Question

In the Haber process:
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$
$30L$ of $H_2$ and $30L$ of $N_2$ were taken for reaction which yielded only $50\%$ of expected product. What will be the composition of the gaseous mixture in the end?

• A. 20 L $N{H}_3$, 25 L $N_2$ and 20 L $H_2$
• B. 10 L $N{H}_3$, 25 L $N_2$ and 15 L $H_2$
• C. 20 L $N{H}_3$, 10 L $N_2$ and 30 L $H_2$
• D. 20 L $N{H}_3$, 25 L $N_2$ and 15 L $H_2$

Answer 1

The correct option is A 20 L $N{H}_3$, 25 L $N_2$ and 20 L $H_2$

For the following reaction

$N_2+3H_2\to 2N{H}_3$

We are given 30 L of $H_2$ and 30 L of $N_2$

From the equation it is clear that

for 3 volume of $H_2$ - 1 volume of $N_2$ required

thus, 30 volume of $H_2$ will require 10 volume of $N_2$

again, from 3 volume of $H_2$ - 2 volume of $N{H}_3$ is formed

thus, from 30 volume of $H_2$ ,20 volume of $N{H}_3$ is formed

so, from the given amount of $H_2$ and $N_2$(i.e 30 L each) it is clear that

10 L of $N_2$ react with 30 L of $H_2$ to give 20 L of $N{H}_3$

but, according to question the yeild of $N{H}_3$ is 50% of the expected i.e 10 L of $N{H}_3$ is produced.

now again from the equation

2 volume of $N{H}_3$ is formed from 1 volume of $N_2$

thus, 10 volume of $N{H}_3$ will be formed from 5 volume of $N_2$

i.e 10 L of $N{H}_3$ will be formed from 5L of $N_2$

2 volume of $N{H}_3$ is formed from 3 volume of $H_2$

thus, 10 volume of $N{H}_3$ will be formed from 15 volume of $H_2$

i.e 10 L of $N{H}_3$ will be formed from 15 L of $H_2$

thus, overall 5L of $N_2$ and 15 L of $H_2$ is used to produce 10 L of $N{H}_3$

hence the remaining amount of $N_2$ = 30 -5 = 25 L

remaining amount of $H_2$ = 30-15 = 15 L

and amount of $N{H}_3$ produced is 10 L

thus option B is correct.