Q. In the reaction, $N_{2}O_{4} \rightleftharpoons 2{NO}_{2},\ \alpha$ is that part of $N_{2}O_{4}$ which dissociates, then the number of moles at equilibrium will be
A) 1
B) 3
C) $(1 + \alpha)$
D) $(1 - \alpha)^{2}$
Explanation
(c)
$N_{2}O_{4} \rightleftharpoons 2{NO}_{2}$
1 0 initially
$(1 - \alpha)\ \ \ 2\alpha$ at equilibrium
Total moles at equilibrium $= (1 - \alpha) + 2\alpha$
$= 1 + \alpha$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.