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NEET 2012
Exam
NEET
Year
2012
Subject
Chemistry
Chapter
Chemical Equilibrium

NEET 2012 · Chemistry · Chemical Equilibrium

Given that the equilibrium constant for the reaction, 2 SO _ 2 (g)+...

Given that the equilibrium constant for the reaction, 2 \mathrm{SO}}_{2}(g)+ \mathrm{O}}_{2}(g) ⇌\rightleftharpoons2 \mathrm{SO}}_{3}(g) has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature? \mathrm{SO}}_{3}(g) \rightleftharpoons \mathrm{SO}}_{2}(g)+\frac{1}{2} \mathrm{O}}_{2}(g)

Options

  1. A1.8×10−31.8 \times 10^{-3}
  2. B3.6×10−33.6 \times 10^{-3}
  3. C6.0×10−26.0 \times 10^{-2}
  4. D1.3×10−51.3 \times 10^{-5}

Correct answer

C. 6.0×10−26.0 \times 10^{-2}

Step-by-step solution

2 \mathrm{SO}}_{2}(g)+ \mathrm{O}}_{2}(g) ⇌\rightleftharpoons2 \mathrm{SO}}_{3}(g) Equilibrium constant for this reaction, \begin{gathered} K=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}\\ \mathrm{SO}}_{3}(g) \rightleftharpoons \mathrm{SO}}_{2}(g)+\frac{1}{2} \mathrm{O}}_{2}(g) \end{gathered} Equilibrium constant for this reaction, K^{\prime}=\frac{\left[\mathrm{\text{ SO }}_2\right]\left[\mathrm{O}_2\right]^{\frac{1}{2}}}{\left[\mathrm{SO}_3\right]}On squaring Eq. (ii) both sides, we have \left(K^{\prime}\right)^2=[SO2]2[O2][SO3]2\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}=1K=1278∴K′=1278=0.003597=5.99×10−2≈6×10−2\begin{aligned} & =\frac{1}{K} \\ & =\frac{1}{278} \\ \therefore \quad K^{\prime} & =\sqrt{\frac{1}{278}} \\ & =\sqrt{0.003597} \\ & =5.99 \times 10^{-2} \\ & \approx 6 \times 10^{-2}\end{aligned}$$
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