Question

When $1$ mole of $N_{2}$ and $1$ mole of $H_{2}$ is enclosed in $5L$ 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. $\frac{2 x}{3}$
• B. $\frac{2 \left(1 + x\right)}{3}$
• C. $\frac{2 \left(1 - x\right)}{3}$
• D. $\frac{\left(1 - x\right)}{2}$

Answer 1

Answer: (C)

$N_{2}+3H_{2}\rightleftharpoons2NH_{3}110$
$t=0\Rightarrow $ $1$ mol $1$ mol $0$
Eq. $\Rightarrow \left(1 - y\right)\left(1 - 3 y\right)2y$
Given $(1-3 y)=x\left(\right.$ moles of $\left.H _{2}\right)$
$y=\frac{1 - x}{3}$
$\therefore $ Moles of $\left(NH\right)_{3}=2y=\frac{2 \left(1 - x\right)}{3}$