Question

Given the data N2(g) + O2(g) → 2 NO(g) ΔH = +180.7 kJ 2 NO(g) + O2(g) → 2 NO2(g) ΔH = −113.1 kJ 2 N2O(g) → 2 N2(g) + O2(g) ΔH = −163.2 kJ use Hess's law to calculate ΔH for the reaction N2O(g) + NO2(g) → 3 NO(g).

Answer 1

N2(g) + O2(g) → 2 NO(g) ΔH = +180.7 kJ

2 NO(g) + O2(g) → 2 NO2(g) ΔH = −113.1 kJ

2 N2O(g) → 2 N2(g) + O2(g) ΔH = −163.2 kJ

get

N2O(g) + NO2(g) → 3 NO(g)

2 N2O(g) → 2 N2(g) + O2(g) ΔH = −163.2 kJ wont move since we need N2O in the left

invert (2) since we need NO2 ni the other side

2 NO2(g) → 2 NO(g) + O2(g) ΔH = +113.1 kJ

We have

2 N2O(g) → 2 N2(g) + O2(g) ΔH = −163.2 kJ

2 NO2(g) → 2 NO(g) + O2(g) ΔH = +113.1 kJ

N2(g) + O2(g) → 2 NO(g) ΔH = +180.7 kJ

Multilpy (3) by 2 so we can cancel N2

2 N2O(g) → 2 N2(g) + O2(g) ΔH = −163.2 kJ

2 NO2(g) → 2 NO(g) + O2(g) ΔH = +113.1 kJ

2N2(g) + 2O2(g) → 4 NO(g) ΔH = 2*180.7 = 361.4 kJ

add all

2 N2O(g) + 2 NO2(g) + 2N2(g) + 2O2(g)  → 2 N2(g) + O2(g) +  2 NO(g) + O2(g) +  4 NO(g)

Cancel common terms

2 N2O(g) + 2 NO2(g) → 6NO(g) ΔH = −163.2 +113.1 + 361.4 = 311.3 kJ

divide by 2 so we get the equation we need

N2O(g) + NO2(g) → NO(g) ΔH = 311.3/2 kJ = 155.65 kJ

then

ΔH = 155.65 kJ