Answer to Question #259482 in General Chemistry for sissy

Question

Given the following enthalpy change values for the reactions below:

A + B → 2C ; Δ_rH = -39.2 kJ mol^-1

C + 2D → E + F  ; Δ_rH = -296.8 kJ mol^-1

½ E + G → H ;  Δ_rH = 474.9 kJ mol^-1

Calculate Δ_rH for the following reaction (in kJ mol^-1) using Hess's Law:

A + B + 4D + 4G → 4H + 2F

Answer 1

2A + D → F + 2G

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