The reaction $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{Br}^{-} $$ 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. 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 \(25^{\circ} \mathrm{C}\) .

Step by step solution

01

Use the Arrhenius equation

The Arrhenius equation is given by: \(k = Ae^{-E_{a}/RT}\) where: - k is the rate constant - A is the frequency factor - \(E_{a}\) is the activation energy (in J/mol) - R is the gas constant (8.314 J/mol·K) - T is the temperature (in K) Taking the natural logarithm of both sides, we get: \[\ln(k) = \ln(A) - \frac{E_{a}}{R} \cdot \frac{1}{T}\]

02

Find the activation energy (Ea)

Since we're given the slope of the graph as \(-1.10 \times 10^{4}\) K, we can directly relate it to the slope of the equation above: Slope = \(-\frac{E_{a}}{R}\) We need to find the activation energy, so we can rearrange this to: \(E_{a} = -R \cdot (-1.10 \times 10^{4})\) Substitute the value of R and calculate Ea: \(E_{a} = (8.314\,\text{J/mol·K}) \times (1.10 \times 10^{4}\,\text{K})\) \(E_{a} = 92154\,\text{J/mol}\) So, the activation energy of this reaction is \(92154\,\text{J/mol}\).

03

Find the frequency factor A

The y-intercept of the graph is given as 33.5. This corresponds to the term \(\ln(A)\) in the equation. We can find A using the following equation: \(A = e^{\ln(A)}\) Plug in the given y-intercept value: \(A = e^{33.5}\) Calculate the value of A: \(A \approx 3.49 \times 10^{14}\,\text{s}^{-1}\) The frequency factor A is \(3.49 \times 10^{14}\,\text{s}^{-1}\).

04

Calculate the rate constant k at 25ºC

Now, we will use the Arrhenius equation to find the rate constant at 25ºC. First, convert the temperature to Kelvin: \(T = 25 + 273.15 = 298.15\,\text{K}\) Now, substitute the values of A, Ea, R, and T into the equation: \(k = (3.49 \times 10^{14}\,\text{s}^{-1}) \cdot e^{(-92154\,\text{J/mol})/(8.314\,\text{J/mol·K} \cdot 298.15\,\text{K})}\) Calculate the value of k: \(k \approx 1.64\times 10^{-2}\,\text{s}^{-1}\) The rate constant k at 25°C is \(1.64\times 10^{-2}\,\text{s}^{-1}\). In conclusion: a. The activation energy for this reaction is \(92154\,\text{J/mol}\). b. The frequency factor A is \(3.49 \times 10^{14}\,\text{s}^{-1}\). c. The rate constant k at 25°C is \(1.64\times 10^{-2}\,\text{s}^{-1}\).

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