Question

For the reaction:

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

K = 1.8×10^-7 at a certain temperature. If at equilibrium [O₂] = 0.062 M, calculate the equilibrium O₃ concentration.

Short answer

The concentration of O₃ at equilibrium is$6.5\times {10}^{-6^{}}M$.

Step-by-step solution

Equilibrium constant expression

The equilibrium constant, K expression in terms of concentrations for the given reaction, is shown below:

$\text{K =}\frac{{\left[{\text{O}}_{\text{3}}\right]}^{\text{2}}}{{\left[{\text{O}}_{\text{2}}\right]}^{\text{3}}}$..........(1)

The equilibrium constant, K value, is given as $1.8\times {10}^{-7}$.

The equilibrium concentration of oxygen is given as 0.062 M.

Concentration of ozone at equilibrium

Substitute$1.8\times {10}^{-7}$for K and 0.062 for [$O_2$] into Equation (1) to calculate the [$O_3$].

$1.8\times {10}^{-7}=\frac{{\left[O_3\right]}^2}{{(0.062)}^3}$

Rearrange to $O_3$:

$$\begin{gathered} \left[O_3\right]=\sqrt{1.8\times {10}^{-7}\times {(0.062)}^3} \\ \left[O_3\right]=6.5\times {10}^{-6}M \end{gathered}$$

Therefore, the concentration of O₃ at equilibrium is$6.5\times {10}^{-6}M$.