📚 International A-level Chemistry Example Responses: CH05 Unit 5 Calculation Questions | 国际A-Level化学样题解析：CH05 单元5 计算题型
Unit 5 of International A-level Chemistry presents a variety of numerical problems that test your ability to apply theoretical concepts to quantitative situations. This article walks through common calculation question types, with example responses that highlight the essential steps and common pitfalls.
国际A-Level化学第五单元涉及多种计算题型，考查将理论知识应用于定量分析的能力。本文精选常见计算题型，通过样题解析展示关键解题步骤与常见错误，助你扎实掌握解题方法。
1. Overview of Unit 5 Calculation Questions | 单元5计算题型概述
Unit 5 covers thermodynamics, equilibrium, redox chemistry, and acid-base systems. Calculation questions often require multi-step reasoning, including unit conversions, correct use of formulas, and interpretation of data from tables or graphs.
第五单元涵盖热力学、化学平衡、氧化还原与酸碱体系。计算题常需多步推理，涉及单位换算、公式的正确运用以及图表数据的解读。
Typical topics include: ideal gas calculations, equilibrium constants (Kc and Kp), Gibbs free energy, Hess’s law, Born-Haber cycles, entropy changes, electrode potentials, titration curves, and buffer solutions.
典型主题包括：理想气体计算、平衡常数(Kc与Kp)、吉布斯自由能、赫斯定律、玻恩-哈伯循环、熵变、电极电势、滴定曲线及缓冲溶液。
In each example response, always show your working clearly, check significant figures, and include units.
每道样题解析中，务必清晰展示步骤、检查有效数字并标明单位。
2. Using the Ideal Gas Equation | 理想气体状态方程的应用
The ideal gas equation, pV = nRT, is fundamental for relating pressure, volume, temperature and amount of gas. Common tasks include calculating molar mass or density of a gas.
理想气体状态方程 pV = nRT 是联系压力、体积、温度与物质的量的基础公式，常用于计算气体摩尔质量或密度。
Example: A 0.250 g sample of a volatile liquid is vaporised, producing 85.0 cm³ of gas at 100 °C and 101 kPa. Calculate its molar mass.
示例：将0.250 g易挥发液体气化，在100 °C、101 kPa下得到85.0 cm³气体，求其摩尔质量。
Solution: Convert to SI units – P = 101×10³ Pa, V = 85.0×10⁻⁶ m³, T = 373 K. n = PV/RT = (101×10³ × 85.0×10⁻⁶)/(8.31 × 373) = 2.77×10⁻³ mol. M = mass/n = 0.250 g / 2.77×10⁻³ mol ≈ 90.3 g mol⁻¹.
解答：换算SI单位 – P=101×10³ Pa，V=85.0×10⁻⁶ m³，T=373 K。n=PV/RT=(101×10³×85.0×10⁻⁶)/(8.31×373)=2.77×10⁻³ mol。M=质量/n=0.250 g/2.77×10⁻³ mol ≈ 90.3 g mol⁻¹。
Always ensure temperature is in Kelvin and pressure in Pascals when using R = 8.31 J K⁻¹ mol⁻¹.
使用R=8.31 J K⁻¹ mol⁻¹时，务必确保温度单位为开尔文，压力单位为帕斯卡。
3. Equilibrium Constants (Kc and Kp) | 平衡常数 (Kc 和 Kp) 计算
Kc and Kp calculations require a balanced equation and careful use of concentration or partial pressure values at equilibrium. Remember that solids and pure liquids do not appear in the expression.
Kc与Kp的计算需配平方程式，并准确使用平衡时的浓度或分压。注意固体和纯液体不写入表达式。
For Kp, partial pressure = mole fraction × total pressure. Mole fraction = moles of substance / total moles of gas.
对于Kp，分压=摩尔分数×总压。摩尔分数=该气体物质的量/气体总物质的量。
Example: For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), at equilibrium the flask contains 0.20 mol SO₂, 1.60 mol O₂, 0.80 mol SO₃ at total pressure 200 kPa. Calculate Kp.
示例：反应2SO₂(g)+O₂(g)⇌2SO₃(g)，平衡时容器含0.20 mol SO₂、1.60 mol O₂、0.80 mol SO₃，总压200 kPa，求Kp。
Total moles = 0.20+1.60+0.80 = 2.60 mol. Mole fractions: SO₂ = 0.20/2.60 = 0.0769, O₂ = 1.60/2.60 = 0.615, SO₃ = 0.80/2.60 = 0.308. Partial pressures: pSO₂ = 15.4 kPa, pO₂ = 123 kPa, pSO₃ = 61.6 kPa. Kp = (pSO₃)² / [(pSO₂)² × pO₂] = (61.6)² / [(15.4)² × 123] ≈ 1.30 kPa⁻¹.
总物质的量=2.60 mol。摩尔分数：SO₂=0.0769，O₂=0.615，SO₃=0.308。分压：pSO₂=15.4 kPa，pO₂=123 kPa，pSO₃=61.6 kPa。Kp=(pSO₃)²/[(pSO₂)²×pO₂] ≈ 1.30 kPa⁻¹。
Include units for Kp based on the expression; exponents can make units kPa⁻¹ or kPa etc.
Kp需带单位，根据表达式指数不同，单位可为kPa⁻¹或kPa等。
4. Gibbs Free Energy and Equilibrium | 吉布斯自由能与平衡
The relationship ΔG = -RT ln K links thermodynamics to equilibrium. If ΔG is negative, K > 1, products are favoured. Use ΔG = ΔH – TΔS to find ΔG first.
关系式ΔG = -RT ln K将热力学与平衡联系起来。若ΔG为负，K>1，产物占优。先用ΔG=ΔH-TΔS求ΔG。
Example: For a reaction, ΔH = -92 kJ mol⁻¹, ΔS = -199 J K⁻¹ mol⁻¹ at 298 K. Calculate K.
示例：某反应ΔH=-92 kJ mol⁻¹，ΔS=-199 J K⁻¹ mol⁻¹，温度298 K，求K。
First ensure consistent units: ΔH = -92000 J mol⁻¹, ΔS in J. ΔG = -92000 – (298 × -199) = -92000 + 59302 = -32698 J mol⁻¹. ln K = -ΔG/RT = 32698/(8.314×298) ≈ 13.19. K = e¹³·¹⁹ ≈ 5.4 × 10⁵.
首先统一单位：ΔH=-92000 J mol⁻¹。ΔG=-92000-(298×-199)= -92000+59302=-32698 J mol⁻¹。ln K=-ΔG/RT=32698/(8.314×298)≈13.19。K≈5.4×10⁵。
Note that T must be in Kelvin and R = 8.314 J K⁻¹ mol⁻¹.
注意T须为开尔文，R=8.314 J K⁻¹ mol⁻¹。
5. Hess’s Law and Enthalpy Cycles | 赫斯定律与焓变循环
Hess’s law allows calculation of enthalpy changes that cannot be measured directly. Use standard enthalpies of formation or combustion to construct a cycle.
赫斯定律可用于计算难以直接测量的焓变。利用标准生成焓或燃烧焓构建循环。
Example: Calculate ΔH for: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) given ΔcH of C(s) = -394 kJ mol⁻¹, H₂(g) = -286 kJ mol⁻¹, C₂H₅OH(l) = -1367 kJ mol⁻¹.
示例：求反应2C(s)+3H₂(g)+½O₂(g)→C₂H₅OH(l)的ΔH，已知ΔcH：C(s)=-394 kJ mol⁻¹，H₂(g)=-286 kJ mol⁻¹，C₂H₅OH(l)=-1367 kJ mol⁻¹。
Using a combustion cycle: ΔH = Σ ΔcH(reactants) – Σ ΔcH(products) = [2(-394) + 3(-286)] – (-1367) = (-788 -858) + 1367 = -279 kJ mol⁻¹.
使用燃烧焓循环：ΔH=ΣΔcH(反应物)-ΣΔcH(产物)=[2(-394)+3(-286)]-(-1367)=(-788-858)+1367=-279 kJ mol⁻¹。
Always check the direction of arrows in your cycle; energy released in combustion means values are negative.
务必核对循环中箭头的方向；燃烧放热故数值为负。
6. Born-Haber Cycle Calculations | 玻恩-哈伯循环计算
Born-Haber cycles apply Hess’s law to ionic compounds. They relate lattice energy to ionisation energies, electron affinities, and enthalpies of atomisation and formation.
玻恩-哈伯循环将赫斯定律应用于离子化合物，关联晶格能与电离能、电子亲和能、原子化焓和生成焓。
Example: For NaCl, ΔfH = -411 kJ mol⁻¹. Given: Na(s) → Na(g) ΔatH = +108 kJ mol⁻¹, Na(g) → Na⁺(g) + e⁻ IE = +496 kJ mol⁻¹, ½Cl₂(g) → Cl(g) ΔatH = +121 kJ mol⁻¹, Cl(g) + e⁻ → Cl⁻(g) EA = -349 kJ mol⁻¹. Calculate lattice energy.
示例：对NaCl，ΔfH=-411 kJ mol⁻¹。已知：Na(s)→Na(g) ΔatH=+108，Na(g)→Na⁺(g)+e⁻ IE=+496，½Cl₂(g)→Cl(g) ΔatH=+121，Cl(g)+e⁻→Cl⁻(g) EA=-349 kJ mol⁻¹。求晶格能。
Cycle: ΔfH = ΔatH(Na) + IE(Na) + ΔatH(Cl) + EA(Cl) + U. So U = ΔfH – [ΔatH(Na)+IE(Na)+ΔatH(Cl)+EA(Cl)] = -411 – (108+496+121-349) = -411 – 376 = -787 kJ mol⁻¹. Lattice energy is exothermic.
循环：ΔfH=ΔatH(Na)+IE(Na)+ΔatH(Cl)+EA(Cl)+U。U=ΔfH-[ΔatH(Na)+IE(Na)+ΔatH(Cl)+EA(Cl)]=-411-(108+496+121-349)=-411-376=-787 kJ mol⁻¹，晶格能为放热。
Lattice energy is always negative for stable ionic solids; magnitude indicates strength of ionic bonding.
稳定离子固体的晶格能总为负值；其大小反映离子键强度。
7. Entropy Changes and Feasibility | 熵变与反应可行性
Total entropy change ΔS_total = ΔS_system + ΔS_surroundings. ΔS_surroundings = -ΔH/T. A reaction is feasible when ΔS_total > 0.
总熵变ΔSₜₒₜₐₗ=ΔSₛᵧₛₜₑₘ+ΔSₛᵤᵣᵣₒᵤₙₔᵢₙ₉ₛ。ΔSₛᵤᵣᵣₒᵤₙₔᵢₙ₉ₛ=-ΔH/T。当ΔSₜₒₜₐₗ>0时反应可行。
Example: For the decomposition of CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹, ΔS_system = +161 J K⁻¹ mol⁻¹ at 298 K. Find ΔS_total and the temperature at which the reaction becomes feasible.
示例：CaCO₃(s)→CaO(s)+CO₂(g)，ΔH=+178 kJ mol⁻¹，ΔSₛᵧₛₜₑₘ=+161 J K⁻¹ mol⁻¹(298 K)。求ΔSₜₒₜₐₗ及反应可行温度。
ΔS_surr = -178000 J mol⁻¹ / 298 K = -597 J K⁻¹ mol⁻¹. ΔS_total = 161 – 597 = -436 J K⁻¹ mol⁻¹ (not feasible at 298 K). For feasibility, set ΔS_total = 0: 161 + (-178000/T) = 0 → T = 178000/161 ≈ 1106 K. Above this temperature, reaction is feasible.
ΔSₛᵤᵣᵣ=-178000/298=-597 J K⁻¹ mol⁻¹，ΔSₜₒₜₐₗ=161-597=-436 J K⁻¹ mol⁻¹(298 K不可行)。设ΔSₜₒₜₐₗ=0，161+(-178000/T)=0→T≈1106 K，高于此温度反应可行。
Always convert ΔH to J when combining with ΔS in J K⁻¹.
与ΔS(J K⁻¹)结合计算时，务必把ΔH换算为J。
8. Electrode Potentials and Cell EMF | 电极电势与电池电动势
Cell EMF is calculated as E_cell = E_right – E_left under standard conditions. The more positive half-cell undergoes reduction.
标准条件下，电池电动势E_cell=E_right-E_left。电势更正的一极发生还原。
Example: A cell is set up with Zn²⁺/Zn (-0.76 V) and Cu²⁺/Cu (+0.34 V). Write the cell diagram and calculate EMF.
示例：由Zn²⁺/Zn(-0.76 V)和Cu²⁺/Cu(+0.34 V)组成的电池，书写电池图示并计算EMF。
Cell: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). E_cell = +0.34 – (-0.76) = +1.10 V.
电池图示：Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s)。E_cell=+0.34-(-0.76)=+1.10 V。
Remember the double line represents the salt bridge; the half-cell with the more negative potential is placed on the left for a spontaneous reaction.
双竖线表示盐桥；自发反应时，电势更负的半电池置于左侧。
9. Acid-Base Titrations and pH Curves | 酸碱滴定与pH曲线
Titration calculations involve finding the equivalence point and selecting a suitable indicator. pH curves differ for strong acid-strong base, weak acid-strong base, etc.
滴定计算涉及确定等当点与选择合适指示剂。强酸强碱、弱酸强碱等的pH曲线形状不同。
Example: 25.0 cm³ of 0.100 mol dm⁻³ CH₃COOH (Ka = 1.74×10⁻⁵) is titrated with 0.100 mol dm⁻³ NaOH. Calculate pH after adding 12.5 cm³ of NaOH.
示例：以0.100 mol dm⁻³ NaOH滴定25.0 cm³ 0.100 mol dm⁻³ CH₃COOH (Ka=1.74×10⁻⁵)，加入12.5 cm³ NaOH后求pH。
At half-neutralisation, [HA] = [A⁻], so pH = pKa = -log(1.74×10⁻⁵) ≈ 4.76. 12.5 cm³ is exactly half of the 25 cm³ required for equivalence, so this buffer region has pH = pKa.
半中和点时，[HA]=[A⁻]，故pH=pKa=-log(1.74×10⁻⁵)≈4.76。12.5 cm³恰为中和所需体积的一半，此缓冲区域pH=pKa。
For other points, use the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]).
其他点使用Henderson-Hasselbalch方程：pH=pKa+log([A⁻]/[HA])。
10. Buffer Calculations | 缓冲溶液计算
Buffer solutions resist changes in pH. They consist of a weak acid and its conjugate base. Use Ka to calculate pH or the required ratio of concentrations.
缓冲溶液能抵抗pH变化，由弱酸及其共轭碱组成。利用Ka计算pH或所需浓度比。
Example: A buffer is made from 0.20 mol dm⁻³ ethanoic acid and 0.15 mol dm⁻³ sodium ethanoate. Ka = 1.74×10⁻⁵. Find pH.
示例：由0.20 mol dm⁻³乙酸和0.15 mol dm⁻³乙酸钠配制缓冲液，Ka=1.74×10⁻⁵，求pH。
pH = pKa + log([salt]/[acid]) = 4.76 + log(0.15/0.20) = 4.76 + log(0.75) = 4.76 – 0.125 ≈ 4.64.
pH=pKa+log([盐]/[酸])=4.76+log(0.15/0.20)=4.76+log(0.75)=4.76-0.125≈4.64。
When a small amount of strong acid or base is added, the ratio [A⁻]/[HA] changes, but pH shifts only slightly. Calculations require stoichiometric adjustments to the moles of acid and conjugate base.
加入少量强酸或强碱时，[A⁻]/[HA]比值变化，但pH仅微小变动。计算需调整酸与共轭碱的物质的量。
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