For the reaction, $C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g)$ the $\frac{K_{p}}{K_{c}}$ is equal to

\sqrt{R} T

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\frac{1}{R T}

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Solution

The correct option is D $\frac{1}{R T}$
$K_{p} = K_{c} (R T)^{Δ n}$ ;
Here,
$Δ n =$ Moles of gaseous product $-$ Moles of gaseous reactant

$Δ n = 1 - 2 = - 1$

$K_{p}$ = $K_{c}$ (RT) $^{- 1}$

$∴ \frac{K_{p}}{K_{c}} = \frac{1}{R T}$

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