Q6 / 90 · JEE Main 2024 (04 Apr Shift 2) · Chemistry

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 :

  • A
    4.9
  • B
    49
  • C
    41.6
  • Correct answer
    D
    416

Solution

$\begin{aligned} & \mathrm{K}_{\mathrm{C}}^{\prime}=\left(\frac{1}{\mathrm{~K}_{\mathrm{C}}}\right)^2=\left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \\ & \mathrm{~K}_{\mathrm{C}}^{\prime}=416.49\end{aligned}$

Hence, the correct answer is option: 'D'.

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