OCR化学 焓变 盖斯定律 键焓 熵变 自由能
OCR 化学 A（H432）把焓变分成两块：模块 3 的 3.2.1 讲焓变的测量与盖斯定律，模块 5 的 5.2.2 讲晶格能、熵变与吉布斯自由能。两块内容最后都汇到 H432/01 那张 100 分、2 小时 15 分钟的卷子里，这张卷占整门课 37% 的分值。焓变的题目不像有机推断那样靠背结构，它更像做账，每一步的能量都要有清楚的去向。很多学生以为算出 q 就等于算出 ΔH，实际上从 q 到 ΔH 中间还隔着溶液质量、摩尔数和正负号三道关。常见的错误是把 q 除以反应物的总质量，而该除的是真正被加热的溶液质量。下面按考纲顺序把常考环节拆开，每个环节给一个可复算的算例。H432 的数学要求写在 Ofqual 的理科科目条件里，化学卷面至少 20% 的分数落在数学技能上，所以这些算例不是装饰。
Enthalpy carries a large share of the marks in OCR Chemistry A (H432), and it sits in two separate modules that end up in the same 100-mark paper. Students who treat the calculations as arithmetic usually lose marks on the sign, the mass of solution, or the number of moles, all of which sit between a thermometer reading and a final answer. What follows works through the definitions, the calorimetry procedure used in PAG 3, the two Hess cycles, mean bond enthalpies, Born-Haber cycles, and the entropy and free energy work in 5.2.2; each section includes a calculation you can reproduce and the mistake examiners report most often at that step.
焓变的符号与焓变图
焓变 ΔH 是反应在恒压下吸收或放出的热量，符号只看体系本身。放热反应的 ΔH 为负值，因为产物回到环境时焓更低；吸热反应的 ΔH 为正值。焓变图把反应物和产物的焓画成两条水平线，箭头方向表示吸热或放热。很多学生以为图的纵轴是温度，其实纵轴是焓，温度只在量热实验里出现。图上反应物线与产物线的高度差就是 ΔH，与箭头画在哪一侧无关。活化能是反应物线到过渡态峰顶的高度，催化剂只压低这一段，不改变两端的高度差。常见的错误是把箭头从产物指回反应物之后，仍然照抄原来的正负号。
A negative delta H means the system has lost enthalpy to the surroundings, while a positive value means it has gained enthalpy from them; the sign belongs to the system, not to the thermometer. On an enthalpy profile the vertical axis is enthalpy, so the gap between the reactant and product lines is the enthalpy change, and the peak shows the activation energy that a catalyst can lower without touching that gap. Reversing the arrow reverses the sign, which is the step where a surprising number of candidates copy the forward value straight across.
标准条件与四类标准焓变
标准条件在 OCR 考纲里写明是 100 kPa 与一个说明的温度，通常取 298 K。压力变了，焓变值会跟着变，所以数据表上的数字必须标条件才有意义。标准反应焓 ΔrH 对应一个写明的方程式，系数变了数值跟着变。标准生成焓 ΔfH 是一摩尔化合物由元素在标准状态下生成时的焓变，元素的 ΔfH 按定义取零。标准燃烧焓 ΔcH 是一摩尔物质在氧气中完全燃烧时的焓变，产物必须是最高价氧化物，碳变二氧化碳、氢变液态水。标准中和焓 ΔneuH 是酸碱反应生成一摩尔水时的焓变。阅卷时最常见的失分是把生成焓和燃烧焓的定义混着写，少写一摩尔或多写标准状态。
Standard conditions are 100 kPa and a stated temperature, usually 298 K, and every value in the data sheet depends on those two numbers. The four standard enthalpy changes differ in what counts as one mole: one mole of compound formed from elements for delta f H, one mole of substance burned completely for delta c H, and one mole of water produced for neutralisation. Definitions lose marks far more often than calculations do, and the missing words are almost always one mole or standard states.
量热法 用 q = mcΔT 求中和焓
量热实验的核心公式是 q = mcΔT，c 取 OCR 数据表给的 4.18 J g⁻¹ K⁻¹。溶液的密度按 1.00 g cm⁻³ 处理，所以 50.0 cm³ 的混合液按 50.0 g 计算。ΔT 一律取正值，符号在最后一步才补上。算完 q 之后除以被研究物质的摩尔数，得到摩尔焓变。以 25.0 cm³ 的 1.00 mol dm⁻³ 盐酸与同体积同浓度氢氧化钠混合为例，温度升高 6.8 K，q 等于 50.0 乘以 4.18 再乘以 6.8，约 1421 J。生成的水是 0.0250 mol，所以 ΔH 约负 56.8 kJ mol⁻¹，与文献值负 57.1 很接近。这一步最容易出错的地方是把体积当成质量，或者忘记除以 1000 把焦换算成千焦。
Calorimetry starts from q equals mc delta T, with c taken as 4.18 from the OCR data sheet and the solution's mass estimated from its volume at 1.00 g per cubic centimetre. A worked case: 25.0 cubic centimetres each of 1.00 mol per cubic decimetre hydrochloric acid and sodium hydroxide, a 6.8 K rise, so q is about 1421 J; dividing by the 0.0250 mol of water formed gives roughly minus 56.8 kJ per mole. The two errors that cost marks here are using the volume as the mass and forgetting to convert joules into kilojoules before dividing.
量热误差来源与实验改进
量热法测出的 ΔH 几乎总比数据表值小，因为热量在温度计读完数之前已经散掉一部分。用保温杯加盖子、减少开口时间、把温度计插到液体中部，都能减少对流和蒸发带走的热。用酒精灯加热水时，火焰周围的空气也被加热，灯体本身的质量损失不等于真正参与燃烧的量。这类实验最好重复三次取平均，并让火焰与容器距离固定。改用弹式量热计可以解决完全燃烧的问题，学校实验室通常没有这件器材。答题时把误差写成「热量散失」四个字通常只拿一分，要指出是哪一步散失、用什么办法减少。常见的错误是只列改进措施，不说这些措施减少了哪一种热损失。
Measured enthalpy changes almost always come out less exothermic than the data book value, because heat escapes before the thermometer has finished reading; lagging, a lid and a steady thermometer position reduce that loss but never remove it. A spirit burner also heats the surrounding air, so the mass it loses is larger than the mass that actually burned, which pushes the calculated value in the same direction. Answers that just say heat loss without naming the step and the fix tend to stay at one mark.
盖斯定律与生成焓循环
盖斯定律说，反应的焓变只取决于始态和终态，与走哪条路径无关。这句话的用处是把测不到的焓变拆成几步测得到的焓变相加。用生成焓做循环时，把反应物和产物都拆回元素，元素那一层是共同的起点。于是 ΔrH 等于产物的生成焓之和减去反应物的生成焓之和，元素的项直接消掉。算例是甲烷燃烧，生成焓分别是甲烷负 74.8、二氧化碳负 393.5、液态水负 285.8，代进去得到负 890.3 kJ mol⁻¹。很多学生以为要先把方程式配平再乘生成焓，其实配平系数就是乘数，顺序不影响结果。错得最多的是把水的状态写成气态，数值会差出四十多千焦。
Hess's law says the enthalpy change depends only on the starting and finishing materials, so an unmeasurable step can be replaced by any route that can be measured. With formation enthalpies the route goes down to the elements and back up, which is why the elements cancel and the answer is the sum for products minus the sum for reactants. For methane the three values give minus 890.3 kJ per mole, and writing water as a gas instead of a liquid shifts the answer by more than forty kilojoules.
燃烧焓循环的箭头方向
燃烧焓循环走的是相反的路线，反应物和产物都烧成同样的氧化物，氧化物那一层是共同终点。因为燃烧焓本身都是负值，这个循环的减法方向容易写反。正确的式子是反应物的燃烧焓之和减去产物的燃烧焓之和。判断方向的稳妥办法是画箭头，箭头朝下时把数值带上符号相加，再让两条路线相等。一氧化碳的燃烧焓可以由碳的两次燃烧求出：碳到二氧化碳负 393.5，碳到一氧化碳负 110.5，相减得到负 283.0 kJ mol⁻¹。常见的错误是看到产物有两个就先把系数乘上去再加，方向在那一步已经错了。写完循环后回代一遍数值，正负号写反立刻能看出来。
Combustion cycles run the other way: everything is burned to the same oxides, so the oxides form the shared bottom of the diagram. Because combustion enthalpies are already negative, the arithmetic reverses and the answer is reactants minus products; drawing arrows and equating the two routes is safer than memorising the formula. Carbon burning to carbon dioxide and to carbon monoxide gives minus 393.5 and minus 110.5, so the difference of minus 283.0 kJ per mole is the combustion of carbon monoxide.
平均键焓与反应焓
断键吸热、成键放热，所以反应焓等于断开的键能之和减去形成的键能之和。键能数据是平均键焓，来自一组化合物的平均，不是某一次具体反应的真实值。氢与氯反应生成氯化氢，氢氢键 436、氯氯键 243、氢氯键 432，代入得到负 185 kJ mol⁻¹，与数据表的负 184 只差一点。差别来自平均键焓本身的性质，遇到键环境差别大的分子误差会更明显。平均键焓只对气态物质成立，液态或固态要先加汽化焓再算。常见的错误是把键能当成可以随便套用的常数，忽略了它只是平均值。
Breaking bonds costs energy and making bonds releases it, so the reaction enthalpy is the energy of the bonds broken minus the energy of the bonds formed. Mean bond enthalpies are averages over many compounds, which is why hydrogen and chlorine giving minus 185 kJ per mole sits just one kilojoule from the data book value while other molecules drift further. They apply to gaseous species only, so a liquid or solid needs its enthalpy of vaporisation added first.
晶格能与 Born-Haber 循环
晶格生成焓是一摩尔固态离子化合物由其气态离子生成时的焓变，符号为负。Born-Haber 循环把生成焓拆成五步：金属原子化、金属电离、非金属原子化、非金属得电子、气态离子结合成晶体。循环的等式是生成焓等于这五步之和，其中金属电离与非金属原子化吸热，其余放热。以氯化钠为例，各步数值相加应回到负 411 kJ mol⁻¹ 的生成焓。很多学生以为晶格能一定放热，其实晶格解离焓是它的相反数，符号为正。这一步最容易丢分的是漏掉非金属的原子化焓，或者把二分之一用在错误的物种上。
Lattice enthalpy of formation is the enthalpy change when one mole of a solid ionic compound forms from its gaseous ions, and it is strongly exothermic. A Born-Haber cycle splits the formation enthalpy into atomisation, ionisation, electron affinity and lattice steps, with the endothermic steps adding and the exothermic ones subtracting; for sodium chloride the cycle returns minus 411 kJ per mole. Marks go missing when the non-metal atomisation step is skipped, or when the half is applied to the wrong species in a Group 2 compound.
电子亲和能与第二电子亲和能
第一电子亲和能是一摩尔气态原子得到一摩尔电子形成负一价气态离子时的焓变，对氧和卤素都是负值。第二电子亲和能是把第二个电子加到负一价离子上的焓变，符号为正。原因很直接，已经带负电的离子会排斥再来的电子，需要外界做功。氧的第二电子亲和能因此是正值，写循环时不能照抄第一项的符号。这一项在氧化物的 Born-Haber 循环里最容易写错，尤其是氧化镁这类二价金属氧化物。判卷时看到氧的第二电子亲和能写成负值，末尾算出的晶格能一定对不上公布值。
The first electron affinity is exothermic for oxygen and the halogens, because the incoming electron is attracted to the nucleus of a neutral atom. The second electron affinity is endothermic, since the electron is being forced onto an ion that already carries a negative charge, and oxygen is the standard example in the data sheet. Oxide cycles collapse whenever that second value is copied with a negative sign, and the lattice enthalpy at the end never matches the published figure.
溶解焓与离子水合焓
溶解焓是一摩尔固态离子化合物溶于水形成无限稀释溶液时的焓变。它可以拆成两步：先让晶格解离成气态离子，再让气态离子被水分子包围。晶格解离吸热，水合放热，两者相加得到溶解焓。水合放热的多少取决于离子的电荷密度，电荷越高、半径越小，放出的热量越多。同一族从上到下离子半径变大，水合焓变小，晶格解离焓也变小，两者的相对变化决定了溶解焓的符号。很多学生以为只要水合焓大就一定易溶，其实晶格解离那一项同样在变。常见的错误是把水合焓当成一摩尔离子对的值，实际上它按每一摩尔离子单独给出。
Enthalpy of solution is the change when one mole of a solid ionic compound dissolves to infinite dilution, and it is the sum of an endothermic lattice breakdown and an exothermic hydration of the gaseous ions. Hydration releases more energy as charge density rises, so a doubly charged magnesium ion pulls water in far more strongly than a potassium ion, and the balance between that term and the lattice term decides the sign. A frequent slip is treating the hydration enthalpy as belonging to a pair of ions rather than to each ion separately.
熵变与总熵变
熵是能量分散程度的量度，也可以理解成粒子排列方式的数目。固体熔化成液体、液体汽化成气体，熵都增大。温度升高、气体摩尔数增加，熵同样增大。反应的总熵变等于体系熵变加上环境熵变，环境熵变由负的焓变除以温度给出。放热反应让环境变热，环境熵增大，所以 ΔH 为负时环境项一定为正。算例是碳酸钙分解，体系熵变约正 160.6 J K⁻¹ mol⁻¹，焓变为正 178 kJ mol⁻¹，在 298 K 下环境熵变约负 597 J K⁻¹ mol⁻¹。这一步最容易错的是单位，熵用焦每开每摩尔，焓用千焦每摩尔，两者必须先统一。
Entropy measures how widely energy is spread, and it rises as a solid becomes a liquid, a liquid becomes a gas, temperature increases, or the number of moles of gas increases. Total entropy change is the system term plus the surroundings term, and the surroundings term equals minus delta H divided by T, so an exothermic reaction always pushes total entropy up. Units are the trap: entropies come in joules per kelvin per mole while enthalpies come in kilojoules, and the factor of a thousand has to be applied before the two can be added.
吉布斯自由能与可行性判断
吉布斯自由能变等于焓变减去温度乘以熵变，符号为负时反应在热力学上可行。温度只出现在熵变这一项上，所以焓与熵同号时，温度就成了开关。碳酸钙分解的焓变为正 178 kJ mol⁻¹，体系熵变为正 160.6 J K⁻¹ mol⁻¹，令 ΔG 等于零解出约 1108 K。低于这个温度分解不利，高于它才有推动力。热力学可行不等于反应会快，动力学上仍可能需要高温或催化剂。阅卷时看到只答 ΔG 为负所以反应发生、却没提温度条件的答案，通常拿不到完整分数。写可行性题时把焓、熵、温度三项逐个交代，比背结论稳。
Gibbs free energy is enthalpy minus temperature times entropy, and a negative value marks a reaction that is thermodynamically feasible. When enthalpy and entropy have the same sign, temperature decides the outcome: setting delta G to zero for calcium carbonate gives about 1108 K, below which the decomposition is not favoured. Feasibility is not rate, so a reaction can be spontaneous on paper and still need a high temperature or a catalyst in the laboratory.
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