Question

In a mixture of $N_2$ and $H_2$ in the ratio $1:3$ at $30atm$ and ${300}^{o}C$, the % of ${NH}_3$ at equilibrium $17.8$. Calculate $K_p$ for $N_2+3H_2\rightleftharpoons 2{NH}_3$.

Answer 1

$N_2+3H_2\rightleftharpoons 2{NH}_3{300}^{0}C$

$a$ $3a$ $0$

$a(1-\alpha )$ $3a(1-\alpha )$ $2a\alpha$

Assuming $17.8$% in weight

$\frac{\left(2a\alpha \right)\left(17\right)}{\left(2a\alpha \right)17+a(1-\alpha )\left(28\right)+3a(1-\alpha )\left(2\right)}=\frac{17.8}{100}$

$\frac{34\alpha }{28-28\alpha +6-6\alpha +34\alpha }=\frac{178}{100}$

$\frac{34\alpha }{34}=\frac{17.8}{100}\alpha =0.178$

$K_P=\frac{{{PNH}^2}_3}{P_{N2}{P^3}_{H2}}\Rightarrow \frac{{\left(\frac{2a\alpha }{a(4-2\alpha )}P\right)}^2}{\frac{a(1-\alpha )}{a(4-2\alpha )}P{\left(\frac{a(1-2)3}{a(4-2\alpha )}P\right)}^3}$ ($\because$ partial pressure $=$ mole fraction $\times$ total pressure)

$\Rightarrow \frac{\frac{{4\alpha }^2}{{(4-2\alpha )}^2}X{P}^2}{\frac{1-\alpha }{(4-2\alpha )}\times \frac{{(1-\alpha )}^3}{{(4-2\alpha )}^3}\times 27\times P^4}\Rightarrow \frac{{4\alpha }^2}{{(1-\alpha )}^4{(4-2\alpha )}^2\times 27\times {\left(30\right)}^2}$

$\Rightarrow 0.0086\times {10}^{-4}$

$K_P\Rightarrow 86\times {10}^{-8}$