Q. $3O_2(g) \rightleftharpoons 2O_3(g)$ for the given reaction at 298 K, $K_c$ is found to be $3.0 \times 10^{-59}$. If the concentration of $O_2$ at equilibrium is 0.040 M then concentration of $O_3$ in M is
A) $1.2 \times 10^{21}$
B) $4.38 \times 10^{-32}$
C) $1.9 \times 10^{-63}$
D) $2.4 \times 10^{31}$
Explanation
[Solution] $[O_3]^2 = K_c[O_2]^3 = 3.0 \times 10^{-59} \times (0.040)^3 = 1.92 \times 10^{-63}$, so $[O_3] = 4.38 \times 10^{-32}$ M. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.