Question

The reaction $$ \left(\mathrm{CH}_{\mathrm{3}}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{Br}^{-} $$ in a certain solvent is first order with respect to \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and zero order with respect to \(\mathrm{OH}^{-}\). In several experiments, the rate constant \(k\) was determined at different temperatures. \(\mathrm{A}\) plot of \(\ln (k)\) versus \(1 / T\) was constructed resulting in a straight line with a slope value of \(-1.10 \times 10^{4} \mathrm{~K}\) and \(y\) -intercept of 33.5. Assume \(k\) has units of \(\mathrm{s}^{-1}\). a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor \(A\). c. Calculate the value of \(k\) at \(25^{\circ} \mathrm{C}\).

Short answer

a. The activation energy for this reaction is \(91.5 \mathrm{~kJ} \ \mathrm{mol}^{-1}\). b. The frequency factor is \(3.6 \times 10^{14} \ \mathrm{s}^{-1}\). c. The rate constant at \(25^{\circ} \mathrm{C}\) is \(3.74 \times 10^{-2} \ \mathrm{s}^{-1}\).

Step-by-step solution

Find the Activation Energy

We can rewrite the \(\ln(k)\) equation as follows: \[\ln(k) = \ln(A) - \frac{E_a}{R}\frac{1}{T}\] Comparing it to the equation of a straight line, \(y = mx + b\), we can see that: - \(\ln(k)\) corresponds to \(y\) - \(\frac{1}{T}\) corresponds to \(x\) - \(-\frac{E_a}{R}\) corresponds to the slope (\(m\)) - \(\ln(A)\) corresponds to the y-intercept (\(b\)) Now, we have the slope value, which is \(-1.10 \times 10^4 \ \mathrm{K}\). Equating the slope with \(-\frac{E_a}{R}\), we can solve for \(E_a\): -\(\frac{E_a}{R} = -1.10 \times 10^4 \ \mathrm{K}\) Now, we can multiply both sides by the negative gas constant: \(E_a = 1.10 \times 10^4 \ \mathrm{K} \times 8.314 \ \mathrm{J} \ \mathrm{mol}^{-1} \ \mathrm{K}^{-1}\) \(E_a = 91.5 \times 10^3 \ \mathrm{J} \ \mathrm{mol}^{-1}\) or \(91.5 \mathrm{~kJ} \ \mathrm{mol}^{-1}\) The activation energy for this reaction is \(91.5 \mathrm{~kJ} \ \mathrm{mol}^{-1}\).

Find the Frequency Factor

We have determined the activation energy and have been given the y-intercept. We can now use the given y-intercept (33.5) to find the frequency factor \(A\): \(\ln(A) = 33.5\) To find \(A\), we can exponentiate both sides of the equation: \(A = e^{33.5}\) \(A = 3.6 \times 10^{14} \ \mathrm{s}^{-1}\) The frequency factor for this reaction is \(3.6 \times 10^{14} \ \mathrm{s}^{-1}\).

Calculate the Rate Constant at \(25^{\circ} \mathrm{C}\)

Now we can calculate the value of the rate constant \(k\) at \(25^{\circ} \mathrm{C}\). First, we need to convert the temperature to Kelvin: \(T = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \ \mathrm{K}\) Using the Arrhenius equation and the values found for the activation energy and frequency factor: \(k = 3.6 \times 10^{14} \ \mathrm{s}^{-1} \ e^{\frac{-91.5 \times 10^3 \ \mathrm{J} \ \mathrm{mol}^{-1}}{8.314 \ \mathrm{J} \ \mathrm{mol}^{-1} \ \mathrm{K}^{-1} \times 298.15 \ \mathrm{K}}}\) \(k = 3.6 \times 10^{14} \ \mathrm{s}^{-1} \ e^{-36.81}\) \(k = 3.6 \times 10^{14} \ \mathrm{s}^{-1} \times 1.04 \times 10^{-16}\) \(k = 3.74 \times 10^{-2} \ \mathrm{s}^{-1}\) The rate constant at \(25^{\circ} \mathrm{C}\) is \(3.74 \times 10^{-2} \ \mathrm{s}^{-1}\). In summary: a. The activation energy for this reaction is \(91.5 \mathrm{~kJ} \ \mathrm{mol}^{-1}\). b. The frequency factor is \(3.6 \times 10^{14} \ \mathrm{s}^{-1}\). c. The rate constant at \(25^{\circ} \mathrm{C}\) is \(3.74 \times 10^{-2} \ \mathrm{s}^{-1}\).

Key concepts

Activation Energy

Activation energy, denoted as \(E_a\), is a fundamental concept in chemical kinetics. It represents the minimum energy that reacting species must possess for a chemical reaction to occur. Think of it as a barrier that the reactants need to overcome to transform into products. Without the required energy, the reaction simply can’t proceed.
In the context of the Arrhenius equation, the activation energy is part of the relationship that helps us understand how temperature affects the rate of reaction. The equation is:\[k = Ae^{-\frac{E_a}{RT}}\]Here, \(k\) is the rate constant, \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.
Higher activation energy means that molecules require more encouragement (energy input) to react, typically resulting in slower reactions. Conversely, a lower \(E_a\) corresponds to faster reactions at a given temperature because the molecules need less energy to get started. Knowing \(E_a\) is essential for predicting how a reaction rate changes with temperature, which is particularly useful in chemical engineering and various industrial processes.

Frequency Factor

The frequency factor, sometimes called the pre-exponential factor or simply \(A\), is another crucial part of the Arrhenius equation.It reflects the number of times that reactants approach each other in the right orientation to react per unit time. Essentially, it's all about collision frequency and orientation compatibility. Even if two molecules collide, they might not react unless they align correctly.
The frequency factor embodies two things:

• The frequency of collision between reacting molecules.
• The proportion of collisions with the correct orientation for a successful reaction.

In practical terms, \(A\) can be derived from the y-intercept of the linear Arrhenius plot, where \(\ln(A)\) is plotted as a function of \(1/T\). A higher \(A\) value suggests a high probability of reactive collisions whenever the reactants come together, facilitating a faster rate constant. Engineers and chemists often look at \(A\) to understand or manipulate reaction conditions in a lab or industrial settings.

Rate Constant

The rate constant, symbolized as \(k\), is a central element in the rate equation of a reaction. It's not just a number; it connects the concentration of reactants to the speed (rate) of a chemical process.
Through the Arrhenius equation \(k = Ae^{-\frac{E_a}{RT}}\), \(k\) is shown to depend on both the temperature and the inherent properties of the reaction system captured by \(A\) and \(E_a\).
A higher rate constant means a faster reaction, assuming constant conditions like concentration.
Here are some key aspects of the rate constant:

• It develops units based on the reaction order. For first order reactions, like in our exercise, it has units of \(\text{s}^{-1}\).
• It varies with temperature. As in our exercise, changes can be predicted with the Arrhenius equation by knowing \(A\) and \(E_a\).

Understanding \(k\) is crucial in scenarios ranging from simple laboratory experiments to large-scale industrial processes, where controlling the reaction’s speed might mean the difference between success and failure.