For the reaction 2CH4 (g) + 3 Cl2 (g) → 2 CHCl3 (l) + 3 H2 (g), ΔH° = -118.6 kJ. ΔH°f = -134.1 kJ/mol for CHCl3 (l). Find ΔH°f for CH4 (g).

Answer 1

2CH4(g) + 3Cl2(g) ==> 2CHCl3(l) + 3H2(g) ... ∆Hº = -118.6 kJ

∆H_f products = 2 mol x -134.1 kJ/mol + 0 = -268.2 kJ

∆H_f reactants = ∆H_f CH₄ + 0 = ∆H_f CH₄

∆Hrxn = -118.6 kJ = ∑products - ∑reactants

-118.6 = -268.2 - ∆H_f CH₄

∆H_f CH₄ = -118.6 + 268.2

∆H_f CH₄ = 149.6 kJ/mol