Problem 103 - Chemistry The Central Science In Si Units

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{tabular}{cc} \hline 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{tabular} 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 reaction is second order. (b) The rate constant is 0.1625 \(\mathrm{M}^{-1} \mathrm{s}^{-1}\).

Step 1: Determine the Reaction Order

First, we need to determine the order of the reaction by analyzing plots of concentration versus time. We plot \([\mathrm{C}_{5} \mathrm{H}_{6}]\), \(\ln [\mathrm{C}_{5} \mathrm{H}_{6}]\), and \(\frac{1}{[\mathrm{C}_{5} \mathrm{H}_{6}]}\) against time. For a first-order reaction, the plot of \(\ln [\mathrm{C}_{5} \mathrm{H}_{6}]\) vs. time should be linear. For a second-order reaction, the \(\frac{1}{[\mathrm{C}_{5} \mathrm{H}_{6}]}\) plot would be linear. We calculate and plot the data to see which is linear.

Step 2: Calculate Natural Logarithm and Reciprocal Values

Using the initial concentration \([\mathrm{C}_{5} \mathrm{H}_{6}]=0.0400\,\text{M}\), we calculate \(\ln [\mathrm{C}_{5} \mathrm{H}_{6}]\) and \(\frac{1}{[\mathrm{C}_{5} \mathrm{H}_{6}]}\) as follows:\\(\begin{align*}\text{Time (s)} & : \ln[\mathrm{C}_{5} \mathrm{H}_{6}] & \frac{1}{[\mathrm{C}_{5} \mathrm{H}_{6}]} \0.0 & : \ln(0.0400) & 25.0 \50.0 & : \ln(0.0300) & 33.3 \100.0 & : \ln(0.0240) & 41.7 \150.0 & : \ln(0.0200) & 50.0 \200.0 & : \ln(0.0174) & 57.5 \\end{align*}\)

Step 3: Analyze the Graphs

Plot each of the calculated values against time. By analyzing the graphs, we observe that the plot for \(\frac{1}{[\mathrm{C}_{5} \mathrm{H}_{6}]}\) vs. time is a straight line, indicating a second-order reaction. This linear relationship confirms the reaction order as second order.

Step 4: Calculate the Rate Constant

For a second-order reaction, the integrated rate law is \(\frac{1}{[A]} = kt + \frac{1}{[A]_0}\). From the slope of the \(\frac{1}{[\mathrm{C}_{5} \mathrm{H}_{6}]}\) vs. time graph, we can determine the rate constant \(k\). Using two points from the plot, say \((0.0 \,\text{s}, 25.0)\) and \((200.0 \,\text{s}, 57.5)\), the slope \(k\) is calculated as: \ \( k = \frac{57.5 - 25.0}{200 - 0} = \frac{32.5}{200} = 0.1625 \mathrm{\, M^{-1} s^{-1}} \).

Key Concepts

Reaction Order Determination

Determining the reaction order is a crucial step in understanding chemical kinetics. It involves assessing how the concentration of reactants influences the rate of reaction. In this case, with cyclopentadiene (1C_{53H_{6}2}) reacting to form dicyclopentadiene, plots are used to reveal the reaction order.

Three plots are commonly compared: concentration 1C1 versus time, natural logarithm of concentration 1Ln [C_{53H_{6}2]} versus time, and reciprocal of concentration 11/[C_{53H_{6}2]} versus time.

• If the 1Ln [C_{53H_{6}2]} plot is linear, the reaction is first-order.
• If the 11/[C_{53H_{6}2]} plot is linear, the reaction is second-order.

In this exercise, the linearity of the 11/[C_{53H_{6}2]} vs. time plot indicates that the reaction is second-order. This means that the rate of reaction depends on the square of the concentration of cyclopentadiene. Knowing the order allows us to predict how changing conditions affect reaction rates.

Integrated Rate Laws

Integrated rate laws are equations that relate the concentration of reactants to time. They are tailored to different reaction orders - zero, first, and second. For a second-order reaction like the one in this exercise, the integrated rate law is especially useful.

The second-order rate equation is given by: \[ \frac{1}{[A]} = kt + \frac{1}{[A]_0} \]where:

• \([A]\) is the concentration of the reactant at time \(t\)
• \([A]_0\) is the initial concentration
• \(k\) is the rate constant

This equation describes how the concentration changes over time and can be rearranged to predict concentrations at different times or determine the rate constant (\(k\)). In many lab settings, this relationship helps interpret how quickly reactions proceed under various settings by integrating observed data.

Rate Constant Calculation

The rate constant (\(k\)) is vital in kinetic studies as it quantifies the speed of a reaction. To find \(k\) for a second-order reaction as in this exercise, you use the slope of a linear plot of 1B/[C_{53H_{6}2]} against time.

The slope of this plot is equivalent to \(k\). Given points like (0.0 s, 25.0) and (200.0 s, 57.5), the calculation follows:\[ k = \frac{57.5 - 25.0}{200 - 0} = \frac{32.5}{200} = 0.1625 \, \text{M}^{-1}\text{s}^{-1} \]This formula shows the change in reciprocal concentration over time, directly correlating to the rate constant.

With \(k\) known, you can predict the reaction rate in particular scenarios, which is valuable for process optimization and industrial applications. It reflects how efficiently and swiftly a reaction unfolds at given concentrations.