Question

The value of Kc for the reaction $3O_2\left(g\right)\rightleftharpoons 2O_3\left(g\right)$ is $1.7\times {10}^{-56}$ at ${25}^{o}C$. Do you expect pure air at ${25}^{o}C$ to contain much $O_3$ (ozone) when $O_2$ and $O_3$ are in equilibrium ? If the equilibrium concentration of $O_2$ in air at ${25}^{o}C$ is $8\times {10}^{3}M$, what is the equilibrium concentration of $O_3$?

• A. Yes,$\sim 3\times {10}^{-32}mol/L$
• B. Yes,$\sim 6\times {10}^{-32}mol/L$
• C. Yes,$\sim 9\times {10}^{-32}mol/L$
• D. None of these

Answer 1

The correct option is C Yes,$\sim 9\times {10}^{-32}mol/L$

$3O_2\left(g\right)=2O_3\left(g\right)$

Then, $K_c=\frac{[O_3{]}^2}{[O_2{]}^3}$

$1.7\times {10}^{-56}=\frac{[O_3{]}^2}{[8\times {10}^{-3}{]}^3}$

$\left[O_3\right]=8.7\times {10}^{-32}mol/L$ ~ $9\times {10}^{-32}mol/L$