AP Chemistry

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:

Moles are the bridge between the lab (grams you weigh) and the equation (particles that react).

The mole is the hub: convert between mass, particles, gas volume, and concentration

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

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.

Relative atomic mass is the abundance-weighted average of the isotopes

The mass spectrum of chlorine: two isotopes with different abundances

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

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.

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.

Finding the empirical formula: mass to moles, divide by the smallest, read the ratio

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:

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.

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.

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.

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

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.

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.

Successive ionisation energies reveal the shell structure through big jumps

The photoelectron spectrum of neon: one peak per subshell, height set by electron count

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

Electronegativity rises across a period and falls down a group

• 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.

Periodic trends: how radius, ionization energy, and electronegativity change across and down

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.

• 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.

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 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.

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.

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

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

Metallic bonding: positive ions in a sea of delocalised electrons

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

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.

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.

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

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.

• 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 ($\text{CO}_2$).

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

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; stronger for larger, more polarizable electron clouds.
• Dipole–dipole 偶极-偶极: between polar molecules.
• Hydrogen bonding 氢键: a strong dipole force when H is bonded to N, O, or F.

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 $-162\,{}^{\circ}\text{C}$.

A solid's properties reflect the particles and forces holding it: ionic, covalent-network (like diamond, very hard, high-melting), metallic, and molecular solids (held by weak IMFs, soft, low-melting). Matching a solid's properties to its structure is a common exam task.

The four solid structures: giant ionic, simple molecular, giant covalent, and metallic

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

An ideal gas obeys

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

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

Always use kelvin for $T$ and match the units of $R$ to your pressure and volume.

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

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.

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:

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.

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.

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

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.

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.

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.

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

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.

• 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.

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.

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

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 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.

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

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:

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.

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

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:

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.

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

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

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).

• 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.

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

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

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

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.

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

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

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.

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.

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

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

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 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.

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.

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.

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

• 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.

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 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

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.

The heat to change a substance's temperature is

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

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

By energy conservation the reaction released this $3.3\ \text{kJ}$, so it is exothermic ($q_{\text{rxn}}=-3.3\ \text{kJ}$).

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.

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.

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

This is an approximation, since bond enthalpies are averages.

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 ($2\times431=862$):

exothermic, because the strong H–Cl bonds formed release more than the reactant bonds cost.

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

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:

a large release, as expected for a combustion.

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

• 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.

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

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

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.

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

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.

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

The NO$_2$ coefficient of $2$ becomes the power; N$_2$O$_4$ (coefficient $1$) is just to the first power.

• $K\gg 1$: 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."

$K$ 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.

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:

so $[\text{HI}]=2x=0.156\ \text{M}$. Taking the square root of both sides works here because the expression is a perfect square.

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.

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

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.

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

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

Adding NaCl (a common ion) would raise $[\text{Cl}^-]$, forcing $s$ down – far less AgCl would dissolve.

• 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.

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

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

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

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

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 $K_b$. 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

The weak acid is far less acidic than a strong acid of the same concentration (which would be pH $1.0$).

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.

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).

A titration curve has a steep jump near the equivalence point

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.

$\text{p}K_a=-\log K_a$: 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.

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

For a buffer, the pH follows from the acid-to-base ratio:

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

A little more conjugate base than acid pushes the pH just above $\text{p}K_a$, as the equation predicts.

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.

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.

• $\text{pH}=-\log[\text{H}^+]$ 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.

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

Every substance has a positive absolute entropy $S^\circ$. For a reaction,

Predict its sign from the change in gas moles: making more gas raises entropy ($\Delta S>0$).

Gibbs free energy 吉布斯自由能 combines enthalpy and entropy:

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

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 ($\Delta S=0.120\ \text{kJ}$):