Q. In the reactions, $A + 2B \rightleftharpoons 2C$, if 2 moles of $A$, 3.0 moles of $B$ and 2.0 moles of $C$ are placed in a 2 L flask and the equilibrium concentration of $C$ is 0.5 mol/L, the equilibrium constant $(K_{c})$ for the reactions is
A) 0.21
B) 0.50
C) 0.75
D) 0.025
Explanation
(a)
$A + 2B\ \ \rightleftharpoons \ \ 2C$
2 3 2 initial moles
$(2 - 0.5)(3 - 0.5)(2 \times 0.5)$ at equilibrium
Molar concentration of $A = \frac{1.5}{2}$
Molar concentration of $B = \frac{2.5}{2}$
Molar concentration of $C = \frac{1}{2}$
$K = \frac{\lbrack C\rbrack^{2}}{\lbrack A\rbrack\lbrack B\rbrack^{2}} = \frac{1 \times 1 \times 2 \times 2 \times 2 \times 2}{2 \times 2 \times 1.5 \times 2.5 \times 2.5}$
$= \frac{2}{1.5 \times 2.5 \times 2.5} = 0.21$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.