Question

For the reaction $2N{H}_3\left(g\right)\to N_2\left(g\right)+3H_2\left(g\right)$, what is the $\%$ of $N{H}_3$ converted if the mixture diffuses twice as fast as that of $S{O}_2$ under similar conditions :

• A. $3.125\%$
• B. $31.25\%$
• C. $6.25\%$
• D. $62.5\%$

Answer 1

Answer: (C)

$6.25\%$

	$N{H}_3$	$N_2$	$H_2$
Initial moles	1	0	0
Equilibrium moles	$1-2x$	$x$	$3x$
Equilibrium mole fraction	$\frac{1-2x}{1+2x}$	$\frac{x}{1+2x}$	$\frac{3x}{1+2x}$

The rate of diffusion is inversely proportional to the square root of molar mass.
The mixture diffuses twice as fast as that of $S{O}_2$ under similar conditions.
$\frac{r}{r^{\prime }}=\sqrt{\frac{M^{\prime }}{M}}$
$\frac{2}{1}=\sqrt{\frac{64}{M}}$
$M=16$ $g/mol$
$16=\frac{17(1-2x)+28\left(x\right)+2\left(3x\right)}{1+2x}$
$16+32x=17-34x+28x+6x$
$32x=1$
$x=0.03125$
The percentage of ammonia converted
$\frac{2x}{1}\times 100=\frac{2\times 0.03125}{1}\times 100=6.25$%