Question

Given

$N_2\left(g\right)+3H_2\left(g\right)\to 2{NH}_3\left(g\right):△H^{\theta }=92.4kJ{mol}^{-1}$

What is the standard enthalpy of formation of ${NH}_3$ gas?

Answer 1

$N_2\left(g\right)+3H_2\left(g\right)\to 2{NH}_3\left(g\right):△H^{\theta }=92.4kJ{mol}^{-1}$

Enthalpy of formation of $N{H}_3$ means heat released in the formation of 1 mole of $N{H}_3$ .

So, let us divide both sides by 2 in the above reaction,

$\frac{1}{2}$$N_2\left(g\right)+3/2H_2\left(g\right)\to {NH}_3\left(g\right)$

Therefore,

$△H^{\theta }$ = $\frac{-92.4}{2}$

= 46.2 KJ /mol