Question

$3 O _{2}( g ) \rightleftharpoons 2 O _{3}( g )$
for the above 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. $2.4 \times 10^{31}$

Answer 1

Answer: (B)

$3 O _{2}( g ) \rightleftharpoons 2 O _{3}( g )$
$K _{ c }=\frac{\left[ O _{3}\right]^{2}}{\left[ O _{2}\right]^{3}}$
$3 \times 10^{-59}=\frac{\left[ O _{3}\right]^{2}}{\left(4 \times 10^{-2}\right)^{3}}$
${\left[ O _{3}\right]^{2}=3 \times 10^{-59} \times 64 \times 10^{-6} }$
$=19.2 \times 10^{-64}$
$=4.38 \times 10^{-32} M$