The standard enthalpy of formation for ${\mathrm{NH}}_3\left(\mathrm{g}\right)$ is $-46.1\mathrm{kJ}\times {\mathrm{mol}}^{-1}$. Calculate $∆\mathrm{H}°$ for the reaction:

$\frac{1}{2}{\mathrm{N}}_2+\frac{3}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{NH}}_3\left(\mathrm{g}\right);\mathrm{\Delta H}=-46.2{\mathrm{kJmol}}^{-1}$

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right);\mathrm{\Delta H}=-46.2\times 2=-92.4\mathrm{kJ}$

As a result, the sign of the reverse reaction will be opposite, i.e.,$∆\mathrm{H}$=92.4 kJ.