At temperature T, compound (AB_ 2(~g) ) dissociates as (AB_ 2(~g) leftharpoons AB_ (g) +(1)/(2) ~B_ 2(~g) ) having degree of dissociation x (small compared to unity). The correct expression for x in terms of (K_ p ) and

Correct answer: C. ( [3] 2 ~K_ p ^2 p ) ( array lcc & AB_2(g) & leftharpoons & AB(g) & + & (1)/(2) ~B_2(g) t=0 & p_0 & t=t_ eq & p_0(1-x) & & p_0 x & & (p_0 x)/(2) array ( p=p _0 x-p_0 x+p_0 x+ p_0 x 2 ) gathered p=p_0(1+(x)/(2)) p_0=(p)/((1+(x)/(2))) K_p= (p_ A B )(p_ B_2 )^ 1 / 2 (p_ AB_2 ) K_p= (p_0 x)((p_0 x)/(2))^ 1 / 2 p_0(1-x) gathered K_ p = px (1+(x)/(2)) ( p 1+(x)/(2) × x 2 )^ 1 / 2 p(1-x) (1+(x)/(2)) Since x 1 aligned & K_p= p^ 1 / 2 x^ 3 / 2 2^ 1 / 2 & x= [3] 2 ~K_ p ^2 p aligned

JEE Main 202529 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Equilibrium

JEE Main 2025 · 29 January · Morning Shift · Chemistry · Chemical Equilibrium

At temperature T, compound (AB_ 2(~g) ) dissociates as (AB_ 2(~g)...

At temperature T, compound (\mathrm{AB}_{2(\mathrm{~g})}) dissociates as (\mathrm{AB}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{AB}_{(\mathrm{g})}+\frac{1}{2} \mathrm{~B}_{2(\mathrm{~g})}) having degree of dissociation

x

(small compared to unity). The correct expression for

x

in terms of (\mathrm{K}_{\mathrm{p}}) and p is

Options

A(\sqrt[4]{\frac{2 K_p}{p}})
B(\sqrt[3]{\frac{2 K_{\mathrm{p}}}{\mathrm{p}}})
C(\sqrt[3]{\frac{2 \mathrm{~K}_{\mathrm{p}}^2}{\mathrm{p}}})
D(\sqrt{K_p})

Correct answer

C. (\sqrt[3]{\frac{2 \mathrm{~K}_{\mathrm{p}}^2}{\mathrm{p}}})

Step-by-step solution

(

\begin{array}{lcc} & \mathrm{AB}_2(\mathrm{g}) & \rightleftharpoons & \mathrm{AB}(\mathrm{g}) & + & \frac{1}{2} \mathrm{~B}_2(\mathrm{g}) \\ t=0 & p_0 & \\ t=t_{\text{eq }} & p_0(1-x) & & p_0 x & & \frac{p_0 x}{2} \end{array}

(

\mathrm{p}=\mathrm{p}

_0 \mathrm{x}-\mathrm{p}_0 \mathrm{x}+\mathrm{p}_0 \mathrm{x}+\frac{\mathrm{p}_0 \mathrm{x}}{2})

\begin{gathered}p=p_0\left(1+\frac{x}{2}\right) \\ p_0=\frac{p}{\left(1+\frac{x}{2}\right)} \\ K_p=\frac{\left(p_{A B}\right)\left(p_{B_2}\right)^{1 / 2}}{\left(p_{\mathrm{AB}_2}\right)} \\ K_p=\frac{\left(p_0 x\right)\left(\frac{p_0 x}{2}\right)^{1 / 2}}{p_0(1-x)}\end{gathered} \mathrm{K}_{\mathrm{p}}=\frac{\frac{\mathrm{px}}{\left(1+\frac{x}{2}\right)}\left(\frac{\mathrm{p}}{1+\frac{x}{2}} \times \frac{\mathrm{x}}{2}\right)^{1 / 2}}{\frac{\mathrm{p}(1-\mathrm{x})}{\left(1+\frac{x}{2}\right)}}

Since

x \ll 1

\begin{aligned} & K_p=\frac{\mathrm{p}^{1 / 2} \mathrm{x}^{3 / 2}}{2^{1 / 2}} \\ & \mathrm{x}=\sqrt[3]{\frac{2 \mathrm{~K}_{\mathrm{p}}^2}{\mathrm{p}}} \end{aligned}

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