Question

When 1 mole of $N_2$ and $1$ mole of $H_2$ is enclosed in $3L$ vessel and the reaction is allowed to attain equilibrium .it is found that at equilibrium there is 'x' mole of $H_2$ The number of moles of ${NH}_3$ formed would be?

• A. $\frac{2x}{3}$
• B. $\frac{2(1+x)}{3}$
• C. $\frac{2(1-x)}{3}$
• D. $\frac{(1+x)}{2}$

Answer 1

The correct option is C $\frac{2(1-x)}{3}$

Solution:- (C) $\frac{2(1-x)}{3}$

Let $y$ is the degree of dissociation.

$\underset{\text{Initially}}{\underset{\text{At equillibrium}}{}}\underset{(1-y)}{\underset{1}{N_2}}+\underset{(1-3y)}{\underset{1}{H_2}}⟶\underset{2y}{\underset{0}{2N{H}_3}}$

Given that, no of moles of $H_2$ at equillibrium is $x$.

$\therefore 1-3y=x$

$\Rightarrow y=\frac{(1-x)}{3}$

No. of moles of $N{H}_3$ at equillibrium $=2y=\frac{2(1-x)}{3}$

Hence the correct answer is (C) $\frac{2(1-x)}{3}$.