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Derivation of Kp and Kc

With partial ...

Question

With partial pressure of $S O_{3}$ and $S O_{2}$ taken to be equal to $10^{- 5}$ atm and 10 atm respectively then what must be the partial pressure of $O_{2}$ so that the reaction may remain in equilibrium?

A

10

atm

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B

0.01

atm

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C

100

atm

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D

0.1

atm

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Solution

The correct option is B $0.01$ atm
The value of the equilibrium constant is $K_{P} = 1.0 \times 10^{- 5} .$
The expression for the equilibrium constant is $K_{P} = \frac{P_{S O_{3}}}{P_{S O_{2}} \times P_{O_{2}}^{\frac{1}{2}}} .$
Substitute values in the above expression.
$1.0 \times 10^{- 5} = \frac{10^{- 5}}{10 \times P_{O_{2}}^{\frac{1}{2}}}$

Hence, $P_{O_{2}} = 0.01 a t m .$

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