Question

question 3 of 15
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° = kj

Explanation:

Step1: Reverse the first reaction

$$2NH_{3}(g)+3N_{2}O(g)\to4N_{2}(g)+3H_{2}O(l)\quad\Delta H^{\circ}=- 1010\ kJ$$
Reversed:
$$4N_{2}(g)+3H_{2}O(l)\to2NH_{3}(g)+3N_{2}O(g)\quad\Delta H^{\circ}=+ 1010\ kJ$$

Step2: Divide the second reaction by 2

$$4NH_{3}(g)+3O_{2}(g)\to2N_{2}(g)+6H_{2}O(l)\quad\Delta H^{\circ}=1531\ kJ$$
Divided by 2:
$$2NH_{3}(g)+\frac{3}{2}O_{2}(g)\to N_{2}(g)+3H_{2}O(l)\quad\Delta H^{\circ}=\frac{1531}{2}\ kJ = 765.5\ kJ$$

Step3: Add the two modified reactions

Reversed first reaction:
$$4N_{2}(g)+3H_{2}O(l)\to2NH_{3}(g)+3N_{2}O(g)\quad\Delta H^{\circ}=+ 1010\ kJ$$
Divided second reaction:
$$2NH_{3}(g)+\frac{3}{2}O_{2}(g)\to N_{2}(g)+3H_{2}O(l)\quad\Delta H^{\circ}=765.5\ kJ$$
Adding them:
$$4N_{2}(g)+3H_{2}O(l)+2NH_{3}(g)+\frac{3}{2}O_{2}(g)\to2NH_{3}(g)+3N_{2}O(g)+N_{2}(g)+3H_{2}O(l)$$
Simplify:
$$3N_{2}(g)+\frac{3}{2}O_{2}(g)\to3N_{2}O(g)$$
Divide by 3:
$$N_{2}(g)+\frac{1}{2}O_{2}(g)\to N_{2}O(g)$$
For the enthalpy change:
$$\Delta H=\frac{1010 + 765.5}{3}\ kJ$$
$$\Delta H=\frac{1775.5}{3}\ kJ\approx592\ kJ$$

Answer:

$592$