Question

For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene \(\left( {{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{ = CH - CH = C}}{{\bf{H}}_{\bf{2}}}} \right)\) has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:

The isomerization of cyclobutene to butadiene is first-order, and the rate constant has been measured as \({\bf{2}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 150 \({\bf{^\circ C}}\) in a 0.53-L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at 150 \({\bf{^\circ C}}\) with an initial pressure of 55 torr.

Short answer

The isomerization can be said as the change in the bonding and the position of the atoms in the molecule, but the number of atoms is same. The partial pressure of cyclobutene after 30 minutes at 150⁰C is 78.65torr.

Step-by-step solution

Reaction Rate

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 the 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.

Numerical Explanation

Time Taken = 30minutes = 1800 seconds

Rate constant of the cyclobutene calculated = \({\bf{2}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{{\bf{s}}^{{\bf{ - 1}}}}\)

Initial Pressure = 55torr

The partial pressure of cyclobutene is:

\(\begin{align}k &= {\bf{ }}\frac{{2.303}}{t}Log{\bf{ }}\frac{P}{{{P_0}}}\\1800s &= {\bf{ }}\frac{{2.303}}{{0.0002s}}Log{\bf{ }}\frac{P}{{55}}\\Log{\bf{ }}\frac{P}{{55}} &= {\bf{ }}\frac{{0.0002}}{{2.303}} \times 1800\\\frac{P}{{55}} &= AntiLog{\bf{ }}(0.156)\end{align}\)

\(\begin{align}\frac{P}{{55}} &= 1.43\\P &= 55 \times 1.43\\P &= 78.65torr\end{align}\)

Hence, the partial pressure of cyclobutene after 30 minutes\({\bf{150^\circ C}}\) is 78.65torr.