Q. 2 moles of $PCl_{5}$ were heated in a closed vessel of 2 L capacity. At equilibrium 40% of $PCl_{5}$ is dissociated into $PCl_{3}$and $Cl_{2}$. The value of equilibrium constant is
A) 0.266
B) 0.366
C) 2.66
D) 3.66
Explanation
(a)
$PCl_{5} \rightleftharpoons {PCl}_{3} + {Cl}_{2}$
$0\ \ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ 0$ initially
$\frac{2 \times 60}{100}\frac{\ 2 \times 40}{100}\frac{2 \times 40}{100}$ at equilibrium
Volume of container = 2L
$K_{c} = \frac{\frac{2 \times 40}{100 \times 2} \times \frac{2 \times 40}{100 \times 2}}{\frac{2 \times 60}{100 \times 2}} = 0.266$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.