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Derivation of Kp and Kc

At 700 K, the...

Question

At 700 K, the equilibrium constant $K_{p},$ for the reaction
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$
is $1.8 \times 10^{- 3} k P a .$ What is the numerical value of $K_{C}$ for this reaction at the same temperature?
$R = 0.0821 L a t m m o l^{- 1} K^{- 1}$

3.09 \times 10^{- 7} m o l e l i t r e^{- 1}

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9.03 \times 10^{- 7} m o l e l i t r e^{- 1}

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5.05 \times 10^{- 9} m o l e l i t r e^{- 1}

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5.05 \times 10^{- 5} m o l e l i t r e^{- 1}

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Solution

The correct option is A $3.09 \times 10^{- 7} m o l e l i t r e^{- 1}$
We know the relationship
$K_{P} = K_{C} (R T)^{△ n}$
Here $K_{P} = 1.80 \times 10^{- 3}$
$K_{P} = \frac{1.8 \times 10^{- 3}}{101.3} a t m$
$= 1.78 \times 10^{- 5} a t m$
$R = 0.0821$ litre atm $K^{- 1} m o l^{- 1}$
$△ n = 3 - 2 = 1$
$T = 700 K$
$K_{C} = \frac{K_{P}}{(R T)^{△ n}}$
putting the values,
$K_{C} = \frac{1.78 \times 10^{- 5}}{0.0821 \times 700}$
$= 3.09 \times 10^{- 7} m o l e l i t r e^{- 1} .$

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