Question

In a mixture of $N_2$ and $H_2$ in the ratio 1:3 at 30 atm and ${300}^{0}C$, the % of ${NH}_3$ at equilibrium is 17.8.
Calculate $K_P$ for ${N_2+{3H}_2\rightleftharpoons 2NH}_3$.

• A. $K_P=2.7\times {10}^{-2}{atm}^{-2}$
• B. $K_P=90.11\times {10}^{-4}{atm}^{-2}$
• C. $K_P=2.42\times {10}^{-3}{atm}^{-2}$
• D. $K_P=7.29\times {10}^{-4}{atm}^{-2}$

Answer 1

The correct option is D $K_P=7.29\times {10}^{-4}{atm}^{-2}$

${N_2+{3H}_2\rightleftharpoons 2NH}_3$
1 3 0 Moles before dissociation
$(1-x)$ $(3-3x)$ $2x$ Moles after dissociation

$\therefore Totalmolesatequilibrium=4-2x$
and$Molesof{NH}_{3}atequilibrium=2x$
$\because$ % ${NH}_3$ in mixture $=17.8$
$\therefore \frac{2x}{4-2x}=\frac{17.8}{100}$
$\therefore x=0.302$
$K_P=\frac{{\left(n_{{NH}_3}\right)}^2}{n_{N_2}\times {\left(n_{H_2}\right)}^3}\times {\left(\frac{P}{\sum n}\right)}^{△n}$
$=\frac{4x^2}{(1-x){(3-3x)}^3}\times {\left(\frac{P}{4-2x}\right)}^{-2}$
$=\frac{4x^2({4-2x)}^2}{P^2(1-x){(3-3x)}^3}$
Now $P=30atm$ and $x=0.302$
$K_P=7.29\times {10}^{-4}{atm}^{-2}$