JEE MAIN - Chemistry (2024 - 4th April Evening Shift - No. 17)
The equilibrium constant for the reaction
$$\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$$
is $$\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}$$. The value of $$\mathrm{K}_{\mathrm{c}}$$ for the reaction given below is $$2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$$ is :
49
416
41.6
4.9
Explanation
The reaction
$$2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3$$
can be formed from the given reaction by reverting it and multiplying coefficients by 2.
Thus,
$$\begin{aligned} \mathrm{K}_c^{\prime} & =\mathrm{K}_{\mathrm{c}}^{-2}=\frac{1}{\mathrm{~K}_{\mathrm{c}}^2}=\left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \\ & =416 \end{aligned}$$
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