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Kp value for $2 {SO}_{2 (g)} + O_{2 (g)} \Leftrightarrow 2 {SO}_{3 (g)} is 5.0 {atm}^{- 1}$ What is the equilibrium partial pressure of $O_{2}$ if the equilibrium pressures of ${SO}_{2} and {SO}_{3}$ are equal ?

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0.1 atm

0.3

0.2

0.4

answer is A.

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Detailed Solution

$2 {SO}_{2 (g)} + O_{2 (g)} \Leftrightarrow 2 {SO}_{3 (g)}$

$K_{P} = \frac{P_{{SO}_{3}^{2}}}{P_{{SO}_{2}^{2}} : P_{O_{2}}} (∴ P_{{SO}_{3}} = P_{{SO}_{2}})$

$\frac{5.0}{1} = \frac{1}{P_{O_{2}}}$

$\Rightarrow P_{O_{2}} = \frac{1}{5} = 0.2 atm .$

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