Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for \(\frac{{\bf{1}}}{{{\bf{64}}}}\) of the initial dose to remain in the athlete’s body?

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 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 first order of Steroids is:

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

The half-life period of the steroid taken by an athlete = 42days

\(\begin{align}Half - life{\bf{ }}period{\bf{ }} &= {\bf{ }}\frac{{0.693}}{k}\\42days{\bf{ }} &= {\bf{ }}\frac{{0.693}}{k}\\k{\bf{ }} &= {\bf{ }}\frac{{0.693}}{{42days}}\\k &= 0.0165day{s^{ - 1}}\end{align}\)

Rate constant of the steroids calculated = 0.0165 days^-1

Therefore, the time period taken it \(\frac{1}{{64}}\)for the initial dose is:

\(\begin{align}k &= {\bf{ }}\frac{{2.303}}{t}Log{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}}\\t &= {\bf{ }}\frac{{2.303}}{k}Log{\bf{ }}\frac{{{{\left( A \right)}_0}}}{{1/64{{\left( A \right)}_0}}}\\t &= {\bf{ }}\frac{{2.303}}{{0.0165}}Log{\bf{ }}64\\t &= {\bf{ }}\frac{{2.303}}{{0.0165}} \times 1.81\\t &= {\bf{ }}252.63days\end{align}\)