Column – I(Reaction) Column – IIKPKCA A2(g)+3B2(g)↔2AB3(g)1 (RT)−2B A2(g)+B2(g)↔2AB(g)2 (RT)0C A(s)+32B2(g)↔AB3(g)3 (RT)1/2D AB2(g)↔AB(g)+12B2(g)4 (RT)−1/2

A Δn=2−(1+3)=−2⇒KPKC=(RT)Δn=(RT)−2B Δn=2−(1+1)=0⇒KPKC=(RT)0C Δn=1−32=−12⇒KPKC=(RT)−1/2D (1+12)−1=12⇒KPKC=(RT)1/2

Column – I(Reaction) Column – IIKPKCA A2(g)+3B2(g)↔2AB3(g)1 (RT)−2B A2(g)+B2(g)↔2AB(g)2 (RT)0C A(s)+32B2(g)↔AB3(g)3 (RT)1/2D AB2(g)↔AB(g)+12B2(g)4 (RT)−1/2

A Δn=2−(1+3)=−2⇒KPKC=(RT)Δn=(RT)−2B Δn=2−(1+1)=0⇒KPKC=(RT)0C Δn=1−32=−12⇒KPKC=(RT)−1/2D (1+12)−1=12⇒KPKC=(RT)1/2

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Column – I
(Reaction)
Column – II
$(\frac{K_{P}}{K_{C}})$
$(A) A_{2 (g)} + 3 B_{2 (g)} \leftrightarrow 2 A B_{3 (g)}$ $(1) {(R T)}^{- 2}$
$(B) A_{2 (g)} + B_{2 (g)} \leftrightarrow 2 A B_{(g)}$ $(2) {(R T)}^{0}$
$(C) A_{(s)} + \frac{3}{2} B_{2 (g)} \leftrightarrow A B_{3 (g)}$ $(3) {(R T)}^{1 / 2}$
$(D) A B_{2 (g)} \leftrightarrow A B_{(g)} + \frac{1}{2} B_{2 (g)}$ $(4) {(R T)}^{- 1 / 2}$

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$A - 1, B - 2, C - 3, D - 4$

$A - 1, B - 2, C - 4, D - 3$

$A - 2, B - 3, C - 4, D - 1$

$A - 3, B - 1, C - 2, D - 4$

answer is B.

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Detailed Solution

$(A) Δ n = 2 - (1 + 3) = - 2 \Rightarrow \frac{K_{P}}{K_{C}} = {(R T)}^{Δ n} = {(R T)}^{- 2}$

$(B) Δ n = 2 - (1 + 1) = 0 \Rightarrow \frac{K_{P}}{K_{C}} = {(R T)}^{0}$

$(C) Δ n = 1 - \frac{3}{2} = - \frac{1}{2} \Rightarrow \frac{K_{P}}{K_{C}} = {(R T)}^{- 1 / 2}$

$(D) (1 + \frac{1}{2}) - 1 = \frac{1}{2} \Rightarrow \frac{K_{P}}{K_{C}} = {(R T)}^{1 / 2}$

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Column – I (Reaction)	Column – II $(\frac{K_{P}}{K_{C}})$
$(A) A_{2 (g)} + 3 B_{2 (g)} \leftrightarrow 2 A B_{3 (g)}$	$(1) {(R T)}^{- 2}$
$(B) A_{2 (g)} + B_{2 (g)} \leftrightarrow 2 A B_{(g)}$	$(2) {(R T)}^{0}$
$(C) A_{(s)} + \frac{3}{2} B_{2 (g)} \leftrightarrow A B_{3 (g)}$	$(3) {(R T)}^{1 / 2}$
$(D) A B_{2 (g)} \leftrightarrow A B_{(g)} + \frac{1}{2} B_{2 (g)}$	$(4) {(R T)}^{- 1 / 2}$

Column – I(Reaction) Column – IIKPKCA A2(g)+3B2(g)↔2AB3(g)1 (RT)−2B A2(g)+B2(g)↔2AB(g)2 (RT)0C A(s)+32B2(g)↔AB3(g)3 (RT)1/2D AB2(g)↔AB(g)+12B2(g)4 (RT)−1/2

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