Given the standard enthalpy changes for the following two reactions:

Question

Given the standard enthalpy changes for the following two reactions:

(1) N₂(g) + 2O₂(g)2NO₂(g)...... ΔH° = 66.4 kJ

(2) 2N₂O(g)2N₂(g) + O₂(g)......ΔH° = -164.2 kJ

what is the standard enthalpy change for the reaction:

(3) 2N₂O(g) + 3O₂(g)4NO₂(g)......ΔH° = ?

 kJ

Answer 1

You can use Hess' Law to solve this problem.

Target equation is 2N₂O(g) + 3O₂(g) ==> 4NO₂(g)

(1) N₂(g) + 2O₂(g)2NO₂(g)...... ΔH° = 66.4 kJ

(2) 2N₂O(g)2N₂(g) + O₂(g)......ΔH° = -164.2 kJ

Copy eq. (2): 2N₂O(g)2N₂(g) + O₂(g)......ΔH° = -164.2 kJ

Copy eq. (1) and multiply by 2: 2N₂(g) + 4O₂(g)4NO₂(g)...... ΔH° = 132.8 kJ

Add them together

2N₂O(g) + 2N₂(g) + 4O₂(g) ==> 2N₂(g) + O₂(g) + 4NO₂(g) and cancel like items to obtain..

2N₂O(g) + 3O₂(g) ==> 4NO₂(g) TARGET EQUATION

∆H = -164.2 kJ + 132.8 kJ

∆H = -31.4 kJ