Q. 1.6 moles of $PCl_{5}(g)$ is placed in $4\ dm^{3}$ closed vessel. When the temperature is raised to 500 K, it decomposes and at equilibrium 1.2 moles of $PCl_{5}(g)$ remains. What is the $K_{c}$ value for the decomposition of $PCl_{5}(g)$ to $PCl_{3}(g)\ and\ {Cl}_{2}(g)\$at 500 K?
Explanation
(c)
$PCl_{5}\ \ \ \ \rightleftharpoons \ \ \ \ \ PCl_{3}\ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Cl_{2}$
1.6 0 0 initially
(1.6-$x$) $x$ $x$ At equilibrium
$\frac{1.6 - x}{4}$ $\frac{x}{4}$ $\frac{x}{4}$
$\because$ $1.6 - x\ \ \ \ \ \ 1.2$
$x\ \ \ \ \ \ \ 0.4$
$\left\lbrack PCl_{5} \right\rbrack\ \ \ \ \ \ \ \ \frac{1.2}{4}$
$\left\lbrack PCl_{3} \right\rbrack\ \ \ \ \ \ \ \ \frac{0.4}{4}$
$\left\lbrack Cl_{2} \right\rbrack\ \ \ \ \ \ \ \ \ \ \frac{0.4}{4}$
$K_{c}\ \ \ \ \ \ \frac{\left\lbrack PCl_{3} \right\rbrack\left\lbrack Cl_{2} \right\rbrack}{\left\lbrack PCl_{5} \right\rbrack}\$
$\frac{\frac{\ 0.4}{4}\ \ \ \ \ \ \ \ \ \ \frac{\ 0.4}{4}}{\frac{1.2}{4}}\$
$\frac{0.1\ \ \ \ \ \ \ \ \ \ \ \ 0.1}{0.3}$
0.033
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.