Q. Consider the reaction $\mathrm{X_2Y(g)}\rightleftharpoons\mathrm{X_2(g)}+\frac12\mathrm{Y_2(g)}$ The equation representing correct relationship between the degree of dissociation (x) of $\mathrm{X_2Y(g)}$ with its equilibrium constant $K_p$ is ______ . Assume $x$ to be very very small.
A) $x=\sqrt[3]{\frac{2K_p}P}$
B) $x=\sqrt[3]{\frac{2K_p^2}P}$
C) $x=\sqrt[3]{\frac{K_p}P}$
D) $x=\sqrt[3]{\frac{K_p}{2P}}$
Explanation
[Solution by Dr. Abubacker] For small $x$: $p_{X_2Y}\approx P$, $p_{X_2}\approx xP$, $p_{Y_2}\approx\frac{xP}2$. $K_p=\frac{xP\left(\frac{xP}2\right)^{1/2}}P=\frac{x^{3/2}P^{1/2}}{\sqrt2}$. Then $x^3=\frac{2K_p^2}P$, so $x=\sqrt[3]{\frac{2K_p^2}P}$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.