Question

Given the following enthalpy change values for the reactions below: A + 2B → C ; ΔrH = -228.3 kJ mol-1, E + F → 2C + D ; ΔrH = 493.9 kJ mol-1, ½ E → G + 2B ; ΔrH = 314.3 kJ mol-1. Calculate ΔrH for the following reaction (in kJ mol-1) using Hess's Law: 2A + D → F + 2G

Answer 1

2A + D → F + 2G

∆H= (-228.3)+493.9 - 314.3 = -48.7 kJ/mol