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Gibbs Free Energy Calculations | 吉布斯自由能计算

2026-08-29 10:42 · 本文部分内容由AI辅助创作

📚 Gibbs Free Energy Calculations | 吉布斯自由能计算

In A-Level Chemistry, Gibbs free energy is the key quantity that determines whether a chemical reaction is thermodynamically feasible under constant temperature and pressure. It combines enthalpy, entropy and temperature into a single criterion, allowing chemists to predict the direction of spontaneous change without analysing the surroundings in detail. This article focuses on the calculations Cambridge International A-Level Chemistry students must master: the Gibbs equation, standard Gibbs free energy changes, feasibility temperature, sign analysis, and the link between ΔG° and the equilibrium constant K.

在 A-Level 化学中，吉布斯自由能是判断化学反应在恒温恒压下是否热力学可行的关键物理量。它把焓、熵和温度整合为单一判据，使化学家无需详细分析环境即可预测自发变化的方向。本文聚焦剑桥国际 A-Level 化学学生必须掌握的吉布斯自由能计算：吉布斯方程、标准吉布斯自由能变化、可行性温度、符号分析，以及 ΔG° 与平衡常数 K 之间的联系。

1. Defining Gibbs Free Energy | 定义吉布斯自由能

e functions (or fixed for a given process), the value of G depends only on the current state of the system, not on the route taken to reach it. The change ΔG measures the maximum non-expansion work that a process can deliver at constant temperature and pressure.

吉布斯自由能，符号为 G，是一个热力学状态函数，定义为 G = H − TS。由于 H、T、S 都是状态函数（或对给定过程为定值），G 的数值只取决于系统当前的状态，而与到达该状态的路径无关。其变化量 ΔG 衡量在恒温恒压下过程所能提供的最大非体积功。

G = H − T S

mically feasible, ΔG > 0 means it is not feasible under those conditions, and ΔG = 0 corresponds to a system at equilibrium. Note that 'feasible' ics only; it says nothing about the rate, which depends on activation energy and kinetics.

对于恒温恒压下的反应，ΔG 的符号就是可行性判据：ΔG < 0 表示正向反应热力学可行；ΔG > 0 表示在该条件下不可行；ΔG = 0 对应系统处于平衡状态。注意，"可行"只描述热力学层面，并不涉及速率；速率取决于活化能和动力学。

2. The Gibbs Equation and Its Units | 吉布斯方程及其单位

Applying the definition to a process at constant temperature gives the Gibbs equation, which is the workhorse of every A-Level calculation on this topic. Because T is constant, the enthalpy and entropy terms can be treated separately, and their values are usually taken from standard data tables.

把定义应用于恒温过程就得到吉布斯方程，它是本专题所有 A-Level 计算的核心工具。由于 T 恒定，焓项与熵项可以分别处理，其数值通常取自标准数据表。

ΔG = ΔH − T ΔS

⁻¹. Before substituting into the equation you must convert ΔS to kJ mol⁻¹ K⁻¹ by dividing by 1000, or convert ΔH to J mol⁻¹ by multiplying by 1000. Temperature must always be in kelvin: T/K = θ/°C + 273.

单位是最常见的失分点。焓变通常以 kJ mol⁻¹ 列出，而熵变以 J mol⁻¹ K⁻¹ 列出。代入方程前，必须把 ΔS 除以 1000 换算成 kJ mol⁻¹ K⁻¹，或把 ΔH 乘以 1000 换算成 J mol⁻¹。温度必须使用开尔文：T/K = θ/°C + 273。

QuantitySymbolTypical unitConversion
Enthalpy changeΔHkJ mol⁻¹× 1000 → J mol⁻¹
Entropy changeΔSJ mol⁻¹ K⁻¹÷ 1000 → kJ mol⁻¹ K⁻¹
TemperatureTKθ/°C + 273
Gibbs changeΔGkJ mol⁻¹must match ΔH unit

3. Standard Gibbs Free Energy Change | 标准吉布斯自由能变化

The standard Gibbs free energy change, ΔG°, refers to a reaction carried out under standard conditions: all gases at 100 kPa, all solutions at 1 mol dm⁻³, and a stated temperature, usually 298 K. Under these conditions the standard equation applies directly, with ΔH° and ΔS° taken from data tables.

标准吉布斯自由能变化 ΔG° 指反应在标准条件下进行：所有气体压力为 100 kPa，所有溶液浓度为 1 mol dm⁻³，并指明温度，通常为 298 K。在此条件下，标准方程可直接应用，ΔH° 与 ΔS° 均取自数据表。

ΔG° = ΔH° − T ΔS°

s the number of moles of electrons transferred and F is the Faraday constant, 96 500 C mol⁻¹. A positive E°cell therefore corresponds to a negative ΔG° and a feasible cell reaction, which is exactly the criterion used in electrochemistry questions.

ΔG° 还与电化学电池的标准电极电势相关：ΔG° = −nFE°cell，其中 n 为转移电子的物质的量，F 为法拉第常数 96 500 C mol⁻¹。因此 E°cell 为正值对应 ΔG° 为负值，电池反应可行——这正是电化学题目所用的判据。

4. Combining Enthalpy and Entropy Data | 结合焓变与熵变数据

Standard enthalpy changes can be obtained from standard enthalpies of formation using Hess's law, and standard entropy changes from standard molar entropies of products minus reactants. Once ΔH° and ΔS° are known, a single substitution gives ΔG° at any chosen temperature.

标准焓变可用标准生成焓通过赫斯定律求出，标准熵变则由产物的标准摩尔熵之和减去反应物的标准摩尔熵之和得到。一旦得到 ΔH° 与 ΔS°，只需一次代入即可求出任意指定温度下的 ΔG°。

ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants)

ΔS° = ΣS°(products) − ΣS°(reactants)

andard molar entropies are not zero: S° for a perfect crystalline solid at 0 K would be zero, but at 298 K every substance has a positive, measurable S° value. Ignoring the entropy of a solid reactant is a frequent error in multi-step questions.

请记住，处于标准状态的单质 ΔH°f = 0，但其标准摩尔熵不为零：0 K 时完美晶体的 S° 才为零，而在 298 K 下每种物质都有正值且可测的 S°。在多步计算中忽略固态反应物的熵是常见错误。

5. Predicting Feasibility from Signs | 用符号判断可行性

A large number of exam questions ask you to predict feasibility without any numerical data, simply by reasoning about the signs of ΔH and ΔS and the size of T. There are four possible combinations, and each has a distinct temperature dependence.

大量考题要求你无需数值数据，仅通过推理 ΔH、ΔS 的符号以及 T 的大小来预测可行性。共有四种组合，每种都有不同的温度依赖关系。

ΔHΔSSign of ΔGFeasibility
negativepositivealways negativefeasible at all temperatures
positivenegativealways positivenever feasible
negativenegativenegative only at low Tfeasible below a threshold T
positivepositivenegative only at high Tfeasible above a threshold T

The crucial insight is that T magnifies the entropy term. At low temperature, ΔH dominates and the sign of ΔG tends to follow the sign of ΔH; at high temperature, the −TΔS term dominates, so reactions with positive ΔS become feasible once T is large enough.

关键在于 T 会放大熵项。低温时 ΔH 占主导，ΔG 的符号趋于跟随 ΔH 的符号；高温时 −TΔS 项占主导，因此 ΔS 为正的反应在 T 足够大后变得可行。

6. Feasibility Temperature | 可行性温度

n for the reaction being just at the boundary between feasible and non-feasible. Rearranging the Gibbs equation under this condition gives a very compact expression.

阈值温度可通过令 ΔG = 0 求得，这正是反应恰好处于可行与不可行边界的条件。在该条件下整理吉布斯方程，可得一个非常简洁的表达式。

ΔG = 0 → ΔH = T ΔS → T = ΔH / ΔS

When you use this expression, ΔH and ΔS must be in matching units: if ΔH is in kJ mol⁻¹, then ΔS must be in kJ mol⁻¹ K⁻¹. The answer comes out in kelvin. For a reaction with positive ΔH and positive ΔS, the reaction becomes feasible at temperatures above this value; for negative ΔH and negative ΔS, it is feasible below it.

使用该式时，ΔH 与 ΔS 的单位必须匹配：若 ΔH 用 kJ mol⁻¹，则 ΔS 必须用 kJ mol⁻¹ K⁻¹。所得结果单位为开尔文。对于 ΔH 为正、ΔS 为正的反应，温度高于该值时反应可行；对于 ΔH 为负、ΔS 为负的反应，则低于该值时可行。

A subtle point worth stating in exam answers is that ΔH° and ΔS° are usually assumed to be independent of temperature over a modest range. This approximation is why the threshold temperature calculated this way is described as an estimate rather than an exact value.

考试作答时值得指出一个细节：在不算大的温度区间内，通常假设 ΔH° 与 ΔS° 不随温度变化。正因为这一近似，用此法算出的阈值温度只能称为估算值，而非精确值。

7. Worked Example: Thermal Decomposition of Calcium Carbonate | 例题：碳酸钙的热分解

l⁻¹ K⁻¹. First convert the entropy term: +160.5 J mol⁻¹ K⁻¹ ÷ 1000 = +0.1605 kJ mol⁻¹ K⁻¹.

考虑分解反应 CaCO₃(s) → CaO(s) + CO₂(g)，已知 ΔH° = +178 kJ mol⁻¹，ΔS° = +160.5 J mol⁻¹ K⁻¹。先换算熵项：+160.5 J mol⁻¹ K⁻¹ ÷ 1000 = +0.1605 kJ mol⁻¹ K⁻¹。

T = ΔH° / ΔS° = 178 / 0.1605 ≈ 1109 K

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So the decomposition becomes thermodynamically feasible at approximately 1110 K, or about 837 °C. This explains why lime kilns must be operated at high temperature, and why the reaction does not occur on its own at room temperature even though the products are entropically favoured.

因此该分解反应在约 1110 K（约 837 °C）以上热力学可行。这解释了为什么石灰窑必须在高温下操作，以及为什么尽管产物在熵上有利，室温下该反应并不会自发发生。

ol⁻¹, a strongly positive value that confirms the reaction is not feasible at room temperature. This temperature-dependent check is a reliable way to earn full method marks.

在 298 K 下计算得 ΔG° = 178 − 298 × 0.1605 = 178 − 47.8 = +130.2 kJ mol⁻¹，为强正值，说明室温下反应不可行。这种随温度变化的验算方法是稳妥的满分答题思路。

8. Gibbs Free Energy and the Equilibrium Constant | 吉布斯自由能与平衡常数

and this leads to one of the most powerful relationships in physical chemistry. The standard Gibbs free energy change is related to the equilibrium constant K by the equation below, where R is the gas constant, 8.314 J mol⁻¹ K⁻¹.

当反应处于平衡时 ΔG = 0，由此可导出物理化学中最有力的关系式之一。标准吉布斯自由能变化与平衡常数 K 的关系如下式所示，其中 R 为气体常数，8.314 J mol⁻¹ K⁻¹。

ΔG° = −R T ln K

ucts. If ΔG° is positive, then K < 1 and the equilibrium mixture is dominated by reactants. If ΔG° = 0, then K = 1 and neither side is favoured.

符号带来的推论很重要。若 ΔG° 为负，则 ln K 为正，K > 1，说明平衡位置偏向产物；若 ΔG° 为正，则 K < 1，平衡混合物以反应物为主；若 ΔG° = 0，则 K = 1，两侧都不占优势。

Note the practical warning: R must be in J mol⁻¹ K⁻¹ in this equation, while ΔG° is so often quoted in kJ mol⁻¹. Convert ΔG° to J mol⁻¹ (multiply by 1000) before dividing by RT, otherwise the answer will be wrong by a factor of 1000 in the exponential.

注意一个实用提醒：该式中 R 必须取 J mol⁻¹ K⁻¹，而 ΔG° 常以 kJ mol⁻¹ 给出。除以 RT 之前必须把 ΔG° 换算成 J mol⁻¹（乘以 1000），否则指数上的结果会相差 1000 倍。

9. Worked Example: Calculating K from ΔG° | 例题：由 ΔG° 求 K

ol⁻¹ at 298 K. Convert to J mol⁻¹: −20 000 J mol⁻¹. Then rearrange the relationship to isolate ln K and substitute the values carefully.

设某反应在 298 K 下 ΔG° = −20.0 kJ mol⁻¹。换算为 J mol⁻¹：−20 000 J mol⁻¹。然后整理关系式以分离 ln K，并仔细代入数值。

ln K = −ΔG° / (R T) = 20 000 / (8.314 × 298) ≈ 8.07

K = e⁸·⁰⁷ ≈ 3.2 × 10³

n K = −30 000 / (8.314 × 298) ≈ −12.1 and K = e⁻¹²·¹ ≈ 5.5 × 10⁻⁶, showing that the reaction barely proceeds at all.

由于 K 远大于 1，平衡明显偏右，与 ΔG° 为负相一致。若 ΔG° 为 +30.0 kJ mol⁻¹，则 ln K = −30 000 / (8.314 × 298) ≈ −12.1，K = e⁻¹²·¹ ≈ 5.5 × 10⁻⁶，说明该反应几乎不进行。

Remember also that K responds to temperature: since ΔG° depends on T, a change in temperature changes K. This is exactly the quantitative version of Le Chatelier's principle, and it explains why the value of K quoted in a data book is always tied to a stated temperature.

还要记住 K 会随温度变化：由于 ΔG° 依赖于 T，温度改变会改变 K。这正是勒夏特列原理的定量版本，也解释了为什么数据手册中给出的 K 值总是与指定温度挂钩。

10. Non-Standard Conditions and Reaction Quotient | 非标准条件与反应商

ere Q is the reaction quotient with the same form as K but using instantaneous concentrations or partial pressures.

在真实体系中，浓度和分压很少恰好处于标准状态。此时实际自由能变化为 ΔG = ΔG° + RT ln Q，其中 Q 为反应商，其形式与 K 相同，但使用瞬时浓度或分压。

ive and the reverse reaction is spontaneous; if Q = K, then ΔG = 0 and the system is at equilibrium. Everything in this section reduces to the comparison of Q with K.

该方程的价值在于预测方向。若 Q < K，则 ΔG 为负，正反应自发进行；若 Q > K，则 ΔG 为正，逆反应自发进行；若 Q = K，则 ΔG = 0，系统处于平衡。本节内容最终归结为 Q 与 K 的比较。

When writing Q, pure solids and pure liquids are omitted, since their activity is taken as 1. This is the same rule you apply when writing an expression for K, and it is essential for matching the model answer in structured questions.

书写 Q 时，纯固体和纯液体不写入表达式，因其活度视为 1。这与书写 K 表达式时的规则相同，在结构化问答题中是与标准答案一致的关键。

11. Graphical Methods and Exam Technique | 图像法与考试技巧

he y-axis and T on the x-axis. The intercept on the y-axis is ΔH, and the gradient is −ΔS.

以 ΔG 对 T 作图得到一条直线，因为吉布斯方程形如 y = c + m x，其中 ΔG 在 y 轴、T 在 x 轴。y 轴截距为 ΔH，斜率为 −ΔS。

ΔG = ΔH + (−ΔS) × T

A negative gradient indicates a positive ΔS; a positive gradient indicates a negative ΔS. Reading gradient and intercept from a graph is a common exam task, so always label the axes and show the units.

直线与 x 轴的交点即为可行性温度，因为该处 ΔG = 0。斜率为负说明 ΔS 为正，斜率为正说明 ΔS 为负。从图像读取斜率和截距是常见考题，因此务必标注坐标轴名称并写出单位。

itute into ΔG = ΔH − TΔS, give the sign of the result with the unit kJ mol⁻¹, and finish with a sentence explaining what that sign means at that temperature. This four-step pattern covers almost every mark scheme.

当题目问反应是否可行时，请按以下结构作答：写出 ΔH 与 ΔS 的数值及符号，完成单位换算，代入 ΔG = ΔH − TΔS，给出结果符号与单位 kJ mol⁻¹，最后用一句话说明该符号在该温度下的含义。这个四步模式几乎覆盖所有评分标准。

12. Common Mistakes and Revision Checklist | 常见错误与复习清单

he same calculation. Another frequent problem is dropping the negative sign when writing ΔG° = −RT ln K, which makes the link between the sign of ΔG° and the size of K appear to be reversed.

有些错误在评分标准中反复出现。用摄氏度而非开尔文最具破坏性，其次是同一计算中混用 kJ 与 J。另一常见问题是写 ΔG° = −RT ln K 时漏掉负号，导致 ΔG° 的符号与 K 的大小之间的关系看起来正好相反。

of the system; the −TΔS term already accounts for the entropy change of the surroundings through ΔSsurroundings = −ΔH/T. Adding an extra surroundings term leads to double counting.

学生也常把体系熵变与宇宙总熵变混淆。在吉布斯方程中，ΔS 是体系的熵变；−TΔS 项已通过 ΔSsurroundings = −ΔH/T 计入了环境的熵变。额外再加一项环境熵变就会重复计算。