JEE Main 2022 · 24 Jun · Shift 1 · Chemistry Q64

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ChemistrySingle correct+4 / -1Question 64 of 90

For a reaction at equilibrium
$\mathrm{A(g)\rightleftharpoons B(g)+\tfrac12C(g)}$
the relation between dissociation constant ( $K$ ), degree of dissociation ( $\alpha$ ) and equilibrium pressure ( $p$ ) is given by :

K=\dfrac{\alpha^{\frac12}p^{\frac32}}{\left(1+\frac32\alpha\right)^{\frac12}(1-\alpha)}

K=\dfrac{\alpha^{\frac32}p^{\frac12}}{(2+\alpha)^{\frac12}(1-\alpha)}

K=\dfrac{(\alpha p)^{\frac32}}{\left(1+\frac32\alpha\right)^{\frac12}(1-\alpha)}

K=\dfrac{(\alpha p)^{\frac32}}{(1+\alpha)(1-\alpha)^{\frac12}}

Answer and worked solution

Official answer: B

Moles at equilibrium (start with 1 mol A): A $1-\alpha$ , B $\alpha$ , C $\frac\alpha2$ ; total $\dfrac{2+\alpha}2$ .

Partial pressures. $p_A=\dfrac{2(1-\alpha)}{2+\alpha}p$ , $p_B=\dfrac{2\alpha}{2+\alpha}p$ , $p_C=\dfrac{\alpha}{2+\alpha}p$ .

Constant.
$K=\frac{p_Bp_C^{1/2}}{p_A}=\frac{\alpha}{1-\alpha}\left(\frac{\alpha p}{2+\alpha}\right)^{1/2}=\frac{\alpha^{3/2}p^{1/2}}{(2+\alpha)^{1/2}(1-\alpha)}.$

(B)

Topics:Law of mass action and Kc, Kp

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JEE Main 2022 · 24 Jun · Shift 1, Chemistry question 64