Question

Ammonia dissociates into $N_2$ and $H_2$ such that degree of dissociation $\alpha$ is very less than $1$ and equilibrium pressure is $P_0$ than the value of $\alpha$ is:

[if $K_p$ for 2$N{H}_3$(g) $\rightleftharpoons$ $N_2$(g)+3$H_2\left(g\right)$ is $27\times {10}^{-8}{P}_0^2$]

• A. ${10}^{-2}$
• B. $4\times {10}^{-4}$
• C. $0.02$
• D. cannot be determined

Answer 1

Answer: (A)

${10}^{-2}$

$2N{H}_3\to N_2+3H_2$

Initial mole 1 0 0

at equilibrium $1-2\alpha$ $\alpha$ $3\alpha$

Given equilibrium pressure is $P_0$

Given $K_P=27\times {10}^{-8}{P}_0^2$

Total Moles $=1-2\alpha +\alpha +3\alpha =1+2\alpha$

$X_{N{H}_3}=\frac{1-2\alpha }{1+2\alpha }$

$X_{N_2}=\frac{\alpha }{1+2\alpha }$

$X_{H_2}=\frac{3\alpha }{1+2\alpha }$

$27\times {10}^{-8}{P}_0^2=\frac{\frac{\alpha P_0}{1+2\alpha }\times (\frac{3\alpha P_0}{1+2\alpha }{)}^3}{(\frac{1-2\alpha }{1+2\alpha }{)}^2{P}_0^2}$

as $\alpha$ is very less than 1 so

${10}^{-8}={\alpha }^4$

$\alpha ={10}^{-2}$