Problem 103 - Chemistry: The Central Science

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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: \[ -\frac{\mathrm{d}\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]}{\mathrm{d}t} = k \left[\mathrm{C}_{5} \mathrm{H}_{6}\right] \]

Step 3: 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 \(k = 0.00575 \mathrm{s}^{-1} \).