Problem 52 - Chemistry The Molecular Science

Question

\(p\) -Methylphenyl acetate reacts with imidazole to produce \(p\) -methylphenol and acetyl imidazole. The rate constants for this second-order reaction at two temperatures are given in the table. \begin{tabular}{cc} \hline\(T\left({ }^{\circ} \mathrm{C}\right)\) & \(k\left(\mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 10.0 & \(2.34 \times 10^{-2}\) \\ 60.0 & \(1.52 \times 10^{-1}\) \\ \hline \end{tabular} (a) Calculate the activation energy and frequency factor for this reaction. (b) Estimate the rate constant for this reaction at a temperature of \(100.0^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

Activation energy is approximately 52.5 kJ/mol, frequency factor is approximately 1.03 x 10^8 L/mol s, and rate constant at 100 °C is about 4.93 L/mol s.

Step 1: Calculate ln(k) Values

First, we need to determine the natural logarithms of the given rate constants: \[\ln(2.34 \times 10^{-2}) = -3.75\]\[\ln(1.52 \times 10^{-1}) = -1.88\]

Step 2: Convert Temperatures to Kelvin

Convert the given Celsius temperatures to Kelvin:\[T_1 = 10.0 + 273.15 = 283.15 \text{ K}\]\[T_2 = 60.0 + 273.15 = 333.15 \text{ K}\]

Step 3: Use Arrhenius Equation

The Arrhenius equation in linear form is:\[\ln(k) = \ln(A) - \frac{E_a}{R} \cdot \frac{1}{T}\]Where:- \(k\) is the rate constant- \(A\) is the frequency factor- \(E_a\) is the activation energy \( (\text{J/mol}) \)- \(R\) is the ideal gas constant \(8.314 \, \text{J/mol K}\)

Step 4: Calculate Activation Energy (Ea)

We can rewrite the equation for two temperatures as:\[\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]Substitute the values:\[\ln\left(\frac{1.52 \times 10^{-1}}{2.34 \times 10^{-2}}\right) = -\frac{E_a}{8.314} \cdot \left(\frac{1}{333.15} - \frac{1}{283.15}\right)\]Solve for \(E_a\):\[1.87 = -\frac{E_a Ea\_a}{8.314} \times (-5.043 \times 10^{-4})\]\[E_a = 2.87 \times 8.314 \times 1.985 \times 10^{4} \approx 52471.14 \, \text{J/mol} \]

Step 5: Calculate Frequency Factor (A)

Using the linear form of the Arrhenius equation and the derived \(E_a\), calculate \(A\): \[\ln(k_1) = \ln(A) - \frac{E_a}{R \times T_1} \]\[-3.75 = \ln(A) - \frac{52471.14}{8.314 \times 283.15}\] Find \(\ln(A)\):\[ \ln(A) = -3.75 + 22.239 \approx 18.489\] \[A \approx e^{18.489} \approx 1.03 \times 10^{8} \, \text{L/mol s} \]

Step 6: Estimate Rate Constant at 100 °C

Now estimate the rate constant at 100 °C (373.15 K) using the Arrhenius equation: \[\ln(k) = \ln(A) - \frac{E_a}{R \times T}\]\[\ln(k) = 18.489 - \frac{52471.14}{8.314 \times 373.15}\]\[\ln(k) = 18.489 - 16.893\]\[k \approx e^{1.596} \approx 4.93 \, \text{L/mol s} \]

Key Concepts

Activation Energy

Activation energy, often denoted as \( E_a \), is the minimum amount of energy that reactant molecules must have for a chemical reaction to occur. This concept stems from the energy barrier that needs to be overcome for a chemical transformation to happen.
A useful way to envision activation energy is to think about pushing a boulder over a hill. The hill represents the energy barrier. Only when enough energy is provided to push the boulder up the hill, it can roll down the other side. Similarly, in chemical reactions, reactants require sufficient energy to reach an "activated" state. Once in this state, the reaction can proceed to form products.
Understanding activation energy:

• It determines the rate of a reaction—lower activation energies result in faster reactions.
• Enzymes in living organisms function to lower the activation energy threshold, allowing biochemical reactions to occur more rapidly and at lower temperatures.

Arrhenius Equation

The Arrhenius equation is a mathematical expression that describes how changes in temperature affect the rate constants of chemical reactions. It is given in its exponential form as:\[k = A \, e^{-\frac{E_a}{RT}}\]Where:

• \( k \) is the rate constant.
• \( A \) is the frequency factor or pre-exponential factor, representing the number of times reactants approach the activation energy per unit time.
• \( E_a \) is the activation energy (J/mol).
• \( R \) is the universal gas constant, approximately 8.314 J/mol K.
• \( T \) is the temperature in Kelvin.

This equation highlights the exponential relationship between the rate constant \( k \) and temperature \( T \). As the temperature increases, molecules have more kinetic energy, meaning more can surpass the activation energy barrier, leading to an increase in the reaction rate.
Moreover, by transforming the Arrhenius equation to its linearized form, \( \ln(k) = \ln(A) - \frac{E_a}{R} \frac{1}{T} \), we can easily find \( E_a \) and \( A \) by plotting \( \ln(k) \) against \( \frac{1}{T} \), forming a straight line.

Temperature Dependence of Reaction Rates

Temperature plays a critical role in determining the rates at which chemical reactions occur. Usually, an increase in temperature results in an increase in reaction rates. This occurs because higher temperatures provide more energy to the molecules, which leads to more collisions and more energy to overcome the activation energy barrier.
Consider the following aspects:

• With every 10°C rise in temperature, the rate of a chemical reaction generally doubles. This is a rough rule of thumb but underscores the sensitivity of reactions to temperature changes.
• Molecules at higher temperatures move faster, increasing both the frequency and the energy of collisions between them.

Let's apply this to the reaction of \(p\)-methylphenyl acetate with imidazole: As the solution indicates, the rate constant increases significantly between 10°C and 60°C, affirming the temperature dependence described by the Arrhenius equation.
In summary, temperature dependence is crucial for understanding how and why chemical processes can be accelerated or slowed down by simply adjusting the thermal conditions under which they occur.