Q. For the reaction, $N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)$ , the value of $K_{c}$ at 800$℃$ is 0.1. When the equilibrium concentration of both the reactants is 0.5 mole, what is the value of $K_{p}$ at the same temperature?
A) 0.5
B) 0.1
C) 0.01
D) 0.025
Explanation
(b)
$N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)$
$K_{c} = 0.1,\ K_{p} = K_{c}(RT)^{\mathrm{\Delta}n}$
$\mathrm{\Delta}n =$Number of gaseous product – Number of gaseous reactants
$\mathrm{\Delta}n = 2 - 2 = 0$
$\ \ \ \ \ \mathrm{\Delta}n = 0$
$\therefore$ $K_{p} = K_{c} = 0.1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.