5 In surface catalysis, a reactant or intermediate binds to, or forms a covalent bond with, the surface. This introduces elementary reactions involving these new bound reaction intermediate(s).
    日本語

    学習目標 5.11.A: 触媒の反応への影響と反応メカニズムの変化との関係について説明する。

    • 5.11.A.1 触媒が反応速度を増加させるためには、触媒の添加により有効衝突の数が増加するか、あるいは元の反応座標に対して低い活性化エネルギーを持つ反応経路を提供しなければならない。
    • 5.11.A.2 触媒を含む反応メカニズムにおいて、触媒の正味濃度は一定である。しかし、触媒はしばしば反応の律速段階で消費され、その後メカニズムの次の段階で再生されることがある。
    • 5.11.A.3 一部の触媒は、反応物に結合することで反応を加速させる。反応物はより有利な配向をとるか、あるいは低い活性化エネルギーで反応する。これには、触媒が反応物に結合した新しい反応中間体が存在することが多い。多くの酵素がこの機能を持つ。
    • 5.11.A.4 一部の触媒は、触媒と反応物との間に共有結合を形成する。例として酸塩基触媒があり、反応物または中間体がプロトンを獲得または失う。これにより、その中間体を含む新しい反応中間体および新しい素反応が生じる。
    • 5.11.A.5 表面触媒では、反応物または中間体が表面に結合するか、または表面と共有結合を形成する。これにより、これらの新しい結合した反応中間体を含む素反応が生じる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    A catalyst 催化剂 speeds a reaction by providing a new pathway with a lower activation energy, without being consumed. It does not change $\Delta H$ or the equilibrium position – only how fast equilibrium is reached. On an energy profile, a catalyst lowers the peak(s).

    日本語

    A catalyst 催化剂 speeds a reaction by providing a new pathway with a lower activation energy, without being consumed. It does not change $\Delta H$ or the equilibrium position – only how fast equilibrium is reached. On an energy profile, a catalyst lowers the peak(s).

    A catalyst gives a route with lower activation energy; the enthalpy change is unchanged
    A catalyst gives a route with lower activation energy; the enthalpy change is unchanged
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    catalyst/ˈkætəlɪst/ 触媒
    5.11

    Exam tips

    • Rate rises with concentration, temperature, surface area, and a catalyst — explain each with collision theory (more, or more energetic, successful collisions).
    • Temperature works mainly by getting more particles above the activation energy, not just more collisions.
    • Reaction orders come from experiment, not the balanced coefficients — see how the rate changes when one concentration is varied.
    • On an energy profile the hill height is the activation energy and the reactant–product gap is $\Delta H$.
    • A catalyst lowers the activation energy (a new pathway) but leaves $\Delta H$ and the equilibrium position unchanged.
  • 6

    Thermochemistry · ⁨熱化学⁩

    Watch lesson · ⁨レッスンを視聴⁩
    6.1

    Endothermic and Exothermic Processes

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 6.1.A: Explain the relationship between experimental observations and energy changes associated with a chemical or physical transformation.

    • 6.1.A.1 Temperature changes in a system indicate energy changes.
    • 6.1.A.2 Energy changes in a system can be described as endothermic and exothermic processes such as the heating or cooling of a substance, phase changes, or chemical transformations.
    • 6.1.A.3 When a chemical reaction occurs, the energy of the system either decreases (exothermic reaction), increases (endothermic reaction), or remains the same. For exothermic reactions, the energy lost by the reacting species (system) is gained by the surroundings, as heat transfer from or work done by the system. Likewise, for endothermic reactions, the system gains energy from the surroundings by heat transfer to or work done on the system.
    • 6.1.A.4 The formation of a solution may be an exothermic or endothermic process, depending on the relative strengths of intermolecular/interparticle interactions before and after the dissolution process.
    日本語

    学習目標 6.1.A: 化学的または物理的変換に伴うエネルギー変化と実験観察結果との関係について説明する。

    • 6.1.A.1 システムにおける温度の変化は、エネルギーの変化を示している。
    • 6.1.A.2 システムのエネルギー変化は、物質の加熱や冷却、相転移、または化学変換など、吸熱過程および発熱過程として記述できる。
    • 6.1.A.3 化学反応が起こると、システムのエネルギーは減少する(発熱反応)、増加する(吸熱反応)、または一定のままとなる。発熱反応の場合、反応種(システム)が失ったエネルギーは、熱伝達またはシステムによる仕事として環境に与えられる。同様に、吸熱反応の場合、システムは熱伝達またはシステムに対する仕事によって環境からエネルギーを得る。
    • 6.1.A.4 溶液の形成は、溶解前後の分子間力・粒子間相互作用の相対的な強さに応じて、発熱過程または吸熱過程となり得る。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Thermochemistry 热化学 tracks energy in reactions. A process is exothermic 放热 if it releases energy to the surroundings (feels hot, $\Delta H<0$) and endothermic 吸热 if it absorbs energy (feels cold, $\Delta H>0$). Breaking bonds costs energy; forming bonds releases it – the net decides the sign.

    日本語
    An instant cold pack: endothermic processes absorb heat from the surroundings
    An instant cold pack: endothermic processes absorb heat from the surroundings

    Thermochemistry 热化学 tracks energy in reactions. A process is exothermic 放热 if it releases energy to the surroundings (feels hot, $\Delta H<0$) and endothermic 吸热 if it absorbs energy (feels cold, $\Delta H>0$). Breaking bonds costs energy; forming bonds releases it – the net decides the sign.

    An exothermic reaction warms the surroundings; an endothermic one cools them
    An exothermic reaction warms the surroundings; an endothermic one cools them
    Explore · ⁨探索⁩

    Compare endothermic and exothermic profiles · ⁨吸熱反応と発熱反応のプロファイル比較⁩

    An exothermic reaction releases energy (products lower than reactants, $\Delta H<0$); an endothermic one absorbs it. The hump is the activation energy. · ⁨発熱反応 はエネルギーを放出し(生成物のエネルギーが反応物より低く、$\Delta H<0$)、吸熱反応 はエネルギーを吸収します。山頂は活性化エネルギーです。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Thermochemistry/ˈθɜːməkemɪstri/ 熱化学
    exothermic/eɡzəˈðɜːmɪk/ 発熱反応
    endothermic/ˌendəʊˈθɜːmɪk/ 吸熱反応
    6.2

    Energy Diagrams

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 6.2.A: Represent a chemical or physical transformation with an energy diagram.

    • 6.2.A.1 A physical or chemical process can be described with an energy diagram that shows the endothermic or exothermic nature of that process.
    日本語

    学習目標 6.2.A: エネルギー図を用いて、化学的または物理的変換を表す。

    • 6.2.A.1 物理的または化学的過程は、その過程が吸熱的か発熱的かを示すエネルギー図によって記述できる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    An energy diagram plots energy from reactants to products. Reactants above products means exothermic; below means endothermic. The vertical gap between them is the enthalpy change $\Delta H$.

    Energy diagrams: exothermic products sit below the reactants, endothermic above
    Energy diagrams: exothermic products sit below the reactants, endothermic above
    6.3

    Heat Transfer and Thermal Equilibrium

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 6.3.A: Explain the relationship between the transfer of thermal energy and molecular collisions.

    • 6.3.A.1 The particles in a warmer body have a greater average kinetic energy than those in a cooler body.
    • 6.3.A.2 Collisions between particles in thermal contact can result in the transfer of energy. This process is called "heat transfer," "heat exchange," or "transfer of energy as heat."
    • 6.3.A.3 Eventually, thermal equilibrium is reached as the particles continue to collide. At thermal equilibrium, the average kinetic energy of both bodies is the same, and hence, their temperatures are the same.
    日本語

    学習目標 6.3.A: 熱エネルギーの移動と分子衝突との関係について説明する。

    • 6.3.A.1 高温物体の粒子は、低温物体の粒子よりも平均運動エネルギーが大きい。
    • 6.3.A.2 熱的に接触した粒子間の衝突により、エネルギーの移動が生じることがある。このプロセスは「熱伝達」、「熱交換」、または「熱としてのエネルギー移動」と呼ばれる。
    • 6.3.A.3 最終的に、粒子が衝突し続けることで熱平衡に達する。熱平衡状態では、両物体の平均運動エネルギーが等しくなるため、温度も等しくなる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Heat 热量 flows from hot to cold until objects reach thermal equilibrium 热平衡 (equal temperature). Energy is conserved: the heat lost by the hot object equals the heat gained by the cold one.

    日本語

    Heat 热量 flows from hot to cold until objects reach thermal equilibrium 热平衡 (equal temperature). Energy is conserved: the heat lost by the hot object equals the heat gained by the cold one.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Heat/hiːt/ 熱
    thermal equilibrium/ˈθɜːml ˌiːkwɪˈlɪbrɪəm/ 熱的平衡
    6.4

    Heat Capacity and Calorimetry

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 6.4.A: Calculate the heat $q$ absorbed or released by a system undergoing heating/cooling based on the amount of the substance, the heat capacity, and the change in temperature.

    • 6.4.A.1 The heating of a cool body by a warmer body is an important form of energy transfer between two systems. The amount of heat transferred between two bodies may be quantified by the heat transfer equation:

      • Equation: $q = mc\Delta T$.

      Calorimetry experiments are used to measure the transfer of heat.

    • 6.4.A.2 The first law of thermodynamics states that energy is conserved in chemical and physical processes.

    • 6.4.A.3 The transfer of a given amount of thermal energy will not produce the same temperature change in equal masses of matter with differing specific heat capacities.

    • 6.4.A.4 Heating a system increases the energy of the system, while cooling a system decreases the energy of the system.

    • 6.4.A.5 The specific heat capacity of a substance and the molar heat capacity are both used in energy calculations.

    • 6.4.A.6 Chemical systems change their energy through three main processes: heating/cooling, phase transitions, and chemical reactions.

    • 6.4.A.7 In calorimetry experiments involving dissolution, temperature changes of the mixture within the calorimeter can be used to determine the direction of energy flow. If the temperature of the mixture increases, thermal energy is released by the dissolution process (exothermic). If the temperature of the mixture decreases, thermal energy is absorbed by the dissolution process (endothermic).

    日本語

    学習目標 6.4.A: 物質の量、比熱容、および温度変化に基づき、加熱/冷却を受けるシステムが吸収または放出する熱 $q$ を計算する。

    • 6.4.A.1 高温物体による低温物体の加熱は、2つのシステム間のエネルギー移動の重要な形態である。2つの物体間で移動する熱の量は、熱伝達式によって定量化できる:

      • 式: $q = mc\Delta T$。

      熱量測定法の実験は、熱の移動を測定するために用いられる。

    • 6.4.A.2 熱力学第一法則は、化学反応や物理変化においてエネルギーが保存されることを示しています。

    • 6.4.A.3 同一量の熱エネルギーを移動させても、比熱容が異なる等質量の物質では、同じ温度変化を生じない。

    • 6.4.A.4 システムを加熱するとシステムのエネルギーが増加し、冷却するとシステムのエネルギーが減少する。

    • 6.4.A.5 物質の比熱容とモル熱容は、どちらもエネルギー計算に用いられる。

    • 6.4.A.6 化学システムは、主に3つのプロセスを通じてエネルギーを変化させる:加熱/冷却、相転移、および化学反応。

    • 6.4.A.7 溶解を含む熱量測定法の実験において、 calorimeter内の混合物の温度変化を利用してエネルギーの流れの方向を決定できる。混合物の温度が上昇すれば、溶解過程によって熱エネルギーが放出され(発熱)、混合物の温度が低下すれば、溶解過程によって熱エネルギーが吸収される(吸熱)。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    The heat to change a substance's temperature is

    $$q=mc\,\Delta T,$$
    where $c$ is the specific heat 比热容 (energy per gram per degree). Calorimetry 量热法 measures a reaction's heat by tracking the temperature change of surrounding water: the heat the water gains equals the heat the reaction releases.

    Worked example. A reaction in a coffee-cup calorimeter warms $100\ \text{g}$ of water by $8.0\,{}^{\circ}\text{C}$ ($c=4.18\ \text{J/(g}\,{}^{\circ}\text{C)}$). The heat absorbed by the water is

    $$q=mc\,\Delta T=100\times4.18\times8.0=3.3\times10^{3}\ \text{J}.$$
    By energy conservation the reaction released this $3.3\ \text{kJ}$, so it is exothermic ($q_{\text{rxn}}=-3.3\ \text{kJ}$).

    日本語

    The heat to change a substance's temperature is

    $$q=mc\,\Delta T,$$
    where $c$ is the specific heat 比热容 (energy per gram per degree). Calorimetry 量热法 measures a reaction's heat by tracking the temperature change of surrounding water: the heat the water gains equals the heat the reaction releases.

    Calorimetry: measure the temperature change of a known mass of solution
    Calorimetry: measure the temperature change of a known mass of solution

    Worked example. A reaction in a coffee-cup calorimeter warms $100\ \text{g}$ of water by $8.0\,{}^{\circ}\text{C}$ ($c=4.18\ \text{J/(g}\,{}^{\circ}\text{C)}$). The heat absorbed by the water is

    $$q=mc\,\Delta T=100\times4.18\times8.0=3.3\times10^{3}\ \text{J}.$$
    By energy conservation the reaction released this $3.3\ \text{kJ}$, so it is exothermic ($q_{\text{rxn}}=-3.3\ \text{kJ}$).

    Explore · ⁨探索⁩

    Heat different materials · ⁨異なる材料に熱を加える⁩

    $Q=mc\Delta T$: a high specific heat (like water's) means a lot of energy for a small temperature rise. Compare materials for the same heat input. · ⁨$Q=mc\Delta T$:比熱 が大きい(例:水など)場合、少量の温度上昇に対して多くのエネルギーが必要です。同じ熱量入力で材料を比較してください。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    specific heat/spəˈsɪfɪk hiːt/ 比熱
    Calorimetry/ˌkælɔːˈrɪmətri/ 熱量計測定法
    6.5

    Energy of Phase Changes

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 6.5.A: Explain changes in the heat $q$ absorbed or released by a system undergoing a phase transition based on the amount of the substance in moles and the molar enthalpy of the phase transition.

    • 6.5.A.1 Energy must be transferred to a system to cause a substance to melt (or boil). The energy of the system therefore increases as the system undergoes a solid-to-liquid (or liquid-to-gas) phase transition. Likewise, a system releases energy when it freezes (or condenses). The energy of the system decreases as the system undergoes a liquid-to-solid (or gas-to-liquid) phase transition. The temperature of a pure substance remains constant during a phase change.
    • 6.5.A.2 The energy absorbed during a phase change is equal to the energy released during a complementary phase change in the opposite direction. For example, the molar enthalpy of condensation of a substance is equal to the negative of its molar enthalpy of vaporization. Similarly, the molar enthalpy of fusion can be used to calculate the energy absorbed when melting a substance and the energy released when freezing a substance.
    日本語

    学習目標 6.5.A: 物質量(モル数)および相転移のモルエンタルピーに基づき、相転移を受けるシステムが吸収または放出する熱 $q$ の変化について説明する。

    • 6.5.A.1 物質を融解(または沸騰)させるには、システムへエネルギーを供給する必要がある。したがって、システムが固体から液体(または液体から気体)への相転移を行う際、システムのエネルギーは増加する。同様に、システムが凝固(または凝縮)する際はエネルギーを放出する。システムが液体から固体(または気体から液体)への相転移を行う際、システムのエネルギーは減少する。純物質の相転移中は温度は一定である。
    • 6.5.A.2 相転移中に吸収されるエネルギーは、逆方向の補完的な相転移中に放出されるエネルギーと等しい。例えば、ある物質の凝縮のモルエンタルピーは、その蒸発のモルエンタルピーの負の値に等しい。同様に、融解のモルエンタルピーを用いて、物質を融解する際に吸収されるエネルギーや、凝固する際に放出されるエネルギーを計算できる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    During a phase change 相变 (melting, boiling) the temperature stays constant while heat goes into breaking intermolecular forces, not raising kinetic energy. The energy needed is $q=n\,\Delta H_{\text{fus}}$ (melting) or $q=n\,\Delta H_{\text{vap}}$ (boiling) – the flat steps on a heating curve.

    日本語

    During a phase change 相变 (melting, boiling) the temperature stays constant while heat goes into breaking intermolecular forces, not raising kinetic energy. The energy needed is $q=n\,\Delta H_{\text{fus}}$ (melting) or $q=n\,\Delta H_{\text{vap}}$ (boiling) – the flat steps on a heating curve.

    Steam from boiling water: phase changes absorb or release energy at constant temperature
    Steam from boiling water: phase changes absorb or release energy at constant temperature
    Explore · ⁨探索⁩

    Heat through a phase change · ⁨相転移中の熱の移動を見る⁩

    During a phase change the temperature holds flat while energy breaks bonds — the latent heat. Watch the plateaus at melting and boiling. · ⁨相転移 の間、温度は一定のままエネルギーが結合破壊に使われます(潜熱)。融解と沸騰での水平部分(Plateau)を確認してください。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    phase change/feɪz tʃeɪndʒ/ 相転移
    6.6

    Introduction to Enthalpy of Reaction

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 6.6.A: Calculate the heat $q$ absorbed or released by a system undergoing a chemical reaction in relationship to the amount of the reacting substance in moles and the molar enthalpy of reaction.

    • 6.6.A.1 The enthalpy change of a reaction gives the amount of heat energy released (for negative values) or absorbed (for positive values) by a chemical reaction at constant pressure.
    • 6.6.A.2 When the products of a reaction are at a different temperature than their surroundings, they exchange energy with the surroundings to reach thermal equilibrium. Thermal energy is transferred to the surroundings as the reactants convert to products in an exothermic reaction. Thermal energy is transferred from the surroundings as the reactants convert to products in an endothermic reaction.
    • 6.6.A.3 The chemical potential energy of the products of a reaction is different from that of the reactants because of the breaking and forming of bonds. The energy difference results in a change in the kinetic energy of the particles, which manifests as a temperature change.
      • Exclusion Statement: The technical distinctions between enthalpy and internal energy will not be assessed on the AP Exam. Most reactions studied at the AP level are carried out at constant pressure, where the enthalpy change of the process is equal to the heat (and by extension, the energy) of reaction.
    日本語

    学習目標 6.6.A: 化学反応に伴う系が吸収または放出する熱量 $q$を、反応物物の量(モル)および反応のモルエンタルピーとの関係において計算する。

    • 6.6.A.1 反応のエンタルピー変化は、定圧下での化学反応によって放出される(負の値の場合)または吸収される(正の値の場合)熱エネルギーの量を表す。
    • 6.6.A.2 反応生成物の温度が周囲環境と異なる場合、熱平衡に達するために周囲環境とエネルギー交換を行う。発熱反応では、反応物が生成物に変化する際に熱エネルギーが周囲環境へ伝わる。吸熱反応では、反応物が生成物に変化する際に熱エネルギーが周囲環境から伝わる。
    • 6.6.A.3 化学反応の生成物の化学ポテンシャルエネルギーは、結合の切断および形成により反応物とは異なる。このエネルギー差は粒子の運動エネルギーの変化を引き起こし、温度変化として現れる。
      • 除外事項: エンタルピーと内部エネルギーの技術的な区別については、AP試験では評価されない。APレベルで研究されるほとんどの反応は定圧下で行われ、その際のエンタルピー変化は反応の熱量(および結果としてエネルギー)に等しい。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    The enthalpy of reaction 反应焓 $\Delta H_{\text{rxn}}$ is the heat released or absorbed at constant pressure. Because enthalpy is a state function 状态函数, $\Delta H$ depends only on the initial and final states, not the path taken – the key that makes the next three methods work.

    日本語

    The enthalpy of reaction 反应焓 $\Delta H_{\text{rxn}}$ is the heat released or absorbed at constant pressure. Because enthalpy is a state function 状态函数, $\Delta H$ depends only on the initial and final states, not the path taken – the key that makes the next three methods work.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    enthalpy of reaction/enˈθælpi ɒv rɪˈækʃn/ 反応エンタルピー
    state function/steɪt ˈfʌŋkʃn/ 状態関数
    6.7

    Bond Enthalpies

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 6.7.A: Calculate the enthalpy change of a reaction based on the average bond energies of bonds broken and formed in the reaction.

    • 6.7.A.1 During a chemical reaction, bonds are broken and/or formed, and these events change the potential energy of the system.
    • 6.7.A.2 The average energy required to break all of the bonds in the reactant molecules can be estimated by adding up the average bond energies of all the bonds in the reactant molecules. Likewise, the average energy released in forming the bonds in the product molecules can be estimated. If the energy released is greater than the energy required, the reaction is exothermic. If the energy required is greater than the energy released, the reaction is endothermic.
    日本語

    学習目標 6.7.A: 反応中に切断・形成される結合の平均結合エネルギーに基づいて、反応のエンタルピー変化を計算する。

    • 6.7.A.1 化学反応中、結合が切断され、かつ/または形成され、これらの事象が系のポテンシャルエネルギーを変化させる。
    • 6.7.A.2 反応物分子のすべての結合を切断するために必要な平均エネルギーは、反応物分子のすべての結合の平均結合エネルギーを合計することで推定できる。同様に、生成物分子の結合を形成する際に放出される平均エネルギーも推定できる。放出されるエネルギーが必要とするエネルギーより大きい場合、反応は発熱である。必要とするエネルギーが放出されるエネルギーより大きい場合、反応は吸熱である。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    One way to estimate $\Delta H$: sum the energy to break all reactant bonds, then subtract the energy released forming product bonds:

    $$\Delta H \approx \sum (\text{bonds broken}) - \sum (\text{bonds formed}).$$
    This is an approximation, since bond enthalpies are averages.

    Breaking bonds takes in energy; making bonds releases it
    Breaking bonds takes in energy; making bonds releases it

    Worked example. Estimate $\Delta H$ for $\text{H}_2+\text{Cl}_2\rightarrow2\text{HCl}$ using bond enthalpies H–H $=436$, Cl–Cl $=242$, H–Cl $=431\ \text{kJ/mol}$. Break both reactant bonds ($436+242=678$) and form two H–Cl bonds ($2\times431=862$):

    $$\Delta H\approx 678-862=-184\ \text{kJ},$$
    exothermic, because the strong H–Cl bonds formed release more than the reactant bonds cost.

    6.8

    Enthalpy of Formation

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 6.8.A: Calculate the enthalpy change for a chemical or physical process based on the standard enthalpies of formation.

    • 6.8.A.1 Tables of standard enthalpies of formation can be used to calculate the standard enthalpies of reactions.
      • Equation: $\Delta H^{\circ}_{reaction} = \Sigma \Delta H^{\circ}_{f\ products} - \Sigma \Delta H^{\circ}_{f\ reactants}$
    日本語

    学習目標 6.8.A: 標準生成エンタルピーに基づいて、化学的または物理的プロセスのエンタルピー変化を計算する。

    • 6.8.A.1 標準生成エンタルピーの表を用いて、反応の標準エンタルピーを計算することができる。
      • 式: $\Delta H^{\circ}_{reaction} = \Sigma \Delta H^{\circ}_{f\ products} - \Sigma \Delta H^{\circ}_{f\ reactants}$

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    The standard enthalpy of formation 生成焓 $\Delta H_f^\circ$ is the enthalpy to make one mole of a compound from its elements (zero for an element in its standard state). Then

    $$\Delta H_{\text{rxn}}^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}).$$

    Worked example. Find $\Delta H_{\text{rxn}}^\circ$ for burning methane, $\text{CH}_4+2\text{O}_2\rightarrow\text{CO}_2+2\text{H}_2\text{O}$, given $\Delta H_f^\circ$: CH$_4=-75$, CO$_2=-394$, H$_2$O$=-286\ \text{kJ/mol}$ (O$_2=0$). Products minus reactants:

    $$\Delta H_{\text{rxn}}^\circ=[-394+2(-286)]-[-75+0]=-966+75=-891\ \text{kJ},$$
    a large release, as expected for a combustion.

    日本語

    The standard enthalpy of formation 生成焓 $\Delta H_f^\circ$ is the enthalpy to make one mole of a compound from its elements (zero for an element in its standard state). Then

    $$\Delta H_{\text{rxn}}^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}).$$

    Formation makes a compound from its elements; combustion burns it in oxygen
    Formation makes a compound from its elements; combustion burns it in oxygen

    Worked example. Find $\Delta H_{\text{rxn}}^\circ$ for burning methane, $\text{CH}_4+2\text{O}_2\rightarrow\text{CO}_2+2\text{H}_2\text{O}$, given $\Delta H_f^\circ$: CH$_4=-75$, CO$_2=-394$, H$_2$O$=-286\ \text{kJ/mol}$ (O$_2=0$). Products minus reactants:

    $$\Delta H_{\text{rxn}}^\circ=[-394+2(-286)]-[-75+0]=-966+75=-891\ \text{kJ},$$
    a large release, as expected for a combustion.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    standard enthalpy of formation/ˈstændəd enˈθælpi ɒv fɔːˈmeɪʃn/ 標準生成エンタルピー
    6.9

    Hess's Law

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 6.9.A: Represent a chemical or physical process as a sequence of steps.

    • 6.9.A.1 Many processes can be broken down into a series of steps. Each step in the series has its own energy change.

    Learning Objective 6.9.B: Explain the relationship between the enthalpy of a chemical or physical process and the sum of the enthalpies of the individual steps.

    • 6.9.B.1 Because total energy is conserved (first law of thermodynamics), and each individual reaction in a sequence transfers thermal energy to or from the surroundings, the net thermal energy transferred in the sequence will be equal to the sum of the thermal energy transfers in each of the steps. These thermal energy transfers are the result of potential energy changes among the species in the reaction sequence; thus, at constant pressure, the enthalpy change of the overall process is equal to the sum of the enthalpy changes of the individual steps.
    • 6.9.B.2 The following are essential principles of Hess's law:
      • i. When a reaction is reversed, the enthalpy change stays constant in magnitude but becomes reversed in mathematical sign.
      • ii. When a reaction is multiplied by a factor $c$, the enthalpy change is multiplied by the same factor $c$.
      • iii. When two (or more) reactions are added to obtain an overall reaction, the individual enthalpy changes of each reaction are added to obtain the net enthalpy change of the overall reaction.
      • Exclusion Statement: The concept of state functions will not be assessed on the AP Exam.
    日本語

    学習目標 6.9.A: 化学的または物理的プロセスを一連の段階として表現する。

    • 6.9.A.1 多くのプロセスは一連の段階に分解できる。一連の各段階には独自のエネルギー変化がある。

    学習目標 6.9.B: 化学的または物理的プロセスのエンタルピーと個々の段階のエンタルピーの総和との関係を説明する。

    • 6.9.B.1 全エネルギーが保存される(熱力学第一法則)ため、連鎖反応の各個々の反応が周囲に対して熱エネルギーを放出または吸収する場合、連鎖全体での正味に移動した熱エネルギーは、各ステップでの熱エネルギー移動量の総和に等しくなります。これらの熱エネルギーの移動は、反応連鎖中の種間のポテンシャルエネルギーの変化によるものです。したがって、一定圧力下では、全体のプロセスのエンタルピー変化は、個々のステップのエンタルピー変化の総和に等しくなります。
    • 6.9.B.2 以下はヘスの法則の必須原則である:
      • i. 反応を逆方向にする場合、エンタルピー変化の絶対値は一定であるが、数学的な符号は逆になる。
      • ii. 反応をある係数 $c$倍する場合、エンタルピー変化も同じ係数 $c$倍される。
      • iii. 2つ(またはそれ以上)の反応を加えて全体反応を得る場合、各反応の個別のエンタルピー変化を加算して、全体反応の正味エンタルピー変化を求める。
      • 除外事項: 状態関数の概念については、AP試験では評価されない。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    Hess's law: the route does not matter

    Hess's law 盖斯定律: if a reaction is the sum of several steps, its $\Delta H$ is the sum of the steps' $\Delta H$ values. Reverse a step and flip the sign; scale a step and scale its $\Delta H$. This lets you find a hard-to-measure $\Delta H$ by combining known reactions – a frequent exam calculation.

    日本語
    Hess's law: the route does not matter

    Hess's law 盖斯定律: if a reaction is the sum of several steps, its $\Delta H$ is the sum of the steps' $\Delta H$ values. Reverse a step and flip the sign; scale a step and scale its $\Delta H$. This lets you find a hard-to-measure $\Delta H$ by combining known reactions – a frequent exam calculation.

    Hess's law: the direct and indirect routes give the same total enthalpy change
    Hess's law: the direct and indirect routes give the same total enthalpy change
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Hess's law/ˈhesɪz lɔː/ ヘスの法則
    6.9

    Exam tips

    • An exothermic reaction has $\Delta H<0$ (feels hot); endothermic has $\Delta H>0$ — always give $\Delta H$ a sign and units.
    • In calorimetry use $q=mc\,\Delta T$ with the mass of the water/solution; the reaction releases what the water gains.
    • Bond enthalpies: $\Delta H\approx\sum(\text{bonds broken})-\sum(\text{bonds made})$ — breaking is endothermic, making is exothermic (the classic sign trap).
    • Hess's law: the total $\Delta H$ is the sum of the steps' — reverse a step and flip the sign, scale a step and scale $\Delta H$.
    • During a phase change the temperature stays constant while energy goes into the forces between particles.
  • 7

    Equilibrium · ⁨平衡⁩

    Watch lesson · ⁨レッスンを視聴⁩
    7.1

    Introduction to Equilibrium

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 7.1.A: Explain the relationship between the occurrence of a reversible chemical or physical process, and the establishment of equilibrium, to experimental observations.

    • 7.1.A.1 Many observable processes are reversible. Examples include evaporation and condensation of water, absorption and desorption of a gas, or dissolution and precipitation of a salt. Some important reversible chemical processes include the transfer of protons in acid-base reactions and the transfer of electrons in redox reactions.
    • 7.1.A.2 When equilibrium is reached, no observable changes occur in the system. Reactants and products are simultaneously present, and the concentrations or partial pressures of all species remain constant.
    • 7.1.A.3 The equilibrium state is dynamic. The forward and reverse processes continue to occur at equal rates, resulting in no net observable change.
    • 7.1.A.4 Graphs of concentration, partial pressure, or rate of reaction versus time for simple chemical reactions can be used to understand the establishment of chemical equilibrium.
    日本語

    学習目標 7.1.A: 可逆的な化学的または物理的プロセスの発生と平衡の確立との関係を、実験的观察と関連付けて説明する。

    • 7.1.A.1 多くの観察可能なプロセスは可逆である。例として、水の蒸発および凝縮、ガスの吸着および脱離、塩の溶解および沈殿などが挙げられる。重要な可逆的化学プロセスには、酸塩基反応におけるプロトンの移動や酸化還元反応における電子の移動が含まれる。
    • 7.1.A.2 平衡に達すると、系では observable な変化は生じない。反応物と生成物は同時に存在し、すべての種の濃度または分圧は一定に保たれる。
    • 7.1.A.3 平衡状態は動的である。正反応および逆反応は等しい速度で継続して起こり、正味の観測可能な変化は生じない。
    • 7.1.A.4 単純な化学反応に対する、時間に対する濃度、分圧、または反応速度のプロットを用いて、化学平衡の確立を理解することができる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    Dynamic equilibrium

    A reversible reaction 可逆反应 runs both ways. Chemical equilibrium 化学平衡 is reached when the forward and reverse rates become equal, so concentrations stop changing. Equilibrium is dynamic – both reactions continue, but at matching rates, so nothing appears to change.

    The rates graph above explains why concentrations settle; this next graph shows what the concentrations do – reactants fall and products rise until, once the rates match, both hold constant (they need not be equal).

    日本語
    Cobalt chloride solutions: equilibrium systems shift when conditions change (Le Chatelier)
    Cobalt chloride solutions: equilibrium systems shift when conditions change (Le Chatelier)
    Dynamic equilibrium

    A reversible reaction 可逆反应 runs both ways. Chemical equilibrium 化学平衡 is reached when the forward and reverse rates become equal, so concentrations stop changing. Equilibrium is dynamic – both reactions continue, but at matching rates, so nothing appears to change.

    Dynamic equilibrium: the forward and reverse rates become equal
    Dynamic equilibrium: the forward and reverse rates become equal

    The rates graph above explains why concentrations settle; this next graph shows what the concentrations do – reactants fall and products rise until, once the rates match, both hold constant (they need not be equal).

    Reactant and product concentrations change, then hold constant once equilibrium is reached
    Reactant and product concentrations change, then hold constant once equilibrium is reached
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    reversible reaction/rɪˈvɜːsɪbl rɪˈækʃn/ 可逆反応
    Chemical equilibrium/ˈkemɪkl ˌiːkwɪˈlɪbrɪəm/ 化学平衡
    7.2

    Direction of Reversible Reactions

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 7.2.A: Explain the relationship between the direction in which a reversible reaction proceeds and the relative rates of the forward and reverse reactions.

    • 7.2.A.1 If the rate of the forward reaction is greater than the reverse reaction, then there is a net conversion of reactants to products. If the rate of the reverse reaction is greater than that of the forward reaction, then there is a net conversion of products to reactants. An equilibrium state is reached when these rates are equal.
    日本語

    学習目標 7.2.A: 可逆反応がどの方向に進むかについて、正反応および逆反応の相対的な速度との関係を説明する。

    • 7.2.A.1 正反応の速度が逆反応の速度より大きい場合、反応物から生成物への正味の転換が起こる。逆に、逆反応の速度が正反応の速度より大きい場合、生成物から反応物への正味の転換が起こる。これらの速度が等しくなったとき、平衡状態に達する。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    At equilibrium the amounts of reactants and products are fixed but usually not equal. Whether the mixture favors products or reactants depends on the reaction; you compare the current state to the equilibrium condition to predict which way it shifts.

    7.3

    The Reaction Quotient and Equilibrium Constant

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 7.3.A: Represent the reaction quotient $Q_c$ or $Q_p$, for a reversible reaction, and the corresponding equilibrium expressions $K_c = Q_c$ or $K_p = Q_p$.

    • 7.3.A.1 The reaction quotient $Q_c$ describes the relative concentrations of reaction species at any time. For gas phase reactions, the reaction quotient may instead be written in terms of partial pressures as $Q_p$. The reaction quotient tends toward the equilibrium constant such that at equilibrium $K_c = Q_c$ and $K_p = Q_p$. As examples, for the reaction

      $$a\,\mathrm{A} + b\,\mathrm{B} \rightleftarrows c\,\mathrm{C} + d\,\mathrm{D}$$
      the law of mass action indicates that the equilibrium expression for $(K_c, Q_c)$ is

      • Equation: $K_c = \dfrac{[\mathrm{C}]^c [\mathrm{D}]^d}{[\mathrm{A}]^a [\mathrm{B}]^b}$

      and that for $(K_p, Q_p)$ is

      • Equation: $K_p = \dfrac{(P_\mathrm{C})^c (P_\mathrm{D})^d}{(P_\mathrm{A})^a (P_\mathrm{B})^b}$
      • Exclusion statement: Conversion between $K_c$ and $K_p$ will not be assessed on the AP Exam. Students should be aware of the conceptual differences and pay attention to whether $K_c$ or $K_p$ is used in an exam question.
      • Exclusion statement: Equilibrium calculations on systems where a dissolved species is in equilibrium with that species in the gas phase will not be assessed on the AP Exam.
    • 7.3.A.2 The reaction quotient does not include substances whose concentrations (or partial pressures) are independent of the amount, such as for solids and pure liquids.

    日本語

    学習目標 7.3.A: 可逆反応に対する反応係数$Q_c$または$Q_p$、および対応する平衡式$K_c = Q_c$または$K_p = Q_p$を表記する。

    • 7.3.A.1 反応係数$Q_c$は、任意の時刻における反応化学種の相対的な濃度を記述する。気相反応の場合、反応係数は分圧を用いて$Q_p$として表すこともある。反応係数は平衡定数に漸近し、平衡において$K_c = Q_c$および$K_p = Q_p$となる。例として、以下の反応については

      $$a\,\mathrm{A} + b\,\mathrm{B} \rightleftarrows c\,\mathrm{C} + d\,\mathrm{D}$$
      質量作用の法則により、$(K_c, Q_c)$に対する平衡式は

      • 式: $K_c = \dfrac{[\mathrm{C}]^c [\mathrm{D}]^d}{[\mathrm{A}]^a [\mathrm{B}]^b}$

      および$(K_p, Q_p)$に対する平衡式は

      • 式: $K_p = \dfrac{(P_\mathrm{C})^c (P_\mathrm{D})^d}{(P_\mathrm{A})^a (P_\mathrm{B})^b}$
      • 除外事項: AP試験では、$K_c$ と $K_p$ の間の変換は評価されません。学生は概念上の違いを理解し、問題文で $K_c$ または $K_p$ のどちらが使われているかに注意を払うべきです。
      • 除外事項: 溶解した化学種が気相中の同種化学種と平衡にある系に関する平衡計算は、AP試験の評価対象外とする。
    • 7.3.A.2 反応係数には、固体や純液体のように、量に依存せず一定の濃度(または分圧)を持つ物質は含まれない。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    The reaction quotient 反应商 $Q$ has the same form as the equilibrium expression but uses current concentrations:

    $$Q=\frac{[\text{products}]}{[\text{reactants}]}\ \text{(each raised to its coefficient)}.$$
    At equilibrium $Q$ equals the equilibrium constant 平衡常数 $K$. Comparing them predicts direction: $Q shifts forward (toward products); $Q>K$ shifts reverse; $Q=K$ means already at equilibrium.

    日本語

    The reaction quotient 反应商 $Q$ has the same form as the equilibrium expression but uses current concentrations:

    $$Q=\frac{[\text{products}]}{[\text{reactants}]}\ \text{(each raised to its coefficient)}.$$
    At equilibrium $Q$ equals the equilibrium constant 平衡常数 $K$. Comparing them predicts direction: $Q shifts forward (toward products); $Q>K$ shifts reverse; $Q=K$ means already at equilibrium.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    reaction quotient/rɪˈækʃn ˈkwəʊʃənt/ 反応商
    equilibrium constant/ˌiːkwɪˈlɪbrɪəm ˈkɒnstənt/ 平衡定数
    7.4

    Calculating the Equilibrium Constant

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 7.4.A: Calculate $K_c$ or $K_p$ based on experimental observations of concentrations or pressures at equilibrium.

    • 7.4.A.1 Equilibrium constants can be determined from experimental measurements of the concentrations or partial pressures of the reactants and products at equilibrium.
    日本語

    学習目標 7.4.A: 平衡時の濃度または圧力の実験的観測値に基づき、$K_c$または$K_p$を計算する。

    • 7.4.A.1 平衡定数は、平衡時における反応物および生成物の濃度または分圧の実験測定値から決定できる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    Write $K$ from the balanced equation (pure solids and liquids are left out). From equilibrium concentrations (or partial pressures for $K_p$), plug in and compute. $K_c$ uses molarities; $K_p$ uses pressures.

    Worked example. For $\text{N}_2\text{O}_4\rightleftharpoons2\text{NO}_2$, the equilibrium concentrations are $[\text{N}_2\text{O}_4]=0.20\ \text{M}$ and $[\text{NO}_2]=0.10\ \text{M}$. Then

    $$K_c=\frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}=\frac{(0.10)^2}{0.20}=0.050.$$
    The NO$_2$ coefficient of $2$ becomes the power; N$_2$O$_4$ (coefficient $1$) is just to the first power.

    7.5

    Magnitude of the Equilibrium Constant

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 7.5.A: Explain the relationship between very large or very small values of $K$ and the relative concentrations of chemical species at equilibrium.

    • 7.5.A.1 Some equilibrium reactions have very large $K$ values and proceed essentially to completion. Others have very small $K$ values and barely proceed at all.
    日本語

    学習目標 7.5.A: $K$ の非常に大きい値や非常に小さい値と、平衡状態における化学種間の相対的な濃度の関係について説明する。

    • 7.5.A.1 一部の平衡反応では$K$の値が非常に大きく、ほぼ完全に進行する。他の反応では$K$の値が非常に小さく、ほとんど進行しない。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    • $K\gg 1$: products strongly favored (reaction nearly complete).
    • $K\ll 1$: reactants favored (little reaction).
    • $K\approx 1$: significant amounts of both.

    The size of $K$ tells you where the equilibrium "sits."

    7.6

    Properties of the Equilibrium Constant

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 7.6.A: Represent a multistep process with an overall equilibrium expression, using the constituent $K$ expressions for each individual reaction.

    • 7.6.A.1 When a reaction is reversed, $K$ is inverted.
    • 7.6.A.2 When the stoichiometric coefficients of a reaction are multiplied by a factor $c$, $K$ is raised to the power $c$.
    • 7.6.A.3 When reactions are added together, the $K$ of the resulting overall reaction is the product of the $K$'s for the reactions that were summed.
    • 7.6.A.4 Since the expressions for $K$ and $Q$ have identical mathematical forms, all valid algebraic manipulations of $K$ also apply to $Q$.
    日本語

    学習目標 7.6.A: 各単独反応に対する構成要素の$K$表現を用いて、全体としての平衡式で多段階プロセスを表記する。

    • 7.6.A.1 反応を逆に行うと、$K$は逆数になる。
    • 7.6.A.2 反応の化学量論係数を定数 $c$ で倍率したとき、$K$ は $c$ 乗される。
    • 7.6.A.3 反応を加算すると、得られる全体の反応の$K$は、加算された各反応の$K$の積となる。
    • 7.6.A.4 $K$ と $Q$ の式は数学的に同一の形式を持つため、$K$ に対する有効な代数的変形はすべて $Q$ にも適用される。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    $K$ changes in predictable ways: reversing a reaction inverts $K$ ($1/K$); multiplying coefficients by $n$ raises $K$ to the $n$th power; adding reactions multiplies their $K$ values. Only temperature changes the value of $K$ itself.

    7.7

    Calculating Equilibrium Concentrations

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 7.7.A: Identify the concentrations or partial pressures of chemical species at equilibrium based on the initial conditions and the equilibrium constant.

    • 7.7.A.1 The concentrations or partial pressures of species at equilibrium can be predicted given the balanced reaction, initial concentrations, and the appropriate $K$.
    • 7.7.A.2 When $Q < K$, the reaction will proceed with a net consumption of reactants and generation of products. When $Q > K$, the reaction will proceed with a net consumption of products and generation of reactants. When $Q = K$, the system is at dynamic equilibrium; both forward and reverse reactions proceed at the same rate, and the proportion of reactants and products remains constant.
    日本語

    学習目標 7.7.A: 初期条件および平衡定数に基づき、平衡時における化学種の濃度または分圧を特定する。

    • 7.7.A.1 平衡時における化学種の濃度または分圧は、balanced reaction(バランス反応)、初期濃度、および適切な$K$が与えられた場合に予測可能である。
    • 7.7.A.2 $Q < K$の場合、反応は試薬の純粋な消費と生成物の純粋な生成を伴って進行する。$Q > K$の場合、反応は生成物の純粋な消費と試薬の純粋な生成を伴って進行する。$Q = K$の場合、系は動的不平衡状態にあります。正反応と逆反応が同じ速度で進行し、試薬と生成物の比率は一定に保たれます。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    Use an ICE table (Initial, Change, Equilibrium): fill in starting amounts, express the change with $x$ using the mole ratios, then substitute into the $K$ expression and solve for $x$. When $K$ is small, the approximation "$x$ is negligible" often simplifies the algebra.

    Worked example. For $\text{H}_2+\text{I}_2\rightleftharpoons2\text{HI}$ with $K_c=50$, start with $0.100\ \text{M}$ each of H$_2$ and I$_2$. The ICE table gives equilibrium $[\text{H}_2]=[\text{I}_2]=0.100-x$ and $[\text{HI}]=2x$. Substitute:

    $$K_c=\frac{(2x)^2}{(0.100-x)^2}=50\;\Rightarrow\;\frac{2x}{0.100-x}=\sqrt{50}=7.07\;\Rightarrow\;x=0.078,$$
    so $[\text{HI}]=2x=0.156\ \text{M}$. Taking the square root of both sides works here because the expression is a perfect square.

    7.8

    Representations of Equilibrium

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 7.8.A: Represent a system undergoing a reversible reaction with a particulate model.

    • 7.8.A.1 Particulate representations can be used to describe the relative numbers of reactant and product particles present prior to and at equilibrium, and the value of the equilibrium constant.
    日本語

    学習目標 7.8.A: 可逆反応を行う系を粒子モデルで表すこと。

    • 7.8.A.1 粒子による表現は、平衡前および平衡時における試薬と生成物の粒子数の相対的な数、および平衡定数の値を記述するために使用できます。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    A particulate picture at equilibrium shows a fixed mix of reactant and product particles. Graphs of concentration vs time level off (never reaching zero) once equilibrium is reached – both curves flatten at the same moment.

    7.9

    Introduction to Le Chatelier's Principle

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 7.9.A: Identify the response of a system at equilibrium to an external stress, using Le Châtelier's principle.

    • 7.9.A.1 Le Châtelier's principle can be used to predict the response of a system to stresses such as addition or removal of a chemical species, change in temperature, change in volume/pressure of a gas-phase system, or dilution of a reaction system.
    • 7.9.A.2 Le Châtelier's principle can be used to predict the effect that a stress will have on experimentally measurable properties such as pH, temperature, and color of a solution.
    日本語

    学習目標 7.9.A: ル・シャトリエの原理を用いて、外部ストレスに対する平衡状態にある系の応答を特定すること。

    • 7.9.A.1 ル・シャトリエの原理は、化学種添加または除去、温度変化、気相系における体積/圧力の変化、あるいは反応系の希釈といったストレスに対する系の応答を予測するために使用できます。
    • 7.9.A.2 ル・シャトリエの原理は、pH、温度、溶液の色といった実験的に測定可能な性質に対して、ストレスが及ぼす影響を予測するために使用できます。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    Le Chatelier's principle

    Le Chatelier's principle 勒沙特列原理: if you disturb a system at equilibrium, it shifts to partly counteract the change. Adding a reactant shifts forward; removing product shifts forward; increasing pressure (by reducing volume) shifts toward the side with fewer gas moles; raising temperature shifts in the endothermic direction.

    日本語
    Le Chatelier's principle

    Le Chatelier's principle 勒沙特列原理: if you disturb a system at equilibrium, it shifts to partly counteract the change. Adding a reactant shifts forward; removing product shifts forward; increasing pressure (by reducing volume) shifts toward the side with fewer gas moles; raising temperature shifts in the endothermic direction.

    Le Chatelier's principle: the equilibrium shifts to oppose the change made
    Le Chatelier's principle: the equilibrium shifts to oppose the change made
    An ammonia plant: industrial equilibria like Haber's process are driven by pressure, temperature and catalysts
    An ammonia plant: industrial equilibria like Haber's process are driven by pressure, temperature and catalysts
    Explore · ⁨探索⁩

    Disturb an equilibrium · ⁨平衡系に乱れを与える⁩

    Le Chatelier's principle: an equilibrium shifts to oppose a change. Add reactant, change pressure or temperature and watch the position of equilibrium move. · ⁨ルシャトリエの原理 :平衡系は変化に反対する方向に移動します。反応物を加えたり、圧力や温度を変えたりすると、平衡位置が動くのが見えます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Le Chatelier's principle/lə ˈtʃeɪtlɪəz ˈprɪnsɪpl/ ル・シャトリエの原理
    7.10

    The Reaction Quotient and Le Chatelier's Principle

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 7.10.A: Explain the relationships between $Q$, $K$, and the direction in which a reversible reaction will proceed to reach equilibrium.

    • 7.10.A.1 A disturbance to a system at equilibrium causes $Q$ to differ from $K$, thereby taking the system out of equilibrium. The system responds by bringing $Q$ back into agreement with $K$, thereby establishing a new equilibrium state.
    • 7.10.A.2 Some stresses, such as changes in concentration, cause a change in $Q$ only. A change in temperature causes a change in $K$. In either case, the concentrations or partial pressures of species redistribute to bring $Q$ and $K$ back into equality.
    日本語

    学習目標 7.10.A: $Q$、$K$、および可逆反応が平衡に達するために進行する方向との関係を説明する。

    • 7.10.A.1 平衡状態にある系への乱れにより、$Q$ が $K$ と異なり、系が平衡から外れる。系は応答して、$Q$ を $K$ に一致させ、新たな平衡状態を確立する。
    • 7.10.A.2 濃度の変化のようなある応力に対しては、$Q$のみが変化する。温度の変化は$K$を変化させる。いずれの場合も、化学種浓度または分圧は再分配され、$Q$と$K$が等しくなるようにする。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    You can justify every Le Chatelier shift with $Q$ versus $K$: a disturbance changes $Q$, and the system reacts to bring $Q$ back to $K$. This quantitative view backs up the qualitative rule and is the fuller answer on the exam.

    Explore · ⁨探索⁩

    Compare Q with K · ⁨QとKを比較する⁩

    If the reaction quotient $Q the forward reaction is favoured; $Q>K$ favours the reverse. Perturb the system and watch it move back toward $Q=K$. · ⁨もし 反応商 $Q 则 forward reaction is favored; $Q>K$ favours the reverse. Perturb the system and watch it move back toward $Q=K$.⁩

    7.11 7.12

    Introduction to Solubility Equilibria

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 7.11.A: Calculate the solubility of a salt based on the value of $K_{sp}$ for the salt.

    • 7.11.A.1 The dissolution of a salt is a reversible process whose extent can be described by $K_{sp}$, the solubility-product constant.
    • 7.11.A.2 The solubility of a substance can be calculated from the $K_{sp}$ for the dissolution process. This relationship can also be used to predict the relative solubility of different substances.
    • 7.11.A.3 The solubility rules (see 4.7.A.5) can be quantitatively related to $K_{sp}$, in which $K_{sp}$ values $>1$ correspond to soluble salts.
    • 7.11.A.4 The molar solubility of one or more species in a saturated solution can be used to calculate the $K_{sp}$ of a substance.
    日本語

    学習目標 7.11.A: $K_{sp}$の値に基づき、塩の溶解度を計算する。

    • 7.11.A.1 塩の溶解は可逆過程であり、その程度は$K_{sp}$(溶解度積定数)によって記述される。
    • 7.11.A.2 物質の溶解度は、溶解反応に対する$K_{sp}$から計算できる。この関係式は、異なる物質間の相対的な溶解度の予測にも用いられる。
    • 7.11.A.3 溶解性規則(4.7.A.5 を参照)は、$K_{sp}$ と定量的に関係付けられ、$K_{sp}$ が $>1$ の場合、可溶性塩に対応します。
    • 7.11.A.4 飽和溶液中の一種以上の種のモル溶解度は、その物質の $K_{sp}$ を計算するために使用できます。
    English

    Learning Objective 7.12.A: Identify the solubility of a salt, and/or the value of $K_{sp}$ for the salt, based on the concentration of a common ion already present in solution.

    • 7.12.A.1 The solubility of a salt is reduced when it is dissolved into a solution that already contains one of the ions present in the salt. The impact of this "common-ion effect" on solubility can be understood qualitatively using Le Châtelier's principle or calculated from the $K_{sp}$ for the dissolution process.
    日本語

    学習目標 7.12.A: 溶液内に既に存在する共通イオンの濃度に基づき、塩の溶解度および/または$K_{sp}$の値を特定する。

    • 7.12.A.1 塩を、その塩に含まれるイオンのいずれかを含む溶液に溶解させると、溶解度は低下する。「共通イオン効果」が溶解度に与える影響は、ル・シャトリエの原理を用いて定性的に理解したり、溶解反応の$K_{sp}$から計算したりして説明できる。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    For a slightly soluble salt, the solubility product 溶度积 $K_{sp}$ is the equilibrium constant for its dissolving:

    $$\text{M}_a\text{X}_b(s)\rightleftharpoons a\,\text{M}^{b+}+b\,\text{X}^{a-},\qquad K_{sp}=[\text{M}^{b+}]^a[\text{X}^{a-}]^b.$$
    A smaller $K_{sp}$ means less soluble. The common-ion effect – adding an ion already in the salt – pushes the equilibrium back and lowers solubility.

    Worked example. Silver chloride has $K_{sp}=1.8\times10^{-10}$. If its molar solubility is $s$, then $[\text{Ag}^+]=[\text{Cl}^-]=s$, so $K_{sp}=s^2$ and

    $$s=\sqrt{1.8\times10^{-10}}=1.3\times10^{-5}\ \text{M}.$$
    Adding NaCl (a common ion) would raise $[\text{Cl}^-]$, forcing $s$ down – far less AgCl would dissolve.

    日本語

    For a slightly soluble salt, the solubility product 溶度积 $K_{sp}$ is the equilibrium constant for its dissolving:

    $$\text{M}_a\text{X}_b(s)\rightleftharpoons a\,\text{M}^{b+}+b\,\text{X}^{a-},\qquad K_{sp}=[\text{M}^{b+}]^a[\text{X}^{a-}]^b.$$
    A smaller $K_{sp}$ means less soluble. The common-ion effect – adding an ion already in the salt – pushes the equilibrium back and lowers solubility.

    Worked example. Silver chloride has $K_{sp}=1.8\times10^{-10}$. If its molar solubility is $s$, then $[\text{Ag}^+]=[\text{Cl}^-]=s$, so $K_{sp}=s^2$ and

    $$s=\sqrt{1.8\times10^{-10}}=1.3\times10^{-5}\ \text{M}.$$
    Adding NaCl (a common ion) would raise $[\text{Cl}^-]$, forcing $s$ down – far less AgCl would dissolve.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    solubility product/ˌsɒljuːˈbɪlɪti ˈprɒdʌkt/ 難溶性塩のイオン積
    7.11 7.12

    Exam tips

    • At equilibrium the forward and reverse rates are equal, but the amounts are usually not.
    • Leave pure solids and liquids out of the $K$ expression; a large $K$ favours products, a small $K$ favours reactants.
    • Apply Le Chatelier: add reactant → shift forward; raise pressure → shift toward fewer gas moles; raise temperature → shift in the endothermic direction.
    • A catalyst does not shift the position of equilibrium — it only speeds the approach.
    • Use an ICE table for equilibrium concentrations, and the small-$x$ approximation when $K$ is small.
  • 8

    Acids and Bases · ⁨酸と塩基⁩

    Watch lesson · ⁨レッスンを視聴⁩
    8.1

    Introduction to Acids and Bases

    Syllabus · ⁨シラバス⁩
    Learning ObjectiveEssential Knowledge

    8.1.A
    Calculate the values of $\mathrm{pH}$ and $\mathrm{pOH}$, based on $K_w$ and the concentration of all species present in a neutral solution of water.

    • 8.1.A.1 The concentrations of hydronium ion and hydroxide ion are often reported as $\mathrm{pH}$ and $\mathrm{pOH}$, respectively.
      • Equation: $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$
      • Equation: $\mathrm{pOH} = -\log[\mathrm{OH^-}]$
      • The terms "hydrogen ion" and "hydronium ion" and the symbols $\mathrm{H^+}(aq)$ and $\mathrm{H_3O^+}(aq)$ are often used interchangeably for the aqueous ion of hydrogen. Hydronium ion and $\mathrm{H_3O^+}(aq)$ are preferred, but $\mathrm{H^+}(aq)$ is also accepted on the AP Exam.
    • 8.1.A.2 Water autoionizes with an equilibrium constant $K_w$.
      • Equation: $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$ at 25°C
    • 8.1.A.3 In pure water, $\mathrm{pH} = \mathrm{pOH}$ is called a neutral solution. At 25°C, $\mathrm{p}K_w = 14.0$ and thus $\mathrm{pH} = \mathrm{pOH} = 7.0$.
      • Equation: $\mathrm{p}K_w = 14 = \mathrm{pH} + \mathrm{pOH}$ at 25°C
    • 8.1.A.4 The value of $K_w$ is temperature dependent, so the pH of pure, neutral water will deviate from 7.0 at temperatures other than 25°C.

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    By the Brønsted–Lowry definition, an acid 酸 is a proton ($\text{H}^+$) donor and a base 碱 is a proton acceptor. When an acid donates a proton it becomes its conjugate base 共轭碱; a base gaining a proton becomes its conjugate acid 共轭酸. Water is amphoteric – it can act as either.

    日本語
    Litmus paper: acid–base indicators change colour with H⁺ concentration (pH)
    Litmus paper: acid–base indicators change colour with H⁺ concentration (pH)

    By the Brønsted–Lowry definition, an acid 酸 is a proton ($\text{H}^+$) donor and a base 碱 is a proton acceptor. When an acid donates a proton it becomes its conjugate base 共轭碱; a base gaining a proton becomes its conjugate acid 共轭酸. Water is amphoteric – it can act as either.

    A strong acid is fully dissociated; a weak acid is only partly dissociated
    A strong acid is fully dissociated; a weak acid is only partly dissociated
    pH test strips: acid–base strength and concentration set the pH you measure
    pH test strips: acid–base strength and concentration set the pH you measure
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    acid/ˈæsɪd/ 酸
    base/beɪs/ 底
    conjugate base/ˈkɒndʒuːɡeɪt beɪs/ 共役塩基
    conjugate acid/ˈkɒndʒuːɡeɪt ˈæsɪd/ 共役酸
    8.1

    The Autoionization of Water

    Syllabus · ⁨シラバス⁩
    Learning ObjectiveEssential Knowledge

    8.1.A
    Calculate the values of $\mathrm{pH}$ and $\mathrm{pOH}$, based on $K_w$ and the concentration of all species present in a neutral solution of water.

    • 8.1.A.1 The concentrations of hydronium ion and hydroxide ion are often reported as $\mathrm{pH}$ and $\mathrm{pOH}$, respectively.
      • Equation: $\mathrm{pH} = -\log[\mathrm{H_3O^+}]$
      • Equation: $\mathrm{pOH} = -\log[\mathrm{OH^-}]$
      • The terms "hydrogen ion" and "hydronium ion" and the symbols $\mathrm{H^+}(aq)$ and $\mathrm{H_3O^+}(aq)$ are often used interchangeably for the aqueous ion of hydrogen. Hydronium ion and $\mathrm{H_3O^+}(aq)$ are preferred, but $\mathrm{H^+}(aq)$ is also accepted on the AP Exam.
    • 8.1.A.2 Water autoionizes with an equilibrium constant $K_w$.
      • Equation: $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$ at 25°C
    • 8.1.A.3 In pure water, $\mathrm{pH} = \mathrm{pOH}$ is called a neutral solution. At 25°C, $\mathrm{p}K_w = 14.0$ and thus $\mathrm{pH} = \mathrm{pOH} = 7.0$.
      • Equation: $\mathrm{p}K_w = 14 = \mathrm{pH} + \mathrm{pOH}$ at 25°C
    • 8.1.A.4 The value of $K_w$ is temperature dependent, so the pH of pure, neutral water will deviate from 7.0 at temperatures other than 25°C.

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Even pure water conducts a tiny current, because water molecules react with each other in a reaction called autoionization 自偶电离:

    $$2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-.$$
    One water molecule passes a proton to another, making a hydronium ion 水合氢离子 ($\text{H}_3\text{O}^+$, the ion we loosely write as $\text{H}^+$) and a hydroxide ion. This equilibrium has its own constant, the ion-product constant of water 水的离子积:
    $$K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0\times10^{-14}$$
    at 25 °C. Defining $\text{pOH}=-\log[\text{OH}^-]$ and taking $-\log$ of the $K_w$ expression gives the rule the rest of this topic leans on:
    $$\text{pH} + \text{pOH} = \text{p}K_w = 14.0$$
    (again at 25 °C). Know one of pH or pOH and you get the other by subtracting from 14.

    In pure water the two ions form in equal numbers, so $[\text{H}_3\text{O}^+]=[\text{OH}^-]$ and $\text{pH}=\text{pOH}=7.0$. This is what neutral 中性 means. Acidic means $[\text{H}_3\text{O}^+]>[\text{OH}^-]$ (pH below 7); basic is the reverse.

    Worked example. A solution has $[\text{H}_3\text{O}^+]=2.0\times10^{-3}\ \text{M}$. Find $[\text{OH}^-]$. Because $K_w$ always holds in water,

    $$[\text{OH}^-]=\frac{K_w}{[\text{H}_3\text{O}^+]}=\frac{1.0\times10^{-14}}{2.0\times10^{-3}}=5.0\times10^{-12}\ \text{M}.$$
    The hydroxide concentration is far smaller than the hydronium, confirming the solution is acidic.

    $K_w$ is temperature dependent: autoionization is endothermic, so heating water shifts it forward and raises $K_w$. At 50 °C, $K_w>1.0\times10^{-14}$, so neutral water has $\text{pH}=\text{pOH}<7.0$. It is still neutral, because the ion concentrations are still equal – neutral means equal ions, not a pH of exactly 7.

    日本語

    Even pure water conducts a tiny current, because water molecules react with each other in a reaction called autoionization 自偶电离:

    $$2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-.$$
    One water molecule passes a proton to another, making a hydronium ion 水合氢离子 ($\text{H}_3\text{O}^+$, the ion we loosely write as $\text{H}^+$) and a hydroxide ion. This equilibrium has its own constant, the ion-product constant of water 水的离子积:
    $$K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0\times10^{-14}$$
    at 25 °C. Defining $\text{pOH}=-\log[\text{OH}^-]$ and taking $-\log$ of the $K_w$ expression gives the rule the rest of this topic leans on:
    $$\text{pH} + \text{pOH} = \text{p}K_w = 14.0$$
    (again at 25 °C). Know one of pH or pOH and you get the other by subtracting from 14.

    In pure water the two ions form in equal numbers, so $[\text{H}_3\text{O}^+]=[\text{OH}^-]$ and $\text{pH}=\text{pOH}=7.0$. This is what neutral 中性 means. Acidic means $[\text{H}_3\text{O}^+]>[\text{OH}^-]$ (pH below 7); basic is the reverse.

    Worked example. A solution has $[\text{H}_3\text{O}^+]=2.0\times10^{-3}\ \text{M}$. Find $[\text{OH}^-]$. Because $K_w$ always holds in water,

    $$[\text{OH}^-]=\frac{K_w}{[\text{H}_3\text{O}^+]}=\frac{1.0\times10^{-14}}{2.0\times10^{-3}}=5.0\times10^{-12}\ \text{M}.$$
    The hydroxide concentration is far smaller than the hydronium, confirming the solution is acidic.

    $K_w$ is temperature dependent: autoionization is endothermic, so heating water shifts it forward and raises $K_w$. At 50 °C, $K_w>1.0\times10^{-14}$, so neutral water has $\text{pH}=\text{pOH}<7.0$. It is still neutral, because the ion concentrations are still equal – neutral means equal ions, not a pH of exactly 7.

    pH strips: water autoionizes so pure water is pH 7; acids and bases shift H+ and OH-
    pH strips: water autoionizes so pure water is pH 7; acids and bases shift H+ and OH-
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    autoionization/ˌɔːtəʊˌaɪənaɪˈzeɪʃn/ 自己解離
    hydronium ion/haɪˈdrəʊnɪəm ˈaɪɒn/ ヒドロニウムイオン
    ion-product constant of water/ˈaɪɒn ˈprɒdʌkt ˈkɒnstənt ɒv ˈwɔːtə/ 水のイオン積定数
    neutral/ˈnjuːtrəl/ 中性
    8.2

    pH and pOH of Strong Acids and Bases

    Syllabus · ⁨シラバス⁩
    Learning ObjectiveEssential Knowledge

    8.2.A
    Calculate $\mathrm{pH}$ and $\mathrm{pOH}$ based on concentrations of all species in a solution of a strong acid or a strong base.

    • 8.2.A.1 Molecules of a strong acid (e.g., $\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$, $\mathrm{HClO_4}$, $\mathrm{H_2SO_4}$, and $\mathrm{HNO_3}$) will completely ionize in aqueous solution to produce hydronium ions and the conjugate base of the acid. As such, the concentration of $\mathrm{H_3O^+}$ in a strong acid solution is equal to the initial concentration of the strong acid, and thus the $\mathrm{pH}$ of the strong acid solution is easily calculated.
    • 8.2.A.2 When dissolved in solution, strong bases (e.g., group I and II hydroxides) completely dissociate to produce hydroxide ions. As such, the concentration of $\mathrm{OH^-}$ in a strong base solution is equal to the initial concentration of a group I hydroxide and double the initial concentration of a group II hydroxide, and thus the $\mathrm{pOH}$ (and $\mathrm{pH}$) of the strong base solution is easily calculated.

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    The pH pH值 scale measures acidity: $\text{pH}=-\log[\text{H}^+]$, with $\text{pH}+\text{pOH}=14$ at 25 °C. A strong acid 强酸 or strong base 强碱 dissociates completely, so its ion concentration equals its concentration – read pH directly. Lower pH means more acidic.

    Worked example. Find the pH of $0.010\ \text{M}$ HCl. Because HCl is a strong acid it is fully dissociated, so $[\text{H}^+]=0.010\ \text{M}$ and

    $$\text{pH}=-\log(0.010)=2.0.$$
    For a strong base like $0.010\ \text{M}$ NaOH, $\text{pOH}=2.0$, so $\text{pH}=14-2.0=12.0$.

    日本語

    The pH pH值 scale measures acidity: $\text{pH}=-\log[\text{H}^+]$, with $\text{pH}+\text{pOH}=14$ at 25 °C. A strong acid 强酸 or strong base 强碱 dissociates completely, so its ion concentration equals its concentration – read pH directly. Lower pH means more acidic.

    The pH scale: pH is the negative log of the hydrogen-ion concentration
    The pH scale: pH is the negative log of the hydrogen-ion concentration

    Worked example. Find the pH of $0.010\ \text{M}$ HCl. Because HCl is a strong acid it is fully dissociated, so $[\text{H}^+]=0.010\ \text{M}$ and

    $$\text{pH}=-\log(0.010)=2.0.$$
    For a strong base like $0.010\ \text{M}$ NaOH, $\text{pOH}=2.0$, so $\text{pH}=14-2.0=12.0$.

    Explore · ⁨探索⁩

    Move along the pH scale

    pH measures hydrogen-ion concentration on a log scale: each unit is a tenfold change. Strong acids sit low, strong bases high, 7 is neutral.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    pH/piː eɪtʃ/ pH
    strong acid/strɒŋ ˈæsɪd/ 強酸
    strong base/strɒŋ beɪs/ 強塩基
    8.3

    Weak Acid and Base Equilibria

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 8.3.A: Explain the relationship among $\mathrm{pH}$, $\mathrm{pOH}$, and concentrations of all species in a solution of a monoprotic weak acid or weak base.

    • 8.3.A.1 Weak acids react with water to produce hydronium ions. However, only a small percentage of molecules of a weak acid will ionize in this way. Thus, the concentration of $\mathrm{H_3O^+}$ is much less than the initial concentration of the molecular acid, and the vast majority of the acid molecules remain un-ionized.
    • 8.3.A.2 A solution of a weak acid involves equilibrium between an un-ionized acid and its conjugate base. The equilibrium constant for this reaction is $K_a$, often reported as $\mathrm{p}K_a$. The pH of a weak acid solution can be determined from the initial acid concentration and the $\mathrm{p}K_a$.
      • Equation: $K_a = \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$
      • Equation: $\mathrm{p}K_a = -\log K_a$
    • 8.3.A.3 Weak bases react with water to produce hydroxide ions in solution. However, ordinarily just a small percentage of the molecules of a weak base in solution will ionize in this way. Thus, the concentration of $\mathrm{OH^-}$ in the solution does not equal the initial concentration of the base, and the vast majority of the base molecules remain un-ionized.
    • 8.3.A.4 A solution of a weak base involves equilibrium between an un-ionized base and its conjugate acid. The equilibrium constant for this reaction is $K_b$, often reported as $\mathrm{p}K_b$. The $\mathrm{pH}$ of a weak base solution can be determined from the initial base concentration and the $\mathrm{p}K_b$.
      • Equation: $K_b = \dfrac{[\mathrm{OH^-}][\mathrm{HB^+}]}{[\mathrm{B}]}$
      • Equation: $\mathrm{p}K_b = -\log K_b$
    • 8.3.A.5 The percent ionization of a weak acid (or base) can be calculated from its $\mathrm{p}K_a$ ($\mathrm{p}K_b$) and the initial concentration of the acid (base). The percent ionization can also be calculated from the initial concentration of the acid (base) and the equilibrium concentration of any of the species in the equilibrium expression.
    • 8.3.A.6 For any conjugate acid-base pair, the acid ionization constant and base ionization constant are related by $K_w$:
      • Equation: $K_w = K_a \times K_b$
      • Equation: $\mathrm{p}K_w = \mathrm{p}K_a + \mathrm{p}K_b$
    日本語

    学習目標 8.3.A: $\mathrm{pH}$、$\mathrm{pOH}$ 、および単プロトン弱酸または弱塩基の溶液中の全化学種の濃度の間の関係を説明すること。

    • 8.3.A.1 弱酸は水と反応してヒドリオンを生じますが、このように電離する弱酸の分子はごく一部に過ぎません。したがって、$\mathrm{H_3O^+}$ の濃度は分子状酸の初期濃度よりはるかに小さく、酸の大部分は未電離のままです。
    • 8.3.A.2 弱酸の溶液には、未電離酸とその共役塩基との間に平衡が存在します。この反応の平衡定数は $K_a$ であり、通常は $\mathrm{p}K_a$ として報告されます。弱酸溶液の pH は、初期酸濃度と $\mathrm{p}K_a$ を用いて決定できます。
      • 式: $K_a = \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$
      • 式: $\mathrm{p}K_a = -\log K_a$
    • 8.3.A.3 弱塩基は水と反応して溶液中にヒドロキシドイオンを生じますが、溶液中の弱塩基の分子の多くは、このように電離することは一般的ではありません。したがって、溶液中の $\mathrm{OH^-}$ の濃度は塩基の初期濃度とは異なり、塩基の大部分は未電離のままです。
    • 8.3.A.4 弱塩基の溶液における平衡は、非イオン化塩基とその共役酸との間に存在する。この反応の平衡定数は $K_b$ で表され、通常は $\mathrm{p}K_b$ として報告される。弱塩基溶液の $\mathrm{pH}$ は、初期塩基濃度と $\mathrm{p}K_b$ から決定できる。
      • 式: $K_b = \dfrac{[\mathrm{OH^-}][\mathrm{HB^+}]}{[\mathrm{B}]}$
      • 式: $\mathrm{p}K_b = -\log K_b$
    • 8.3.A.5 弱酸(または塩基)の電離率は、その $\mathrm{p}K_a$ ($\mathrm{p}K_b$) と酸(または塩基)の初期濃度から計算できる。また、平衡式に含まれるいかなる種の平衡濃度と酸(または塩基)の初期濃度からも電離率を計算できる。
    • 8.3.A.6 いかなる共役酸塩基対についても、酸電離定数と塩基電離定数は $K_w$ の関係にある:
      • 式: $K_w = K_a \times K_b$
      • 式: $\mathrm{p}K_w = \mathrm{p}K_a + \mathrm{p}K_b$

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    A weak acid 弱酸 only partly dissociates, described by an equilibrium constant $K_a$ (larger $K_a$ = stronger weak acid); a weak base has $K_b$. For any conjugate acid–base pair the two are linked through water, $K_a\times K_b = K_w$, so $\text{p}K_a+\text{p}K_b=14$ – a stronger acid always has a weaker conjugate base. Because dissociation is small, find $[\text{H}^+]$ with an ICE table and the $K_a$ expression, often using the small-$x$ approximation.

    Worked example. Find the pH of $0.10\ \text{M}$ acetic acid, $K_a=1.8\times10^{-5}$. Let $x=[\text{H}^+]$ at equilibrium; the ICE table gives $K_a=\dfrac{x^2}{0.10-x}\approx\dfrac{x^2}{0.10}$ (small-$x$ approximation), so

    $$x=\sqrt{K_a\times0.10}=\sqrt{1.8\times10^{-6}}=1.3\times10^{-3}\ \text{M}\;\Rightarrow\;\text{pH}=-\log(1.3\times10^{-3})=2.9.$$
    The weak acid is far less acidic than a strong acid of the same concentration (which would be pH $1.0$).

    日本語

    A weak acid 弱酸 only partly dissociates, described by an equilibrium constant $K_a$ (larger $K_a$ = stronger weak acid); a weak base has $K_b$. For any conjugate acid–base pair the two are linked through water, $K_a\times K_b = K_w$, so $\text{p}K_a+\text{p}K_b=14$ – a stronger acid always has a weaker conjugate base. Because dissociation is small, find $[\text{H}^+]$ with an ICE table and the $K_a$ expression, often using the small-$x$ approximation.

    Worked example. Find the pH of $0.10\ \text{M}$ acetic acid, $K_a=1.8\times10^{-5}$. Let $x=[\text{H}^+]$ at equilibrium; the ICE table gives $K_a=\dfrac{x^2}{0.10-x}\approx\dfrac{x^2}{0.10}$ (small-$x$ approximation), so

    $$x=\sqrt{K_a\times0.10}=\sqrt{1.8\times10^{-6}}=1.3\times10^{-3}\ \text{M}\;\Rightarrow\;\text{pH}=-\log(1.3\times10^{-3})=2.9.$$
    The weak acid is far less acidic than a strong acid of the same concentration (which would be pH $1.0$).

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    weak acid/wiːk ˈæsɪd/ 弱酸
    8.4

    Acid-Base Reactions and Buffers

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 8.4.A: Explain the relationship among the concentrations of major species in a mixture of weak and strong acids and bases.

    • 8.4.A.1 When a strong acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation: $\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \rightarrow \mathrm{H_2O}(l)$. The pH of the resulting solution may be determined from the concentration of excess reagent.
    • 8.4.A.2 When a weak acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation: $\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftarrows \mathrm{A^-}(aq)\ \mathrm{H_2O}(l)$. If the weak acid is in excess, then a buffer solution is formed, and the $\mathrm{pH}$ can be determined from the Henderson-Hasselbalch (H–H) equation (see 8.9.A.1). If the strong base is in excess, then the $\mathrm{pH}$ can be determined from the moles of excess hydroxide ion and the total volume of solution. If they are equimolar, then the (slightly basic) $\mathrm{pH}$ can be determined from the equilibrium represented by the equation: $\mathrm{A^-}(aq) + \mathrm{H_2O}(l) \rightleftarrows \mathrm{HA}(aq) + \mathrm{OH^-}(aq)$.
    • 8.4.A.3 When a weak base and a strong acid are mixed, they will react quantitatively in a reaction represented by the equation: $\mathrm{B}(aq) + \mathrm{H_3O^+}(aq) \rightleftarrows \mathrm{HB^+}(aq) + \mathrm{H_2O}(l)$. If the weak base is in excess, then a buffer solution is formed, and the $\mathrm{pH}$ can be determined from the H–H equation. If the strong acid is in excess, then the $\mathrm{pH}$ can be determined from the moles of excess hydronium ion and the total volume of solution. If they are equimolar, then the (slightly acidic) $\mathrm{pH}$ can be determined from the equilibrium represented by the equation: $\mathrm{HB^+}(aq) + \mathrm{H_2O}(l) \rightleftarrows \mathrm{B}(aq) + \mathrm{H_3O^+}(aq)$.
    • 8.4.A.4 When a weak acid and a weak base are mixed, they will react to an equilibrium state whose reaction may be represented by the equation: $\mathrm{HA}(aq) + \mathrm{B}(aq) \rightleftarrows \mathrm{A^-}(aq) + \mathrm{HB^+}(aq)$.
    日本語

    学習目標 8.4.A: 弱酸・強酸および弱塩基・強塩基の混合物における主要な種の濃度の関係を説明せよ。

    • 8.4.A.1 強酸と強塩基を混合すると、方程式で表される反応に従って定量反応を起こす: $\mathrm{H^+}(aq) + \mathrm{OH^-}(aq) \rightarrow \mathrm{H_2O}(l)$ 。得られた溶液のpHは、過剰試薬の濃度から決定できる。
    • 8.4.A.2 弱酸と強塩基を混合すると、化学量論的に反応し、その反応は次の式で表される:$\mathrm{HA}(aq) + \mathrm{OH^-}(aq) \rightleftarrows \mathrm{A^-}(aq)\ \mathrm{H_2O}(l)$。弱酸が過剰である場合、緩衝溶液が形成され、$\mathrm{pH}$ はヘンダーソン=ハ塞尔バルフ(H–H)式によって決定できる(8.9.A.1 を参照)。強塩基が過剰である場合、$\mathrm{pH}$ は過剰の水酸化物イオンのモル数と溶液の全容積から決定できる。両者が等モルである場合、(わずかに塩基性な)$\mathrm{pH}$ は、次の式で表される平衡系から決定できる:$\mathrm{A^-}(aq) + \mathrm{H_2O}(l) \rightleftarrows \mathrm{HA}(aq) + \mathrm{OH^-}(aq)$。
    • 8.4.A.3 弱塩基と強酸を混合すると、方程式で表される反応に従って定量反応を起こす: $\mathrm{B}(aq) + \mathrm{H_3O^+}(aq) \rightleftarrows \mathrm{HB^+}(aq) + \mathrm{H_2O}(l)$ 。弱塩基が過剰の場合、緩衝溶液が形成され、 $\mathrm{pH}$ はH–H式から決定できる。強酸が過剰の場合、 $\mathrm{pH}$ は過剰ヒドロニウムイオンの物質量と溶液の全容積から決定できる。等モル量の場合、(わずかに酸性である) $\mathrm{pH}$ は、方程式で表される平衡から決定できる: $\mathrm{HB^+}(aq) + \mathrm{H_2O}(l) \rightleftarrows \mathrm{B}(aq) + \mathrm{H_3O^+}(aq)$ 。
    • 8.4.A.4 弱酸と弱塩基を混合すると、方程式で表される反応が起こり、平衡状態に至る: $\mathrm{HA}(aq) + \mathrm{B}(aq) \rightleftarrows \mathrm{A^-}(aq) + \mathrm{HB^+}(aq)$ 。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    A buffer 缓冲溶液 resists pH change. It contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable amounts. Added acid is neutralized by the conjugate base, and added base by the weak acid, so pH barely moves.

    日本語

    A buffer 缓冲溶液 resists pH change. It contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable amounts. Added acid is neutralized by the conjugate base, and added base by the weak acid, so pH barely moves.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    buffer/ˈbʌfə/ バッファー
    8.5

    Acid-Base Titrations

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 8.5.A: Explain results from the titration of a mono- or polyprotic acid or base solution, in relation to the properties of the solution and its components.

    • 8.5.A.1 An acid-base reaction can be carried out under controlled conditions in a titration. A titration curve, plotting $\mathrm{pH}$ against the volume of titrant added, is useful for summarizing results from a titration.
    • 8.5.A.2 At the equivalence point for titrations of monoprotic acids or bases, the number of moles of titrant added is equal to the number of moles of analyte originally present. This relationship can be used to obtain the concentration of the analyte. This is the case for titrations of strong acids/bases and weak acids/bases.
    • 8.5.A.3 For titrations of weak acids/bases, it is useful to consider the point halfway to the equivalence point, that is, the half-equivalence point. At this point, there are equal concentrations of each species in the conjugate acid-base pair, for example, for a weak acid $[\mathrm{HA}] = [\mathrm{A^-}]$. Because $\mathrm{pH} = \mathrm{p}K_a$ when the conjugate acid and base have equal concentrations, the $\mathrm{p}K_a$ can be determined from the $\mathrm{pH}$ at the half-equivalence point in a titration.
    • 8.5.A.4 At the equivalence point, pH is determined by the major species in solution. Strong acid and strong base titrations result in neutral pH at the equivalence point. However, in titrations of weak acids (weak bases), the conjugate base of the weak acid (conjugate acid of the weak base) is present at the equivalence point and can undergo proton-transfer reactions with the surrounding water, producing basic (acidic) solutions.
    • 8.5.A.5 For polyprotic acids, titration curves can be used to determine the number of acidic protons. In doing so, the major species present at any point along the curve can be identified, along with the $\mathrm{p}K_a$ associated with each proton in a weak polyprotic acid.
      • Exclusion statement: Computation of the concentration of each species present in the titration curve for polyprotic acids will not be assessed on the AP Exam. Such computations for titration of monoprotic acids are within the scope of the course (see 8.4.A.2 and 8.4.A.3), as is qualitative reasoning regarding what species are present in large versus small concentrations at any point in a titration of a polyprotic acid.
    日本語

    学習目標 8.5.A: 単プロトン酸・塩基溶液および多プロトン酸・塩基溶液の滴定結果を、溶液とその成分の性質に関連付けて説明せよ。

    • 8.5.A.1 酸塩基反応は、滴定という制御条件下で行うことができる。添加した滴定剤の体積に対して $\mathrm{pH}$ をプロットした滴定曲線は、滴定の結果を要約するために有用である。
    • 8.5.A.2 単プロトン酸または塩基の滴定における等量点では、添加された滴定剤の物質量は、もともと存在していた分析対象物の物質量に等しい。この関係を用いて分析対象物の濃度を求めることができる。これは強酸・強塩基の滴定および弱酸・弱塩基の滴定の両方に当てはまる。
    • 8.5.A.3 弱酸/弱塩基の滴定においては、等量点への途中の半分、すなわち半等量点を考えることが有用である。この時点では、共役酸塩基対の各種の濃度が等しくなる。例えば、弱酸 $[\mathrm{HA}] = [\mathrm{A^-}]$ の場合などである。共役酸と共役塩基の濃度が等しいとき $\mathrm{pH} = \mathrm{p}K_a$ となるため、 $\mathrm{p}K_a$ は滴定中の半等量点における $\mathrm{pH}$ から決定できる。
    • 8.5.A.4 等量点におけるpHは、溶液中の主要な種によって決定される。強酸と強塩基の滴定では等量点で中性のpHとなる。しかし、弱酸(弱塩基)の滴定では、等量点に弱酸の共役塩基(弱塩基の共役酸)が存在し、周囲の水とプロトン移動反応を起こして、塩基性(酸性)の溶液を生じる。
    • 8.5.A.5 多プロトン酸については、滴定曲線を用いて酸性プロトンの数を決定できる。この際、曲線上の任意の地点で存在する主要な種を特定し、さらに弱多プロトン酸の各プロトンに対応する $\mathrm{p}K_a$ も特定する。
      • 除外事項: 多プロトン酸の滴定曲線における各種の濃度の計算はAP試験の対象外とする。単プロトン酸の滴定に対する此类の計算は課程の範囲内である(8.4.A.2 および 8.4.A.3 参照)。また、多プロトン酸の滴定の任意の地点でどの種の濃度が大きいのか小さいのかに関する定性的な推論も出題範囲である。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    An acid–base titration curve

    A titration curve 滴定曲线 plots pH as base (or acid) is added. Key points: the equivalence point 等当点 (moles of acid = moles of base – a steep jump), and the half-equivalence point, where $\text{pH}=\text{p}K_a$ (half the weak acid is converted, so a buffer is at its center).

    An acid-base indicator 酸碱指示剂 is itself a weak acid whose protonated and deprotonated forms have different colours, so its colour responds to pH. You pick an indicator whose colour change happens near the titration's equivalence-point pH, so the colour flips just as the reaction completes.

    日本語
    An acid–base titration curve

    A titration curve 滴定曲线 plots pH as base (or acid) is added. Key points: the equivalence point 等当点 (moles of acid = moles of base – a steep jump), and the half-equivalence point, where $\text{pH}=\text{p}K_a$ (half the weak acid is converted, so a buffer is at its center).

    An acid-base indicator 酸碱指示剂 is itself a weak acid whose protonated and deprotonated forms have different colours, so its colour responds to pH. You pick an indicator whose colour change happens near the titration's equivalence-point pH, so the colour flips just as the reaction completes.

    A titration curve has a steep jump near the equivalence point
    A titration curve has a steep jump near the equivalence point
    Explore · ⁨探索⁩

    Titrate to the equivalence point

    A titration curve rises gently, then sharply at the equivalence point where moles of acid and base match. The steep jump pinpoints that volume.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    titration curve/taɪˈtreɪʃn kɜːv/ 滴定曲線
    equivalence point/ɪˈkwɪvələns pɔɪnt/ 当量点
    acid-base indicator/ˈæsɪd beɪs ˈɪndɪkeɪtə/ 酸塩基指示薬
    8.6

    Molecular Structure of Acids and Bases

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 8.6.A: Explain the relationship between the strength of an acid or base and the structure of the molecule or ion.

    • 8.6.A.1 The protons on a molecule that will participate in acid-base reactions, and the relative strength of these protons, can be inferred from the molecular structure.
      • i. Strong acids (such as $\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$, $\mathrm{HClO_4}$, $\mathrm{H_2SO_4}$, and $\mathrm{HNO_3}$) have very weak conjugate bases that are stabilized by electronegativity, inductive effects, resonance, or some combination thereof.
      • ii. Carboxylic acids are one common class of weak acid.
      • iii. Strong bases (such as group I and II hydroxides) have very weak conjugate acids.
      • iv. Common weak bases include nitrogenous bases such as ammonia as well as carboxylate ions.
      • v. Electronegative elements tend to stabilize the conjugate base relative to the conjugate acid, and so increase acid strength.
    日本語

    学習目標 8.6.A: 酸または塩基の強さと、分子またはイオンの構造との関係を説明せよ。

    • 8.6.A.1 酸塩基反応に関与する分子上のプロトンや、これらのプロトンの相対的な強さは、分子構造から推察できる。
      • i. 強酸( $\mathrm{HCl}$ 、 $\mathrm{HBr}$ 、 $\mathrm{HI}$ 、 $\mathrm{HClO_4}$ 、 $\mathrm{H_2SO_4}$ 、 $\mathrm{HNO_3}$ など)は非常に弱い共役塩基を持ち、これらは電気陰性度、誘起効果、共鳴効果、あるいはそれらの組み合わせにより安定化されている。
      • ii. カルボン酸は、一般的な弱酸の一種である。
      • iii. 強塩基(例:第I族および第II族水酸化物)は、非常に弱い共役酸を持つ。
      • iv. 一般的な弱塩基には、アンモニアなどの窒素含有塩基、およびカルボキシレートイオンが含まれる。
      • v. 電気陰性度の高い元素は、共役酸に対して共役塩基を安定化させる傾向があり、そのため酸性を強める。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    Strength has structural roots. An acid is stronger when its conjugate base is more stable – for example, more electronegative atoms or resonance spreading the negative charge stabilize it. For oxoacids, more oxygen atoms on the central atom means a stronger acid.

    8.7

    pH and pKa

    Syllabus · ⁨シラバス⁩
    Learning ObjectiveEssential Knowledge

    8.7.A
    Explain the relationship between the predominant form of a weak acid or base in solution at a given $\mathrm{pH}$ and the $\mathrm{p}K_a$ of the conjugate acid or the $\mathrm{p}K_b$ of the conjugate base.

    • 8.7.A.1 The protonation state of an acid or base (i.e., the relative concentrations of $\mathrm{HA}$ and $\mathrm{A^-}$) can be predicted by comparing the $\mathrm{pH}$ of a solution to the $\mathrm{p}K_a$ of the acid in that solution. When solution $\mathrm{pH} < \mathrm{acid}\ \mathrm{p}K_a$, the acid form has a higher concentration than the base form. When solution $\mathrm{pH} > \mathrm{acid}\ \mathrm{p}K_a$, the base form has a higher concentration than the acid form.
    • 8.7.A.2 Acid-base indicators are substances that exhibit different properties (such as color) in their protonated versus deprotonated state, making that property respond to the $\mathrm{pH}$ of a solution.
    • 8.7.A.3 To ensure accurate results in a titration experiment, acid-base indicators should be selected that have a $\mathrm{p}K_a$ close to the $\mathrm{pH}$ at the equivalence point.

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    $\text{p}K_a=-\log K_a$: a smaller $\text{p}K_a$ means a stronger acid. Comparing pH to $\text{p}K_a$ tells you the dominant form: below $\text{p}K_a$ the acid form dominates; above it, the conjugate base form dominates.

    8.8

    Properties of Buffers

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 8.8.A: Explain the relationship between the ability of a buffer to stabilize $\mathrm{pH}$ and the reactions that occur when an acid or a base is added to a buffered solution.

    • 8.8.A.1 A buffer solution contains a large concentration of both members in a conjugate acid-base pair. The conjugate acid reacts with added base and the conjugate base reacts with added acid. These reactions are responsible for the ability of a buffer to stabilize $\mathrm{pH}$.
    日本語

    学習目標 8.8.A: バッファーが $\mathrm{pH}$ を安定化する能力と、バッファー溶液に酸または塩基を加えた際に行われる反応との間の関係性を説明する。

    • 8.8.A.1 バッファー溶液には、共役酸塩基対の両成分が大量に含まれている。共役酸は加えた塩基と反応し、共役塩基は加えた酸と反応する。これらの反応が、バッファーが $\mathrm{pH}$ を安定化する能力を生み出している。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    How a buffer resists pH change

    A buffer works best when the weak acid and conjugate base concentrations are similar (pH near $\text{p}K_a$). Choose a buffer whose $\text{p}K_a$ is close to the target pH. Diluting a buffer barely changes its pH, because the acid-to-base ratio stays the same.

    A buffer resists pH change by mopping up added acid or base
    A buffer resists pH change by mopping up added acid or base
    8.9

    The Henderson-Hasselbalch Equation

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 8.9.A: Identify the $\mathrm{pH}$ of a buffer solution based on the identity and concentrations of the conjugate acid-base pair used to create the buffer.

    • 8.9.A.1 The $\mathrm{pH}$ of the buffer is related to the $\mathrm{p}K_a$ of the acid and the concentration ratio of the conjugate acid-base pair. This relation is a consequence of the equilibrium expression associated with the dissociation of a weak acid, and is described by the Henderson-Hasselbalch equation. Adding small amounts of acid or base to a buffered solution does not significantly change the ratio of $[\mathrm{A^-}]/[\mathrm{HA}]$ and thus does not significantly change the solution $\mathrm{pH}$. The change in $\mathrm{pH}$ on addition of acid or base to a buffered solution is therefore much less than it would have been in the absence of the buffer.
      • Equation: $\mathrm{pH} = \mathrm{p}K_a + \log\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}$
      • Exclusion statement: Computation of the change in pH resulting from the addition of an acid or a base to a buffer will not be assessed on the AP Exam.
      • Exclusion statement: Derivation of the Henderson-Hasselbalch equation will not be assessed on the AP Exam.
    日本語

    学習目標 8.9.A: バッファー作成に用いた共役酸塩基対の正体および濃度に基づいて、バッファー溶液の $\mathrm{pH}$ を特定する。

    • 8.9.A.1 バッファーの $\mathrm{pH}$ は、酸の $\mathrm{p}K_a$ および共役酸塩基対の濃度比と関連している。この関係性は、弱酸の解離に関連する平衡定数式の帰結であり、ヘンダーソン=ハ塞尔バルフの式によって記述される。バッファー溶液に少量の酸または塩基を加えても、$[\mathrm{A^-}]/[\mathrm{HA}]$ の比は大きく変化しないため、溶液の $\mathrm{pH}$ も大きく変化しない。したがって、バッファー溶液に酸または塩基を加えたときの $\mathrm{pH}$ の変化は、バッファーがない場合に比べてはるかに小さい。
      • 式: $\mathrm{pH} = \mathrm{p}K_a + \log\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}$
      • 除外事項: バッファーに酸または塩基を加えることによるpHの変化を計算する問題は、AP試験では出題されない。
      • 除外事項: ヘンダーソン=ハ塞尔バルフの式の導出に関する問題は、AP試験では出題されない。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    For a buffer, the pH follows from the acid-to-base ratio:

    $$\text{pH}=\text{p}K_a+\log\frac{[\text{A}^-]}{[\text{HA}]}.$$
    When the concentrations are equal, the log is zero and $\text{pH}=\text{p}K_a$. Use it to design a buffer or find its pH quickly.

    Worked example. A buffer holds $0.20\ \text{M}$ acetic acid ($\text{p}K_a=4.74$) and $0.30\ \text{M}$ acetate. Its pH is

    $$\text{pH}=4.74+\log\frac{0.30}{0.20}=4.74+\log(1.5)=4.74+0.18=4.92.$$
    A little more conjugate base than acid pushes the pH just above $\text{p}K_a$, as the equation predicts.

    8.10

    Buffer Capacity

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 8.10.A: Explain the relationship between the buffer capacity of a solution and the relative concentrations of the conjugate acid and conjugate base components of the solution.

    • 8.10.A.1 Increasing the concentration of the buffer components (while keeping the ratio of these concentrations constant) keeps the $\mathrm{pH}$ of the buffer the same but increases the capacity of the buffer to neutralize added acid or base.
    • 8.10.A.2 When a buffer has more conjugate acid than base, it has a greater buffer capacity for addition of added base than acid. When a buffer has more conjugate base than acid, it has a greater buffer capacity for addition of added acid than base.
    日本語

    学習目標 8.10.A: 溶液のバッファー容量と、その溶液の共役酸および共役塩基成分の相対的な濃度の間の関係を説明すること。

    • 8.10.A.1 バッファー成分の濃度を増加させる(これらの濃度の比を一定に保ったまま)と、バッファーの $\mathrm{pH}$ は同じままに保たれつつ、加えた酸または塩基を中和するバッファーの容量が増加します。
    • 8.10.A.2 バッファー中に共役酸が塩基よりも多い場合、加えた塩基に対するバッファー容量は加えた酸に対するものより大きくなります。逆に、共役塩基が酸よりも多い場合、加えた酸に対するバッファー容量は加えた塩基に対するものより大きくなります。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Buffer capacity 缓冲容量 is how much acid or base a buffer can absorb before its pH changes sharply. It is greatest when the components are concentrated and in roughly equal amounts. Once one component is used up, the buffer fails.

    日本語

    Buffer capacity 缓冲容量 is how much acid or base a buffer can absorb before its pH changes sharply. It is greatest when the components are concentrated and in roughly equal amounts. Once one component is used up, the buffer fails.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Buffer capacity/ˈbʌfə kəˈpæsɪti/ 緩衝能
    8.11

    pH and Solubility

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 8.11.A: Identify the qualitative effect of changes in pH on the solubility of a salt.

    • 8.11.A.1 The solubility of a salt is pH sensitive when one of the constituent ions is a weak acid, a weak base, or the hydroxide ion. These effects can be understood qualitatively using Le Châtelier's principle.
      • Exclusion statement: Computations of solubility as a function of pH will not be assessed on the AP Exam.
    日本語

    学習目標 8.11.A: pH の変化が塩の溶解度に及ぼす定性的な効果を特定すること。

    • 8.11.A.1 塩の構成イオンのいずれかが弱酸、弱塩基、またはヒドロキシドイオンである場合、その塩の溶解度は pH に敏感です。これらの効果はル・シャトリエの原理を用いて定性的に理解できます。
      • 除外事項: pH を関数とする溶解度の計算については AP試験では評価されません。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    The solubility of a salt with a basic anion rises in acidic solution: the added $\text{H}^+$ reacts with the anion, removing it from the solubility equilibrium and pulling more solid to dissolve (Le Chatelier applied to $K_{sp}$). So pH can control whether an ionic solid dissolves.

    8.11

    Exam tips

    • Neutral means equal ions, not pH 7. In pure water $[\text{H}_3\text{O}^+]=[\text{OH}^-]$; since $K_w$ rises with temperature, neutral pH drifts below 7 above 25 °C. Use $K_w=[\text{H}_3\text{O}^+][\text{OH}^-]=1.0\times10^{-14}$ (25 °C) to convert between $[\text{OH}^-]$ and $[\text{H}_3\text{O}^+]$.
    • $\text{pH}=-\log[\text{H}^+]$ is logarithmic — each unit is a ten-fold change in $[\text{H}^+]$.
    • A strong acid/base dissociates completely (so $[\text{H}^+]$ = concentration); a weak one only partly, needing $K_a$.
    • A buffer contains a weak acid and its conjugate base together; it resists pH change by mopping up added acid or base.
    • On a titration curve the equivalence point is the steep jump; the half-equivalence point has $\text{pH}=\text{p}K_a$.
    • A smaller $\text{p}K_a$ means a stronger acid.
  • 9

    Thermodynamics and Electrochemistry · ⁨熱力学と電気化学⁩

    Watch lesson · ⁨レッスンを視聴⁩
    9.1

    Introduction to Entropy

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 9.1.A: Identify the sign and relative magnitude of the entropy change associated with chemical or physical processes.

    • 9.1.A.1 Entropy increases when matter becomes more dispersed. For example, the phase change from solid to liquid or from liquid to gas results in a dispersal of matter as the individual particles become freer to move and generally occupy a larger volume. Similarly, for a gas, the entropy increases when there is an increase in volume (at constant temperature), and the gas molecules are able to move within a larger space. For reactions involving gas-phase reactants or products, the entropy generally increases when the total number of moles of gas-phase products is greater than the total number of moles of gas-phase reactants.
    • 9.1.A.2 Entropy increases when energy is dispersed. According to kinetic molecular theory (KMT), the distribution of kinetic energy among the particles of a gas broadens as the temperature increases. As a result, the entropy of the system increases with an increase in temperature.
    日本語

    学習目標 9.1.A: 化学的または物理的過程に伴うエントロピー変化の符号および相対的な大きさを特定する。

    • 9.1.A.1 物質がより分散する際、エントロピーは増加する。例えば、固体から液体、あるいは液体から気体への相転移では、個々の粒子が自由に動くようになり、一般的により大きな体積を占めるようになるため、物質が分散する。同様に、気体については、温度一定で体積が増加すると、気体分子がより広い空間内で移動できるようになるため、エントロピーは増加する。気体反応物または生成物を伴う反応において、気体生成物のモル数が気体反応物のモル数の合計より大きい場合、エントロピーは一般的に増加する。
    • 9.1.A.2 エネルギーが分散する際、エントロピーは増加する。運動論分子理論(KMT)によると、温度が上昇するにつれて、気体粒子間での運動エネルギーの分布は広がる。その結果、系のエントロピーは温度の上昇とともに増加する。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Entropy 熵 $S$ measures the dispersal of energy and matter – loosely, the number of ways to arrange a system. Entropy increases when a substance goes solid → liquid → gas, when a solid dissolves, when gas moles increase, or when temperature rises. More disorder means higher entropy.

    日本語
    Household batteries: electrochemical cells convert chemical free energy into electrical work
    Household batteries: electrochemical cells convert chemical free energy into electrical work

    Entropy 熵 $S$ measures the dispersal of energy and matter – loosely, the number of ways to arrange a system. Entropy increases when a substance goes solid → liquid → gas, when a solid dissolves, when gas moles increase, or when temperature rises. More disorder means higher entropy.

    Entropy rises from solid to liquid to gas
    Entropy rises from solid to liquid to gas
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Entropy/ˈentrəpi/ エントロピー
    9.2

    Absolute Entropy and Entropy Change

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 9.2.A: Calculate the standard entropy change for a chemical or physical process based on the absolute entropies (standard molar entropies) of the species involved in the process.

    • 9.2.A.1 The entropy change for a process can be calculated from the absolute entropies of the species involved before and after the process occurs.
      • Equation: $\Delta S^{\circ}_{reaction} = \Sigma S^{\circ}_{products} - \Sigma S^{\circ}_{reactants}$
    日本語

    学習目標 9.2.A: 過程に関与する各種の絶対エントロピー(標準モルエントロピー)に基づいて、化学的または物理的過程の標準エントロピー変化を計算する。

    • 9.2.A.1 過程のエントロピー変化は、過程前後に関与する各種の絶対エントロピーから計算できる。
      • 式: $\Delta S^{\circ}_{reaction} = \Sigma S^{\circ}_{products} - \Sigma S^{\circ}_{reactants}$

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    Every substance has a positive absolute entropy $S^\circ$. For a reaction,

    $$\Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants}).$$
    Predict its sign from the change in gas moles: making more gas raises entropy ($\Delta S>0$).

    9.3

    Gibbs Free Energy and Thermodynamic Favorability

    Syllabus · ⁨シラバス⁩
    Learning ObjectiveEssential Knowledge

    9.3.A
    Explain whether a physical or chemical process is thermodynamically favored based on an evaluation of $\Delta G^{\circ}$.

    • 9.3.A.1 The Gibbs free energy change for a chemical process in which all the reactants and products are present in a standard state (as pure substances, as solutions of 1.0 M concentration, or as gases at a pressure of 1.0 atm (or 1.0 bar)) is given the symbol $\Delta G^{\circ}$.

    • 9.3.A.2 The standard Gibbs free energy change for a chemical or physical process is a measure of thermodynamic favorability. Historically, the term "spontaneous" has been used to describe processes for which $\Delta G^{\circ} < 0$. The phrase "thermodynamically favored" is preferred instead so that common misunderstandings (equating "spontaneous" with "suddenly" or "without cause") can be avoided. When $\Delta G^{\circ} < 0$ for the process, it is said to be thermodynamically favored.

    • 9.3.A.3 The standard Gibbs free energy change for a physical or chemical process may also be determined from the standard Gibbs free energy of formation of the reactants and products.

      • Equation: $\Delta G^{\circ}_{reaction} = \Sigma \Delta G^{\circ}_{f\ products} - \Sigma \Delta G^{\circ}_{f\ reactants}$
    • 9.3.A.4 In some cases, it is necessary to consider both enthalpy and entropy to determine if a process will be thermodynamically favored. The freezing of water and the dissolution of sodium nitrate are examples of such phenomena.

    • 9.3.A.5 Knowing the values of $\Delta H^{\circ}$ and $\Delta S^{\circ}$ for a process at a given temperature allows $\Delta G^{\circ}$ to be calculated directly.

      • Equation: $\Delta G^{\circ} = \Delta H^{\circ} - T\,\Delta S^{\circ}$
    • 9.3.A.6 In general, the temperature conditions for a process to be thermodynamically favored ($\Delta G^{\circ} < 0$) can be predicted from the signs of $\Delta H^{\circ}$ and $\Delta S^{\circ}$ as shown in the table below:

      $\Delta H^{\circ}$ $\Delta S^{\circ}$ Symbols $\Delta G^{\circ} < 0$, favored at:
      $< 0$ $> 0$ $<\ >$ all $T$
      $> 0$ $< 0$ $>\ <$ no $T$
      $> 0$ $> 0$ $>\ >$ high $T$
      $< 0$ $< 0$ $<\ <$ low $T$

      In cases where $\Delta H^{\circ} < 0$ and $\Delta S^{\circ} > 0$, no calculation of $\Delta G^{\circ}$ is necessary to determine that the process is thermodynamically favored ($\Delta G^{\circ} < 0$). In cases where $\Delta H^{\circ} > 0$ and $\Delta S^{\circ} < 0$, no calculation of $\Delta G^{\circ}$ is necessary to determine that the process is thermodynamically unfavored ($\Delta G^{\circ} > 0$).

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Gibbs free energy 吉布斯自由能 combines enthalpy and entropy:

    $$\Delta G = \Delta H - T\Delta S.$$
    A process is thermodynamically favorable 热力学有利 when $\Delta G<0$. So exothermic ($\Delta H<0$) and entropy-increasing ($\Delta S>0$) reactions are always favorable; when the two oppose, temperature decides.

    Worked example. A reaction has $\Delta H=+40\ \text{kJ/mol}$ and $\Delta S=+120\ \text{J/(mol K)}$. It is endothermic (unfavorable enthalpy) but entropy-increasing, so it becomes favorable only when hot enough. Setting $\Delta G=\Delta H-T\Delta S<0$ and matching units ($\Delta S=0.120\ \text{kJ}$):

    $$T>\frac{\Delta H}{\Delta S}=\frac{40}{0.120}=333\ \text{K}\;(60\,{}^{\circ}\text{C}).$$

    日本語

    Gibbs free energy 吉布斯自由能 combines enthalpy and entropy:

    $$\Delta G = \Delta H - T\Delta S.$$
    A process is thermodynamically favorable 热力学有利 when $\Delta G<0$. So exothermic ($\Delta H<0$) and entropy-increasing ($\Delta S>0$) reactions are always favorable; when the two oppose, temperature decides.

    Whether a reaction is favorable, from the signs of the enthalpy and entropy change
    Whether a reaction is favorable, from the signs of the enthalpy and entropy change

    Worked example. A reaction has $\Delta H=+40\ \text{kJ/mol}$ and $\Delta S=+120\ \text{J/(mol K)}$. It is endothermic (unfavorable enthalpy) but entropy-increasing, so it becomes favorable only when hot enough. Setting $\Delta G=\Delta H-T\Delta S<0$ and matching units ($\Delta S=0.120\ \text{kJ}$):

    $$T>\frac{\Delta H}{\Delta S}=\frac{40}{0.120}=333\ \text{K}\;(60\,{}^{\circ}\text{C}).$$

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Gibbs free energy/ɡɪbz friː ˈenədʒi/ ギブズ自由エネルギー
    thermodynamically favorable/ˌθɜːməʊdaɪˈnæmɪkli ˈfeɪvərəbl/ 熱力学的に有利である
    9.4

    Thermodynamic and Kinetic Control

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 9.4.A: Explain, in terms of kinetics, why a thermodynamically favored reaction might not occur at a measurable rate.

    • 9.4.A.1 Many processes that are thermodynamically favored do not occur to any measurable extent, or they occur at extremely slow rates.
    • 9.4.A.2 Processes that are thermodynamically favored, but do not proceed at a measurable rate, are under "kinetic control." High activation energy is a common reason for a process to be under kinetic control. The fact that a process does not proceed at a noticeable rate does not mean that the chemical system is at equilibrium. If a process is known to be thermodynamically favored, and yet does not occur at a measurable rate, it is reasonable to conclude that the process is under kinetic control.
    日本語

    学習目標 9.4.A: 運動論の観点から、なぜ熱力学的に優先される反応が測定可能な速度で発生しないのかを説明する。

    • 9.4.A.1 熱力学的に優先される多くの過程は、測定可能な程度では発生せず、または極めて遅い速度でしか進行しません。
    • 9.4.A.2 熱力学的に優先されるが、測定可能な速度で進行しない過程は「運動論的制御」下にあります。高い活性化エネルギーは、過程が運動論的制御下にある一般的な理由です。 noticeableな速度で進行しないという事実は、化学系が平衡状態にあることを意味するものではありません。ある過程が熱力学的に優先されることが分かっているにもかかわらず、測定可能な速度で進行しない場合、その過程が運動論的制御下にあると結論づけることは妥当です。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    $\Delta G<0$ says a reaction can happen, not that it will happen fast. A reaction can be thermodynamically favorable yet kinetically slow because of a high activation energy (diamond → graphite). Thermodynamics gives the direction; kinetics gives the speed.

    9.5

    Free Energy and Equilibrium

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 9.5.A: Explain whether a process is thermodynamically favored using the relationships between $K$, $\Delta G^{\circ}$, and $T$.

    • 9.5.A.1 The phrase "thermodynamically favored" ($\Delta G^{\circ} < 0$) means that the products are favored at equilibrium ($K > 1$) under standard conditions.
    • 9.5.A.2 The equilibrium constant is related to free energy by the equations
      • Equation: $K = e^{-\Delta G^{\circ}/RT}$
      • Equation: $\Delta G^{\circ} = -RT \ln K$
    • 9.5.A.3 Connections between $K$ and $\Delta G^{\circ}$ can be made qualitatively through estimation. When $\Delta G^{\circ}$ is near zero, the equilibrium constant will be close to 1. When $\Delta G^{\circ}$ is much larger or much smaller than $RT$, the value of $K$ deviates strongly from 1.
    • 9.5.A.4 Processes with $\Delta G^{\circ} < 0$ favor products (i.e., $K > 1$) and those with $\Delta G^{\circ} > 0$ favor reactants (i.e., $K < 1$).
    日本語

    学習目標 9.5.A: $K$、$\Delta G^{\circ}$、および $T$ の関係を用いて、ある過程が熱力学的に優先されるかどうかを説明する。

    • 9.5.A.1 「熱力学的に優先される」($\Delta G^{\circ} < 0$)とは、標準条件下で平衡において生成物が優先される($K > 1$)ことを意味します。
    • 9.5.A.2 平衡定数は、以下の式によって自由エネルギーと関連付けられています。
      • 式: $K = e^{-\Delta G^{\circ}/RT}$
      • 式: $\Delta G^{\circ} = -RT \ln K$
    • 9.5.A.3 $K$ と $\Delta G^{\circ}$ の間には、推定を通じて定性的な関連付けを行うことができる。$\Delta G^{\circ}$ がゼロに近い場合、平衡定数は 1 に近づく。$\Delta G^{\circ}$ が $RT$ よりもはるかに大きい、あるいははるかに小さい場合、$K$ の値は 1 から大きく逸脱する。
    • 9.5.A.4 $\Delta G^{\circ} < 0$ の過程は生成物を優先し(つまり $K > 1$)、$\Delta G^{\circ} > 0$ の過程は反応物を優先します(つまり $K < 1$)。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    Free energy links to the equilibrium constant:

    $$\Delta G^\circ = -RT\ln K.$$
    So $\Delta G^\circ<0$ gives $K>1$ (products favored), and $\Delta G^\circ>0$ gives $K<1$. At equilibrium $\Delta G=0$.

    A battery converts free energy of a spontaneous redox reaction into electrical work
    A battery converts free energy of a spontaneous redox reaction into electrical work
    9.6

    Free Energy of Dissolution

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 9.6.A: Explain the relationship between the solubility of a salt and changes in the enthalpy and entropy that occur in the dissolution process.

    • 9.6.A.1 The free energy change ($\Delta G^{\circ}$) for dissolution of a substance reflects a number of factors: the breaking of the intermolecular interactions that hold the solid together, the reorganization of the solvent around the dissolved species, and the interaction of the dissolved species with the solvent. It is possible to estimate the sign and relative magnitude of the enthalpic and entropic contributions to each of these factors. However, making predictions for the total change in free energy of dissolution can be challenging due to the cancellations among the free energies associated with the three factors cited.
    日本語

    学習目標 9.6.A: 塩の溶解度と、溶解過程で生じるエンタルピーおよびエントロピーの変化との関係を説明する。

    • 9.6.A.1 物質の溶解に伴う自由エネルギー変化($\Delta G^{\circ}$)は、複数の要因を反映している:固体を保持する分子間相互作用の切断、溶質種周囲での溶媒の再組織化、および溶質種と溶媒との相互作用。これらの各要因に対するエンタルピー寄与とエントロピー寄与の符号および相対的な大きさを推定することは可能である。しかし、引用された3つの要因に関連する自由エネルギー間の打ち消し合いのため、溶解の全自由エネルギー変化を予測することは困難である場合がある。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    Whether a salt dissolves depends on the free-energy change of dissolving. Dissolving often increases entropy (ordered solid → dispersed ions) but may cost enthalpy; the sign of $\Delta G$ (and thus $K_{sp}$) follows from $\Delta H - T\Delta S$.

    9.7

    Coupled Reactions

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 9.7.A: Explain the relationship between external sources of energy or coupled reactions and their ability to drive thermodynamically unfavorable processes.

    • 9.7.A.1 An external source of energy can be used to make a thermodynamically unfavorable process occur. Examples include:
      • 9.7.A.1.i Electrical energy to drive an electrolytic cell or charge a battery.
      • 9.7.A.1.ii Light to drive the overall conversion of carbon dioxide to glucose in photosynthesis.
    • 9.7.A.2 A desired product can be formed by coupling a thermodynamically unfavorable reaction that produces that product to a favorable reaction (e.g., the conversion of $ATP$ to $ADP$ in biological systems). In the coupled system, the individual reactions share one or more common intermediates. The sum of the individual reactions produces an overall reaction that achieves the desired outcome and has $\Delta G^{\circ} < 0$.
    日本語

    学習目標 9.7.A: 外部エネルギー源または連成反応が、熱力学的に不利なプロセスを進行させる能力との関係を説明せよ。

    • 9.7.A.1 外部エネルギー源を用いることで、熱力学的に不利なプロセスを進行させることができる。例として以下が挙げられる:
      • 9.7.A.1.i 電気分解セルを駆動したり、バッテリーに充電したりするための電気エネルギー。
      • 9.7.A.1.ii 光合成における二酸化炭素からグルコースへの全体変換を駆動するための光。
    • 9.7.A.2 目的の生成物を得るために、その生成物を生じる熱力学的に不利な反応を有利な反応に連成させることができる(例:生物系における $ATP$ から $ADP$ への変換)。連成系では、個々の反応が1つ以上の共通中間体を共有する。個々の反応の合計は、目的の結果を実現し、$\Delta G^{\circ} < 0$ を持つ全体反応を生み出す。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    An unfavorable reaction ($\Delta G>0$) can be driven by coupling it to a favorable one ($\Delta G<0$) that shares a common intermediate, as long as the sum has $\Delta G<0$. This is how cells use ATP to power otherwise unfavorable processes.

    9.8

    Galvanic and Electrolytic Cells

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 9.8.A: Explain the relationship between the physical components of an electrochemical cell and the overall operational principles of the cell.

    • 9.8.A.1 Each component of an electrochemical cell (electrodes, solutions in the half-cells, salt bridge, voltage/current measuring device) plays a specific role in the overall functioning of the cell. The operational characteristics of the cell (galvanic vs. electrolytic, direction of electron flow, reactions occurring in each half-cell, change in electrode mass, evolution of a gas at an electrode, ion flow through the salt bridge) can be described at both the macroscopic and particulate levels.
    • 9.8.A.2 Galvanic, sometimes called voltaic, cells involve a thermodynamically favored reaction, whereas electrolytic cells involve a thermodynamically unfavored reaction. Visual representations of galvanic and electrolytic cells are tools of analysis to identify where half-reactions occur and in what direction current flows.
    • 9.8.A.3 For all electrochemical cells, oxidation occurs at the anode and reduction occurs at the cathode.
      • Exclusion statement: Labeling an electrode as positive or negative will not be assessed on the AP Exam.
    日本語

    学習目標 9.8.A: 電気化学セルの物理的構成要素とセル全体の動作原理との関係を説明せよ。

    • 9.8.A.1 電気化学セルの各構成要素(電極、半セル内の溶液、塩橋、電圧/電流計測装置)は、セル全体の機能において特定の役割を果たす。セルの動作特性(ガルバニ vs 電気分解、電子の流れの方向、各半セルで起こる反応、電極質量の変化、電極における気体発生、塩橋を介したイオンの流れ)は、巨視レベルおよび粒子レベルの両方で記述できる。
    • 9.8.A.2 ガルバニセル(ボルタセルとも呼ばれる)は熱力学的に有利な反応を伴い、一方、電気分解セルは熱力学的に不利な反応を伴う。ガルバニセルおよび電気分解セルの図示は、半反応が発生する場所や電流の流向を特定するための分析ツールである。
    • 9.8.A.3 すべての電気化学セルにおいて、アノードでは酸化が起こり、カソードでは還元が起こる。
      • 除外事項: AP試験において、電極を正極または負極としてラベル付けすることを評価することはない。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    The galvanic cell

    Redox reactions can move electrons through a wire:

    • A galvanic (voltaic) cell 原电池 uses a favorable reaction ($\Delta G<0$) to produce electricity – a battery.
    • An electrolytic cell 电解池 uses electricity to force an unfavorable reaction ($\Delta G>0$).

    In both, oxidation happens at the anode 阳极 and reduction at the cathode 阴极.

    日本語
    The galvanic cell

    Redox reactions can move electrons through a wire:

    A galvanic cell with a salt bridge and voltmeter
    A galvanic cell with a salt bridge and voltmeter
    • A galvanic (voltaic) cell 原电池 uses a favorable reaction ($\Delta G<0$) to produce electricity – a battery.
    • An electrolytic cell 电解池 uses electricity to force an unfavorable reaction ($\Delta G>0$).

    In both, oxidation happens at the anode 阳极 and reduction at the cathode 阴极.

    Commercial batteries: galvanic cells convert chemical free energy into electrical work
    Commercial batteries: galvanic cells convert chemical free energy into electrical work
    Explore · ⁨探索⁩

    Transfer electrons in a cell · ⁨セル内での電子移動を確認する⁩

    In a galvanic cell a spontaneous redox reaction drives electrons through a wire, doing electrical work; oxidation at one electrode, reduction at the other. · ⁨ガルバニセル では、自発的な酸化還元反応により導線を通じて電子が流れ、電気的仕事を行います。一方の電極で酸化、他方の電極で還元が起こります。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    galvanic (voltaic) cell/ɡælˈvænɪk sel/ ガルバニ(ボルタ)電池
    electrolytic cell/ɪˌlektrəˈlɪtɪk sel/ 電解セル
    anode/ˈænəʊd/ アノード
    cathode/ˈkæθəʊd/ カソード
    9.9

    Cell Potential and Free Energy

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 9.9.A: Explain whether an electrochemical cell is thermodynamically favored, based on its standard cell potential and the constituent half-reactions within the cell.

    • 9.9.A.1 Electrochemistry encompasses the study of redox reactions that occur within electrochemical cells. The reactions are either thermodynamically favored (resulting in a positive voltage) or thermodynamically unfavored (resulting in a negative voltage and requiring an externally applied potential for the reaction to proceed).
    • 9.9.A.2 The standard cell potential of electrochemical cells can be calculated by identifying the oxidation and reduction half-reactions and their respective standard reduction potentials.
    • 9.9.A.3 $\Delta G^{\circ}$ (standard Gibbs free energy change) is proportional to the negative of the cell potential for the redox reaction from which it is constructed. Thus, a cell with a positive $E^{\circ}$ involves a thermodynamically favored reaction, and a cell with a negative $E^{\circ}$ involves a thermodynamically unfavored reaction.
      • Equation: $\Delta G^{\circ} = -nFE^{\circ}$
    日本語

    学習目標 9.9.A: 標準セル電位およびセル内の構成半反応に基づき、電気化学セルが熱力学的に有利かどうかを説明せよ。

    • 9.9.A.1 電気化学は、電気化学セル内で起こる酸化還元反応の研究を含む。これらの反応は、熱力学的に有利(正の電圧产生)か、あるいは熱力学的に不利(負の電圧产生であり、反応が進行するためには外部印加電位が必要)かのいずれかである。
    • 9.9.A.2 電気化学セルの標準セル電位は、酸化半反応および還元半反応を特定し、それぞれの標準還元電位を用いて計算できる。
    • 9.9.A.3 $\Delta G^{\circ}$(標準ギブズ自由エネルギー変化)は、そこから構成される酸化還元反応のセル電位の負の値に比例する。したがって、正の $E^{\circ}$ を持つセルは熱力学的に有利な反応を伴い、負の $E^{\circ}$ を持つセルは熱力学的に不利な反応を伴う。
      • 式: $\Delta G^{\circ} = -nFE^{\circ}$

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    The cell potential 电池电势 $E^\circ_{\text{cell}}$ measures the driving force in volts, found from standard reduction potentials ($E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$). It links to free energy by

    $$\Delta G^\circ = -nFE^\circ_{\text{cell}},$$
    where $n$ is moles of electrons and $F$ the Faraday constant. A positive $E^\circ_{\text{cell}}$ means a favorable (galvanic) reaction.

    Worked example. A cell pairs a copper cathode ($\text{Cu}^{2+}+2e^-\rightarrow\text{Cu}$, $E^\circ=+0.34\ \text{V}$) with a zinc anode ($\text{Zn}^{2+}+2e^-\rightarrow\text{Zn}$, $E^\circ=-0.76\ \text{V}$). The cell potential is

    $$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}=0.34-(-0.76)=1.10\ \text{V}.$$
    It is positive, so the reaction is spontaneous and the cell (a Daniell cell) acts as a battery.

    日本語

    The cell potential 电池电势 $E^\circ_{\text{cell}}$ measures the driving force in volts, found from standard reduction potentials ($E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$). It links to free energy by

    $$\Delta G^\circ = -nFE^\circ_{\text{cell}},$$
    where $n$ is moles of electrons and $F$ the Faraday constant. A positive $E^\circ_{\text{cell}}$ means a favorable (galvanic) reaction.

    The electrochemical series of standard electrode potentials
    The electrochemical series of standard electrode potentials

    Worked example. A cell pairs a copper cathode ($\text{Cu}^{2+}+2e^-\rightarrow\text{Cu}$, $E^\circ=+0.34\ \text{V}$) with a zinc anode ($\text{Zn}^{2+}+2e^-\rightarrow\text{Zn}$, $E^\circ=-0.76\ \text{V}$). The cell potential is

    $$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}=0.34-(-0.76)=1.10\ \text{V}.$$
    It is positive, so the reaction is spontaneous and the cell (a Daniell cell) acts as a battery.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    cell potential/sel pəˈtenʃl/ 電池起電力
    9.10

    Cell Potential Under Nonstandard Conditions

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 9.10.A: Explain the relationship between deviations from standard cell conditions and changes in the cell potential.

    • 9.10.A.1 In a real system under nonstandard conditions, the cell potential will vary depending on the concentrations of the active species. The cell potential is a driving force toward equilibrium; the farther the reaction is from equilibrium, the greater the magnitude of the cell potential.

    • 9.10.A.2 Equilibrium arguments such as Le Châtelier's principle do not apply to electrochemical systems, because the systems are not in equilibrium.

    • 9.10.A.3 The standard cell potential $E^{\circ}$ corresponds to the standard conditions of $Q = 1$. As the system approaches equilibrium, the magnitude (i.e., absolute value) of the cell potential decreases, reaching zero at equilibrium (when $Q = K$). Deviations from standard conditions that take the cell further from equilibrium than $Q = 1$ will increase the magnitude of the cell potential relative to $E^{\circ}$. Deviations from standard conditions that take the cell closer to equilibrium than $Q = 1$ will decrease the magnitude of the cell potential relative to $E^{\circ}$. In concentration cells, the direction of spontaneous electron flow can be determined by considering the direction needed to reach equilibrium.

    • 9.10.A.4 Algorithmic calculations using the Nernst equation are insufficient to demonstrate an understanding of electrochemical cells under nonstandard conditions. However, students should qualitatively understand the effects of concentration on cell potential and use conceptual reasoning, including the qualitative use of the Nernst equation:

      • Equation: $E = E^{\circ} - (RT/nF) \ln Q$

      to solve problems.

    日本語

    学習目標 9.10.A: 標準セル条件からの逸脱とセル電位の変化との間の関係性を説明する。

    • 9.10.A.1 非標準条件下の実際の系において、セル電位は活性種浓度に応じて変化する。セル電位は平衡に向かう駆動力であり、反応が平衡からどれだけ離れているかによって、セル電位の大きさ(絶対値)は大きくなる。

    • 9.10.A.2 ラ・シュトリエルの法則などの平衡に関する議論は、系が平衡状態にないため、電気化学系には適用できない。

    • 9.10.A.3 標準セル電位 $E^{\circ}$ は、$Q = 1$ という標準条件に対応する。系が平衡に近づくにつれて、セル電位の大きさ(絶対値)は減少し、平衡時($Q = K$ のとき)にはゼロに達する。標準条件からの逸脱が、$Q = 1$ よりも平衡から遠ざかる方向にセルを進めると、$E^{\circ}$ に対してセル電位の大きさが増加する。一方、標準条件からの逸脱が、$Q = 1$ よりも平衡に近づく方向にセルを進めると、$E^{\circ}$ に対してセル電位の大きさは減少する。濃度セルにおいては、平衡に到達するために必要な電子の流れの方向を考慮することで、自発的な電子流の方向を決定できる。

    • 9.10.A.4 ネルンスト式を用いたアルゴリズム計算のみでは、非標準条件下の電気化学セルに対する理解を示すことは不十分である。しかし、学生は濃度がセル電位に及ぼす影響を定性的に理解し、ネルンスト式の定性的な使用を含む概念的推論を用いるべきである。

      • 式: $E = E^{\circ} - (RT/nF) \ln Q$

      これを使って問題を解く。

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English

    Away from standard conditions, the potential shifts with concentration – the Nernst equation 能斯特方程 (qualitatively): as reactants are consumed, $Q$ rises and $E_{\text{cell}}$ falls, reaching zero at equilibrium (a dead battery). Changing a concentration shifts $E_{\text{cell}}$ the way Le Chatelier predicts.

    日本語

    Away from standard conditions, the potential shifts with concentration – the Nernst equation 能斯特方程 (qualitatively): as reactants are consumed, $Q$ rises and $E_{\text{cell}}$ falls, reaching zero at equilibrium (a dead battery). Changing a concentration shifts $E_{\text{cell}}$ the way Le Chatelier predicts.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Nernst equation/nɜːnst ɪˈkweɪʒn/ ネルンストの式
    9.11

    Electrolysis and Faraday's Law

    Syllabus · ⁨シラバス⁩
    English

    Learning Objective 9.11.A: Calculate the amount of charge flow based on changes in the amounts of reactants and products in an electrochemical cell.

    • 9.11.A.1 Faraday's laws can be used to determine the stoichiometry of the redox reaction occurring in an electrochemical cell with respect to the following:
      • 9.11.A.1.i Number of electrons transferred
      • 9.11.A.1.ii Mass of material deposited on or removed from an electrode (as in electroplating)
      • 9.11.A.1.iii Current
      • 9.11.A.1.iv Time elapsed
      • 9.11.A.1.v Charge of ionic species
      • Equation: $I = q/t$
    日本語

    学習目標 9.11.A: 電気化学セルにおける反応物と生成物の量の変化に基づいて、電荷移動量を計算する。

    • 9.11.A.1 ファラデーの法則は、次の点に関する電気化学セルで起こる酸化還元反応の化学量論を決定するために使用できる:
      • 9.11.A.1.i 移動した電子の数
      • 9.11.A.1.ii 電極に析出または除去された物質の質量(メッキなど)
      • 9.11.A.1.iii 電流
      • 9.11.A.1.iv 経過時間
      • 9.11.A.1.v イオン種の電荷
      • 式: $I = q/t$

    Source: College Board AP Course and Exam Description · ⁨出典: College Board AP コースおよび試験説明書⁩

    English
    Electrolysis

    In electrolysis 电解, the charge passed determines how much substance is deposited or produced – Faraday's law 法拉第定律. Convert current × time to charge, charge to moles of electrons ($F=96{,}485$ C/mol), then use the half-reaction's electron ratio to get moles (and mass) of product.

    Worked example. A current of $2.0\ \text{A}$ flows for $30\ \text{minutes}$ through copper(II) sulfate ($\text{Cu}^{2+}+2e^-\rightarrow\text{Cu}$). How much copper is deposited? The charge is $Q=It=2.0\times1800=3600\ \text{C}$, giving $3600/96485=0.0373\ \text{mol}$ of electrons. Since each Cu needs $2$ electrons, $0.0187\ \text{mol}$ of Cu forms, a mass of $0.0187\times63.5=1.2\ \text{g}$.

    日本語
    Electrolysis

    In electrolysis 电解, the charge passed determines how much substance is deposited or produced – Faraday's law 法拉第定律. Convert current × time to charge, charge to moles of electrons ($F=96{,}485$ C/mol), then use the half-reaction's electron ratio to get moles (and mass) of product.

    Electrolysis: ions move to the electrodes and are discharged
    Electrolysis: ions move to the electrodes and are discharged

    Worked example. A current of $2.0\ \text{A}$ flows for $30\ \text{minutes}$ through copper(II) sulfate ($\text{Cu}^{2+}+2e^-\rightarrow\text{Cu}$). How much copper is deposited? The charge is $Q=It=2.0\times1800=3600\ \text{C}$, giving $3600/96485=0.0373\ \text{mol}$ of electrons. Since each Cu needs $2$ electrons, $0.0187\ \text{mol}$ of Cu forms, a mass of $0.0187\times63.5=1.2\ \text{g}$.

    Explore · ⁨探索⁩

    Electrolyse a molten salt · ⁨溶融塩の電気分解⁩

    Electrolysis uses current to force a non-spontaneous reaction: positive ions gain electrons at the cathode, negative ions lose them at the anode. Charge sets the amount deposited. · ⁨電気分解 は電流を用いて自発的でない反応を強制します。陽極では負イオンが電子を失い、陰極では正イオンが電子を得ます。流した電荷量によって析出する量が決まります。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    electrolysis/ɪlekˈtrɒləsɪs/ 電気分解
    Faraday's law/ˈfærədeɪz lɔː/ ファラデーの法則
    9.11

    Exam tips

    • A process is thermodynamically favourable when $\Delta G<0$; combine enthalpy and entropy with $\Delta G=\Delta H-T\Delta S$ (match units — kJ vs J).
    • Entropy increases solid→liquid→gas and when more gas moles are produced.
    • Favourable does not mean fast — a high activation energy can make a $\Delta G<0$ reaction extremely slow (kinetic control).
    • In electrochemistry a positive $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$ means a spontaneous (galvanic) cell; oxidation is at the anode, reduction at the cathode.
    • In electrolysis the charge passed ($Q=It$) fixes the amount deposited (Faraday's law).

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