Chemical Thermodynamics: Edge Cases and Subtle Traps (5)

Question

Calculate $\Delta H$ for the reaction $C(s) + 2H_2(g) \to CH_4(g)$ given:

(i) $C(s) + O_2(g) \to CO_2(g)$, $\Delta H_1 = -393.5$ kJ/mol (ii) $H_2(g) + \tfrac{1}{2}O_2(g) \to H_2O(l)$, $\Delta H_2 = -285.8$ kJ/mol (iii) $CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)$, $\Delta H_3 = -890.4$ kJ/mol

Solution — Step by Step

Why This Works

Hess’s law: $\Delta H$ is a state function. The total enthalpy change of a reaction depends only on initial and final states, not on the path. So we can manipulate intermediate reactions freely — multiply, reverse, add — and as long as they algebraically sum to the target, the corresponding $\Delta H$ values sum the same way.

This is why standard enthalpies of formation work: $\Delta H_{rxn} = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants})$.

Alternative Method

Direct from formation enthalpies: $\Delta H_{rxn} = \Delta H_f^\circ(CH_4) - \Delta H_f^\circ(C) - 2\Delta H_f^\circ(H_2) = \Delta H_f^\circ(CH_4) - 0 - 0$. The given $\Delta H_3$ is essentially the heat of combustion, used here in the reversed form. Same result.

Common Mistake

The five most common Hess’s law traps:

1. Forgetting to flip the sign of $\Delta H$ when reversing a reaction.
2. Multiplying the species but not $\Delta H$ when scaling.
3. Mixing units (kJ vs J).
4. Confusing $\Delta H$ (enthalpy) with $\Delta U$ (internal energy): related by $\Delta H = \Delta U + \Delta n_g RT$ for gas-phase reactions.
5. Counting heat of formation of elements as nonzero: $\Delta H_f^\circ = 0$ for elements in their standard state ($O_2(g)$, $C(graphite)$, $H_2(g)$, etc.). $C(diamond)$ is not standard, so its $\Delta H_f$ is nonzero.

The fix: write the target equation, write each given reaction, plan the algebraic combination on paper, and double-check signs at each step.