For the following gas phase equilibrium reaction at constant temperature, if the total pressure is atm and the pressure equilibrium constant is 9 atm, then the degree of dissociation is given as The value of is ______.(nearest integer)

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Solution

Q.
For the following gas phase equilibrium reaction at constant temperature,

${NH}_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$

if the total pressure is $\sqrt{3}$ atm and the pressure equilibrium constant $(K_{p})$ is 9 atm, then the degree of dissociation is given as ${(x \times 10^{- 2})}^{- 1 / 2} .$ The value of $x$ is ______.(nearest integer) [2026]

Ans.
(125)

$\begin{matrix} {NH}_{3 (g)} & ⇌ & \frac{1}{2} N_{2 (g)} & + & \frac{3}{2} H_{2 (g)} \\ t = 0 & 1 mole & - & - \\ t = t_{e q} & 1 - α & α / 2 & 3 α / 2 \end{matrix}$

$K_{p} = \frac{{(\frac{α}{2})}^{1 / 2} {(\frac{3 α}{2})}^{3 / 2}}{(1 - α)} {[\frac{P_{T}}{1 + α}]}^{1}$ $[∵ P_{T} = \sqrt{3} atm]$

$9 = \frac{{(\frac{α}{2})}^{1 / 2} {(\frac{3 α}{2})}^{3 / 2}}{(1 - α)} \times \frac{{(3)}^{1 / 2}}{1 + α}$

$9 = \frac{9 {(\frac{α}{2})}^{2}}{1 - α^{2}}$

$1 - α^{2} = \frac{α^{2}}{4}$

$\frac{5 α^{2}}{4} = 1$

$α^{2} = 0.8$

$α = {(0.8)}^{1 / 2}$

$α = {[\frac{1}{0.8}]}^{- 1 / 2}$

$α = {[125 \times 10^{- 2}]}^{- 1 / 2}$

$x = 125$

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