Question

In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture under these condition in the end.

Answer 1

Initial, $3H_2+N_2⟶2N{H}_3$

$30L$ $30L$

$30L$ $10L$ $⟶\frac{2}{3}\times 30=20$ {vol.$\propto$mol, PV=nRT}

$N{H}_3⟶50$%$\Rightarrow 20\times \frac{50}{100}=10L$ $N{H}_3$

$N_2=\frac{10}{2}=5L$ left=$2.5L$

$H_2$ left=$30-\frac{10}{2}\times 3=15L$

$N{H}_3=\frac{1}{2}\times 20=10L$ $N{H}_3$