Question

Using the enthalpy changes given below, calculate the enthalpy change for the following equation:

N_2(g) + 2O_2(g) -----> 2 NO₂

using the following equations:

2NO -----> N₂ + O₂ delta H + -180.0 kJ

NO₂ -----> NO + 1/2 O₂ delta H= 112.0 kJ

Answer 1

The given reactions are
$2NO \rightarrow N2 + O2 \: \: \: \delta H = -180.0 kJ \\ NO2 \rightarrow NO + 1/2 O2 \: \: \: \delta H= 112.0 kJ$

Reverse above reactions
$N2 + O2 \rightarrow 2NO \: \: \: \delta H = 180.0 kJ......(1) \\ NO + 1/2 O2 \rightarrow NO2 \: \: \: \delta H= -112.0 kJ......(2)$
Multiply equation (2) with 2
$2NO + O2 \rightarrow 2NO2 \: \: \: \delta H= -112.0 \times 2 = - 224.0 kJ......(3)$
Add equations (1) and (3)
$N2 +2O2 \rightarrow 2NO2 \delta H = 180.0 - 224.0 = -44 kJ$