AQA化学 焓变 BornHaber 熵变 吉布斯自由能
热力学是 AQA A-Level Chemistry 7405 考纲 3.1.8 的一节，卷面上很少只考一个定义，通常把焓变、熵变和吉布斯自由能串成一道计算加解释的题。难点不在公式数量，而在符号约定和数据链条：循环画错方向，或者单位没统一，后面的结论就全反了。下面按考纲顺序把七个考点拆开，每个考点配一个能直接套用的算例和对应的失分点，取自 AQA 7405 specification 与官方数据表的口径。
Thermodynamics occupies section 3.1.8 of the AQA 7405 specification, and exam questions seldom test one definition alone. Enthalpy, entropy and Gibbs free energy are chained into a single calculation, so an early sign error or a missed unit conversion flips the final verdict on whether a reaction happens.
考纲把这一节分成 3.1.8.1 与 3.1.8.2 两块，前者管 Born-Haber 循环和晶格焓，后者管熵与吉布斯自由能。两块在卷面上常常合并出题，先用循环求出某个焓变，再拿这个数去算 ΔG 和转变温度。复习时按这个顺序串，比按教材页码从头背要省时间。
The specification splits the topic into 3.1.8.1 for Born-Haber cycles and lattice enthalpy and 3.1.8.2 for entropy and Gibbs free energy, though a single exam question often uses both. Revising in that order, cycle first and free energy second, mirrors how the marks are usually allocated.
标准生成焓 标准燃烧焓 中和焓的定义边界
标准生成焓指在标准状态下由元素生成一摩尔该化合物时的焓变，元素的 ΔfH 按定义取 0；标准燃烧焓指一摩尔物质在氧气中完全燃烧的焓变，水必须写成液态；中和焓指生成一摩尔水时的焓变，强酸与强碱的稀溶液约为 -57 kJ mol-1。很多学生以为这三个量可以互换，题干给的到底是哪一个，直接决定 Hess 循环里箭头的朝向。标准状态在 AQA 体系里是 100 kPa，温度通常取 298 K，答题时温度条件写在方程式旁更稳。数据表里查不到的数值，要用循环反推，不能自己编。
Standard enthalpy of formation is quoted per mole of compound formed from its elements in their standard states, with elemental values set to zero; combustion is per mole of substance burned completely in oxygen, leaving water as a liquid; neutralisation is per mole of water produced, close to -57 kJ mol-1 for a strong acid and strong base. Which of the three a question names fixes the direction of every arrow in your Hess cycle, so read that line before drawing anything.
用生成焓算反应焓的式子是 ΔH = ΣΔfH(产物) - ΣΔfH(反应物)，系数相乘不要漏。以甲烷完全燃烧为例，ΔfH(CH4) 约 -75，ΔfH(CO2) 约 -393，ΔfH(H2O, l) 约 -286，代入得 (-393) + 2(-286) - (-75) = -890 kJ mol-1。反过来，已知燃烧焓要推生成焓时，把循环画成两条都到 CO2 的路径，再用盖斯定律相减。查表时留意状态符号，水写成气态，数值差约 44 kJ mol-1，整道题的答案就偏了。
The arithmetic is ΔH = ΣΔfH(products) - ΣΔfH(reactants) with every coefficient applied, so methane combustion gives (-393) + 2(-286) - (-75) = -890 kJ mol-1. Running the same cycle in reverse recovers a formation enthalpy from combustion data, and the state symbols matter because liquid and gaseous water differ by about 44 kJ mol-1.
量热法 q=mcΔT 与 ΔH=-q/n 的符号
量热实验里 q = mcΔT，m 用溶液质量，c 取 4.18 J g-1 K-1，ΔT 的数值用摄氏度或开尔文都一样。溶液吸收的热量等于反应放出的热量，所以 ΔH = -q/n，n 是限量试剂的物质的量。常见的错误是把溶质质量当成溶液质量，或者忘记把 cm3 按 1.00 g cm-3 换成克。用聚苯乙烯杯代替玻璃烧杯，就是为了减少向空气和杯壁的散热，但实测值仍然偏小，热量有一部分被温度计和杯壁吃掉。想压住这个误差，可以把温度对时间作图，外推到混合瞬间求 ΔT，AQA 的实验题明确接受这种处理。
The heat evolved comes from q = mcΔT, with m taken as the mass of the solution and c as 4.18 J g-1 K-1 for water. Dividing q by the moles of limiting reagent and reversing the sign gives ΔH, and plotting temperature against time to extrapolate back to the moment of mixing removes much of the heat loss to the cup.
算一遍完整的账：50.0 cm3 的 1.00 mol dm-3 盐酸与 50.0 cm3 同浓度氢氧化钠混合，温度升高 6.8 K，溶液按 100 g 计，q = 100 × 4.18 × 6.8 = 2842 J。限量试剂是酸，n = 0.0500 mol，于是 ΔH = -2842 / 0.0500 = -56.8 kJ mol-1，与教材给的中和焓很接近。若把总体积一半当成质量，或者拿 0.100 mol 做分母，结果就差一倍。两种错法在往年卷面上都出现过，判卷时只看最后的数，中间过程不会替你补分。
A full run: 50.0 cm3 of 1.00 mol dm-3 hydrochloric acid mixed with the same volume of sodium hydroxide raises the temperature by 6.8 K, so q = 100 x 4.18 x 6.8 = 2842 J. With 0.0500 mol of acid as the limiting reagent, ΔH = -2842/0.0500 = -56.8 kJ mol-1, while using the total moles as the denominator halves the answer.
盖斯定律与 BornHaber 循环的每一步符号
盖斯定律说总焓变与路径无关，这条定律让不可直接测量的晶格焓可以拆成几步来算。Born-Haber 循环的常见顺序是：金属升华或原子化，非金属键断裂（用半个键焓），金属电离能，非金属电子亲和能，生成气态离子，最后气态离子结合成晶体放出的就是晶格焓。符号上有两个坑：氧的第一电子亲和能为负（放热），第二电子亲和能为正（吸热），因为 O- 再接受一个电子要克服排斥；AQA 数据表给的是由气态离子生成晶体的晶格焓，所以取负值。画完循环要做一次闭合检查，两条路径的焓变总和应当相等；对不上号，多半是某一步的化学计量数没乘进去。
Hess law makes lattice enthalpy measurable by breaking an impossible step into atomisation, ionisation, electron affinity and bond dissociation, and the loop only balances when the stoichiometry of every half equation is carried through. Two signs trip people up: the first electron affinity of oxygen is negative while the second is positive, and the lattice enthalpy in the AQA data sheet is quoted for gaseous ions forming the solid, so it enters as a negative value.
用盖斯定律推生成焓的例子：碳不完全燃烧生成一氧化碳的 ΔfH 不能直接测，可以借助两次燃烧。C 到 CO2 是 -393 kJ mol-1，CO 到 CO2 是 -283 kJ mol-1，两条路径相减得到 ΔfH(CO) = -393 - (-283) = -110 kJ mol-1。减法方向弄反就会得到 +110，符号一错，后面判断自发性全部反掉。写循环时把每一步的方程式单独列出来，比直接在图上标数字更不容易漏项。
Carbon burning to carbon monoxide cannot be measured directly, so the cycle goes through carbon dioxide: -393 kJ mol-1 for C to CO2 and -283 kJ mol-1 for CO to CO2 leave ΔfH(CO) = -110 kJ mol-1. Subtracting the other way round returns +110, which would invert every later feasibility judgement.
熵变 ΔS 的微观解释与单位换算
熵变的单位是 J K-1 mol-1，与焓变的 kJ mol-1 差了一千倍，这个差别是后面所有计算错误的源头。ΔS 为正意味着微观状态数增加，气体摩尔数变多的反应几乎一定为正，固体变液体变气体的顺序同理。溶解过程的符号要看晶格被破坏和水合放热哪一头更强，不能只凭看起来更乱下结论。算 ΔS 用产物的熵减反应物的熵，系数要乘进去。很多学生会把 ΔS 为正直接当成反应能自发，漏掉了环境熵这一半，判断自发性要回到 ΔS total = ΔS system - ΔH/T 这个式子。
Entropy is measured in J K-1 mol-1, a thousand times smaller than the kJ mol-1 used for enthalpy, and that mismatch causes more lost marks than any conceptual gap. A positive ΔS system says nothing on its own about feasibility, because the surroundings contribute through -ΔH/T and the two must be added before any conclusion is drawn.
碳酸钙分解是常考的熵增例子：CaCO3(s) 变成 CaO(s) 加 CO2(g)，一摩尔固体放出一摩尔气体，ΔS 是较大的正值，约 +160 J K-1 mol-1，也就是 +0.160 kJ K-1 mol-1。写进 ΔG 式子必须用后一个数，用前者会把转变温度算小一千倍。比较两个过程的熵变谁大，先看气体摩尔数的变化，再看液体与固体的比例，最后才轮到分子结构复杂度。
Calcium carbonate decomposing into calcium oxide and carbon dioxide releases one mole of gas per mole of solid, so ΔS is a sizeable positive value, roughly +160 J K-1 mol-1 or +0.160 kJ K-1 mol-1. Substituting the J value into a kJ-based ΔG calculation shrinks the switching temperature by a factor of a thousand.
吉布斯自由能 ΔG=ΔH-TΔS 与转变温度
ΔG = ΔH - TΔS，负值代表热力学可行，等于零代表系统处于平衡，正值代表该温度下不可行。温度必须换成开尔文，K = 摄氏度 + 273。求反应开始自发的温度时令 ΔG = 0，得到 T = ΔH/ΔS，代进去之前把 kJ 和 J 统一，否则答案会差三个数量级。四种符号组合要背成条件反射：焓减熵增任何温度可行，焓增熵减任何温度都不可行，焓减熵减低温可行，焓增熵增高温可行。这一步最容易丢分的地方就是单位，250 J K-1 mol-1 和 -100 kJ mol-1 直接相减，算出来的温度必然是错的。
Setting ΔG equal to zero gives the switching temperature T = ΔH/ΔS, but only after both quantities share the same energy unit, since mixing kJ with J shifts the answer by a factor of a thousand. The four sign combinations then reduce to a short table: feasible at every temperature, at none, only when cold, or only when hot.
设某反应 ΔH = +50 kJ mol-1、ΔS = +100 J K-1 mol-1，令 ΔG = 0 得 T = 50000 / 100 = 500 K，也就是 227 摄氏度以上才自发。若把 ΔS 写成 0.100 kJ K-1 mol-1，算式变成 50 / 0.100，答案仍是 500 K，两种写法只要单位一致都对。题目追问在 400 K 是否自发时，把温度代回去：ΔG = 50 - 400 × 0.100 = +10 kJ mol-1，正值说明该温度下不成立。这类两问的题，第二问往往就是在查第一问的单位有没有换对。
For ΔH = +50 kJ mol-1 and ΔS = +100 J K-1 mol-1, setting ΔG to zero gives T = 50000/100 = 500 K, so the reaction turns feasible above 227 degrees Celsius. Substituting 400 K back gives ΔG = 50 - 400 x 0.100 = +10 kJ mol-1, a positive value that rules the reaction out at that temperature.
考场步骤与循环里的重复计算
拿到题先圈出题干给的是哪一种焓，再定循环方向，不要一上来就画箭头。算 q 时把体积按密度换成质量，算 n 时确认哪个试剂限量；写 Born-Haber 时逐行核对符号，第二电子亲和能和晶格焓是高频错处。答案要带单位也带符号，AQA 的评分方案对 kJ mol-1 与 J K-1 mol-1 分得很清。常见的错误是循环里把同一步算了两遍，例如既用了金属的原子化焓，又用了它的升华焓。有效数字按题给数据取，中间步骤多留一位，最后再收。
A workable order is: identify the enthalpy the question quotes, fix the cycle direction, then fill in numbers. Duplicate steps, such as counting both sublimation and atomisation for the same metal, are the most frequent structural mistake in Born-Haber answers, and they usually survive a careless self-check because the arithmetic itself is correct.
检查单位有个笨办法：把 ΔH 与 TΔS 分别写成带单位的数再相加，单位不一致立刻看得出来。-100 kJ mol-1 与 250 J K-1 mol-1 在 300 K 下的组合是 -100 kJ 加 -75 kJ，得到 -175 kJ mol-1；若把 250 当成 kJ 处理，算出来是 -100 加 -75000，明显不合理。答案的符号也要交代清楚，写放出 175 kJ mol-1 比孤立的一个 175 更安全。把这个检查做成固定动作，热力学部分的计算题失分会少一大半。
Writing both terms with their units exposes the mismatch at once: at 300 K, -100 kJ mol-1 and 250 J K-1 mol-1 combine to -100 minus 75, giving -175 kJ mol-1, whereas treating 250 as kJ produces an absurd figure. Stating the sign in words, as 175 kJ mol-1 released, removes the ambiguity.
解释题的文字也要练。问为什么某个反应在高温才自发，答案要落到 ΔG = ΔH - TΔS 中 TΔS 随温度增大这一项上，而不是重复说高温更好。问晶格焓为什么是负值，答案要回到气态离子结合成晶体放出能量这个定义。评分方案给分是按关键词给的，写出式子再说明符号，比写一整段描述更容易拿满分。
Written explanations are marked on key words rather than length, so an answer about high temperature feasibility should point at the growing TΔS term instead of repeating that heat helps. Likewise, a negative lattice enthalpy is justified by gaseous ions forming the solid and releasing energy, not by saying the bond is strong.
ΔH 与 ΔS 近似恒定的适用边界
把 ΔH 和 ΔS 当成常数代进 ΔG 式子，前提是温度区间不大。严格来说焓变随热容变化，熵变随温度按对数变化，只是在 298 K 到五六百 K 这段范围里变化量小到可以忽略。跨过相变点时这个近似就破了：冰融化、水汽化的潜热要单独计入，热容曲线在相变温度处有折点。题目给出不同温度下的 ΔH 时，分段算再相加，不要硬拿一个数套到底。真正需要积分的场合在 A-Level 不考，知道近似成立的条件就够了。
The assumption that ΔH and ΔS stay constant holds only over a modest temperature range, since enthalpy drifts with heat capacity and entropy varies logarithmically with temperature. Crossing a phase boundary breaks it, because latent heat and the kink in the heat capacity curve have to be handled separately rather than folded into one ΔG expression.
海外直订AQA GCSE Chemistry Student Book AQA GCSE化学学生手册¥588.0📱 去京东购买京东精选
A-Level / GCSE / IB 化学与物理全程辅导，热力学计算题逐题批改
更多咨询请联系16621398022（同微信）
Visit aleveler.com
或截图后用「扫一扫 → 相册」选择本图