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Chemistry · Ch 5 — Thermodynamics

Standard Enthalpy of Combustion

5.5(a)

Standard Enthalpy of Combustion

Standard Enthalpy of Combustion

Combustion reactions are exothermic — they release heat. They are important in industry, rocketry, and everyday life (for example, the burning of cooking gas or fuel in a vehicle).

Definition: The standard enthalpy of combustion is the enthalpy change when one mole of a substance undergoes complete combustion, with all reactants and products in their standard states at the specified temperature.

Important

The standard enthalpy of combustion is always defined per mole of the substance being burned, and the reaction must be complete combustion (usually to CO₂(g) and H₂O(l) for organic compounds).

Example — Butane: Cooking gas in cylinders contains mostly butane (C₄H₁₀). During complete combustion of one mole of butane, 2658 kJ of heat is released.

C4H10(g)+132O2(g)→4CO2(g)+5H2O(l);ΔcH⊖=−2658.0 kJ mol−1\text{C}_4\text{H}_{10}(g) + \frac{13}{2}\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 5\text{H}_2\text{O}(l); \quad \Delta_c H^\ominus = -2658.0\ \text{kJ mol}^{-1}

Example — Glucose: Combustion of glucose gives out 2802.0 kJ per mole.

C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l);ΔcH⊖=−2802.0 kJ mol−1\text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 6\text{H}_2\text{O}(l); \quad \Delta_c H^\ominus = -2802.0\ \text{kJ mol}^{-1}

Our body generates energy from food by the same overall process as combustion, although the final products are produced after a series of complex biochemical reactions involving enzymes.

Note

The overall chemical equation for the metabolic breakdown of glucose in the body is the same as its combustion equation — the energy yield is the same, even though the pathway is different.

Worked Problem: Calculating ΔfH⊖\Delta_f H^\ominus from ΔcH⊖\Delta_c H^\ominus

Problem 5.9: The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO₂(g) and H₂O(l) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, ΔfH⊖\Delta_f H^\ominus, of benzene. Standard enthalpies of formation of CO₂(g) and H₂O(l) are –393.5 kJ mol⁻¹ and –285.83 kJ mol⁻¹ respectively.

Solution:

The formation reaction of benzene is:

6C(graphite)+3H2(g)→C6H6(l);ΔfH⊖=?... (i)6\text{C(graphite)} + 3\text{H}_2(g) \rightarrow \text{C}_6\text{H}_6(l); \quad \Delta_f H^\ominus = ? \quad \text{... (i)}

The enthalpy of combustion of 1 mol of benzene is:

C6H6(l)+152O2(g)→6CO2(g)+3H2O(l);ΔcH⊖=−3267 kJ mol−1... (ii)\text{C}_6\text{H}_6(l) + \frac{15}{2}\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 3\text{H}_2\text{O}(l); \quad \Delta_c H^\ominus = -3267\ \text{kJ mol}^{-1} \quad \text{... (ii)}

The enthalpy of formation of 1 mol of CO₂(g):

C(graphite)+O2(g)→CO2(g);ΔfH⊖=−393.5 kJ mol−1... (iii)\text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g); \quad \Delta_f H^\ominus = -393.5\ \text{kJ mol}^{-1} \quad \text{... (iii)}

The enthalpy of formation of 1 mol of H₂O(l):

H2(g)+12O2(g)→H2O(l);ΔfH⊖=−285.83 kJ mol−1... (iv)\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l); \quad \Delta_f H^\ominus = -285.83\ \text{kJ mol}^{-1} \quad \text{... (iv)}

Multiply equation (iii) by 6 and equation (iv) by 3:

6C(graphite)+6O2(g)→6CO2(g);ΔfH⊖=−2361 kJ mol−16\text{C(graphite)} + 6\text{O}_2(g) \rightarrow 6\text{CO}_2(g); \quad \Delta_f H^\ominus = -2361\ \text{kJ mol}^{-1}

3H2(g)+32O2(g)→3H2O(l);ΔfH⊖=−857.49 kJ mol−13\text{H}_2(g) + \frac{3}{2}\text{O}_2(g) \rightarrow 3\text{H}_2\text{O}(l); \quad \Delta_f H^\ominus = -857.49\ \text{kJ mol}^{-1}

Summing these two equations:

6C(graphite)+3H2(g)+152O2(g)→6CO2(g)+3H2O(l);ΔrH⊖=−3218.49 kJ mol−1... (v)6\text{C(graphite)} + 3\text{H}_2(g) + \frac{15}{2}\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 3\text{H}_2\text{O}(l); \quad \Delta_r H^\ominus = -3218.49\ \text{kJ mol}^{-1} \quad \text{... (v)}

Reverse equation (ii):

6CO2(g)+3H2O(l)→C6H6(l)+152O2(g);ΔH⊖=+3267.0 kJ mol−1... (vi)6\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \rightarrow \text{C}_6\text{H}_6(l) + \frac{15}{2}\text{O}_2(g); \quad \Delta H^\ominus = +3267.0\ \text{kJ mol}^{-1} \quad \text{... (vi)}

Adding equations (v) and (vi): …

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