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 \(25^{\circ} \mathrm{C}\).

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 \(91.5 \mathrm{~kJ} \ \mathrm{mol}^{-1}\).

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 \(3.6 \times 10^{14} \ \mathrm{s}^{-1}\).

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 \(3.74 \times 10^{-2} \ \mathrm{s}^{-1}\).