Problem 101 - Chemistry The Molecular Science

Question

Cyclopropane isomerizes to propene when heated. Rate constants for the reaction cyclopropane \(\longrightarrow\) propene are \(1.10 \times 10^{-4} \mathrm{~s}^{-1}\) at \(470.0^{\circ} \mathrm{C}\) and \(1.02 \times 10^{-3} \mathrm{~s}^{-1}\) at \(510.0^{\circ} \mathrm{C}\) (a) Calculate the activation energy, \(E_{\mathrm{a}},\) for this reaction. (b) Calculate how long it takes at \(500 .{ }^{\circ} \mathrm{C}\) for the concentration of cyclopropane to drop from \(0.10 \mathrm{M}\) to \(0.023 \mathrm{M}\).

Step-by-Step Solution

Answer

(a) 33.2 kJ/mol, (b) 2839 seconds

Step 1: Use the Arrhenius Equation

The Arrhenius equation is given by \[k = A e^{-E_a / (RT)}\]Taking the natural logarithm of both sides, we obtain:\[\ln k = \ln A - \frac{E_a}{RT}\]We can use two temperatures to create the equation for calculating the activation energy:\[\ln \frac{k_2}{k_1} = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]

Step 2: Substitute known values for Activation Energy

Given:- \(k_1 = 1.10 \times 10^{-4} \text{s}^{-1}\) at \(T_1 = 470 + 273 = 743\,\text{K}\)- \(k_2 = 1.02 \times 10^{-3} \text{s}^{-1}\) at \(T_2 = 510 + 273 = 783\,\text{K}\)- \(R = 8.314\,\text{J/mol K}\)Substitute in the equation:\[\ln \frac{1.02 \times 10^{-3}}{1.10 \times 10^{-4}} = -\frac{E_a}{8.314} \left(\frac{1}{783} - \frac{1}{743}\right)\]Solve for \(E_a\).

Step 3: Calculate Activation Energy

Calculate the left side:\[\ln \frac{1.02 \times 10^{-3}}{1.10 \times 10^{-4}} \approx \ln 9.2727 \approx 2.227\]Calculate the right side:\[\frac{1}{783} - \frac{1}{743} \approx -0.000056 \, \text{K}^{-1}\]Substitute into the equation:\[2.227 = -\frac{E_a}{8.314} \times (-0.000056)\]Solve for \(E_a\):\[E_a \approx \frac{2.227 \times 8.314}{0.000056} \approx 33185 \, \text{J/mol} \approx 33.2 \, \text{kJ/mol}\]

Step 4: Use First-order Reaction Integrated Rate Law

The reaction is first-order, so use the equation:\[\ln \frac{[A]_0}{[A]} = kt\]Solve for \(t\):\[t = \frac{\ln ([A]_0/[A])}{k}\]

Step 5: Calculate Time at 500°C

Substitute the given values:- \([A]_0 = 0.10 \, \text{M}\)- \([A] = 0.023 \, \text{M}\)- Find the rate constant \(k\) at 773 K using\[\ln k = \ln A - \frac{E_a}{RT}\]Since \(k\) at 773 K is unknown, use Arrhenius equation:\[k_1 = 1.10 \times 10^{-4} \text{s}^{-1} , T_1 = 743\text{K}\]\[k = k_1 e^{-\frac{E_a}{R} \left( \frac{1}{773}-\frac{1}{743} \right)}\]Solve for \(k\) and use it to find \(t\):\[t = \frac{\ln (0.10/0.023)}{k}\]

Step 6: Calculate Rate Constant and Time

Calculate \(k\) at 773 K:\[\ln k - \ln (1.10 \times 10^{-4}) = -\frac{33185}{8.314} \left(\frac{1}{773}-\frac{1}{743}\right)\]Solve for \(k\):\[k \approx 4.89 \times 10^{-4} \text{s}^{-1}\]Then find \(t\):\[t = \frac{\ln (0.10/0.023)}{4.89 \times 10^{-4}} \approx \frac{1.39}{4.89 \times 10^{-4}} \approx 2839 \, \text{s}\]

Key Concepts

Arrhenius Equation

The Arrhenius Equation is a fundamental expression that describes how the rate constant ( \(k\)) of a chemical reaction changes with temperature. This equation highlights the relationship between the reaction rate and the activation energy (\(E_a\)), which is the minimum energy required to initiate a reaction. The equation is given by:

• \[k = A e^{-E_a / (RT)}\]

Here, \(A\) is the frequency factor or pre-exponential factor, which is indicative of the frequency of collisions with the correct orientation. \(R\) is the universal gas constant, and \(T\) is the absolute temperature in Kelvin. By taking the natural logarithm of both sides, the Arrhenius equation can be linearized, which is useful for deriving values such as the activation energy.

One practical application of this equation is to determine the activation energy using experimental data of rate constants at different temperatures. By plotting \( \ln k\) versus \(1/T\), you can obtain a straight line where the slope is proportional to \(E_a\). In the calculation provided, we used the exponential form of the equation to solve for the activation energy by substituting values for the rate constants at two specific temperatures.

First-Order Reaction

Reactions where the rate is directly proportional to the concentration of one reactant are called first-order reactions. The reaction rate for such processes depends linearly upon the concentration of a single reactant raised to the first power. This means that if you double the concentration of the reactant, the reaction rate also doubles.

The general equation for a first-order reaction is given by

• \[ ext{Rate} = k[A]\]

where \([A]\) is the concentration of the reactant, and \(k\) is the rate constant. Due to their nature, these reactions typically have a characteristic exponential decay of the reactant concentration over time.First-order reactions are characterized by a specific time-dependent relationship which is derived as the integrated rate law:

• \[ ext{{ ext{ln} }} ([A]_0/[A]) = kt\]

where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is again the rate constant. The integration implies that any change in the reactant concentration over time is exponential while the half-life remains constant regardless of concentration.

Integrated Rate Law

The integrated rate law provides a connection between the concentration of reactants and time, making it easier to quantify how reactant concentrations change during a reaction. For first-order reactions, the law is expressed as

• \[ ext{{ ext{ln} }} ([A]_0/[A]) = kt\]

This equation makes it possible to predict the concentration of a reactant at any given time if the initial concentration and the rate constant are known.

To compute time required to reach a certain concentration, like in the exercise, you can rearrange this equation to get

• \[t = \frac{\text{ln}([A]_0/[A])}{k}\]

For our reaction, this form facilitated the determination of time taken for a drop in cyclopropane's concentration at given conditions. Using the determined rate constant at 500°C from previous calculations, we substituted the initial and final concentrations to solve for the time variable. Understanding and using integrated rate laws enable visualization and calculation of the dynamics of chemical reactions effectively.