$l o g \frac{K_{p}}{K_{c}} + l o g R T = 0$ is true relationship for the following reaction.

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

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

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N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}

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both

B

and

C

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Solution

The correct option is C $2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}$
As, $K_{p} = K_{c} {(R T)}^{Δ n}$
Apply logarithm, $log K_{p} = log K_{c} + Δ n log (R T) ⟹ log (\frac{K_{p}}{K_{c}}) - Δ n log (R T) = 0$
Given, $⟹ log (\frac{K_{p}}{K_{c}}) + log (R T) = 0 ⟹ Δ n = - 1$
So, $2 {S O}_{2} + O_{2} ⇌ 2 {S O}_{3}$ is correct because here $Δ n$ $= 2 - (2 + 1) = - 1$

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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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N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

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

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

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