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) + ½o₂(g) → n₂o(g)

δh° =

Explanation:

Step1: Manipulate the given reactions

Let the first reaction be \(2NH_{3}(g)+3N_{2}O(g)\to4N_{2}(g)+3H_{2}O(l)\), \(\Delta H_{1}=- 1010\space kJ\)
Let the second reaction be \(4NH_{3}(g)+3O_{2}(g)\to2N_{2}(g)+6H_{2}O(l)\), \(\Delta H_{2}=1531\space kJ\)

We want to find \(\Delta H\) for \(N_{2}(g)+\frac{1}{2}O_{2}(g)\to N_{2}O(g)\)

First, reverse the first reaction: \(4N_{2}(g)+3H_{2}O(l)\to2NH_{3}(g)+3N_{2}O(g)\), \(\Delta H_{1}^{'}=1010\space kJ\)

Multiply this reversed reaction by \(\frac{1}{3}\): \(\frac{4}{3}N_{2}(g)+H_{2}O(l)\to\frac{2}{3}NH_{3}(g)+N_{2}O(g)\), \(\Delta H_{1}^{''}=\frac{1010}{3}\space kJ\)

Second, take the second reaction \(4NH_{3}(g)+3O_{2}(g)\to2N_{2}(g)+6H_{2}O(l)\), \(\Delta H_{2}=1531\space kJ\) and multiply it by \(\frac{1}{6}\): \(\frac{2}{3}NH_{3}(g)+\frac{1}{2}O_{2}(g)\to\frac{1}{3}N_{2}(g)+H_{2}O(l)\), \(\Delta H_{2}^{'}=\frac{1531}{6}\space kJ\)

Step2: Add the manipulated reactions

Add \(\frac{4}{3}N_{2}(g)+H_{2}O(l)\to\frac{2}{3}NH_{3}(g)+N_{2}O(g)\) (\(\Delta H_{1}^{''}=\frac{1010}{3}\space kJ\)) and \(\frac{2}{3}NH_{3}(g)+\frac{1}{2}O_{2}(g)\to\frac{1}{3}N_{2}(g)+H_{2}O(l)\) (\(\Delta H_{2}^{'}=\frac{1531}{6}\space kJ\))

The \(\frac{2}{3}NH_{3}(g)\) and \(H_{2}O(l)\) cancel out.

We get \(N_{2}(g)+\frac{1}{2}O_{2}(g)\to N_{2}O(g)\)

\(\Delta H=\frac{1010}{3}+\frac{1531}{6}=\frac{2020 + 1531}{6}=\frac{3551}{6}\approx592\space kJ\)

Answer:

\(592\space kJ\)