Question

In a mixture of $N_2$ and $H_2$ in the ratio of $1:3$ at $64$ atmospheric pressure and ${300}^{o}C$, the percentage of ammonia under equilibrium is $33.33$ by volume. Calculate the equilibrium constant of the reaction using the equation.
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ :

• A. $K_P=0.13\times {10}^{-3}at{m}^{-2}$
• B. $K_P=1.3\times {10}^{-3}at{m}^{-2}$
• C. $K_P=2.6\times {10}^{-5}at{m}^{-2}$
• D. None of these

Answer 1

Answer: (B)

$K_P=1.3\times {10}^{-3}at{m}^{-2}$

As given,
$\underset{t=016atm}{N_2}+\underset{48atm}{3H_2}\rightleftharpoons 2N{H}_3$
$t=\infty 16-x48-3x2x\Rightarrow \frac{2x}{64-2x}=\frac{1}{2}\Rightarrow x=8$
$K_P=\frac{{16}^2}{8(24{)}^3}$
$K_P=1.3\times {10}^{-3}at{m}^{-2}$