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Chemical Equilibrium

JEE Main 2024 (Online) 8th April Morning Shift

MCQ+4 / -12024

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


$$\begin{aligned} & \mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_1=1.0 \\ & \mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_2=2.0 \\ & \mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_3=4.0 \end{aligned}$$


The equilibrium constant for the reaction \(\mathrm{X} \rightleftharpoons \mathrm{W}\) is

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