What is the half-life for the decomposition of \({{\bf{O}}_{\bf{3}}}\) when the concentration of \({{\bf{O}}_{\bf{3}}}\)is \({\bf{2}}{\bf{.35 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)M? The rate constant for this second-order reaction is 50.4 L/mol/h.

The reaction involved the effective collision of two reactants to produce the desired products. Reactions can be natural, which occur in the surrounding environment, whereas it can be artificially done in the laboratory to form a desired required product.

The reaction rate can be defined as the reaction speed to produce the products. The reaction rate can be slow, fast or moderate. The reaction can take less than a millisecond to produce products, or it can take years to produce the desired product.

The half-life period can be defined as the time period at which half the concentration of the reactants gets converted into a product.

The half-life period of the second-order decomposition of the O₃ is:

\({\bf{Half - life period = }}\frac{{\left( {\bf{A}} \right){\bf{^\circ }}}}{{{\bf{2k}}}}\)

Where \(\left( {\bf{A}} \right){\bf{^\circ }}\) = initial concentration of reactant

K = Rate constant

Initial concentration of reactant, (A)° =\({\bf{2}}{\bf{.35 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)M

Rate constant for the decomposition of O₃ = 50.4L/mol/h

\(\begin{align}{\bf{Half - life period = }}\frac{{{\bf{2}}{\bf{.35 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{mole/L}}}}{{{\bf{2 \times }}\left( {{\bf{50}}{\bf{.4L/mole/h}}} \right)}}\\{\bf{Half - life period = }}\frac{{{\bf{2}}{\bf{.35 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{mole/L}}}}{{{\bf{100}}{\bf{.8L/mole/h}}}}\\{\bf{Half - life period = 2}}{\bf{.33 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}{\bf{hour}}\end{align}\)

Half-life period = \(2.33 \times {10^{ - 8}}\) hour.

Hence, the half-life period of the second-order decomposition of the O₃ is \(2.33 \times {10^{ - 8}}\) hours which has rate constant is 50.4 L/mol/h.