Question

$40$ of a mixture of $0.2mole$ of $N_2$ and $0.6mole$ 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 volume to the initial volume of gases are :

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

Answer 1

Answer: (A)

$4:5$

$N_2+3H_2\rightleftharpoons 2N{H}_3$
Initially at equilibrium $0.20.60$
$(0.2-a)(0.6-3a)2a$
Total mixture is 0.8; 40% of it reacts, i.e., $\frac{0.8\times 40}{100}$ reacts to give $\frac{0.8\times 40}{100}\times \frac{1}{2}$ mole of $N{H}_3$ or $N{H}_3$ formed is 0.16 mole or $2a=0.16$
$\therefore a=0.08$
$\therefore$ Initial mole $=0.8$
Final mole $=(0.2-0.08)+(0.6-0.24)+0.16$
$=0.12+0.36+0.16=0.64$
$\therefore$ Ratio of final to initial mole $=\frac{0.64}{0.8}=0.8=\frac{4}{5}$