Answer to Question #96558 in General Chemistry for Simon

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

A + 2B =C.......1

E + F = 2C + D........2

1/2 E = G + 2B........3

According to hess law if we do (equation no 1 × 2) - ( equation no 2) + (equation no 3)

In the result we will reach at 2A + D = F + 2G

So ∆rH of these reaction will be -636.2 KJ mol^-1