Question

A mixture of $2$ moles of $N_2$ and $8$ moles of $H_2$ are heated in a $2$ lit vessel, till equilibrium is established. At equilibrium $0.4$ moles of $N_2$ was present. What will be the equilibrium concentration of $H_2$?

• A. $2$ mole/lit
• B. $4$ mole/lit
• C. $1.6$ mole/lit
• D. $1$ mole/lit

Answer 1

Answer: (C)

$1.6$ mole/lit

$H_2+3H_2\rightleftharpoons 2N{H}_3$

$t=0$ $2$ $8$ $-$

t eg. $2.x$ $8-3x$ $2x$

$2-x=0.4$

$\Rightarrow x=1.6$ $\therefore (H_2{)}_{eq}=8-4.8=3.2$

$[H_2{]}_{eq}=\frac{3.2}{2}=1.6$mol/lit