A-Level Chemistry · ⁨A-Level 화학⁩

Atoms 原子 are far too light to weigh in grams, so we compare every mass to one standard. The standard is the unified atomic mass unit 统一原子质量单位 (symbol u), defined as exactly one twelfth of the mass of one carbon-12 atom.

Using this unit, we state masses as simple numbers:

• the relative atomic mass 相对原子质量 $A_r$ of an element is the average mass of its atoms compared with $\tfrac{1}{12}$ of a carbon-12 atom. It is an average over all the isotopes 同位素, weighted by how common each one is.
• the relative isotopic mass 相对同位素质量 is the mass of one atom of a single isotope, compared with $\tfrac{1}{12}$ of a carbon-12 atom.
• the relative molecular mass 相对分子质量 $M_r$ of a molecule 分子 is the sum of the relative atomic masses of all its atoms.
• the relative formula mass 相对式量 is the same idea for a substance that is not made of molecules (such as an ionic compound). Add up the relative atomic masses shown in the formula.

To find $A_r$ from isotope data, multiply each isotope mass by its percentage, add these up, and divide by 100.

Worked example. Chlorine is $75\%$ ${}^{35}\text{Cl}$ and $25\%$ ${}^{37}\text{Cl}$. Find its relative atomic mass.

Worked example. A mass spectrometer shows copper is $69.2\%$ ${}^{63}\text{Cu}$ and $30.8\%$ ${}^{65}\text{Cu}$. Find its relative atomic mass.

Chemists count particles in groups called moles, just as we count eggs in dozens.

One mole 摩尔 (symbol mol) is the amount of substance that contains the same number of particles as there are atoms in exactly 12 g of carbon-12. That number is the Avogadro constant 阿伏伽德罗常量:

So one mole of anything contains $6.02 \times 10^{23}$ particles. The particles may be atoms, molecules or ions 离子 — always say which.

The mass of one mole in grams equals the relative mass ($A_r$ or $M_r$). This is the molar mass 摩尔质量, with units $\text{g mol}^{-1}$. The key equation is:

where $n$ is the amount in moles, $m$ is the mass in grams, and $M$ is the molar mass.

Worked example. How many moles are in $8.0\ \text{g}$ of methane, $\text{CH}_4$? ($A_r$: C $= 12$, H $= 1$.)

The molar mass is $M = 12 + 4(1) = 16\ \text{g mol}^{-1}$, so

A compound 化合物 is a substance made of two or more elements chemically joined.

Charges and formulas of ionic compounds

In an ionic compound 离子化合物 the total positive charge balances the total negative charge, so the compound is neutral overall.

You can predict the charge of many ions from the element's position in the Periodic Table:

Hydrogen forms $\text{H}^+$, and the Group 18 noble gases do not normally form ions.

Some ions you must know by name and formula:

For a metal that can have more than one charge, a Roman numeral shows the oxidation number 氧化数. For example, iron(II) is $\text{Fe}^{2+}$ and iron(III) is $\text{Fe}^{3+}$. To write a formula, balance the charges: iron(III) oxide is $\text{Fe}_2\text{O}_3$, because two $\text{Fe}^{3+}$ balance three $\text{O}^{2-}$.

Equations and state symbols

A chemical equation must be balanced 配平 — the same number of each kind of atom on both sides. Add state symbols 状态符号 to show the state of each species: (s) solid, (l) liquid, (g) gas, and (aq) aqueous 水溶液 (dissolved in water).

An ionic equation 离子方程式 shows only the ions and molecules that actually change. The ions that do not change are spectator ions 旁观离子, and you leave them out. For example, the reaction that forms silver chloride is:

Empirical and molecular formulas

The empirical formula 实验式 is the simplest whole-number ratio of the atoms of each element in a compound. The molecular formula 分子式 shows the actual number of atoms of each element in one molecule.

For example, ethane has empirical formula $\text{CH}_3$ but molecular formula $\text{C}_2\text{H}_6$.

Hydrated and anhydrous solids

Some solids hold water inside their crystals. This water is the water of crystallisation 结晶水. A solid that contains it is hydrated 水合的; the same solid with the water removed is anhydrous 无水的.

For example, hydrated copper(II) sulfate is $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$. Heating it drives off the water to leave anhydrous $\text{CuSO}_4$.

Calculating empirical and molecular formulas

To find the empirical formula from masses (or percentages by mass):

1. divide each element's mass by its $A_r$ to get the moles.
2. divide all the mole values by the smallest one.
3. round to the nearest whole numbers — that ratio is the empirical formula.

To get the molecular formula, you also need $M_r$. Find how many times the empirical formula mass fits into $M_r$, then multiply the formula by that number.

Worked example. A compound is $40.0\%$ carbon, $6.7\%$ hydrogen and $53.3\%$ oxygen by mass. Find its empirical formula. ($A_r$: C $= 12$, H $= 1$, O $= 16$.)

Take $100\ \text{g}$ and divide each mass by its $A_r$ to get moles: C $= 40.0/12 = 3.33$, H $= 6.7/1 = 6.7$, O $= 53.3/16 = 3.33$. Dividing through by the smallest ($3.33$) gives a ratio C : H : O $= 1 : 2 : 1$, so the empirical formula is $\text{CH}_2\text{O}$.

Reacting masses and percentage yield

The numbers in front of each species in a balanced equation give the mole ratio — this is the stoichiometry 化学计量. To find a reacting mass: change the known mass to moles, use the mole ratio to find the moles you want, then change back to mass.

Worked example. What mass of magnesium oxide forms when $4.8\ \text{g}$ of magnesium burns completely? $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$. ($A_r$: Mg $= 24$, O $= 16$.)

Moles of Mg $= 4.8/24 = 0.20\ \text{mol}$. The ratio Mg : MgO is $1 : 1$, so $0.20\ \text{mol}$ of MgO forms. Its molar mass is $24 + 16 = 40\ \text{g mol}^{-1}$, so

In real reactions you usually get less product than the maximum. The percentage yield 产率 compares the amount you actually made with the most you could make:

Limiting and excess reagent

When two reactants are mixed, one usually runs out first. The limiting reagent 限量试剂 is the one that runs out — it decides how much product forms. The other is the excess reagent 过量试剂, because there is more than enough of it. Always base the calculation on the limiting reagent.

To find it: work out the moles of each reactant, divide each by its number in the equation, and the smallest result is the limiting reagent.

Volumes of gases

At the same temperature and pressure, equal volumes of any gases contain equal numbers of molecules. At room temperature and pressure (r.t.p.), one mole of any gas takes up $24.0\ \text{dm}^3$, so:

This is used when burning hydrocarbons 碳氢化合物 (compounds of only carbon and hydrogen), where you compare gas volumes.

Volumes and concentrations of solutions

The concentration 浓度 of a solution is the amount of solute 溶质 in each cubic decimetre of solution 溶液, measured in $\text{mol dm}^{-3}$:

where $c$ is the concentration and $V$ is the volume in $\text{dm}^3$. Remember that $1000\ \text{cm}^3 = 1\ \text{dm}^3$.

Worked example. In a titration, $25.0\ \text{cm}^3$ of sodium hydroxide solution is exactly neutralised by $20.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ hydrochloric acid: $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$. Find the concentration of the sodium hydroxide.

Moles of HCl $= cV = 0.100 \times \dfrac{20.0}{1000} = 2.00 \times 10^{-3}\ \text{mol}$. The ratio is $1 : 1$, so there are $2.00 \times 10^{-3}\ \text{mol}$ of NaOH in $25.0\ \text{cm}^3$:

This is the basis of a titration 滴定, where you find an unknown concentration by reacting it with a solution whose concentration you already know.

Significant figures

Give your answer to a sensible number of significant figures 有效数字 — usually match the data in the question. Do not write more digits than the data supports, and do not round so early that you lose accuracy.

• Convert volumes to $\text{dm}^3$ before using concentration ($25.0\ \text{cm}^3 = 0.0250\ \text{dm}^3$); the missing $\div 1000$ is the most common titration error.
• Use the balancing numbers as the mole ratio between species — never the $M_r$ values.
• Empirical formula: divide each mass/percentage by $A_r$, then by the smallest, then scale to whole numbers; use $M_r$ to reach the molecular formula.
• For gases at r.t.p. use $\text{volume} = \text{moles} \times 24\ \text{dm}^3$; quote the equation you use every time.
• Give the answer to the same significant figures as the data (usually 3) and always include units.

Electronegativity 电负性 is the power of an atom to attract the electrons 电子 in a bond towards itself.

Three factors decide how electronegative an atom is:

• nuclear charge 核电荷: more protons pull the bonding electrons more strongly.
• atomic radius 原子半径: the closer the bond is to the nucleus, the stronger the pull.
• shielding 屏蔽 by inner shells and sub-shells: more inner electrons weaken the pull on the bonding electrons.

So electronegativity rises across a period (more nuclear charge, smaller radius) and falls down a group (larger radius, more shielding). Fluorine is the most electronegative element.

You can use the difference in Pauling electronegativity 鲍林电负性 values to predict the bond type. A large difference gives an ionic bond; a small difference gives a covalent bond.

Ionic bonding 离子键 is the electrostatic attraction 静电引力 between oppositely charged ions 离子 — positive cations 阳离子 and negative anions 阴离子.

It forms when a metal gives electrons to a non-metal. Good examples are sodium chloride ($\text{NaCl}$), magnesium oxide ($\text{MgO}$) and calcium fluoride ($\text{CaF}_2$). The ions pack into a regular giant lattice 晶格, held together by the attraction in every direction.

Metallic bonding 金属键 is the electrostatic attraction between positive metal ions and a "sea" of delocalised electrons 离域电子.

The outer electrons are free to move through the whole metal. This explains why metals conduct electricity and are strong.

Covalent bonding 共价键 is the electrostatic attraction between the nuclei of two atoms and a shared pair of electrons.

Simple molecules with covalent bonds include $\text{H}_2$, $\text{O}_2$, $\text{N}_2$, $\text{Cl}_2$, $\text{HCl}$, $\text{CO}_2$, $\text{NH}_3$, $\text{CH}_4$, $\text{C}_2\text{H}_6$ and $\text{C}_2\text{H}_4$. A double bond shares two pairs; a triple bond (as in $\text{N}_2$) shares three pairs.

Atoms in Period 3 and below can expand the octet 扩展八隅体 — hold more than eight electrons in their outer shell. Examples are $\text{SO}_2$, $\text{PCl}_5$ and $\text{SF}_6$.

A coordinate bond 配位键 (also called a dative covalent bond) is a covalent bond where both shared electrons come from the same atom. For example, when ammonia and hydrogen chloride gases meet, the lone pair on the nitrogen forms a coordinate bond to $\text{H}^+$, making the ammonium ion $\text{NH}_4^+$. Coordinate bonds also join the two halves of the $\text{Al}_2\text{Cl}_6$ molecule.

Sigma and pi bonds

Covalent bonds form when orbitals 轨道 overlap:

• a sigma bond σ键 forms by the direct, head-on overlap 重叠 of orbitals between the two atoms.
• a pi bond π键 forms by the sideways overlap of two p orbitals, above and below the sigma bond.

A single bond is one sigma bond. A double bond (as in $\text{C}_2\text{H}_4$) is one sigma plus one pi bond. A triple bond (as in $\text{N}_2$ and $\text{HCN}$) is one sigma plus two pi bonds.

Hybridisation

Hybridisation 杂化 mixes orbitals in the same shell to make new, equal orbitals for bonding:

• $\text{sp}$: two equal orbitals, used in a linear molecule.
• $\text{sp}^2$: three equal orbitals, used in a flat molecule like $\text{C}_2\text{H}_4$.
• $\text{sp}^3$: four equal orbitals, used in $\text{CH}_4$.

Bond energy and bond length

• bond energy 键能 is the energy needed to break one mole of a particular covalent bond in the gas state.
• bond length 键长 is the distance between the centres of the two bonded atoms.

A shorter bond is usually stronger (higher bond energy). Triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds. Stronger bonds make a molecule harder to react.

To work out a shape, use VSEPR theory 价层电子对互斥理论: the pairs of electrons around the central atom push apart as far as possible, because like charges repel.

A lone pair 孤对电子 (not in a bond) pushes more strongly than a bonding pair 成键电子对. Each lone pair squeezes the bond angle 键角 by about $2.5°$.

$\text{NH}_3$ has one lone pair and $\text{H}_2\text{O}$ has two, which is why their angles drop below the $109.5°$ of $\text{CH}_4$. You can predict the shapes of similar molecules and ions in the same way.

Worked example. Predict the shape and bond angle of $\text{NH}_3$ and of $\text{H}_2\text{O}$. Nitrogen has 5 outer electrons and forms 3 bonds, leaving 3 bonding pairs and 1 lone pair - four pairs in total, so they start from the tetrahedral $109.5°$. The lone pair repels more strongly and squeezes the angle by about $2.5°$, giving a pyramidal shape at about $107°$. Oxygen forms 2 bonds and keeps 2 lone pairs: still four pairs, but now two squeezes, so the shape is bent at about $104.5°$. Count all the pairs to fix the basic geometry, subtract $2.5°$ for each lone pair, and name the shape from the atoms - $\text{NH}_3$ has four pairs but is pyramidal, not tetrahedral.

Intermolecular forces 分子间作用力 are the forces between molecules 分子. They are much weaker than the ionic, covalent and metallic bonding inside substances.

Bond polarity and dipoles

When two atoms with different electronegativity share a bond, the electrons sit closer to the more electronegative atom. The bond then has a polarity 极性: one end is slightly negative ($\delta-$) and the other slightly positive ($\delta+$). This separation of charge is a dipole 偶极.

If the dipoles in a molecule do not cancel, the whole molecule has a dipole moment 偶极矩 and is polar. If they cancel by symmetry (as in $\text{CO}_2$), the molecule is non-polar.

Van der Waals' forces

Van der Waals' forces 范德华力 is the general name for all intermolecular forces. There are two main types.

The first type is the instantaneous dipole–induced dipole force, also called the London dispersion force 伦敦色散力. Moving electrons make a brief instantaneous dipole 瞬时偶极, which then creates a matching induced dipole 诱导偶极 in a nearby molecule. These forces act between all molecules and get stronger when there are more electrons.

The second type is the permanent dipole–permanent dipole force. It acts between molecules that are always polar, because each one has a permanent dipole 永久偶极.

Hydrogen bonding

Hydrogen bonding 氢键 is a strong, special case of permanent dipole forces. It forms when hydrogen is bonded to a very electronegative atom — nitrogen, oxygen or fluorine — and is attracted to a lone pair on an N, O or F atom in a neighbour. Look for N–H and O–H groups, as in ammonia and water.

Hydrogen bonding explains the strange behaviour of water:

• its high melting and boiling point 沸点, because many hydrogen bonds must be broken.
• its high surface tension 表面张力.
• ice is less dense than liquid water, because hydrogen bonds hold the molecules in an open, spread-out structure, so ice floats.

A dot-and-cross diagram 点叉图 shows the outer electrons of each atom, using dots for one atom and crosses for the other. This makes it clear where each bonding electron came from. You can draw them for ionic, covalent and coordinate bonding, including molecules with an expanded octet or an odd number of electrons.

• For shapes, count bonding pairs and lone pairs, name the shape, then give the exact bond angle (e.g. $\text{NH}_3$: pyramidal, $107^\circ$) — each lone pair lowers the angle by about $2.5^\circ$.
• A dative (coordinate) bond has both electrons from one atom (e.g. $\text{NH}_4^+$, $\text{H}_3\text{O}^+$); draw the arrow from the lone pair.
• Name the intermolecular force precisely: hydrogen bonding needs H bonded to N, O or F; otherwise it is permanent-dipole or induced-dipole (van der Waals). Never call van der Waals forces "bonds".
• Explain a physical property by stating which forces are broken, not just "strong bonds".

Every chemical reaction takes in or gives out energy. This energy change, measured at constant pressure, is the enthalpy change 焓变, with symbol $\Delta H$.

• in an exothermic 放热 reaction the system gives out heat, so the products have less energy than the reactants and $\Delta H$ is negative.
• in an endothermic 吸热 reaction the system takes in heat, so the products have more energy than the reactants and $\Delta H$ is positive.

Reaction pathway diagrams

A reaction pathway diagram 反应路径图 shows the energy of the reactants and products, and the energy "hill" between them. The height of the hill is the activation energy 活化能 — the least energy the particles need before they can react.

• exothermic: products sit lower than reactants ($\Delta H < 0$).
• endothermic: products sit higher than reactants ($\Delta H > 0$).

Standard conditions and types of enthalpy change

Energy values are compared under standard conditions 标准条件: $298\ \text{K}$ and $101\ \text{kPa}$, shown by the symbol $^{\ominus}$. Each substance is in its normal physical state at those conditions.

Energy from breaking and making bonds

During a reaction, old bonds break and new bonds form. Breaking a bond needs energy (endothermic); making a bond releases energy (exothermic). The enthalpy change of the reaction is the difference between the two:

The bond energy 键能 is the energy needed to break one mole of a particular bond in the gas state, so it is always positive. Some bond energies are exact (for one specific molecule); others are averages taken over many different molecules, so calculations using them are only approximate.

Worked example. Use bond energies to find $\Delta H$ for $\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$. Bond energies (kJ mol⁻¹): H–H $= 436$, Cl–Cl $= 242$, H–Cl $= 431$.

Measuring enthalpy change in the lab

When a reaction heats up (or cools down) a known mass of water or solution, the heat transferred is:

where $m$ is the mass, $c$ is the specific heat capacity 比热容 (how much energy raises 1 g by 1 K), and $\Delta T$ is the temperature change. The enthalpy change per mole is then:

The minus sign makes $\Delta H$ negative when the temperature rises (an exothermic reaction).

Worked example. Burning $0.50\ \text{g}$ of methanol ($M_r = 32$) raises the temperature of $100\ \text{g}$ of water by $18\ ^{\circ}\text{C}$. Find the enthalpy change of combustion per mole. ($c = 4.18\ \text{J g}^{-1}\,\text{K}^{-1}$.)

The heat released is $q = mc\Delta T = 100 \times 4.18 \times 18 = 7520\ \text{J}$. The amount burnt is $n = 0.50/32 = 0.0156\ \text{mol}$, so

Hess's law 盖斯定律 says that the total enthalpy change for a reaction is the same, no matter which route you take from reactants to products. This is because energy is conserved.

This lets you draw an energy cycle 能量循环: link the reactants and products by a direct step and by an indirect route, then add the steps so that both routes give the same total.

Hess's law is useful in two ways:

• it lets you find an enthalpy change that you cannot measure directly (for example, the formation of a compound that forms slowly or with side reactions).
• it lets you calculate $\Delta H_r$ from bond energy data, or from formation or combustion data given in the question.

• Define each enthalpy change with its exact standard conditions (e.g. combustion = one mole burned completely in excess oxygen).
• Use $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})$; breaking is endothermic ($+$), making is exothermic ($-$) — getting the sign the wrong way round is the classic error.
• In $q = mc\Delta T$ use the mass of the water/solution, then divide by moles and add the minus sign for an exothermic reaction.
• Draw Hess cycles with arrows the same way round, follow the alternative route, and always give $\Delta H$ a sign and units ($\text{kJ mol}^{-1}$).

The oxidation number 氧化数 (also called the oxidation state) shows how many electrons 电子 an atom has lost or gained compared with the free element. You work it out using simple rules:

• an uncombined element has an oxidation number of $0$.
• a simple ion has an oxidation number equal to its charge (so $\text{Mg}^{2+}$ is $+2$).
• Group 1 is always $+1$, Group 2 is always $+2$.
• hydrogen is $+1$ (but $-1$ in metal hydrides).
• oxygen is $-2$ (but $-1$ in peroxides).
• fluorine is always $-1$.
• the oxidation numbers in a neutral compound add up to $0$; in an ion they add up to the charge.

A Roman numeral shows the size of the oxidation number of an element, for example iron(II) means $+2$ and manganese(VII) in $\text{KMnO}_4$ means $+7$.

Worked example. Find the oxidation number of sulfur in the sulfate ion, $\text{SO}_4^{2-}$.

Each oxygen is $-2$, so the four oxygens give $4 \times (-2) = -8$. In an ion the numbers add up to the charge, here $-2$. If sulfur is $x$:

Worked example. Find the oxidation number of nitrogen in the nitrate ion, $\text{NO}_3^-$.

A redox 氧化还原 reaction is one where electrons move from one species to another.

• oxidation 氧化 is the loss of electrons. The oxidation number goes up.
• reduction 还原 is the gain of electrons. The oxidation number goes down.

A useful memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain.

Oxidation and reduction always happen together, because the electrons lost by one species are gained by another. This electron transfer 电子转移 is why we call it a redox reaction.

Changes in oxidation number help you balance a redox equation:

1. find the element whose oxidation number rises (it is oxidised) and the one whose number falls (it is reduced).
2. the total rise must equal the total fall, because every electron lost is gained somewhere.
3. choose the ratio of the two species so the rise and fall match, then balance the rest of the equation.

Worked example. Balance the reaction of manganate(VII) with iron(II) in acid: $\text{MnO}_4^- + \text{Fe}^{2+} + \text{H}^+ \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} + \text{H}_2\text{O}$.

Manganese falls from $+7$ to $+2$ (a fall of $5$); iron rises from $+2$ to $+3$ (a rise of $1$). To make the total fall equal the total rise, take $5$ iron ions for every $1$ manganate ion. Then balance oxygen with water ($4\,\text{H}_2\text{O}$) and hydrogen with $\text{H}^+$ ($8\,\text{H}^+$):

Disproportionation 歧化 is a special redox reaction in which the same element is both oxidised and reduced at the same time. For example, when chlorine reacts with cold water, some chlorine atoms are reduced (to $\text{Cl}^-$) and others are oxidised (to $\text{ClO}^-$).

• an oxidising agent 氧化剂 takes electrons away from another species. In doing so, it is itself reduced.
• a reducing agent 还原剂 gives electrons to another species. In doing so, it is itself oxidised.

So in any redox reaction, the oxidising agent causes the oxidation of the other species, while the reducing agent causes the reduction.

• Apply the oxidation-number rules in order: O is $-2$ except in peroxides ($-1$); H is $+1$ except in metal hydrides ($-1$).
• In a redox equation the total increase equals the total decrease in oxidation number — use this to fix the ratio, then balance O with $\text{H}_2\text{O}$ and H with $\text{H}^+$.
• An oxidising agent is itself reduced; name both what is oxidised/reduced and the agent for full marks.
• Disproportionation is the same element both oxidised and reduced — show both oxidation-number changes.

A reversible reaction 可逆反应 can go both ways. The forward reaction 正反应 makes products; the reverse reaction 逆反应 turns products back into reactants. We show this with the sign $\rightleftharpoons$.

In a closed system 封闭系统 (nothing enters or leaves), the reaction reaches a dynamic equilibrium 动态平衡 when:

• the rate of the forward reaction equals the rate of the reverse reaction.
• the concentrations of reactants and products stay constant.

It is called dynamic because both reactions are still happening — they just cancel out. A closed system is needed, or products would escape and equilibrium could never be reached.

Le Chatelier's principle 勒夏特列原理 says: if you change a system at equilibrium, the position of equilibrium 平衡 moves to oppose (reduce) that change.

A catalyst lets equilibrium be reached faster, but it does not change the position of equilibrium.

For a reaction at equilibrium, the equilibrium constant 平衡常数 links the amounts of products and reactants. For the reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$:

where the square brackets mean concentration in $\text{mol dm}^{-3}$.

Worked example. At equilibrium the mixture $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ has $[\text{H}_2] = 0.20$, $[\text{I}_2] = 0.20$ and $[\text{HI}] = 1.6\ \text{mol dm}^{-3}$. Find $K_c$.

(no units here, because the concentration powers cancel top and bottom).

Worked example (ICE table). $1.00\ \text{mol}$ each of $\text{H}_2$ and $\text{I}_2$ are sealed in a $1.00\ \text{dm}^3$ flask; at equilibrium $0.20\ \text{mol}$ of $\text{H}_2$ remains. Set out an ICE table (Initial, Change, Equilibrium):

$0.80\ \text{mol}$ of $\text{H}_2$ reacted, so $2\times0.80 = 1.60\ \text{mol}$ HI formed. In the $1.00\ \text{dm}^3$ flask the concentrations equal the moles, so $K_c = \dfrac{1.60^2}{0.20\times0.20} = 64$.

For reactions of gases, we use partial pressure 分压 instead of concentration. The partial pressure of a gas is the share of the total pressure that it provides. It is found from the mole fraction 摩尔分数 (the fraction of all the moles that are that gas):

The constant written with partial pressures is $K_p$.

Worked example. A gas mixture holds $2.0\ \text{mol}$ of $\text{N}_2$ and $6.0\ \text{mol}$ of $\text{H}_2$ at a total pressure of $200\ \text{kPa}$. Find the partial pressure of each gas.

There are $8.0\ \text{mol}$ in total, so

Worked example (a numeric $K_p$). For $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ the equilibrium partial pressures are $p(\text{N}_2)=20$, $p(\text{H}_2)=40$ and $p(\text{NH}_3)=10\ \text{kPa}$. Then

Only temperature changes the value of $K_c$ or $K_p$. Changing concentration or pressure shifts the position of equilibrium but leaves the constant unchanged; adding a catalyst changes neither the constant nor the position — it only makes equilibrium arrive faster.

These two industrial processes are chosen by balancing yield, rate and cost using Le Chatelier's principle.

• the Haber process 哈伯法 makes ammonia: $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ (exothermic). It uses about $450\,°\text{C}$, $200\ \text{atm}$ and an iron catalyst. A low temperature would give more ammonia but too slowly, so a moderate temperature is a compromise.
• the Contact process 接触法 makes sulfur trioxide: $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$ (exothermic). It uses about $450\,°\text{C}$, near $1$–$2\ \text{atm}$ and a vanadium(V) oxide catalyst.

A proton 质子 is simply an $\text{H}^+$ ion. The Brønsted–Lowry theory defines acids and bases by what they do with protons:

• an acid 酸 is a proton donor 质子供体 (it gives away $\text{H}^+$).
• a base 碱 is a proton acceptor 质子受体 (it takes $\text{H}^+$). A base that dissolves in water is called an alkali.

Common acids you must know: hydrochloric acid ($\text{HCl}$), sulfuric acid ($\text{H}_2\text{SO}_4$), nitric acid ($\text{HNO}_3$) and ethanoic acid ($\text{CH}_3\text{COOH}$). Common alkalis: sodium hydroxide ($\text{NaOH}$), potassium hydroxide ($\text{KOH}$) and ammonia ($\text{NH}_3$).

This is about how fully an acid or base splits up in water — not how concentrated it is.

• a strong acid 强酸 or strong base 强碱 is fully dissociated 解离 in water (almost every molecule splits into ions).
• a weak acid 弱酸 or weak base 弱碱 is only partly dissociated (most molecules stay whole).

The pH scale measures how acidic a solution is: pure water is pH 7, acids are below 7, and alkalis are above 7. A strong acid has a lower pH than a weak acid of the same concentration.

You can tell strong and weak acids apart by:

• reaction with a reactive metal: a strong acid fizzes faster.
• pH: measured with a pH meter or universal indicator 通用指示剂.
• electrical conductivity: a strong acid conducts better, because it has more ions.

Neutralisation 中和 happens when the hydrogen ions from an acid react with the hydroxide ions from an alkali:

A salt 盐 is also formed, from the rest of the acid and base.

In a titration 滴定 you add one solution to another and follow the pH. The titration curve 滴定曲线 has a steep, almost vertical jump near the end point. The exact shape depends on whether each reactant is strong or weak.

To pick an indicator 指示剂, choose one whose colour change falls inside that steep jump. For a strong acid with a strong base most indicators work; for a weak acid with a strong base you need one that changes in the higher pH range.

• Define dynamic equilibrium with all three points: forward and reverse rates equal, concentrations constant, closed system.
• Answer Le Chatelier questions by stating the change, the direction of shift, and the reason; a catalyst does not shift the position (it speeds both rates).
• Write $K_c$ as products over reactants, each raised to its balancing number; include state symbols to decide what appears.
• Justify Haber/Contact conditions as a compromise between yield, rate and cost — do not just quote them.
• Brønsted: acid = proton donor, base = proton acceptor; conjugate pairs differ by one $\text{H}^+$.

The rate of reaction 反应速率 is how fast reactants turn into products. We measure it as the change in concentration (or amount) in each unit of time.

To react, particles must collide 碰撞. The collision frequency 碰撞频率 is how often the particles hit each other. But not every collision leads to a reaction:

• an effective collision 有效碰撞 has enough energy and the correct direction, so a reaction happens.
• a non-effective collision 无效碰撞 does not have enough energy, or the particles hit at the wrong angle, so nothing happens.

So the rate depends on the frequency of effective collisions — how many useful collisions happen each second.

Concentration and pressure

If you increase the concentration of a solution (or the pressure of a gas), the particles are packed closer together. They collide more often, so there are more effective collisions each second, and the rate goes up.

You can calculate a rate from experimental data — for example, the volume of gas made divided by the time taken.

Worked example. A reaction gives off carbon dioxide. In the first $30\ \text{s}$, $48\ \text{cm}^3$ of gas is collected. Find the average rate of reaction over this time.

The rate is fastest at the start (the graph is steepest there) because the reactants are most concentrated, then it slows as they are used up.

The activation energy 活化能 ($E_A$) is the minimum energy a collision needs in order to be effective.

The Boltzmann distribution 玻尔兹曼分布 is a graph showing how the energies of the molecules are spread out at one temperature. The curve starts at the origin, rises to a peak, then falls away in a long tail. The total area under the curve is the total number of molecules. Only the molecules to the right of $E_A$ have enough energy to react.

When you raise the temperature:

• the curve flattens and spreads to the right, so a much larger fraction of molecules now have energy greater than $E_A$.
• the molecules also move faster and collide more often.

The first effect is the bigger one. This is why a small rise in temperature gives a large rise in rate.

A catalyst 催化剂 speeds up a reaction but is not used up itself. Catalysis 催化作用 is the name for this action.

A catalyst works by giving the reaction a different reaction mechanism 反应机理 — a new route with a lower activation energy. On the Boltzmann distribution, lowering $E_A$ moves the line to the left, so more molecules now have enough energy. This means more effective collisions each second, and a faster rate.

On a reaction pathway diagram 反应路径图, the catalysed route has a lower energy "hill". The enthalpy change of the reaction, $\Delta H$, is not changed by the catalyst.

There are two types:

• a homogeneous catalyst 均相催化剂 is in the same physical state as the reactants — for example, an acid catalyst dissolved in a solution of liquids.
• a heterogeneous catalyst 多相催化剂 is in a different state from the reactants — for example, solid iron speeding up the reaction of gases in the Haber process.

• Explain rate changes with collision theory — more frequent and/or more energetic effective collisions.
• On a Boltzmann distribution mark $E_A$, shade to its right, and show the curve flatten and shift right at higher temperature; it starts at the origin and never touches the axis.
• A catalyst gives an alternative route with lower $E_A$ and does not change $\Delta H$.
• Say why a small temperature rise gives a large rate rise: a much greater proportion of molecules now exceed $E_A$ (the main effect).

Periodicity 周期性 means that properties repeat in a regular pattern as you go across each period 周期 of the Periodic Table. Period 3 (Na to Ar) is the standard example.

The melting point and conductivity follow from the structure and bonding:

• Na, Mg, Al are giant metallic 金属晶体. Melting points rise (Na → Al) because each atom gives more delocalised electrons and the ions get smaller, so the bonding is stronger. They conduct well.
• Si is giant molecular 原子晶体 (giant covalent). It has the highest melting point, because strong covalent bonds must be broken. It barely conducts.
• P, S, Cl, Ar are simple molecular 分子晶体 (or single atoms). Their melting points are low, because the only forces that break are the weak van der Waals' forces 范德华力 between the molecules - the strong covalent bonds inside each molecule are never broken at all. They do not conduct.

Name van der Waals' forces when you explain that drop: it is the marking point, and "the covalent bonds are weak" is the error that loses it. Melting $\text{S}_8$ pulls whole molecules apart from each other; it does not touch a single $\text{S–S}$ bond. Their sizes still follow the molecule: $\text{S}_8$ melts higher than $\text{P}_4$ because it is bigger, with more electrons and so stronger van der Waals' forces, while $\text{Ar}$ (a single atom) is lowest of all.

Reactions with oxygen, chlorine and water

With oxygen:

With chlorine:

With water (only Na and Mg react):

Sodium reacts fast; magnesium reacts only very slowly with cold water.

Oxidation number of the oxides and chlorides

The oxidation number 氧化数 of the Period 3 element in its oxide or chloride rises across the period, because it equals the number of outer-shell (valence shell 价层) electrons 电子 the atom uses in bonding:

The chlorides $\text{NaCl}$, $\text{MgCl}_2$, $\text{AlCl}_3$, $\text{SiCl}_4$, $\text{PCl}_5$ show oxidation numbers $+1$ to $+5$ in the same way.

Oxides with water, and acid–base behaviour

Across the period the oxides 氧化物 change from basic to acidic:

Metal oxides (left) are basic; non-metal oxides (right) are acidic. $\text{Al}_2\text{O}_3$ and its hydroxide 氢氧化物 $\text{Al(OH)}_3$ are amphoteric — they react with both acids and bases:

Chlorides with water

• $\text{NaCl}$ and $\text{MgCl}_2$ are ionic. They simply dissolve, giving a near-neutral solution.
• $\text{SiCl}_4$ and $\text{PCl}_5$ are covalent. They undergo hydrolysis 水解 (react with water) to make acidic solutions and fumes of $\text{HCl}$:

Explaining the trends

These trends follow from the change in bonding and electronegativity 电负性. On the left, the elements are metals with low electronegativity, so their oxides and chlorides are ionic and basic (or neutral). On the right, the elements are non-metals with high electronegativity, so their oxides and chlorides are covalent and acidic. You can use a chloride's or oxide's properties (melting point, conductivity, effect on water) to suggest whether its bonding is ionic or covalent.

Worked example. Predict the pH when $\text{Na}_2\text{O}$, $\text{Al}_2\text{O}_3$ and $\text{SO}_3$ are each added to water. Acid-base character follows the metal to non-metal change across Period 3. $\text{Na}_2\text{O}$ is ionic and basic: it dissolves to give $\text{NaOH}$, so the pH is about 13. $\text{SO}_3$ is covalent and acidic: it reacts to give $\text{H}_2\text{SO}_4$, so the pH is about 1. $\text{Al}_2\text{O}_3$ sits at the changeover - it is amphoteric and essentially insoluble in water, so the pH stays about 7. The trap is that last one: amphoteric does not mean neutral, it means the oxide reacts with acids and with alkalis - it simply does not react with water.

The same idea works for any group. If you know the pattern down a group and across a period, you can:

• predict the properties of an element from its position (for example, a Group 1 element will be a reactive metal forming a $+1$ ion).
• deduce the likely position and identity of an unknown element from its physical and chemical properties.

• Always pair the trend with its reason: atomic radius falls across Period 3 (rising nuclear charge, similar shielding).
• Explain the melting-point pattern by structure: giant metallic (Na→Al), giant covalent (Si, highest), then simple molecular ($\text{P}_4$, $\text{S}_8$, $\text{Cl}_2$) and monatomic (Ar).
• Oxides go basic → amphoteric ($\text{Al}_2\text{O}_3$) → acidic across the period; link to ionic-to-covalent bonding.
• For reactions of oxides/chlorides with water, give the products and the resulting pH.