For the given hypothetical reactions, the equilibrium constants are as follows : \beginaligned & X leftharpoons Y ; K_1=1.0 & Y leftharpoons Z ; K_2=2.0 & Z leftharpoons W ; K_3=4.0 \endaligned The equilibrium constant for the reaction X leftharpoons W is

Chemical Equilibrium PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q12 with Solution

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Correct answer: B

Use the rule for combining equilibrium reactions

When reactions are added, their equilibrium constants are multiplied.

Given:
$\mathrm{X} \rightleftharpoons \mathrm{Y}, \quad K_1=1.0$ $\mathrm{Y} \rightleftharpoons \mathrm{Z}, \quad K_2=2.0$ $\mathrm{Z} \rightleftharpoons \mathrm{W}, \quad K_3=4.0$
Add the reactions

Adding them gives:
$\mathrm{X} \rightleftharpoons \mathrm{Y}$ $\mathrm{Y} \rightleftharpoons \mathrm{Z}$ $\mathrm{Z} \rightleftharpoons \mathrm{W}$
On cancellation of intermediate species $\mathrm{Y}$ and $\mathrm{Z}$ , we get:
$\mathrm{X} \rightleftharpoons \mathrm{W}$
Multiply the equilibrium constants

Therefore,
$K = K_1 K_2 K_3 = (1.0)(2.0)(4.0)=8.0$
Match with the options
$K_{\mathrm{X} \rightleftharpoons \mathrm{W}} = 8.0$
So the correct option is B.