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$3O_2(g)\rightleftharpoons2O_3(g)$ for the given reaction at 298 K, $K_c$ is found to be $3.0\times10^{-59}$. If the concentration of $O_2$ at equilibrium is 0.040 M then concentration of $O_3$ in M is
A
$1.2\times10^{21}$
B
$4.38\times10^{-32}$
C
$1.9\times10^{-63}$
D
$2.4\times10^{31}$
Detailed Solution
$K_c = \frac{[O_3]^2}{[O_2]^3}$
$3 \times 10^{-59} = \frac{[O_3]^2}{(4 \times 10^{-2})^3}$
$[O_3]^2 = 3 \times 10^{-59} \times 64 \times 10^{-6} = 19.2 \times 10^{-64}$
$[O_3] = 4.38 \times 10^{-32}$ M