If $log \frac{k_{c}}{k_{p}} - log \frac{1}{R T} = 0$ , then above is true for the following equilibrium reaction:

{N H}_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)

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C a C O_{3} (s) ⇋ C a O (s) + C O_{2} (g)

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2 {N O}_{2} (g) ⇌ N_{2} O_{4} (g)

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H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

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Solution

The correct options are
A ${N H}_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$
B $C a C O_{3} (s) ⇋ C a O (s) + C O_{2} (g)$
$log \frac{K_{c}}{K_{p}} - log \frac{1}{R T} = 0$
$\Rightarrow \frac{K_{c}}{K_{p}} = \frac{1}{R T}$
$\Rightarrow K_{a} = K_{p} \times (R T)^{- 1}$
$\Rightarrow K_{p} = K_{c} (R T)^{1}$
We know
$K_{p} = K_{c} (R T)^{△ n_{g a s}}$
$∴ △ n_{g a s} = 1$
$A \to △ n_{g a s} = \frac{1}{2} + \frac{3}{2} - 1 = 2 - 1 = 1$
$B \to △ n_{g a s} = 1 - 0 = 1$
$C \to △ n_{g a s} = 1 - 2 = - 1$
$D \to △ n_{g a s} = 2 - 1 - 1 = 0$

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