# Given the following reactions the enthalpy of reaction for 2N2O(g) --> 2NO(g) +N2(g) is __________ kJ? N2 (g) + O2(g) --> 2NO(g) ΔH = +180.7 kJ 2N2O(g) --> O2 (g) + 2N2(g) ΔH = -163.2 kJ

May 26, 2017

17.5 KJ

#### Explanation:

In all this kind of problems the first step is to balance the equations, in this case this is a very simple one, because if you add the two equations without any change you would obtain the asked one.

(1) ${N}_{2 \left(g\right)} + {O}_{2 \left(g\right)} \rightarrow 2 N {O}_{\left(g\right)} \Delta {H}_{1} = + \text{180.7 kJ}$

(2) $2 {N}_{2} {O}_{\left(g\right)} \rightarrow {O}_{2 \left(g\right)} + 2 {N}_{2 \left(g\right)} \Delta {H}_{2} = - \text{163.7 kJ}$

(3) $2 {N}_{2} {O}_{\left(g\right)} \rightarrow 2 N {O}_{\left(g\right)} + {N}_{2 \left(g\right)} \Delta {H}_{2} = \text{? kJ}$

then $\left(1\right) + \left(2\right) = \left(3\right)$

as $\Delta {H}_{1} + \Delta {H}_{2} = \Delta {H}_{3}$

$\therefore \Delta {H}_{3} = + 17.5 K J$