Question

consider the two reactions.
2nh₃(g) + 3n₂o(g) → 4n₂(g) + 3h₂o(l) δh° = -1010 kj
4nh₃(g) + 3o₂(g) → 2n₂(g) + 6h₂o(l) δh° = 1531 kj
using these two reactions, calculate and enter the enthalpy change for the reaction below.
n₂(g) + ½ o₂(g) → n₂o(g)
δh° = kj

Explanation:

Step1: Reverse the first reaction

$$2NH_3(g)+3N_2O(g)\to4N_2(g)+3H_2O(l)\quad\Delta H^{\circ}=- 1010\ \text{kJ}$$
Reversed: $$4N_2(g)+3H_2O(l)\to2NH_3(g)+3N_2O(g)\quad\Delta H^{\circ}=+1010\ \text{kJ}$$

Step2: Divide the second reaction by 2

$$4NH_3(g)+3O_2(g)\to2N_2(g)+6H_2O(l)\quad\Delta H^{\circ}=-1531\ \text{kJ}$$
Divided by 2: $$2NH_3(g)+\frac{3}{2}O_2(g)\to N_2(g)+3H_2O(l)\quad\Delta H^{\circ}=-\frac{1531}{2}\ \text{kJ}=-765.5\ \text{kJ}$$

Step3: Add the two modified reactions

Reaction 1 (reversed): $$4N_2(g)+3H_2O(l)\to2NH_3(g)+3N_2O(g)\quad\Delta H^{\circ}=+1010\ \text{kJ}$$
Reaction 2 (divided by 2): $$2NH_3(g)+\frac{3}{2}O_2(g)\to N_2(g)+3H_2O(l)\quad\Delta H^{\circ}=-765.5\ \text{kJ}$$
Adding them:
$$4N_2(g)+3H_2O(l)+2NH_3(g)+\frac{3}{2}O_2(g)\to2NH_3(g)+3N_2O(g)+N_2(g)+3H_2O(l)$$
Cancel out common terms ($2NH_3(g)$ and $3H_2O(l)$ on both sides):
$$3N_2(g)+\frac{3}{2}O_2(g)\to3N_2O(g)$$
Divide the entire reaction by 3:
$$N_2(g)+\frac{1}{2}O_2(g)\to N_2O(g)$$
For the enthalpy change:
$$\Delta H^{\circ}=\frac{1010 - 765.5}{3}\ \text{kJ}$$
$$\Delta H^{\circ}=\frac{244.5}{3}\ \text{kJ}=81.5\ \text{kJ}$$

Answer:

$81.5$