JEE MAIN - Chemistry Hindi (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
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