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Relation between Kp and Kc

For which rea...

Question

For which reactions at 298 K, the value of $\frac{K_{p}}{K_{c}}$ is maximum and minimum respectively.
(a) $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
(b) $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
(c) $X (g) + Y (g) ⇌ 4 Z (g)$
(d) $A (g) + 3 B (g) ⇌ 7 C (g)$

d, c

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d, b

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c, b

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d, a

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Solution

The correct option is B d, b
We know the relation $K_{p} = K_{c} \times (R T)^{△ n}$
where $△ n = n_{p} - n_{g}$
(a) $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
$△ n = 2 - 1 = 1$
$\Rightarrow K_{p} = K_{c} \times (R T) \Rightarrow \frac{K_{p}}{K_{c}} = R T$

(b) $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
$△ n = 2 - (2 + 1) = - 1$
$K_{p} = K_{c} (R T)^{- 1}$
$\Rightarrow \frac{K_{p}}{K_{c}} = \frac{1}{R T} = = Minimum value$

(c) $X (g) + Y (g) ⇌ 4 Z (g)$
$△ n = 4 - (1 + 1) = 2$
$K_{p} = K_{c} (R T)^{2}$
$\Rightarrow \frac{K_{p}}{K_{c}} = (R T)^{2}$
(d) $A (g) + 3 B (g) ⇌ 7 C (g)$
$△ n = 7 - (1 + 3)$
$= 3$
$K_{p} = K_{c} (R T)^{3}$
$\Rightarrow \frac{K_{p}}{K_{c}} = (R T)^{3} = Maximum value$

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