Hess’s Law: Two Reactions

Question

Given the standard enthalpy changes for the following two reactions:

(1) N₂(g) + 2O₂(g)N₂O₄(g)...... ΔH° = 9.2 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)2N₂O₄(g)......ΔH° = ?

 _______kJ

Answer 1

Use the given equation to find ∆H for the target equation.

TARGET EQUATION: 2N₂O(g) + 3O₂(g) ==> 2N₂O(g) ... ∆H = ?

Given:

(1) N₂(g) + 2O₂(g)==> N₂O₄(g)...... ΔH° = 9.2 kJ

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

Procedure:

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

Copy (1)x2: 2N₂(g) + 4O₂(g)==> 2N₂O₄(g)...... ΔH° = 18.4 kJ

-----------------------------------------------------------------------------------Add them & combine/cancel like terms....

2N₂O(g) + 2N₂(g) + 4O₂(g) ==> 2N₂(g) + O₂(g) + 2N₂O₄(g)

2N₂O(g) + 3O₂(g) ==> 2N₂O₄ = TARGET EQUATION

∆G = -164.2 kJ + 18.4 kJ = -145.8 kJ