Q. In the reaction, $H_{2}(g) + Cl_{2}(g) \rightleftharpoons 2HCl(g)$
A) $K_{p} \neq K_{c}$
B) $K_{p} = K_{c}$
C) $K_{p} > K_{c}$
D) $K_{p} < K_{c}$
Explanation
(b)
$H_{2}(g) + Cl_{2}(g) \rightleftharpoons 2HCl(g)$
We know that,
$K_{p} = K_{c}.(RT)^{\mathrm{\Delta}n_{g}}$
$\mathrm{\Delta}n_{g}$ = no. of moles of gaseous products – no. of moles of gaseous reactants
=2-2=0
$K_{p} = K_{c}.(RT)^{0}$
$K_{p} = K_{c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.