Q43 / 150 · JEE Main 2006 · 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 $$ 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) $$ will be

  • Correct answer
    A
    416
  • B
    $2.40 \times 10^{-3}$
  • C
    $9.8 \times 10^{-2}$
  • D
    $4.9 \times 10^{-2}$

Solution

$\begin{aligned} \mathrm{K}_{\mathrm{c}}^{\prime} & =\left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \\ & =\frac{10^4}{4.9 \times 4.9}=\frac{100 \times 100}{24.01} \\ & =4.1649 \times 100 \\ & =416.49\end{aligned}$

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

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