The enthalpies of formation of gaseous ${\mathrm{N}}_2\mathrm{O}$ and NO at 298 K are 82.0 and 90.0KJ mol^-1 respectively. The enthalpy change of the reaction ${\mathrm{N}}_2\mathrm{O}\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2\mathrm{NO}\left(\mathrm{g}\right)$ is 

$$\begin{gathered} {\mathrm{N}}_2+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{N}}_2\mathrm{O}, ∆{\mathrm{H}}_1=82.0 \mathrm{kJ}/\mathrm{mol}...\left(\mathrm{i}\right) \\ \frac{1}{2}{\mathrm{N}}_2+\frac{1}{2}{\mathrm{O}}_2\to \mathrm{NO}, ∆{\mathrm{H}}_2=90.0 \mathrm{kJ}/\mathrm{mol}...\left(\mathrm{ii}\right) \end{gathered}$$

Reverse equation (i), multiply equation (ii) by 2. Then subtract (ii) from (i).

$$\begin{gathered} \left(\frac{1}{2}{\mathrm{N}}_2+\frac{1}{2}{\mathrm{O}}_2\to \mathrm{NO}, ∆{\mathrm{H}}_2=90.0 \mathrm{kJ}/\mathrm{mol}...\left(\mathrm{ii}\right)\right)\times 2 \\ {\mathrm{N}}_2\mathrm{O}\to {\mathrm{N}}_2+\frac{1}{2}{\mathrm{O}}_2, ∆{\mathrm{H}}_1=-82.0 \mathrm{kJ}/\mathrm{mol}...\left(\mathrm{i}\right) \\ {\mathrm{N}}_2+{\mathrm{O}}_2\to 2\mathrm{NO}, ∆{\mathrm{H}}_2=180.0 \mathrm{kJ}/\mathrm{mol}...\left(\mathrm{ii}\right) \\ \overline{{\mathrm{N}}_2\mathrm{O}+\frac{1}{2}{\mathrm{O}}_2\to 2\mathrm{NO}, ∆\mathrm{H}=98 \mathrm{kJ}/\mathrm{mol}} \end{gathered}$$