Q. For $PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2}$, initial concentration of each reactant and product is 1 M. If $K_{eq} = 0.41$ then
A) More $PCl_{3}$ will form
B) More $Cl_{2}$ will form
C) More $PCl_{5}$ will form
D) No change
Explanation
(c)
$K_{c} = \frac{\lbrack product\rbrack}{\lbrack reactant\rbrack}$
$0.41 = \frac{\lbrack product\rbrack}{\lbrack reactant\rbrack}$
$\Rightarrow \ \ \ \ \ \lbrack product\rbrack = 0.41 \times \lbrack reactant\rbrack$
$\Rightarrow \ \ \ \ \ \lbrack product\rbrack < \lbrack reactant\rbrack$
Therefore, reaction proceeds in backward direction, i.e., more $PCl_{5}$ will be formed.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.