Question

Cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) reacts with itself to form dicyclopentadiene $\left(\mathrm{C}_{10} \mathrm{H}_{12}\right)\(. A \)0.0400 \mathrm{M}\( solution of \)\mathrm{C}_{5} \mathrm{H}_{6}$ was monitored as a function of time as the reaction $2 \mathrm{C}_{5} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12}$ proceeded. The following data were collected: $$ \begin{array}{cc} \hline \text { Time (s) } & {\left[\mathrm{C}_{5} \mathrm{H}_{6}\right](M)} \\\ \hline 0.0 & 0.0400 \\ 50.0 & 0.0300 \\ 100.0 & 0.0240 \\ 150.0 & 0.0200 \\ 200.0 & 0.0174 \\ \hline \end{array} $$ Plot \(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, $\ln \left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\( versus time, and \)1 /\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]$ versus time. (a) What is the order of the reaction? (b) What is the value of the rate constant?

Short answer

(a) The order of the reaction is 1. (b) The value of the rate constant is \(k = 0.00575 \mathrm{s}^{-1} \).

Step-by-step solution

Plot the Data

Create three plots with the given data: 1. Concentration ([C5H6]) versus time. 2. Natural log of concentration (ln[C5H6]) versus time. 3. Reciprocal of concentration (1/[C5H6]) versus time. We need to analyze these graphs to determine which one gives a straight line, which will indicate the order of the reaction.

Identifying the Order of the Reaction

After plotting the data, we will see that the graph of ln([C5H6]) versus time provides a straight line. This implies that the reaction is first-order. In general, for a first-order reaction, we have the rate law: \[ -\frac{\mathrm{d}\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]}{\mathrm{d}t} = k \left[\mathrm{C}_{5} \mathrm{H}_{6}\right] \]

Calculating the Rate Constant

For a first-order reaction, the equation can be expressed in terms of concentration and time: \[ \ln \frac{\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]_{0}}{\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]} = kt\] We can use data from one of the points to calculate the rate constant (k). Using the data at t = 50.0 seconds: \[ \ln \frac{0.0400 \mathrm{M}}{0.0300 \mathrm{M}} = k(50.0 \mathrm{s})\] Calculate k as: \[ k = \frac{\ln \frac{0.0400}{0.0300}}{50.0} = \frac{0.2877}{50.0} = 0.00575 \mathrm{s}^{-1} \] (a) The order of the reaction is 1. (b) The value of the rate constant is \(k = 0.00575 \mathrm{s}^{-1} \).