Question

Given the standard enthalpy changes for the following two reactions:

Answer 1

Answer:

The answer is -297 kJ

Explanation:

The enthalpy change is state function and can be calculated from Hess' s Law

The two reactions can be combined to obtain the third equation to calculate its enthalpy change as the enthalpy change depends upon only the initial and final state irrespective of path taken.

(1) $N_{2}(g) + 2O_{2}(g)--->2NO_{2}(g)$...... ΔH° = 66.4 kJ

(2) $2N_{2}O(g)--->2N_{2}(g) + O_{2}(g)$......ΔH° = -164.2 kJ

The required equation is:

(3) $2N_{2}O(g) + 3O_{2}(g)--->4NO_{2}(g)$......ΔH° = _________?

We will multiply the equation (1) with "2" and then will subtract it from equation "1"

(2) $2N_{2}O(g)--->2N_{2}(g) + O_{2}(g)$......ΔH° = -164.2 kJ

-2[X(1) $N_{2}(g) + 2O_{2}(g)--->2NO_{2}(g)$...... ΔH° = 66.4 kJ ]

......................................................................................................................................

$2N_{2}O(g) + 3O_{2}(g)--->4NO_{2}(g)$

ΔH° = ΔH°(eq 2) - 2ΔH°(eq 1) = -164.2 -(2X66.4) = -297 kJ