Question

A reaction is carried out by 1 mole of $N_2$, 3 mole of $H_2$ shows at equilibrium, the mole fraction of $N{H}_3$ is 0.012 at 500 ${}^{o}C$ and 10 atm pressure. $K_p$ is the equilibrium constant. The pressure (in atm) at which mole % of $N{H}_3$ in equilibrium mixture increased to 10.4 is_________.
[Give your answer in closest integer]

Answer 1

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

$130$
$(1-x)(3-3x)2x$

Given, mole fraction of $N{H}_3=0.012$ and $P=10atm$

$\therefore \frac{2x}{(4-2x)}=0.012$ or $x=0.0237$

$K_p=\frac{{\left(n_{N{H}_3}\right)}^2}{n_{N_2}\times {\left(n_{H_2}\right)}^3}\times {\left(\frac{P}{\sum n}\right)}^{\Delta n}$ $(\because \Delta n=-2)$

$=\frac{{\left(2x\right)}^2}{(1-x){(3-3x)}^3}{\left[\frac{P}{4-2x}\right]}^{-2}$

$=\frac{4x^2{(4-2x)}^2}{(1-x){(3-3x)}^3{P}^2}$ ...(i)

$K_p=\frac{4\times {\left(0.0237\right)}^2\times {[4-2\times \left(0.0237\right)]}^2}{[1-0.0237]{[3-3\times \left(0.0237\right)]}^3\times 100}$

$=1.431\times {10}^{-5}{atm}^{-2}$

Also, if mole% of $N{H}_3$ is to be increased to 10.4, then

$\frac{2x}{4-2x}=\frac{10.4}{100}$
$\therefore x=0.1884$

Again using equation (i)

$K_p=\frac{4x^2{(4-2x)}^2}{(1-x){(3-3x)}^3\cdot P^2}$

$1.431\times {10}^{-5}=\frac{[4\times {\left(0.1884\right)}^2]{[4-2\left(0.1884\right)]}^2}{[1-0.1884]{[3-3\times \left(0.1884\right)]}^3\times P^2}$

$\therefore P=105.41\approx 105atm$