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 $\frac{1}{2}N_{2}+\frac{3}{2}H_{2}\rightleftharpoons NH_{3}$ .

Answer 1

For the reaction:
$\frac{1}{2}N_{2}+\frac{3}{2}H_{2}\rightleftharpoons NH_{3}131-x3-3x2x$
Total moles at equilibrium state $=1-x+3-3x+2x=4-2x$
Given: $NH_{3}$ is $16\%$ in the equilibrium mixture.
So, $\frac{2 x}{4 - 2 x}=0.16$ .
The total pressure given is $20atm$ .
Now, $P_{NH_{3}}=\frac{2 x}{4 - 2 x}\times 20=3.2atm$ .
The remaining pressure will be in $3:1$ ratio.
$P_{N_{2}}=\frac{1 - x}{4 - 2 x}\times 20=4.2atm$
$P_{H_{2}}=\frac{3 - 3 x}{4 - 2 x}\times 20=12.6atm$
$K_{P}=\frac{P_{\left(NH\right)_{3}}}{\sqrt{P_{N_{2}}} \times \left(P_{H_{2}}\right)^{3/2}}$
$K_{P}=\frac{3 . 2}{\sqrt{4 . 2} \times \left(12 . 6\right)^{3/2}}=3.49\times \left(10\right)^{- 2}$
Now, $K_{P}=K_{C}\left(RT\right)^{\Delta n}$
$\Delta n=1-\left(\frac{1}{2} + \frac{3}{2}\right)=-1$
$K_{C}=K_{P}\left(\right.RT\left.\right)$
$K_{C}=3.49\times 10^{- 2}\times 0.0821\times 700=2$