Question

For the reaction 2$N{H}_3$ (g) ⇌ $N_2$(g) + 3$H_2$(g), if the total equilibrium pressure is P and degree of dissociation is α, the equilibrium constant $K_p$ is given by

$\frac{27{\alpha }^4{p}^2}{16{(1-{\alpha }^2)}^2}$

• A. True
• B. False

Answer 1

The correct option is A

True

Initially let us assume n moles of $N{H}_3$ are there.

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

Moles at t = 0 n 0 0

Moles at $t=teq(1-\alpha )$ $n\left(\frac{\alpha }{2}\right)$ $n\left(\frac{3\alpha }{2}\right)$

Total amount of the subtance = n (1 - $\alpha$) + n $\frac{\alpha }{2}+n\frac{3\alpha }{2}$ = n (1 - $\alpha$)

Partial pressure of $N{H}_3$ , $P_{N{H}_3}$ = $x_{N{H}_3}$ P = $\frac{n(1-\alpha )}{n(1+\alpha )}$ P = $\frac{1-\alpha }{1+\alpha }$P

Partial pressure of $N_2$, $P_{N_2}$ = $x_{N_2}$ P = $\frac{n\left(\frac{\alpha }{2}\right)}{n(1+\alpha )}$ P = $\frac{\alpha }{2(1+\alpha )}$P

Partial pressure of $H_2$, $P_{H_2}$ = $x_{H_2}$ P = $\frac{n\left(\frac{3\alpha }{2}\right)}{n(1+\alpha )}$ P = $\frac{3\alpha }{2(1+\alpha )}$P

the expression of $K_p$ is

$K_p$ = $\frac{P_{N_2}{p}^3{H}_2}{P_{N{H}_3}^2}$ = [ $\frac{\alpha }{2(1+\alpha )}$ P ] ${\left[\frac{3\alpha }{2(1+\alpha )}P\right]}^3$ $[\frac{1+\alpha }{1-\alpha }\frac{1}{P}{]}^2$ = $\frac{27{\alpha }^4{P}^2}{16(1-{\alpha }^2{)}^2}$