WBJEE SERIES

Chemistry

Law Of Chemical Equilibrium And Equilibrium Constant

3 previous year questions.

Volume: 3 Ques

Yield: Medium

High-Yield Trend

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2026

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2025

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2016

Chapter Questions
3 MCQs

01

PYQ 2016

medium

chemistry ID: wbjee-20

The equilibrium constant for the reaction is ?K?. Then, the equilibrium constant for the reaction will be

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K

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2K

02

PYQ 2025

medium

chemistry ID: wbjee-20

Consider the following gas phase dissociation, PCl (g) PCl (g) + Cl (g) with equilibrium constant K at a particular temperature and at pressure P. The degree of dissociation () for PCl (g) is PCl (g) PCl (g) + Cl (g)

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Official Solution

Correct Option: (3)

Let the initial moles of PCl5 be 1 and the degree of dissociation be α. We can set up an ICE table based on moles:Reaction:PCl5(g) ⇌ PCl3(g)+Cl2(g)Initial (mol):1 0 0Change (mol):-α +α +αEquil. (mol):1-α α αTotal moles at equilibrium = (1-α) + α + α = 1+α. The total pressure at equilibrium is given as P. Now, calculate the partial pressures using pi = Xi × Ptotal, where Xi is the mole fraction. pPCl5 = ( (1-α) / (1+α) ) P pPCl3 = ( α / (1+α) ) P pCl2 = ( α / (1+α) ) P The expression for the equilibrium constant Kp is: Kp = (pPCl3 × pCl2) / pPCl5 Substitute the partial pressures: Kp = (( αP / (1+α) ) ( αP / (1+α) )) / (( (1-α)P ) / (1+α) ) Kp = (α2P2 / (1+α)2) / (( (1-α)P ) / (1+α) ) Kp = (α2P2 / (1+α)2) × ((1+α) / ((1-α)P)) Kp = (α2P) / ((1+α)(1-α)) Kp = (α2P) / (1-α2) Now, rearrange to solve for α: Kp (1-α2) = α2 P Kp - Kp α2 = α2 P Kp = α2 P + Kp α2 Kp = α2 (P + Kp) α2 = Kp / (P + Kp) α = √( Kp / (Kp + P) ) = ( Kp / (Kp + P) )1/2 This corresponds to option (C).

03

PYQ 2026

medium

chemistry ID: wbjee-20

Consider the gas phase dissociation, PCl (g) PCl (g) + Cl (g) with equilibrium constant at a particular temperature and pressure . The degree of dissociation () for PCl (g) is

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2

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