Consider the two reactions.

$2N{H}_3(g)+3N_{2}O(g)4N_2(g)+3H_{2}O(l)H^{*}=-1010kJ$

$4N{H}_3(g)+3O_2(g)2N_2(g)+6H_{2}O(l)H^{*}=1531kJ$

Using these two reactions, calculate and enter the enthalpy change for the reaction below.

$N_2(g)+\frac{1}{2}{O}_2(g)N_{2}O(g)$

$H^{*}=$