Gibbs Free Energy for GCSE CCEA Chemistry | GCSE CCEA 化学：吉布斯自由能考点精讲

📚 Gibbs Free Energy for GCSE CCEA Chemistry | GCSE CCEA 化学：吉布斯自由能考点精讲

In GCSE CCEA Chemistry, Gibbs free energy is introduced as a way to predict whether a chemical reaction is feasible under given conditions. It combines enthalpy change, entropy change, and temperature into a single quantity, ΔG. Understanding this topic helps you explain why some endothermic reactions happen spontaneously while others do not, and why temperature can switch the direction of feasibility.

在 GCSE CCEA 化学课程中，吉布斯自由能用来预测化学反应在给定条件下是否具有可行性。它将焓变、熵变和温度综合为一个物理量 ΔG。掌握这个主题有助于解释为什么有些吸热反应能自发进行而另一些不能，以及温度为何能改变反应的可行性方向。

1. What is Gibbs Free Energy? | 什么是吉布斯自由能？

Gibbs free energy, symbol G, is a thermodynamic potential that measures the maximum amount of non-expansion work that can be extracted from a closed system at constant temperature and pressure. In GCSE terms, we use the change in Gibbs free energy, ΔG, to decide if a reaction is feasible (can happen on its own) or not.

吉布斯自由能，符号为 G，是一种热力学势，用来衡量在恒温恒压下、封闭体系所能作出的最大非体积功。在 GCSE 层面，我们通过吉布斯自由能的变化量 ΔG 来判断一个反应是否具有可行性（能否自发进行）。

The key idea is simple: if ΔG is negative, the reaction is feasible; if ΔG is positive, the reaction is not feasible under those conditions. A ΔG of zero means the system is at equilibrium.

核心思想很简单：若 ΔG 为负值，反应可行；若 ΔG 为正值，在该条件下反应不可行；若 ΔG = 0，体系处于平衡状态。

The symbol comes from the American scientist Josiah Willard Gibbs, who developed this concept in the 1870s.

这一符号来源于美国科学家约西亚·威拉德·吉布斯，他在 19 世纪 70 年代提出了这一概念。

2. The Gibbs Equation | 吉布斯方程

The change in Gibbs free energy is calculated using the equation:

吉布斯自由能的变化量由以下方程计算：

ΔG = ΔH – TΔS

Where:

其中：

ΔG = change in Gibbs free energy (kJ mol⁻¹ or J mol⁻¹) | 吉布斯自由能变（千焦每摩尔或焦每摩尔）
ΔH = enthalpy change (kJ mol⁻¹ or J mol⁻¹) | 焓变（千焦每摩尔或焦每摩尔）
T = temperature in kelvin (K) | 热力学温度，单位开尔文（K）
ΔS = entropy change (J K⁻¹ mol⁻¹) | 熵变，单位焦每开每摩尔（J K⁻¹ mol⁻¹）

Notice that ΔS is usually given in J K⁻¹ mol⁻¹, while ΔH is often in kJ mol⁻¹. In calculations, you must convert both to the same unit – typically convert ΔH to J mol⁻¹ by multiplying by 1000, or convert ΔS to kJ K⁻¹ mol⁻¹ by dividing by 1000.

请注意，ΔS 通常以 J K⁻¹ mol⁻¹ 为单位，而 ΔH 通常以 kJ mol⁻¹ 为单位。在计算时，必须统一单位——常见做法是将 ΔH 乘以 1000 转换为 J mol⁻¹，或将 ΔS 除以 1000 转换为 kJ K⁻¹ mol⁻¹。

This equation shows that feasibility depends on three factors: the heat transferred (ΔH), the change in disorder (ΔS), and the temperature at which the reaction takes place.

这个方程表明，可行性取决于三个因素：热量传递（ΔH）、无序度的变化（ΔS）以及反应进行的温度。

3. Understanding Entropy ΔS | 理解熵变 ΔS

Entropy, symbol S, is a measure of the disorder or randomness of a system. A positive ΔS means the products are more disordered than the reactants. For example, when a solid dissolves, particles spread out and entropy increases (ΔS > 0). When a gas condenses into a liquid, entropy decreases (ΔS < 0).

熵，符号为 S，是衡量体系无序度或随机程度的物理量。ΔS 为正值表示产物比反应物更无序。例如，固体溶解时，微粒分散开来，熵增加（ΔS > 0）。当气体冷凝为液体时，熵减少（ΔS < 0）。

The units of entropy are J K⁻¹ mol⁻¹. In the Gibbs equation, a larger positive ΔS helps make ΔG more negative, favouring feasibility. A negative ΔS can work against feasibility unless ΔH is sufficiently negative.

熵的单位是 J K⁻¹ mol⁻¹。在吉布斯方程中，较大的正 ΔS 有助于使 ΔG 变得更负，有利于反应进行。负的 ΔS 则对可行性不利，除非 ΔH 足够负。

In GCSE CCEA exams, you may be given ΔS values or asked to explain why a reaction becomes feasible only at higher temperatures due to a large positive ΔS.

在 GCSE CCEA 考试中，你可能会被给出 ΔS 数值，或者需要解释为何一个反应由于具有较大的正 ΔS，仅在较高温度下才变得可行。

4. Temperature in Kelvin | 开尔文温度

The temperature T in the Gibbs equation must be in kelvin. To convert from degrees Celsius to kelvin, add 273:

吉布斯方程中的温度 T 必须以开尔文为单位。将摄氏度转换为开尔文的做法是加上 273：

T (K) = Temperature (°C) + 273

For example, room temperature of 25 °C becomes 298 K. A typical exam question may provide temperature in °C and expect you to convert it before substituting into the equation.

例如，室温 25 °C 转换为 298 K。考试中常见的题目会给出摄氏温度，要求你先转换单位再代入方程。

Always check that you have used kelvin; failure to do so will give the wrong sign or magnitude for ΔG.

务必确认使用了开尔文温度；否则会导致 ΔG 的正负号和大小都出现错误。

5. Unit Consistency in Calculations | 计算中的单位统一

One of the most common mistakes in Gibbs free energy calculations is mixing kJ and J. Always convert ΔH and ΔS to compatible units.

吉布斯自由能计算中最常见的错误之一就是混淆千焦和焦耳。务必将 ΔH 和 ΔS 转换为一致的单位。

For example, if ΔH = –200 kJ mol⁻¹ and ΔS = +150 J K⁻¹ mol⁻¹, convert ΔH to –200 000 J mol⁻¹, or convert ΔS to +0.150 kJ K⁻¹ mol⁻¹. Then perform the calculation:

例如，若 ΔH = –200 kJ mol⁻¹，ΔS = +150 J K⁻¹ mol⁻¹，可将 ΔH 转换为 –200 000 J mol⁻¹，或将 ΔS 转换为 +0.150 kJ K⁻¹ mol⁻¹。然后进行计算：

ΔG = –200 000 J mol⁻¹ – (298 K × 150 J K⁻¹ mol⁻¹) = –200 000 – 44 700 = –244 700 J mol⁻¹ = –244.7 kJ mol⁻¹

The negative ΔG confirms feasibility.

ΔG 为负值，确认反应可行。

An exam tip: write down the units at each step. That helps you see whether you need to multiply or divide by 1000.

考试技巧：每一步都写下单位，这样可以帮你判断是否需要乘以或除以 1000。

6. Feasibility Criteria | 可行性判据

The sign of ΔG tells you whether a reaction is feasible under the specified temperature and pressure:

ΔG 的正负号告诉我们，在指定温度和压力下反应是否可行：

ΔG Value (ΔG 值)	Meaning (含义)
ΔG < 0 (negative)	Reaction is feasible (反应可行)
ΔG > 0 (positive)	Reaction is not feasible; reverse reaction may be feasible (反应不可行；逆反应可能可行)
ΔG = 0	System at equilibrium; no net change (体系处于平衡态；无净变化)

It is important to note that feasibility does not indicate the rate of reaction. A reaction with a negative ΔG might be extremely slow at room temperature and require a catalyst or high temperature to occur at an observable rate.

需要特别注意的是，可行性并不代表反应速率。一个 ΔG 为负的反应在室温下可能极其缓慢，需要催化剂或高温才能在可观察的速率下进行。

7. Using ΔG to Predict the Effect of Temperature | 利用 ΔG 预测温度影响

Because T appears in the term –TΔS, temperature can change the sign of ΔG. Consider four situations:

由于温度 T 出现在 –TΔS 项中，温度可以改变 ΔG 的正负号。思考以下四种情况：

ΔH < 0 and ΔS > 0: ΔG is always negative regardless of temperature. The reaction is feasible at all temperatures.
ΔH < 0 and ΔS > 0：无论温度如何，ΔG 始终为负。反应在任何温度下都可行。
ΔH > 0 and ΔS < 0: ΔG is always positive. The reaction is never feasible.
ΔH > 0 and ΔS < 0：ΔG 始终为正。反应永远不可行。
ΔH < 0 and ΔS < 0: ΔG is negative only at low temperatures. Feasibility is lost when T becomes too large because the –TΔS term becomes positive.
ΔH < 0 and ΔS < 0：ΔG 仅在低温时为负。当 T 过大时，–TΔS 项变为正，反应不再可行。
ΔH > 0 and ΔS > 0: ΔG is negative only at high temperatures. This explains endothermic reactions that are feasible only when hot, such as the thermal decomposition of calcium carbonate.
ΔH > 0 and ΔS > 0：ΔG 仅在高温时为负。这解释了仅在被加热时才可行的吸热反应，例如碳酸钙的热分解。

You may be asked to calculate the temperature at which ΔG becomes zero (the minimum temperature for feasibility of an endothermic reaction with ΔS > 0). Set ΔG = 0, then T = ΔH / ΔS. Remember unit alignment.

你可能需要计算使 ΔG = 0 的温度（即一个 ΔH > 0, ΔS > 0 的反应变得可行的最低温度）。令 ΔG = 0，则 T = ΔH / ΔS。注意单位一致。

8. Worked Example | 典型计算示例

A reaction has ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. Calculate the temperature at which the reaction becomes feasible.

某反应的 ΔH = +178 kJ mol⁻¹，ΔS = +161 J K⁻¹ mol⁻¹。计算反应变得可行的温度。

Step 1: Convert units so they match. ΔH = 178 000 J mol⁻¹. ΔS = 161 J K⁻¹ mol⁻¹.

步骤一：统一单位。ΔH = 178 000 J mol⁻¹，ΔS = 161 J K⁻¹ mol⁻¹。

Step 2: Set ΔG = 0. 0 = ΔH – TΔS → T = ΔH / ΔS.

步骤二：令 ΔG = 0。0 = ΔH – TΔS → T = ΔH / ΔS。

Step 3: T = 178 000 / 161 = 1105.6 K. Convert to °C: 1105.6 – 273 = 832.6 °C.

步骤三：T = 178 000 / 161 = 1105.6 K。转换为摄氏度：1105.6 – 273 = 832.6 °C。

Thus, the reaction becomes feasible at temperatures above approximately 833 °C.

因此，反应在约 833 °C 以上变得可行。

This is typical for thermal decomposition reactions, such as the breakdown of limestone in a blast furnace.

这是热分解反应的典型特征，例如鼓风炉中石灰石的分解。

9. Relating ΔG to Industrial Processes | 将 ΔG 与工业过程联系起来

CCEA GCSE Chemistry often uses industrial examples. The extraction of iron in the blast furnace involves the reaction:

CCEA GCSE 化学常引用工业实例。鼓风炉炼铁涉及以下反应：

CaCO₃(s) → CaO(s) + CO₂(g)

This is endothermic (ΔH > 0) and produces a gas, so ΔS > 0. The reaction becomes feasible only at high temperatures (around 900–1000 °C). The Gibbs equation explains why heating is essential.

此反应吸热（ΔH > 0），同时生成气体，因此 ΔS > 0。该反应仅在高温（约 900–1000 °C）下才变得可行。吉布斯方程解释了为何加热是必需的。

Another example is the formation of ammonia in the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Here ΔH < 0 and ΔS < 0 (fewer moles of gas on product side). Feasibility is better at low temperatures, but the rate is too slow. Therefore, a compromise temperature of about 450 °C is used with a catalyst.

另一个例子是哈伯制氨法中的氨合成：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)。此反应 ΔH < 0，ΔS < 0（产物一侧气体摩尔数减少）。低温更有利于可行性，但速率太慢。因此，工业上采用约 450 °C 的折中温度，并使用催化剂。

10. Common Exam Pitfalls | 常见考试误区

Students often lose marks by:

同学们常因以下原因失分：

Forgetting to convert °C to K. | 忘记将摄氏度转换为开尔文。
Using ΔS in J K⁻¹ mol⁻¹ with ΔH in kJ mol⁻¹ without conversion. | 在计算时未转换单位，直接混合使用 J 和 kJ。
Assuming a negative ΔG means the reaction is fast. | 认为 ΔG 为负就意味着反应速率快。
Incorrectly stating that ΔG must be zero for a reaction to occur. | 错误地认为 ΔG 必须为零才能发生反应。
Not multiplying ΔS by T before subtracting from ΔH. | 未将 ΔS 与 T 相乘就直接从 ΔH 中减去。

To avoid these, always follow a clear method: list your values, check units, apply the equation, and interpret the sign.

为了避免这些错误，请始终遵循清晰的解题步骤：列出数值，检查单位，代入方程，再解释正负号的含义。

11. Practice Calculation with Unit Conversion | 包含单位转换的练习计算

Calculate ΔG at 25 °C for a reaction with ΔH = –92.4 kJ mol⁻¹ and ΔS = –198.3 J K⁻¹ mol⁻¹. Is the reaction feasible at room temperature?

计算 25 °C 下某反应的 ΔG，已知 ΔH = –92.4 kJ mol⁻¹、ΔS = –198.3 J K⁻¹ mol⁻¹。该反应在室温下是否可行？

Solution:

解答：

T = 25 + 273 = 298 K. Convert ΔS: –198.3 J K⁻¹ mol⁻¹ = –0.1983 kJ K⁻¹ mol⁻¹.

T = 25 + 273 = 298 K。转换 ΔS：–198.3 J K⁻¹ mol⁻¹ = –0.1983 kJ K⁻¹ mol⁻¹。

ΔG = –92.4 – (298 × –0.1983) = –92.4 – (–59.1) = –33.3 kJ mol⁻¹.

ΔG 为负值，反应在室温下可行。但请注意，由于 ΔS 为负，升温会使 ΔG 变得不那么负，并最终变为正。你可以进一步计算当 T > ΔH / ΔS 时，反应不再可行。

This illustrates how a reaction feasible at room temperature can become non-feasible at higher temperatures because of a negative entropy change.

这说明了由于熵变为负，一个在室温下可行的反应在更高温度下可能变为不可行。

12. Summary and Key Takeaways | 总结与核心要点

Gibbs free energy combines enthalpy, entropy, and temperature into a single criterion for feasibility: ΔG = ΔH – TΔS. A negative ΔG means the reaction is feasible; a positive ΔG means it is not. Temperature plays a crucial role, especially when ΔS is large. Always check your units, convert °C to K, and do not confuse feasibility with rate. Understanding these principles will help you tackle GCSE CCEA Chemistry questions on energy changes and equilibria with confidence.

吉布斯自由能将焓、熵和温度结合为一个衡量可行性的单一判据：ΔG = ΔH – TΔS。ΔG 为负表示反应可行；ΔG 为正表示不可行。温度起着关键作用，尤其是当 ΔS 数值较大时。务必检查单位，将 °C 转换为 K，切勿混淆可行性概念与反应速率概念。理解这些原理将帮助你自信地应对 GCSE CCEA 化学中关于能量变化和平衡的考题。

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Gibbs Free Energy for GCSE CCEA Chemistry | GCSE CCEA 化学：吉布斯自由能 考点精讲