**We are asked to determine what fraction of the O _{3} will have reacted when the rate falls to one-half of its initial value.**

Recall that the **rate law** only focuses on the reactant concentrations and has a general form of:

$\overline{){\mathbf{rate}}{\mathbf{}}{\mathbf{law}}{\mathbf{=}}{\mathbf{k}}{\left[\mathbf{A}\right]}^{{\mathbf{x}}}{\left[\mathbf{B}\right]}^{{\mathbf{y}}}}$

k = rate constant

A & B = reactants

x & y = reactant orders

**Given: **

$\overline{){\mathbf{Rate}}{\mathbf{=}}{\mathbf{k}}\frac{{\mathbf{\left[}{\mathbf{O}}_{\mathbf{3}}\mathbf{\right]}}^{{\mathbf{2}}}}{\mathbf{\left[}{\mathbf{O}}_{\mathbf{2}}\mathbf{\right]}}}$

[O_{3}] = 1.0 mol

[O_{2}] = 1.0 mol

Volume of Container = 1 L

Consider the following reaction: 2O_{3}(g) → 3O_{2}(g). The rate law for this reaction is as follows: Rate = k[O_{3}]^{2}/[O_{2}]

Suppose that a 1.0-L reaction vessel initially contains 1.0 mol of O_{3} and 1.0 mol of O_{2}. What fraction of the O_{3} will have reacted when the rate falls to one-half of its initial value?

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