Setup the K expression and find the pressure of O_{3}

$\overline{){\mathbf{K}}{\mathbf{=}}\frac{{\mathbf{P}}_{\mathbf{product}}}{{\mathbf{P}}_{\mathbf{reactant}}}}\phantom{\rule{0ex}{0ex}}\mathbf{K}\mathbf{=}\frac{{{\mathbf{P}}_{{\mathbf{O}}_{\mathbf{3}}}}^{\mathbf{2}}}{{{\mathbf{P}}_{{\mathbf{O}}_{\mathbf{2}}}}^{\mathbf{3}}}\phantom{\rule{0ex}{0ex}}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathbf{=}\mathbf{}\frac{{{\mathbf{P}}_{{\mathbf{O}}_{\mathbf{3}}}}^{\mathbf{2}}}{{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{062}\mathbf{)}}^{\mathbf{3}}}$'

For the reaction:

3O_{2}(*g*) ⇌ 2O_{3}(*g*)

*K* = 1.8 x 10 ^{-7} at a certain temperature. If at equilibrium [O _{2}] = 0.062 M, calculate the equilibrium O_{3} concentration.

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