Answer to Question #327090 in General Chemistry for ;ex

Question

Given the following reactions N2 (g) + O2 (g)  2NO (g) ̇ H = +180.7 kJ 2NO (g) + O2(g)  2NO2 (g) ̇H = -113.1 kJ calculate ̇delta H for the following rxn (in kJ). 4NO (g)  2NO2 (g) + N2 (g)

Answer 1

N2(g) + O2(g) -> 2NO(g)

H = +180.7 kJ

2NO(g) + O2(g) -> 2NO2(g)

H = -113.1 kJ

4NO(g) -> 2NO2(g) + N2(g)

ΔH = H(NO2) - 2H(NO) = -113.1 - 2*180.7 = -491.5 kJ