Question

There are two molecules with the formula\({{\bf{C}}_{\bf{3}}}{{\bf{H}}_{\bf{6}}}\). Propene,\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CH = C}}{{\bf{H}}_{\bf{2}}}\), is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anaesthetic:

When heated to 499\({\bf{^\circ C}}\), cyclopropane rearranges (isomerizes) and forms propene with a rate constant of\({\bf{5}}{\bf{.95 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{{\bf{s}}^{{\bf{ - 1}}}}\). What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499.5\({\bf{^\circ C}}\)?

Short answer

The reaction isomerisation of cyclopropane follows first-order reaction. The half-life period of the reaction is 0.324hours. 20% of cyclopropane remains after 0.75h.

Step-by-step solution

Intermolecular Forces

Intermolecular forces are the interaction which are formed by the attraction of the two having opposite charge (partial positive and partial negative charge). The opposite charge are formed by the presence of the electron-negative atom in the molecule. Due to the presence of electron-negative atom, there is and induced partial positive charge is generated on the electron-positive charge (or less electron-negative charge) atom. There will be an attraction between the both oppositely charges to form a bond.

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

Explanation

The half-life period of the first-order reaction is:

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

The half-life period of the reaction does not depend upon the concentration of the reactant.

Rate constant for the isomerisation of cyclopropane\({\bf{ = 5}}{\bf{.95 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{{\bf{s}}^{{\bf{ - 1}}}}\)

\(\begin{align}Half - life{\bf{ }}period{\bf{ }} &= {\bf{ }}\frac{{0.693}}{{5.95{\bf{ }} \times {\bf{ }}{{10}^{ - 4}}{{\sec }^{ - 1}}}}\\Half - life{\bf{ }}period{\bf{ }} &= {\bf{ }}1165\sec onds\end{align}\)

Convert seconds into hours:

\(\begin{align}Half - life{\bf{ }}period{\bf{ }} &= {\bf{ }}\frac{{1165\sec onds}}{{3600\sec onds}} \times 1hour\\Half - life{\bf{ }}period{\bf{ }} &= 0.324hours\end{align}\)

The half-life period of the reaction is 0.324hours.

\(\begin{align}{\bf{Time given = 0}}{\bf{.75 hour }}\\{\bf{ = 2700 seconds}}\end{align}\)

\(\begin{align}k &= {\bf{ }}\frac{{2.303}}{t}Log{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}}\\5.95{\bf{ }} \times {\bf{ }}{10^{ - 4}}{s^{ - 1}} &= {\bf{ }}\frac{{2.303}}{{2700s}}Log{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}}\\5.95{\bf{ }} \times {\bf{ }}{10^{ - 4}} \times \frac{{2700}}{{2.303}} &= {\bf{ }}Log{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}}\\Log{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}} &= 0.7\end{align}\)

\(\begin{align}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}} &= {\bf{ }}AntiLog(0.7)\\\frac{1}{{{{\left( A \right)}_0}}} &= 5.01\\{\left( A \right)_0} &= \frac{1}{{5.01}}\\{\left( A \right)_0} &= 0.2\end{align}\).

The percentage of fraction remained after 0.75 hours is calculated below

\(\begin{align}{\bf {Percentage of Fraction Remained = 0}}{\bf{.2 \times 100}}\\{\bf{ = 20\% }}\end{align}\)