Chemistry 1: Calculate ΔH°f of Propane C₃H₈(g) Using Hess's Law

Exact Chemistry Midterm Problem

Calculate the standard enthalpy of formation ($\Delta H^\circ_f$) of propane gas:

> $$ 3\text{C}(s, \text{graphite}) + 4\text{H}_2(g) \longrightarrow \text{C}_3\text{H}_8(g) $$

Given:

1. $\text{C}_3\text{H}_8 + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O} \quad \Delta H_1^\circ = -2220.0\text{ kJ}$

2. $\text{C} + \text{O}_2 \to \text{CO}_2 \quad \Delta H_2^\circ = -393.5\text{ kJ}$

3. $\text{H}_2 + \frac{1}{2}\text{O}_2 \to \text{H}_2\text{O} \quad \Delta H_3^\circ = -285.8\text{ kJ}$

Step-by-Step Solution

1. Reverse Eq 1: $3\text{CO}_2 + 4\text{H}_2\text{O} \to \text{C}_3\text{H}_8 + 5\text{O}_2 \implies \Delta H_A = +2220.0\text{ kJ}$
2. Multiply Eq 2 by 3: $3\text{C} + 3\text{O}_2 \to 3\text{CO}_2 \implies \Delta H_B = 3(-393.5) = -1180.5\text{ kJ}$
3. Multiply Eq 3 by 4: $4\text{H}_2 + 2\text{O}_2 \to 4\text{H}_2\text{O} \implies \Delta H_C = 4(-285.8) = -1143.2\text{ kJ}$
4. Sum Enthalpies:

$$ \Delta H^\circ_f = +2220.0 - 1180.5 - 1143.2 = \mathbf{-103.7\text{ kJ/mol}} $$