The enthalpy of formation of methane(g) at constant pressure is -18,500 cal/mol at 27°C. The enthalp — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

The reaction for standard formation of methane is: C(s) + 2$H_2$(g) -> $CH_4$(g). The gaseous mole change is δ n_g = 1 - 2 = -1. δ H = δ E + δ n_g * R * T => δ E = δ H - δ n_g * R * T. Using R = 2 cal/(mol K) and T = 300 K: δ E = -18500 - (-1 * 2 * 300) = -17900 cal/mol. However, the book's answer key is (b) (-17,300 cal), which corresponds to taking δ n_g = -2. If δ n_g = -2, then δ E = -18500 - (-2 * 2 * 300) = -17300 cal. This occurs if carbon is incorrectly assumed to be in the gaseous phase.