Q40 / 160 · AP EAMCET 2007 · Chemistry

The equilibrium constant for the reaction $\mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{SO}_3(g)$ is $5 \times 10^{-2}$ atm. The equilibrium constant of the reaction $2 \mathrm{SO}_3(g) \rightleftharpoons 2 \mathrm{SO}_2(g)+\mathrm{O}_2(g)$ would be

  • A
    $100 \mathrm{~atm}$
  • B
    $200 \mathrm{~atm}$
  • Correct answer
    C
    $4 \times 10^2 \mathrm{~atm}$
  • D
    $6.25 \times 10^4 \mathrm{~atm}$

Solution

$\begin{aligned} \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 & \rightleftharpoons \mathrm{SO}_3 \\ K_1 & =\frac{\left[\mathrm{SO}_3\right]}{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{1 / 2}} \\ 2 \mathrm{SO}_3 & \rightleftharpoons 2 \mathrm{SO}_2+\mathrm{O}_2 \\ \mathrm{~K}_2 & =\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}\end{aligned}$ From Eqs. (i) and (ii) $\begin{aligned} K_2 & =\frac{1}{K_1^2} \\ K_2 & =\frac{1}{\left(5 \times 10^{-2}\right)^2}=\frac{1}{25 \times 10^{-4}} \\ & =\frac{100 \times 10^2}{25} \\ & =4 \times 10^2 \mathrm{~atm}\end{aligned}$

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

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