📚 Gibbs Free Energy for OCR A-Level Chemistry | OCR A-Level 化学：吉布斯自由能考点精讲
Gibbs free energy (G) is a thermodynamic potential that allows chemists to predict whether a reaction will occur spontaneously under constant temperature and pressure. For OCR A-Level Chemistry, mastering ΔG is essential because it links the two fundamental driving forces behind chemical change – enthalpy and entropy – into a single decision-making quantity. This article unpacks every major specification point: from the definition and the master equation ΔG = ΔH – TΔS, to its role in equilibrium and electrochemistry, while highlighting common pitfalls students must avoid.
吉布斯自由能 (G) 是一种热力学势能函数，能让化学家在定温定压下判断一个反应是否自发进行。在 OCR A-Level 化学中，掌握 ΔG 至关重要，因为它把化学变化的两大根本驱动力——焓变与熵变——统一成一个判断依据。本文将逐一解析考纲中的所有核心考点：从定义与核心公式 ΔG = ΔH – TΔS，到它在化学平衡和电化学中的延伸应用，同时指出考生必须规避的常见误区。
1. What is Gibbs Free Energy? | 什么是吉布斯自由能？
Gibbs free energy, symbol G, is defined as the energy available within a system to do useful work at a constant temperature and pressure. It combines the system’s enthalpy (H) and entropy (S) into a single state function: G = H – TS. A change in Gibbs free energy (ΔG) for a process is the ultimate indicator of reaction feasibility, sometimes called the ‘chemical driving force’.
吉布斯自由能，符号为 G，定义为系统在定温定压下可用来做有用功的那部分能量。它把系统的焓 (H) 和熵 (S) 综合成一个状态函数：G = H – TS。一个过程的吉布斯自由能变 (ΔG) 是反应可行性的终极判断标准，常被称为“化学驱动力”。
For a reaction to be thermodynamically feasible, the total entropy of the universe must increase – but calculating universal entropy change is impractical. J. Willard Gibbs showed that under constant temperature and pressure, the condition can be simplified: a reaction is feasible if ΔG < 0. This spares us from having to compute the entropy change of the surroundings separately.
一个反应要具有热力学上的可行性，宇宙的总熵必须增加——但要计算宇宙的熵变并不实际。吉布斯证明，在定温定压下，这一条件可以简化为：若 ΔG < 0，反应即自发可行。我们因此无需单独计算环境的熵变。
2. The Key Equation: ΔG = ΔH – TΔS | 关键公式：ΔG = ΔH – TΔS
The central relationship for OCR candidates is:
OCR 考生必须掌握的核心关系式为：
ΔG = ΔH – TΔS
where ΔH is the enthalpy change (kJ mol⁻¹), T is the absolute temperature in kelvin (K), and ΔS is the entropy change, typically expressed in J mol⁻¹ K⁻¹. The term TΔS has units of energy, and it is crucial to ensure unit consistency – either convert ΔH to J mol⁻¹ or TΔS to kJ mol⁻¹ before combining.
其中 ΔH 是焓变（单位 kJ mol⁻¹），T 是热力学温度（单位 K），ΔS 是熵变，通常用 J mol⁻¹ K⁻¹ 表示。TΔS 项具有能量的量纲，务必保持单位一致——计算前要么将 ΔH 换算为 J mol⁻¹，要么将 TΔS 换算为 kJ mol⁻¹。
The equation reveals that spontaneity is a compromise between the tendency to minimise energy (negative ΔH) and the tendency to maximise disorder (positive ΔS). The minus sign is physically meaningful: a large positive ΔS contributes negatively to ΔG, favouring feasibility.
该公式表明，自发性是倾向减少能量（负 ΔH）与倾向增大混乱度（正 ΔS）两者之间的折中。公式中的负号具有物理意义：较大的正 ΔS 会使 ΔG 负得更多，有利于反应自动进行。
3. Understanding the Sign of ΔG | 理解 ΔG 的符号
If ΔG is negative (ΔG < 0), the forward reaction is thermodynamically feasible under the given conditions – it is spontaneous. If ΔG is positive (ΔG > 0), the reaction is not feasible in the forward direction; the reverse reaction would be spontaneous. When ΔG = 0, the system is at equilibrium, with no net tendency to move in either direction.
若 ΔG 为负 (ΔG < 0)，正反应在该条件下具有热力学可行性——它是自发的。若 ΔG 为正 (ΔG > 0)，则正反应不可行，逆反应才是自发的。当 ΔG = 0 时，系统处于平衡状态，无净变化倾向。
This three-state model underpins all applications of ΔG, from predicting reaction direction to calculating equilibrium constants and exam feasibility questions.
这一“三种状态”模型是 ΔG 所有应用的基础，从判断反应方向到计算平衡常数，再到考试中常见的可行性判断题。
4. Temperature Dependence and Feasibility | 温度依赖性与反应可行性
The temperature T explicitly enters the ΔG equation, meaning feasibility is not a fixed property – it can flip as temperature changes. Chemists therefore speak of a temperature window of feasibility rather than a simple ‘can happen’ label.
温度 T 直接出现在 ΔG 的表达式中，这意味着可行性并非一个固有属性——它会随温度变化而翻转。因此，化学家谈论的是一个可行性温度窗口，而非简单的“能发生”标签。
This is particularly important when ΔH and ΔS have the same sign. If both are negative, the negative TΔS term becomes more dominant at high temperatures, potentially making ΔG positive. Conversely, if both are positive, a high T makes the –TΔS term sufficiently negative to drive ΔG below zero.
当 ΔH 和 ΔS 同号时，这一点尤为重要。若两者均为负，负的 TΔS 项在高温下会变得更大（量值），可能使 ΔG 转正。相反，若两者均为正，高温则使 –TΔS 项负得足够大，推动 ΔG 降至零以下。
The temperature at which feasibility switches can be found by setting ΔG = 0:
令 ΔG = 0 即可求得可行性转变温度：
T = ΔH ÷ ΔS
Remember to use consistent units in this calculation – for example, expressing ΔH in J mol⁻¹ to match ΔS in J mol⁻¹ K⁻¹.
计算时务必保持单位一致——例如，将 ΔH 换算为 J mol⁻¹，以匹配 ΔS 的 J mol⁻¹ K⁻¹。
5. Interpreting ΔH and ΔS Combinations | 解读 ΔH 和 ΔS 的组合
The interplay between enthalpy and entropy gives rise to four distinct cases, which are a staple of A-Level exam questions. The table below summarises how the signs of ΔH and ΔS govern feasibility at different temperatures.
焓与熵的相互作用产生了四种典型情形，是 A-Level 考试中的常客。下表总结了 ΔH 和 ΔS 的符号如何决定不同温度下的可行性。
| ΔH | ΔS | ΔG = ΔH – TΔS | Feasibility |
|---|---|---|---|
| Negative (–) | Positive (+) | Always negative | Always feasible |
| Positive (+) | Negative (–) | Always positive | Never feasible |
| Negative (–) | Negative (–) | Negative at low T, positive at high T | Feasible below T = ΔH/ΔS |
| Positive (+) | Positive (+) | Positive at low T, negative at high T | Feasible above T = ΔH/ΔS |
Students often find the negative–negative and positive–positive cases the most challenging, because a simple yes/no answer is impossible without considering temperature.
考生通常觉得“负–负”和“正–正”这两种情况最具挑战性，因为若不讨论温度，就无法给出简单的“是/否”答案。
6. Calculating Gibbs Free Energy Changes | 计算吉布斯自由能变
Calculating ΔG from given enthalpy and entropy data is a routine exam task. The straightforward route uses ΔG = ΔH – TΔS, but the standard free energy change ΔG° can also be determined indirectly via a Hess’s-law-type approach or from equilibrium constants.
根据给定的焓与熵数据计算 ΔG 是考试中的常规题目。最直接的方法是使用 ΔG = ΔH – TΔS，但标准自由能变 ΔG° 也可通过盖斯定律式间接方法或从平衡常数求取。
Worked example: The reaction 2NO₂(g) → N₂O₄(g) has ΔH° = –57.2 kJ mol⁻¹ and ΔS° = –176 J mol⁻¹ K⁻¹. Calculate ΔG° at 298 K.
实例：反应 2NO₂(g) → N₂O₄(g) 的 ΔH° = –57.2 kJ mol⁻¹，ΔS° = –176 J mol⁻¹ K⁻¹。计算 298 K 下的 ΔG°。
Solution: First ensure units match. Express ΔH° as –57200 J mol⁻¹. Then TΔS° = 298 K × (–176 J mol⁻¹ K⁻¹) = –52448 J mol⁻¹. ΔG° = ΔH° – TΔS° = –57200 – (–52448) = –4752 J mol⁻¹ ≈ –4.75 kJ mol⁻¹. The negative sign confirms feasibility at 298 K.
解答：先统一单位。将 ΔH° 写作 –57200 J mol⁻¹。则 TΔS° = 298 K × (–176 J mol⁻¹ K⁻¹) = –52448 J mol⁻¹。ΔG° = ΔH° – TΔS° = –57200 – (–52448) = –4752 J mol⁻¹ ≈ –4.75 kJ mol⁻¹。负号表明 298 K 下反应可行。
Always state your final answer to three significant figures and include units – examiners penalise missing or inconsistent units heavily.
务必以三位有效数字给出最终答案并标明单位——阅卷人会对缺失或单位不一致大力扣分。
7. Gibbs Free Energy and Equilibrium Constants | 吉布斯自由能与平衡常数
A powerful extension of ΔG connects it directly to the position of chemical equilibrium. The standard free energy change ΔG° is related to the equilibrium constant K by the equation:
ΔG 的一个强大延伸是将它直接与化学平衡位置联系起来。标准自由能变 ΔG° 与平衡常数 K 之间的关系为：
ΔG° = –RT ln K
where R is the gas constant (8.314 J mol⁻¹ K⁻¹), T is the temperature in kelvin, and ln K is the natural logarithm of K. This equation reveals that a large negative ΔG° corresponds to a large K (products strongly favoured), while a positive ΔG° gives a small K (reactants favoured).
式中 R 为气体常数 (8.314 J mol⁻¹ K⁻¹)，T 为开氏温度，ln K 为 K 的自然对数。该式表明，大的负 ΔG° 对应大的 K 值（产物强烈占优），而正的 ΔG° 则给出小的 K 值（反应物占优）。
OCR exam questions often require you to rearrange the equation to find K = exp(–ΔG°/RT) or to determine ΔG° from a given K. A classic pitfall is forgetting to convert ΔG° from kJ mol⁻¹ to J mol⁻¹ before calculating, as R uses joules.
OCR 考试的题目常要求你将公式变形为 K = exp(–ΔG°/RT)，或从已知 K 求 ΔG°。一个典型陷阱是，在计算前忘记将 ΔG° 的 kJ mol⁻¹ 换算为 J mol⁻¹，因为 R 的单位为焦耳。
8. Linking ΔG° to Electrochemical Cells | 将 ΔG° 与电化学电池联系起来
Electrochemistry provides another window into Gibbs free energy. For a reaction taking place in an electrochemical cell, the standard free energy change is directly proportional to the standard cell potential E°:
电化学为吉布斯自由能提供了另一扇窗口。对于在电化学电池中发生的反应，标准自由能变与标准电池电动势 E° 直接成正比：
ΔG° = –nFE°
Here n is the number of moles of electrons transferred in the redox reaction, F is the Faraday constant (96485 C mol⁻¹), and E° is the standard cell potential in volts (V). The negative sign ensures that a positive E° (a spontaneous cell) yields a negative ΔG°.
其中 n 为氧化还原反应中转移的电子摩尔数，F 为法拉第常数 (96485 C mol⁻¹)，E° 为标准电池电动势，单位为伏特 (V)。负号保证了正 E°（自发电池）产生负的 ΔG°。
This equation is extremely useful for finding ΔG° from measured cell voltages, or for predicting the maximum voltage a cell can deliver under standard conditions. A typical calculation: for the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), n = 2 and E° = +1.10 V, so ΔG° = –2 × 96485 × 1.10 ≈ –212 kJ mol⁻¹.
该公式在根据实测电池电压求 ΔG°，或预测标准条件下电池可提供的最大电压时极为有用。典型计算：对于反应 Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)，n = 2、E° = +1.10 V，故 ΔG° = –2 × 96485 × 1.10 ≈ –212 kJ mol⁻¹。
9. The Effect of Temperature on Equilibrium | 温度对平衡的影响
Although ΔG° is defined at a specific temperature (often 298 K), the relationship ΔG° = –RT ln K shows that as temperature changes, K also changes. This provides a thermodynamic explanation for Le Chatelier’s principle: for an exothermic reaction (negative ΔH°), increasing temperature makes ΔG° less negative (or more positive), so K decreases, shifting equilibrium to the left. For an endothermic reaction, K increases with temperature.
虽然 ΔG° 是在特定温度（通常为 298 K）下定义的，但关系式 ΔG° = –RT ln K 表明，温度变化时 K 也随之改变。这为勒夏特列原理提供了热力学解释：对于放热反应（ΔH° 为负），升高温度会使 ΔG° 的负值减小（甚至转正），因此 K 减小，平衡左移；对于吸热反应，K 随温度升高而增大。
A more advanced form, the van’t Hoff equation, is not required in OCR but the conceptual link between ΔG°, K and T is explicitly tested. Candidates may be asked to explain how the sign of ΔH° determines the temperature dependence of K, or to calculate K at a non-standard temperature using provided ΔH° and ΔS° values.
更精细的范特霍夫方程虽不在 OCR 考纲内，但 ΔG°、K 和 T 之间的概念联系是明确考查的。考生可能被要求解释 ΔH° 的符号如何决定 K 的温度依赖性，或利用给出的 ΔH° 和 ΔS° 值计算非标准温度下的 K。
10. Standard Conditions and ΔG° | 标准状态与 ΔG°
The superscript ‘°’ in ΔG° signifies standard conditions: pressure of 100 kPa, temperature usually 298 K, and all solutions at a concentration of 1 mol dm⁻³. Standard state is a reference point, not a realistic reaction medium – yet it allows fair comparison between different reactions.
ΔG° 中的上标 ‘°’ 表示标准条件：压力 100 kPa，温度通常为 298 K，所有溶液浓度均为 1 mol dm⁻³。标准态是一个参考基准，并非真实的反应环境——但它使得不同反应之间有了公平比较的可能。
Do not confuse ΔG° with ΔG under non-standard conditions. For any process, ΔG = ΔG° + RT ln Q, where Q is the reaction quotient. This non-standard form is beyond the OCR specification, but understanding that ΔG° refers to a specific set of conditions is essential for proper interpretation of equilibrium and electrochemical data.
不要将 ΔG° 与非标准条件下的 ΔG 混淆。对于任意过程，ΔG = ΔG° + RT ln Q，其中 Q 为反应商。这一非标准形式虽超出 OCR 考纲，但理解 ΔG° 对应一组特定条件是正确解读平衡和电化学数据的关键。
11. Common Misconceptions and Exam Tips | 常见误区与应试技巧
One of the biggest errors is treating ΔG as a measure of reaction rate. A negative ΔG indicates thermodynamic feasibility – it says nothing about how fast the reaction occurs. Many spontaneous reactions are kinetically controlled and may appear not to happen at all without a spark or a catalyst.
最大的误区之一是将 ΔG 视为反应速率的量度。负的 ΔG 仅表明热力学可行性——与反应速率完全无关。许多自发反应都受动力学控制，没有火花或催化剂时可能看起来根本不发生。
Other common mistakes include: forgetting to convert temperature to kelvin; using Celsius in calculations; mixing kJ and J units; treating ΔS as ‘energy’ when it is actually ‘energy per kelvin’; and misapplying the feasibility temperature formula T = ΔH/ΔS when the signs are not suitable.
其他常见错误包括：忘记将温度换算为开尔文；用摄氏温度代入计算；kJ 与 J 单位混用；将 ΔS 当成“能量”，而它实际上是“每开尔文的能量”；在适用符号不当的情况下误用可行性温度公式 T = ΔH/ΔS。
In the exam, always show the rearrangement of the ΔG° = –RT ln K equation clearly if you are solving for K, and round your answer sensibly. If a question asks ‘Explain why this reaction becomes feasible above a certain temperature’, you must reference both the signs of ΔH and ΔS and how they influence the ΔG sign.
考试中，若要从 ΔG° = –RT ln K 求解 K，务必清晰展示公式的变形过程，并合理取舍答案。若题目要求“解释为何该反应在某个温度以上变得可行”，你必须指明 ΔH 和 ΔS 的符号，并说明它们如何影响 ΔG 的符号。
12. Worked Example: Feasibility Temperature | 实例分析：反应可行性温度
Consider the endothermic decomposition of ammonium chloride: NH₄Cl(s) → NH₃(g) + HCl(g). For this reaction, ΔH° = +176 kJ mol⁻¹ and ΔS° = +285 J mol⁻¹ K⁻¹. Determine the temperature above which the reaction is feasible.
考虑氯化铵的吸热分解反应：NH₄Cl(s) → NH₃(g) + HCl(g)。已知 ΔH° = +176 kJ mol⁻¹，ΔS° = +285 J mol⁻¹ K⁻¹。计算该反应可行所需的温度下限。
Set ΔG° = 0, so T = ΔH° ÷ ΔS°. Convert ΔH° to J mol⁻¹: 176 000 J mol⁻¹. Then T = 176 000 ÷ 285 ≈ 618 K. The reaction is thermodynamically feasible above approximately 618 K (345 °C).
令 ΔG° = 0，得 T = ΔH° ÷ ΔS°。将 ΔH° 换算为 J mol⁻¹：176 000 J mol⁻¹。则 T = 176 000 ÷ 285 ≈ 618 K。该反应在约 618 K (345 °C) 以上热力学可行。
This example highlights why heating is needed to drive an endothermic entropy-driven reaction: high T makes the –TΔS term negative enough to overcome the unfavourable positive ΔH.
这个例子说明了为什么加热对驱动这类吸热且熵驱动的反应是必需的：高温使得 –TΔS 项负得足够大，足以克服不利的正 ΔH。
In a full exam answer, also state the assumption that ΔH° and ΔS° are approximately constant over the temperature range considered – a simplification that keeps calculations manageable.
在完整的考试答案中，还须声明假设 ΔH° 和 ΔS° 在考虑的温度区间内近似为常数——这一简化使计算保持在可控范围内。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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