For the given hypothetical reactions, the equilibrium constants are as follows: X rightleftharpoons Y ; K 1 =1.0 Y rightleftharpoons Z ; K 2 =2.0 Z rightleftharpoons W ; K 3 =4.0 The equilibrium constant for the reaction X rightleftharpoons W is \ \ \ \ \ \ .

Chemistry · Chemical Equilibrium

JEE Main 2024 — 8 April, Shift 1 — Question 63

For the given hypothetical reactions, the equilibrium constants are as follows:

$\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_{1}=1.0$

$\mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_{2}=2.0$

$\mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_{3}=4.0$

The equilibrium constant for the reaction $\mathrm{X} \rightleftharpoons \mathrm{W}$ is $\_ \_\_\_\_\_$ .

\mathrm{X \rightleftharpoons Y}, \quad K_1 = 1.0

\mathrm{Y \rightleftharpoons Z}, \quad K_2 = 2.0

\mathrm{Z \rightleftharpoons W}, \quad K_3 = 4.0

Adding the reactions gives: $\mathrm{X \rightleftharpoons W}$ The equilibrium constant for the overall reaction is:

K = K_1 \times K_2 \times K_3 = 1.0 \times 2.0 \times 4.0 = 8

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient