Gibbs Free Energy Exam Essentials | 吉布斯自由能考点精讲

📚 Gibbs Free Energy Exam Essentials | 吉布斯自由能考点精讲

Gibbs free energy is a vital concept in WJEC IGCSE Chemistry that helps us decide whether a chemical reaction can take place on its own. It links two key driving forces – enthalpy and entropy – allowing us to predict spontaneity under constant temperature and pressure. This guide will walk you through everything you need to know for the exam: the equation, units, calculations, and how to interpret temperature effects on feasibility.

吉布斯自由能是 WJEC IGCSE 化学中的一个重要概念，它帮助我们判断化学反应能否自发进行。它联系了两个关键驱动力——焓和熵——让我们能在恒温恒压条件下预测反应的自发性。本指南将带你梳理考试所需的一切：公式、单位、计算，以及如何解读温度对可行性的影响。

1. Introduction to Gibbs Free Energy | 吉布斯自由能简介

Gibbs free energy, symbol G, is a thermodynamic property that combines the enthalpy (ΔH) and entropy (ΔS) of a system. A reaction is said to be spontaneous or feasible if the change in Gibbs free energy (ΔG) for that reaction is negative. It is important to remember that ‘feasible’ does not mean ‘fast’ – a negative ΔG tells us a reaction is energetically possible, but it may still be very slow.

吉布斯自由能，符号为 G，是一个结合了系统焓变 (ΔH) 和熵变 (ΔS) 的热力学性质。如果反应的吉布斯自由能变 (ΔG) 为负值，则称该反应为自发或可行。务必记住，“可行”并不意味着“迅速”——ΔG 为负告诉我们反应在能量上是可能的，但它可能仍然进行得非常缓慢。

In WJEC IGCSE Chemistry, you will use the Gibbs free energy equation to determine feasibility, calculate the temperature at which a reaction becomes spontaneous, and analyse how changes in enthalpy and entropy compete with each other. This concept often appears in the ‘Energy Changes in Reactions’ topic and can be assessed with numerical problems.

在 WJEC IGCSE 化学中，你将运用吉布斯自由能方程来判断可行性、计算反应自发进行的温度，并分析焓变与熵变如何相互竞争。这一概念经常出现在“反应中的能量变化”主题中，并可能以数值计算题的形式进行考查。

2. The Gibbs Equation | 吉布斯方程

The relationship is given by the Gibbs-Helmholtz equation:

这一关系由吉布斯-亥姆霍兹方程给出：

ΔG = ΔH – TΔS

where ΔG is the change in Gibbs free energy (kJ mol⁻¹), ΔH is the change in enthalpy (kJ mol⁻¹), T is the absolute temperature in kelvin (K), and ΔS is the change in entropy (usually given in J K⁻¹ mol⁻¹). The units must be consistent before calculation: if ΔH is in kJ, you will often need to convert ΔS from J to kJ by dividing by 1000.

其中 ΔG 为吉布斯自由能变 (kJ mol⁻¹)，ΔH 为焓变 (kJ mol⁻¹)，T 为绝对温度 (K)，ΔS 为熵变 (通常以 J K⁻¹ mol⁻¹ 给出)。计算前单位必须一致：若 ΔH 单位为 kJ，通常需要将 ΔS 从 J 转换为 kJ，即除以 1000。

In the exam, you are likely to be given ΔH and ΔS values, then asked to calculate ΔG and comment on feasibility. Always write out the full equation, substitute carefully, and pay close attention to the signs of ΔH and ΔS. A surprising number of mistakes come from forgetting the minus sign in the equation or mixing up units.

考试中，你很可能会得到 ΔH 和 ΔS 值，然后要求计算 ΔG 并评论其可行性。务必写出完整方程，仔细代入，并密切关注 ΔH 和 ΔS 的正负号。有相当多的错误源于忘记方程中的减号或混淆了单位。

3. Units and Conversions | 单位与换算

One of the most common pitfalls is unit inconsistency. ΔH is nearly always expressed in kJ mol⁻¹, while ΔS is frequently given in J K⁻¹ mol⁻¹. Since T is in kelvin, multiplying T by ΔS in J gives a value in J. You must convert this to kJ to match ΔH.

最常见的陷阱之一是单位不一致。ΔH 几乎总是以 kJ mol⁻¹ 表示，而 ΔS 常以 J K⁻¹ mol⁻¹ 给出。由于 T 单位为开尔文，将 T 与单位为 J 的 ΔS 相乘会得到焦耳值。你必须将其转换为 kJ 才能与 ΔH 匹配。

The standard conversion is: ΔS (kJ K⁻¹ mol⁻¹) = ΔS (J K⁻¹ mol⁻¹) ÷ 1000. For example, if ΔS = +120 J K⁻¹ mol⁻¹, use 0.120 kJ K⁻¹ mol⁻¹ in the equation. Double-check this conversion before any calculation. Writing the units on each term during substitution will help you spot errors.

标准换算为：ΔS (kJ K⁻¹ mol⁻¹) = ΔS (J K⁻¹ mol⁻¹) ÷ 1000。例如，若 ΔS = +120 J K⁻¹ mol⁻¹，则在方程中使用 0.120 kJ K⁻¹ mol⁻¹。在任何计算前都要仔细核对这一换算。代入时在每个项旁边写出单位有助于发现错误。

Also, remember that temperature must always be in kelvin. If a question gives temperature in °C, add 273 to convert. For standard conditions, 25 °C becomes 298 K. Neglecting to convert Celsius to Kelvin will produce a wildly inaccurate ΔG.

此外，请记住温度必须始终以开尔文为单位。若题目给出的温度是 °C，则加 273 进行转换。对于标准条件，25 °C 转换为 298 K。忽略将摄氏度转换为开尔文会产生极为不准确的 ΔG。

4. How to Calculate ΔG | 如何计算ΔG

Follow a clear stepwise method for every ΔG calculation:

每一次 ΔG 计算都应遵循清晰的逐步法：

Step 1: List the given ΔH (kJ mol⁻¹), ΔS (J K⁻¹ mol⁻¹), and T (°C or K).
步骤1：列出给出的 ΔH (kJ mol⁻¹)、ΔS (J K⁻¹ mol⁻¹) 和 T (°C 或 K)。
Step 2: Convert T to kelvin if necessary (T(K) = t(°C) + 273).
步骤2：如有必要，将 T 转换为开尔文 (T(K) = t(°C) + 273)。
Step 3: Convert ΔS to kJ K⁻¹ mol⁻¹ by dividing by 1000.
步骤3：将 ΔS 除以 1000 转换为 kJ K⁻¹ mol⁻¹。
Step 4: Substitute into ΔG = ΔH – TΔS.
步骤4：代入公式 ΔG = ΔH – TΔS。
Step 5: Calculate and add the correct unit (kJ mol⁻¹).
步骤5：计算并添加正确单位 (kJ mol⁻¹)。

For example, a reaction has ΔH = -200 kJ mol⁻¹ and ΔS = -150 J K⁻¹ mol⁻¹ at 298 K. Convert ΔS: -150 ÷ 1000 = -0.150 kJ K⁻¹ mol⁻¹. Then ΔG = -200 – (298 × -0.150) = -200 + 44.7 = -155.3 kJ mol⁻¹. The negative ΔG confirms the reaction is feasible at this temperature.

例如，某反应 ΔH = -200 kJ mol⁻¹，ΔS = -150 J K⁻¹ mol⁻¹，温度为 298 K。转换 ΔS：-150 ÷ 1000 = -0.150 kJ K⁻¹ mol⁻¹。则 ΔG = -200 – (298 × -0.150) = -200 + 44.7 = -155.3 kJ mol⁻¹。ΔG 为负值证实该反应在此温度下可行。

5. Interpreting ΔG: Feasibility | 解释ΔG：可行性

The sign of ΔG tells us whether the forward reaction is thermodynamically feasible:

ΔG 的正负号告诉我们正向反应在热力学上是否可行：

ΔG < 0: Reaction is feasible (spontaneous). Products are more stable than reactants under the given conditions.
ΔG < 0：反应可行（自发）。在给定条件下，产物比反应物更稳定。
ΔG > 0: Reaction is not feasible. The forward reaction will not occur without an external energy input. The reverse reaction may be feasible.
ΔG > 0：反应不可行。正向反应在无外部能量输入时不会发生。逆向反应可能可行。
ΔG = 0: The system is at equilibrium. The forward and reverse reaction rates are equal, and there is no net change in composition.
ΔG = 0：系统处于平衡状态。正向与逆向反应速率相等，组成没有净变化。

Remember: feasibility depends on temperature. A reaction that is not feasible at room temperature may become feasible at a higher or lower temperature because the TΔS term changes. Always relate your ΔG interpretation back to the question context – mention whether the reaction is feasible under the stated conditions.

记住：可行性取决于温度。在室温下不可行的反应可能在更高或更低温度下变得可行，因为 TΔS 项会变化。始终将你对 ΔG 的解读与题目背景联系起来——提及在该条件下反应是否可行。

6. Effect of Temperature on Spontaneity | 温度对自发性的影响

Because ΔG = ΔH – TΔS, temperature can flip the sign of ΔG, especially when ΔH and ΔS have the same sign. There are four key scenarios you must recognise for the exam:

由于 ΔG = ΔH – TΔS，温度可以改变 ΔG 的正负号，尤其是当 ΔH 和 ΔS 正负号相同时。考试中你必须识别以下四种关键情形：

ΔH negative, ΔS positive: Reaction is feasible at all temperatures. The negative ΔH and positive -TΔS term both drive ΔG to be negative.
ΔH 为负，ΔS 为正：反应在所有温度下均可行。负的 ΔH 和正的 -TΔS 项都使 ΔG 为负。
ΔH positive, ΔS negative: Reaction is never feasible. Both terms oppose spontaneity, so ΔG is always positive.
ΔH 为正，ΔS 为负：反应永远不可行。两项都不利于自发，ΔG 始终为正。
ΔH negative, ΔS negative: Reaction is feasible at low temperatures. At high T, the -TΔS becomes a large positive term, making ΔG positive.
ΔH 为负，ΔS 为负：反应在低温下可行。在高温下，-TΔS 变为很大的正值，使 ΔG 为正。
ΔH positive, ΔS positive: Reaction is feasible at high temperatures. Only when T is large enough does the -TΔS term overcome the positive ΔH.
ΔH 为正，ΔS 为正：反应在高温下可行。仅当 T 足够大时，-TΔS 项才能克服正的 ΔH。

Students often find the last two cases tricky. A helpful way to visualise this is to treat ΔG as a balance between an enthalpy drive and an entropy drive. When they pull in opposite directions, temperature decides the winner.

学生通常会觉得后两种情况棘手。一个有助于形象化的方法是把 ΔG 看作焓驱动与熵驱动之间的平衡。当两者方向相反时，温度将决定胜负。

7. Using ΔG = 0 to Find the ‘Crossover’ Temperature | 利用ΔG = 0 求转变温度

When ΔH and ΔS oppose each other, the temperature at which the reaction just becomes feasible is found by setting ΔG = 0. Rearranging the equation gives:

当 ΔH 与 ΔS 相互对抗时，令 ΔG = 0 可求得反应刚好变得可行的温度。重新整理方程得：

T = ΔH ÷ ΔS

Ensure ΔS is in kJ K⁻¹ mol⁻¹ to match ΔH in kJ mol⁻¹. If ΔS is given in J K⁻¹ mol⁻¹, convert it first. The calculated T is in kelvin. You can then convert to °C if required. This temperature is often called the ‘crossover temperature’ or the temperature at which feasibility changes.

确保 ΔS 单位为 kJ K⁻¹ mol⁻¹，以与单位为 kJ mol⁻¹ 的 ΔH 相匹配。如果 ΔS 以 J K⁻¹ mol⁻¹ 给出，先进行转换。计算出的 T 单位为开尔文。如有需要可转换为 °C。此温度常被称为“转变温度”或可行性发生改变的温度。

For instance, if ΔH = +45 kJ mol⁻¹ and ΔS = +150 J K⁻¹ mol⁻¹ (0.150 kJ K⁻¹ mol⁻¹), then T = 45 ÷ 0.150 = 300 K (27 °C). Above 300 K, the reaction is feasible; below this temperature, it is not. This type of calculation appears regularly in WJEC papers, so practise it thoroughly.

例如，若 ΔH = +45 kJ mol⁻¹，ΔS = +150 J K⁻¹ mol⁻¹ (0.150 kJ K⁻¹ mol⁻¹)，则 T = 45 ÷ 0.150 = 300 K (27 °C)。在 300 K 以上，反应可行；低于此温度则不可行。此类计算经常出现在 WJEC 试卷中，务必充分练习。

8. Practice Calculation Examples | 计算实例练习

Let’s work through a typical exam-style multi-step problem.

我们来演练一道典型的考试风格多步骤题目。

Example: For the reaction A → B, ΔH = -92 kJ mol⁻¹ and ΔS = -200 J K⁻¹ mol⁻¹. (a) Calculate ΔG at 298 K. (b) Determine the temperature below which the reaction is feasible.

例题：对于反应 A → B，ΔH = -92 kJ mol⁻¹，ΔS = -200 J K⁻¹ mol⁻¹。(a) 计算 298 K 时的 ΔG。(b) 确定反应在什么温度以下可行。

Solution (a): Convert ΔS: -200 J K⁻¹ mol⁻¹ = -0.200 kJ K⁻¹ mol⁻¹. ΔG = -92 – (298 × -0.200) = -92 + 59.6 = -32.4 kJ mol⁻¹. The reaction is feasible at 298 K.

解答 (a)：转换 ΔS：-200 J K⁻¹ mol⁻¹ = -0.200 kJ K⁻¹ mol⁻¹。ΔG = -92 – (298 × -0.200) = -92 + 59.6 = -32.4 kJ mol⁻¹。该反应在 298 K 可行。

Solution (b): Since ΔH is negative and ΔS is negative, feasibility is favoured at low temperatures. Set ΔG = 0, T = ΔH ÷ ΔS = (-92) ÷ (-0.200) = 460 K. The reaction is feasible for T < 460 K. Note that the minus signs cancel, giving a positive temperature.

解答 (b)：因为 ΔH 为负且 ΔS 为负，低温有利于可行性。设 ΔG = 0，T = ΔH ÷ ΔS = (-92) ÷ (-0.200) = 460 K。该反应在 T < 460 K 时可行。注意负号相互抵消，得到正的温度值。

Always do a quick sense check: Does the calculation match the prediction from the ΔH/ΔS sign analysis? Here, negative ΔH, negative ΔS means feasibility at low T, so a crossover temperature of 460 K fits the rule.

始终做一个快速合理检查：计算结果是否符合通过 ΔH/ΔS 符号分析所作的预测？在此例中，负 ΔH、负 ΔS 意味着低温可行，因此 460 K 的转变温度符合该规律。

9. Common Mistakes and Exam Tips | 常见错误与考试技巧

Even students who understand the theory can lose marks through small slip-ups. Here are the top mistakes to avoid:

即使理解了理论的学生也可能因小失误而丢分。以下是需要避免的主要错误：

Forgetting to convert ΔS units from J to kJ. Always write ‘/1000’ as a reminder.
忘记将 ΔS 单位从 J 转换为 kJ。始终写下 “/1000” 作为提醒。
Using °C instead of Kelvin. Check that you have added 273 where necessary.
使用 °C 而非开尔文。检查是否已在必要处加上了 273。
Mis-handling signs: when subtracting a negative TΔS, it becomes addition. Write the substitution step carefully to avoid sign errors.
符号处理错误：减去负的 TΔS 时，它变为加法。仔细写出代入步骤以避免符号错误。
Stating a reaction is ‘fast’ because ΔG is negative. Feasibility is about thermodynamics, not kinetics. Always use the word ‘feasible’ rather than ‘will react quickly’.
因 ΔG 为负而表述反应“快速”。可行性关乎热力学而非动力学。始终使用“可行”一词，而非“会快速反应”。
Confusing the direction: ΔG refers to the forward reaction as written. You may need to comment on the reverse reaction’s feasibility.
混淆方向：ΔG 指的是按书写方向的反应。你可能需要对逆反应的可行性做出评论。

Exam tip: Show all working clearly. Even if your final answer is slightly off, well-laid-out steps can gain method marks. Also, always state the final conclusion in words after a calculation, e.g. ‘Since ΔG is negative, the reaction is feasible at this temperature.’

考试技巧：清晰展示所有计算过程。即使最终答案略有偏差，条理清晰的步骤也能获得方法分。此外，计算后始终用文字陈述最终结论，例如：“由于 ΔG 为负，该反应在此温度下可行。”

10. Summary Table for ΔH, ΔS and Feasibility | ΔH、ΔS与可行性总结表

The table below summarises the effect of temperature on spontaneity for all sign combinations. Use it as a quick reference when revising.

下表总结了所有符号组合下温度对自发性的影响。复习时可作为快速参考。

ΔH	ΔS	Feasible at	Example reaction type
Negative	Positive	All temperatures	Combustion
Positive	Negative	Never	Reverse of above
Negative	Negative	Low temperatures	Freezing of water
Positive	Positive	High temperatures	Thermal decomposition

Memorising this table will help you quickly interpret any ΔG scenario. Then practise applying it by predicting feasibility before doing the full calculation. This builds confidence and saves time in the exam.

记住此表将有助于你快速解读任何 ΔG 情景。然后通过在实际计算前预测可行性来练习应用。这能建立信心并在考试中节省时间。

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Gibbs Free Energy Exam Essentials | 吉布斯自由能 考点精讲