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 the aforesaid condition in the end?

• A. $20$ litres ammonia, $25$ litres nitrogen, $15$ litres hydrogen
• B. $20$ litres ammonia, $20$ litres nitrogen, $20$ litres hydrogen
• C. $10$ litres ammonia, $25$ litres nitrogen, $15$ litres hydrogen
• D. $20$ litres ammonia, $10$ litres nitrogen, $30$ litres hydrogen

Answer 1

Answer: (C)

$\underset{\overset{1\,\text{Vol}}{10\,\text{litre}}}{N_2} + \underset{\overset{3\,\text{vol}}{30\,\text{litre}}}{3H_2} \to \underset{\overset{2\,\text{vol}}{20\,\text{litre}}}{2NH_3}$
It is given that only $50\%$ of the expected product is formed
hence only $10$ litres of $NH_3$ is formed
$N_2$ used $= 5$ litres, left $= 30 - 5 = 25$ litres
$H_2$ used $= 15$ litres, left $= 30 -15 = 15$ litres