Q. For which one of the following reactions $K_{p} = K_{c}$?

A) $PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2}$
B) $N_{2} + O_{2} \rightleftharpoons 2NO$
C) $N_{2} + 3H_{2} \rightleftharpoons 2NH_{3}$
D) $2SO_{3} \rightleftharpoons 2SO_{2} + O_{2}$
Explanation

(b)
$K_{p} = K_{c}(RT)^{\mathrm{\Delta}n}$
$K_{p} = K_{c}$ only when $(RT)^{\mathrm{\Delta}n} = 0$
(a)$PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2}$
$\mathrm{\Delta}n = n_{p} - n_{R}$
= 2$-$1
= 1
$\therefore\ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}K_{p} \neq K_{c}$
(b)$N_{2} + O_{2} \rightleftharpoons 2NO$
$\mathrm{\Delta}n = 2 - 2$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K_{p} = K_{c}$
(c)$N_{2} + 3H_{2} \rightleftharpoons 2NH_{3}$
$\mathrm{\Delta}n = 2 - 4$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - 2$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}K_{p} \neq K_{c}$
(d)$2SO_{3} \rightleftharpoons 2SO_{2} + O_{2}$
$\mathrm{\Delta}n = 3 - 2 = 1$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}K_{p} \neq K_{c}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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