Question

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

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

Answer 1

The correct option is A $4:5$

$N_2+3H_2\rightleftharpoons 2N{H}_3$

I 0.2 0.6 0

C -x -3x +2x

E (0.2-x)(0.6-3x) +2x

According to question $÷\frac{40}{100}[0.2-x+0.6-3x]=2x$

$0.4[0.8-4x]=2x$

$0.32-1.6x=2x$

$0.32=3.6x$

$x=0.0888$

Final volume $=(0.2-x)+(0.6-3x)+2x$

$=0.2-0.0888+0.6-3\times 0.0888+2\times 0.0888$

$=0.624$

Initial volume $=0.2+0.6=0.8$

Ratio $=\frac{0.624}{0.8}\simeq 0.78\simeq \frac{4}{5}\to \left(A\right)$