For a reaction at equilibrium A (g) ⇌ B (g) + (1)/(2)C(g) the relation between dissociation constant (K) , degree of dissociation (α) and equilibrium pressure (p) is given by :

Correct answer: A. K = (α(3)/(2)p(1)/(2))/((2 + α)(1)/(2)(1 - α)) cases & A _ (g) & ⇌ & B _ (g) & + & (1)/(2) C _ (g) att = 0 & 1 & & 0 & & 0 ateq . & 1 - α & & α & & (α)/(2) cases total moles at eq. = 1 - α + α + (α)/(2) = 1 + (α)/(2) P _ B = (α)/(1 + (α)/(2))p P _ c = ((α)/(2))/(1 + (α)/(2))p P _ A = (1 - α)/(1 + (α)/(2))p K _ p = P _ B × P _ C ^ (1)/(2) P _ A = (((α)/(1 + (α)/(2))p)(((α)/(2))/(1 + (α)/(2))p)(1)/(2))/(((1 - α)/(1 + (α)/(2))p)) K _ p = α ^ (3)/(2) p ^ (1)/(2) 2 ^ (1)/(2) ( 1 - α ) (1 + (α)/(2)) ^ (1)/(2) k _ p = (α(3)/(2)p(1)/(2))/((1 - α)(2 + α)(1)/(2))

JEE Main 202224 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 24 June
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Equilibrium

JEE Main 2022 · 24 June · Morning Shift · Chemistry · Chemical Equilibrium

For a reaction at equilibrium A (g) ⇌ B (g) + (1)/(2)C(g) the...

For a reaction at equilibrium

A $\left(g\right) ⇌ B \left(g\right) + \dfrac{1}{2}C\left(g\right)$

the relation between dissociation constant \left(K\right) $, degree of dissociation \left(\alpha\right)$ and equilibrium pressure $\left(p\right)$ is given by :

Options

A $K = \dfrac{\alpha\dfrac{3}{2}p\dfrac{1}{2}}{\left(2 + \alpha\right)\dfrac{1}{2}\left(1 - \alpha\right)}$
B $K = \dfrac{\alpha\dfrac{1}{2}p\dfrac{3}{2}}{\left(1 + \dfrac{3}{2}\alpha\right)\dfrac{1}{2}\left(1 - \alpha\right)}$
C $K = \dfrac{\left(\alpha p\right)\dfrac{3}{2}}$ { $\left(1 + \dfrac{3}{2}\alpha\right)\dfrac{1}{2}\left(1 - \alpha\right)}$
D $K = \dfrac{\left(\alpha p\right)\dfrac{3}{2}}$ { $\left(1 + \alpha\right)\left(1 - \alpha\right)\dfrac{1}{2}}$

Correct answer

A. $K = \dfrac{\alpha\dfrac{3}{2}p\dfrac{1}{2}}{\left(2 + \alpha\right)\dfrac{1}{2}\left(1 - \alpha\right)}$

Step-by-step solution

$\begin{cases} & {A}_{\left(g\right)} & ⇌ & {B}_{\left(g\right)} & + & \dfrac{1}{2}{C}_{\left(g\right)} \\ att = 0 & 1 & & 0 & & 0 \\ ateq . & 1 - \alpha & & \alpha & & \dfrac{\alpha}{2}\end{cases}$

total moles at eq. $= 1 - \alpha + \alpha + \dfrac{\alpha}{2} = 1 + \dfrac{\alpha}{2}$

${P}_{B} = \dfrac{\alpha}{1 + \dfrac{\alpha}{2}}p{P}_{c} = \dfrac{\dfrac{\alpha}{2}}{1 + \dfrac{\alpha}{2}}p{P}_{A} = \dfrac{1 - \alpha}{1 + \dfrac{\alpha}{2}}p$

${K}_{p} = \dfrac{{P}_{B} \times {P}_{C}^{\dfrac{1}{2}}}{{P}_{A}} = \dfrac{\left(\dfrac{\alpha}{1 + \dfrac{\alpha}{2}}p\right)\left(\dfrac{\dfrac{\alpha}{2}}{1 + \dfrac{\alpha}{2}}p\right)\dfrac{1}{2}}{\left(\dfrac{1 - \alpha}{1 + \dfrac{\alpha}{2}}p\right)}$ ${K}_{p} = \dfrac{{\alpha}^{\dfrac{3}{2}}{p}^{\dfrac{1}{2}}}{{2}^{\dfrac{1}{2}} ( 1 - \alpha ) {\left(1 + \dfrac{\alpha}{2}\right)}^{\dfrac{1}{2}}}{k}_{p} = \dfrac{\alpha\dfrac{3}{2}p\dfrac{1}{2}}{\left(1 - \alpha\right)\left(2 + \alpha\right)\dfrac{1}{2}}$

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