# Given... N_2(g) + 3H_2(g) rarr 2NH_3(g) + 92.4*kJ... ..what is DeltaH_f^@ NH_3(g)?

Oct 17, 2016

$\Delta {H}_{f}^{\circ}$ $N {H}_{3} \left(g\right)$ $=$ $- 46.2 \cdot k J \cdot m o {l}^{-} 1$.

#### Explanation:

By definition, $\Delta {H}_{f}^{\circ}$ is the enthalpy change associated with the formation of 1 mole of product from its constituent elements in their standard states at $298 \cdot K$.

I think you have quoted the enthalpy change for the reaction:

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \rightarrow 2 N {H}_{3} \left(g\right)$, which of course is TWICE the enthalpy of formation for $N {H}_{3}$, because this reaction formed TWO moles of ammonia.

Finally, we can write:

$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) \rightarrow N {H}_{3} \left(g\right)$ ;DeltaH_"rxn"^@=DeltaH_f^@"ammonia"=-46.2*kJ*mol^-1.