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Effective Equilibrium Constant

The equilibri...

Question

The equilibrium mixture for $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ present in 1L vessel at $600^{o} C$ contains 0.50, 0.12, and 5.0 mole of $S O_{2}, O_{2} and S O_{3}$ respectively.
Calculate $K_{C}$ for the given change at $600^{o} C .$

A

675.23 m o l^{- 1} L

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B

765.32 m o l^{- 1} L

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C

833.33 m o l^{- 1} L

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D

801.23 m o l^{- 1} L

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Solution

The correct option is C $833.33 m o l^{- 1} L$
The equilibrium reaction is $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g) .$
The expression for the equilibrium constant, $K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
Substituting given values in the above expression, we get,
$K_{c} = \frac{(5)^{2}}{(0.5)^{2} \times 0.12} = 833.33 m o l^{- 1} L$

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Effective Equilibrium Constant

Standard XII Chemistry

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