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Thermochemistry

25min Part 2 / Ch3 / Lesson 2
Prerequisites: 1-2-3

Objectives

  • Understand enthalpy of reaction (reaction heat)
  • Write thermochemical equations
  • Apply Hess's law in calculations

Reaction Heat and Enthalpy

Heat changes in chemical reactions:

  • Exothermic reaction: releases heat → ΔH<0\Delta H < 0
  • Endothermic reaction: absorbs heat → ΔH>0\Delta H > 0

ΔH\Delta H: enthalpy change of reaction (unit: kJ/mol)

ExothermicEnthalpy HReactantsProductsΔH < 0↓ heat outEndothermicEnthalpy HReactantsProductsΔH > 0↑ heat in
Energy diagrams for exothermic and endothermic reactions

Thermochemical Equations

A chemical equation with ΔH\Delta H included. State symbols (g, l, s, aq) must be specified.

C(s)+O2(g)→CO2(g)   ΔH=−394 kJ/mol\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}\ \ \ \Delta H = -394\ \mathrm{kJ/mol}

H2(g)+12O2(g)→H2O(l)   ΔH=−286 kJ/mol\mathrm{H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)}\ \ \ \Delta H = -286\ \mathrm{kJ/mol}

In thermochemical equations, fractional coefficients are allowed (they correspond to moles). The ΔH\Delta H value corresponds to the coefficients as written.

Hess’s Law

Hess’s law: regardless of the reaction path, if the initial and final states are the same, the total enthalpy change is the same.

→ You can add and subtract thermochemical equations to find unknown ΔH\Delta H values.

ABCΔH = ?ΔH₁ΔH₂ΔH = ΔH₁ + ΔH₂
Hess's law concept

Worked Example

Example: Hess's law

Find the enthalpy of combustion of CO using these two equations:

(1) C(s)+O2(g)→CO2(g)\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}, ΔH1=−394 kJ/mol\Delta H_1 = -394\ \mathrm{kJ/mol}

(2) C(s)+12O2(g)→CO(g)\mathrm{C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)}, ΔH2=−111 kJ/mol\Delta H_2 = -111\ \mathrm{kJ/mol}

Target: CO(g)+12O2(g)→CO2(g)\mathrm{CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)}, ΔH=?\Delta H = ?

Subtract (2) from (1):

The C(s) cancels, leaving: CO(g)+12O2(g)→CO2(g)\mathrm{CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)}

ΔH=ΔH1−ΔH2=−394−(−111)=−283 kJ/mol\Delta H = \Delta H_1 - \Delta H_2 = -394 - (-111) = -283\ \mathrm{kJ/mol}

Check Your Understanding

Q1 For an exothermic reaction, ΔH is:

Q2 Why does Hess's law work?

Q3 C(s) + O₂(g) → CO₂(g), ΔH = -394 kJ/mol. Heat released when 2 mol CO₂ forms?

Exercises

Q1. Use the following thermochemical equations to find the enthalpy of formation of H2O(l)\mathrm{H_2O(l)}, ΔHf\Delta H_f.

(1) H2(g)+12O2(g)→H2O(g)\mathrm{H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)}, ΔH1=−242 kJ/mol\Delta H_1 = -242\ \mathrm{kJ/mol}

(2) H2O(g)→H2O(l)\mathrm{H_2O(g) \rightarrow H_2O(l)}, ΔH2=−44 kJ/mol\Delta H_2 = -44\ \mathrm{kJ/mol}

Solution

Add equations (1) + (2):

H2(g)+12O2(g)→H2O(l)\mathrm{H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)}

ΔHf=−242+(−44)=−286 kJ/mol\Delta H_f = -242 + (-44) = -286\ \mathrm{kJ/mol}