Q. At $600℃,\ K_{p}$ for the following reaction is 1 atm. $X(g) \rightleftharpoons Y(g) + Z(g)$ At equilibrium, 50% of $X(g)$ is dissociated. The total pressure of the equilibrium system is $p$ atm. What is the partial pressure (in atm) of $X(g)$ at equilibrium?
A) 1
B) 4
C) 2
D) 0.5
Explanation
(a)
$X(g)\ \ \ \ \ \ \ \rightleftharpoons \ \ \ \ \ \ \ \ \ Y(g)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Z(g)$
Initial 1 0 0
At equilibrium 0.5 0.5 0.5
Partial pressure $\frac{0.5}{1.5}\ p$ $\frac{0.5}{1.5}\ p$ $\frac{0.5}{1.5}\ p$
$\because$ $K_{p}\ \ \ \ \ \frac{p_{Y}.p_{Z}}{p_{x}}$
$1\ \ \ \ \ \ \ \ \ \frac{\frac{p}{3}\ \ \ \ \ \ \frac{p}{3}}{\frac{p}{3}}$
$p$ 3 atm
Partial pressure of $X\ \ \ \ \frac{p}{3}\ \ \ \ \ \frac{3}{3}\ \ \ 1\ atm$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.