Question

$40\%$ of a mixture of $0.2$ mole of $N_2$ and $0.6$ mole of $H_2$ react to give $N{H}_3$ according to the equation,
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ at constant temperature and pressure.
Then the ratio of the final volumn to the initial volumn of gases is as-

• A. $4:5$
• B. $5:4$
• C. $7:10$
• D. $8:5$

Answer 1

The correct option is A $4:5$

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)0.20.60$

At Equilibrium $0.2-x$ $0.6-3x$ $2x$

$40\%$ of $N_2=0.2\times 0.4=0.08$

$40\%$ of $H_2=0.6\times 0.4=0.24$

number of moles of $N_2$ remaining $=0.2-0.8=0.12$

number of moles of $H_2$ remaining $=0.6-0.24=0.36$

number of moles of $N{H}_3$ formed $=0.16$

Total number of moles $=0.12+0.36+0.16=0.64$

$\therefore \frac{FinalVolume}{InitialVolume}=\frac{0.64}{0.80}=\frac{4}{5}$