Question

The ammonia in equilibrium with a $1:3,N_2-H_2$ mixture at $20$ $atm$ and $427^{\circ }C$ amounts to $16\%$ . Calculate $K_c$ for $\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\rightleftharpoons N{H}_3$ .

Answer 1

Answer: 2

For the reaction:
$\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\rightleftharpoons N{H}_{3}131-x3-3x2x$
Total moles at equilibrium state $=1-x+3-3x+2x=4-2x$
Given: $N{H}_3$ is $16\%$ in the equilibrium mixture.
So, $\frac{2x}{4-2x}=0.16$ .
The total pressure given is $20atm$ .
Now, $P_{N{H}_3}=\frac{2x}{4-2x}\times 20=3.2atm$ .
The remaining pressure will be in $3:1$ ratio.
$P_{N_2}=\frac{1-x}{4-2x}\times 20=4.2atm$
$P_{H_2}=\frac{3-3x}{4-2x}\times 20=12.6atm$
$K_P=\frac{P_{{\left(NH\right)}_3}}{\sqrt{P_{N_2}}\times {\left(P_{H_2}\right)}^{3/2}}$
$K_P=\frac{3.2}{\sqrt{4.2}\times {\left(12.6\right)}^{3/2}}=3.49\times {\left(10\right)}^{-2}$
Now, $K_P=K_C{\left(RT\right)}^{\Delta n}$
$\Delta n=1-\left(\frac{1}{2}+\frac{3}{2}\right)=-1$
$K_C=K_P(RT)$
$K_C=3.49\times 10^{-2}\times 0.0821\times 700=2$