Question

In a mixture of $N_2$ and $H_2$ initially in a mole ration of 1:3 at 30 atm and ${300}^{o}C,$ the percentage of ammonia by volume under the equilibrium is 17.8. Calculate the equilibrium constant $\left(K_P\right)$ of the mixture, for the reaction, $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

Answer 1

Let the initial moles of $N_2$ and $H_2$ be $1$ and $3.$

$N_2+3H_2⟶2N{H}_3$

Initial $1$ $3$ $0$

at Eqb. $1-x$ $3-3x$ $2x$

$\%$ of volume is same as $\%$ by mole $N{H}_3$

$\frac{2x}{4-2x}=0.178$

$x=\frac{4\times 0.178}{2+(2\times 0.178)}$

$x=0.302$

mole fraction of $H_2$ at eqb.$=\frac{3-3x}{4-2x}=0.6165$

Mole fraction of $N_2$ at eqb.$=1-0.6165=0.178=0.2055$

$K_p=(X_{N{H}_3}.P_T{)}^2/(X_{N_2}\left)\right(X_{H_2}.P_T{)}^3$

$=\frac{(0.178\times 30{)}^2}{(0.2055\times 30)(0.6165\times 30{)}^3}$

$=7.31\times {10}^{-4}at{m}^{-2}$