In which of the following $K_{p}$ is less than $K_{c}$ ?

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

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2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

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2 S O_{2} (g) + O_{2} (g) ⇌ 2 {S O}_{3} (g)

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P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)

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Solution

The correct option is C $2 S O_{2} (g) + O_{2} (g) ⇌ 2 {S O}_{3} (g)$
We have, $K_{P} = K_{C} {(R T)}^{△ n}$
So, for $K_{P} < K_{C} \Rightarrow △ n < 0$
Now, for the reaction,
$2 {S O}_{2} (g) + O_{2} (g) ⇌ 2 {S O}_{3} (g)$
$△ n = 2 - 2 - 1 = - 1$
$\Rightarrow K_{P} = K_{C} {(R T)}^{- 1}$
$\Rightarrow K_{P} = \frac{K_{C}}{R T} \Rightarrow K_{P} < K_{C}$ .

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log \frac{K_{P}}{K_{C}} + log R T = 0

log \frac{K_{p}}{K_{c}} + log R T = 0

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