From the given reaction, $\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2…
From the given reaction,
$\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})} \Delta \mathrm{H}=-92 \cdot 6 \mathrm{~kJ}$, the enthalpy of formation of $\mathrm{NH}_{3}$ is
$-92 \cdot 6 \mathrm{~kJ}$
$-138 \cdot 9 \mathrm{k} \mathrm{J}$
$-185 \cdot 2 \mathrm{~kJ}$
$-46 \cdot 3 \mathrm{~kJ}$
Solution
$\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})} \quad \Delta \mathrm{H}=-92.6 \mathrm{~kJ}$ $\therefore$ Enthalpy of formation of $\mathrm{NH}_{3}=\frac{-92.6}{2}=-46.3 \mathrm{~kJ}$