Using Enthalpies of Formation to Calculate Reaction Enthalpy | 利用生成焓变计算反应焓变

📚 Using Enthalpies of Formation to Calculate Reaction Enthalpy | 利用生成焓变计算反应焓变

In A-Level Chemistry, thermochemistry often requires calculating the enthalpy change of a reaction that is difficult or impossible to measure directly. The standard molar enthalpy of formation provides a reliable and systematic route to such calculations, based on Hess's Law and the fact that enthalpy is a state function.

在 A-Level 化学中，热化学常常需要我们计算那些难以甚至无法直接测定的反应焓变。标准摩尔生成焓为这类计算提供了一条可靠且系统的途径，其依据是赫斯定律以及焓是状态函数这一重要事实。

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1. What Is Standard Molar Enthalpy of Formation? | 什么是标准摩尔生成焓？

The standard molar enthalpy of formation, symbolised as ΔH^°_f, is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions, usually 298 K and 100 kPa. For example, the formation of carbon dioxide from graphite and oxygen can be written as:

标准摩尔生成焓用符号 ΔH^°_f 表示，是指在标准条件（通常为 298 K 和 100 kPa）下，由处于标准态的组成元素单质生成 1 mol 化合物时的焓变。例如，由石墨和氧气生成二氧化碳可写为：

C(s, graphite) + O₂(g) → CO₂(g)　ΔH^° = −393.5 kJ mol⁻¹

This enthalpy change is exactly the standard molar enthalpy of formation of CO₂(g). Notice that the reactants must be elements in their most stable forms under standard conditions, not just any form of the element.

这个焓变正是 CO₂(g) 的标准摩尔生成焓。注意，反应物必须是元素在标准状态下最稳定的形态，而不是任意形态。

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2. The Hess's Law Foundation | 赫斯定律基础

Hess's Law states that the overall enthalpy change for a reaction is independent of the route taken, provided the initial and final states are the same. Since enthalpy is a state function, we can imagine a reaction proceeding by an alternative, artificial route: first breaking all reactants down into their constituent elements, then forming all products from those elements.

赫斯定律指出，只要始态和终态相同，反应的总焓变与反应路径无关。由于焓是状态函数，我们可以设想反应经由另一条人为路径进行：首先将所有反应物分解为其组成元素，然后再由这些元素生成所有产物。

For a general reaction, the enthalpy of the first step is the negative of the sum of the enthalpies of formation of the reactants, and the enthalpy of the second step is the sum of the enthalpies of formation of the products. Adding these gives the enthalpy change of the overall reaction.

对于一般反应而言，第一步的焓变等于反应物生成焓之和的相反数；第二步的焓变等于产物生成焓之和。将两步相加，即可得到总反应的焓变。

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3. The General Equation | 通用计算公式

From the Hess's Law cycle above, the standard enthalpy change of reaction can be expressed using standard enthalpies of formation:

由上述赫斯循环可得到，反应的标准焓变可用标准生成焓表示如下：

ΔH^°_rxn = Σ ΔH^°_f(products) − Σ ΔH^°_f(reactants)

Each ΔH^°_f value must be multiplied by the stoichiometric coefficient of the corresponding substance in the balanced equation. The symbol Σ means 'sum of'. This formula works for reactions at standard conditions and is one of the most frequently used equations in A-Level thermochemistry.

每个 ΔH^°_f 值都必须乘以该物质在配平方程式中的化学计量数。符号Σ表示'求和'。该公式适用于标准条件下的反应，是 A-Level 热化学中最常用的公式之一。

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4. Step-by-Step Calculation Method | 逐步计算法

To apply the formula correctly, follow these steps:

要正确应用这个公式，请按以下步骤操作：

• Write and balance the chemical equation. Pay close attention to state symbols because different states have different ΔH^°_f values.

  写出并配平化学方程式。务必注意状态符号，因为不同状态的物质具有不同的 ΔH^°_f 值。

• Look up the standard molar enthalpy of formation for each reactant and product from the data table provided.

  从题目给出的数据表中查找每个反应物和产物的标准摩尔生成焓。

• Multiply each ΔH^°_f value by the stoichiometric coefficient of that substance in the balanced equation.

  将每个 ΔH^°_f 值乘以该物质在配平方程式中的化学计量数。

• Sum the product values and the reactant values separately.

  分别求出所有产物生成焓之和以及所有反应物生成焓之和。

• Subtract the reactant sum from the product sum. Include the correct sign and units in your final answer.

  用产物生成焓之和减去反应物生成焓之和。最终答案要包含正确的正负号和单位。

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5. Worked Example 1: Combustion of Ethane | 实例一：乙烷的燃烧

Calculate the standard enthalpy change for the complete combustion of ethane:

计算乙烷完全燃烧的标准焓变：

C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(l)

Use the following standard enthalpies of formation:

使用下列标准生成焓数据：

Substance	ΔH°f / kJ mol−1
C2H6(g)	−84.7
CO2(g)	−393.5
H2O(l)	−285.8
O2(g)	0

Using the general equation:

使用通用公式：

ΔH^°_rxn = [2ΔH^°_f(CO₂,g) + 3ΔH^°_f(H₂O,l)] − [ΔH^°_f(C₂H₆,g) + 7/2 ΔH^°_f(O₂,g)]

= [2 × (−393.5) + 3 × (−285.8)] − [(−84.7) + 0]

= (−787.0 − 857.4) + 84.7 = −1559.7 kJ mol⁻¹

The negative sign indicates that the combustion of ethane is highly exothermic. The value −1559.7 kJ mol⁻¹ is the enthalpy change per mole of ethane burned as written in the equation.

负号表明乙烷燃烧是强烈放热的过程。−1559.7 kJ mol⁻¹ 是按方程式所写，每摩尔乙烷燃烧时的焓变。

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6. Worked Example 2: Formation of Calcium Carbonate | 实例二：碳酸钙的生成

Calculate ΔH^°_rxn for the reaction between calcium oxide and carbon dioxide:

计算氧化钙与二氧化碳反应的标准焓变：

CaO(s) + CO₂(g) → CaCO₃(s)

Given: ΔH^°_f(CaCO₃,s) = −1207 kJ mol⁻¹, ΔH^°_f(CaO,s) = −635 kJ mol⁻¹, ΔH^°_f(CO₂,g) = −393.5 kJ mol⁻¹.

已知：ΔH^°_f(CaCO₃,s) = −1207 kJ mol⁻¹，ΔH^°_f(CaO,s) = −635 kJ mol⁻¹，ΔH^°_f(CO₂,g) = −393.5 kJ mol⁻¹。

ΔH^°_rxn = ΔH^°_f(CaCO₃,s) − [ΔH^°_f(CaO,s) + ΔH^°_f(CO₂,g)]

= (−1207) − [(−635) + (−393.5)] = −1207 + 1028.5 = −178.5 kJ mol⁻¹

This reaction is exothermic, which is consistent with the fact that calcium carbonate is more stable than its constituent oxide and carbon dioxide under standard conditions.

该反应放热，这与碳酸钙在标准条件下比氧化钙和二氧化碳的混合物更稳定的事实相符。

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7. Why Are Element Standard Enthalpies of Formation Zero? | 为什么元素单质的标准生成焓为零？

By definition, the standard enthalpy of formation of an element in its most stable form is zero. This is because forming one mole of an element from itself involves no chemical change, so the enthalpy change is zero. Carbon

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