(a) A certain first-order reaction has a rate constant of $2.75 \times 10^{-2} \mathrm{~s}^{-1}\( at \)20^{\circ} \mathrm{C}\(. What is the value of \)k$ at \(60^{\circ} \mathrm{C}\) if $E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{b})\( Another first-order reaction also has a rate constant of \)2.75 \times 10^{-2} \mathrm{~s}^{-1}\( at \)20^{\circ} \mathrm{C}$. What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if $E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{c})$ What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

Short Answer

Expert verified
In summary, for a first-order reaction with a rate constant of \(2.75 \times 10^{-2} ~\mathrm{s}^{-1}\) at \(20^{\circ}\mathrm{C}\) and activation energy of \(75.5 \mathrm{~kJ/mol}\), the value of k at \(60^{\circ} \mathrm{C}\) is approximately \(5.35 \times 10^{-2}~\mathrm{s}^{-1}\). For another first-order reaction with the same rate constant at \(20^{\circ}\mathrm{C}\) and an activation energy of \(125 \mathrm{~kJ/mol}\), the value of k at \(60^{\circ} \mathrm{C}\) is approximately \(1.37 \times 10^{-2}~\mathrm{s}^{-1}\). These calculations are based on the assumptions that the Arrhenius equation is valid, the activation energy and frequency factor A remain constant, and the reaction is a first-order reaction.

Step by step solution

01

Convert temperatures to Kelvin

Before we begin, we need to convert the given temperatures from Celsius to Kelvin. We can do this by adding 273.15 to the Celsius temperature. \(T_1 = 20^{\circ} \mathrm{C} + 273.15 = 293.15~\mathrm{K}\) \(T_2 = 60^{\circ} \mathrm{C} + 273.15 = 333.15~\mathrm{K}\)
02

Use the Arrhenius equation for part (a)

We are given the initial rate constant k1 at T1 and the activation energy Ea. We need to find the rate constant k2 at T2. We can use the Arrhenius equation to relate k1, k2, and Ea as follows: \(\frac{k_2}{k_1} = e^{\frac{-Ea}{R} (\frac{1}{T_2} - \frac{1}{T_1})}\) Now, plug in the given values and solve for k2: \(\frac{k_2}{2.75 \times 10^{-2}~\mathrm{s}^{-1}} = e^{\frac{-75.5 \times 10^3 \mathrm{J/mol}}{8.314 \mathrm{J/mol·K}} (\frac{1}{333.15 \mathrm{K}} - \frac{1}{293.15 \mathrm{K}})}\) After calculating, we get: \(k_2 ≈ 5.35 \times 10^{-2}~\mathrm{s}^{-1}\) So the value of k at \(60^{\circ} \mathrm{C}\) for part (a) is approximately \(5.35 \times 10^{-2}~\mathrm{s}^{-1}\).
03

Use the Arrhenius equation for part (b)

For part (b), we are given a different activation energy. We can use the same equation as in step 2, with the new Ea value: \(\frac{k_2}{2.75 \times 10^{-2}~\mathrm{s}^{-1}} = e^{\frac{-125 \times 10^3 \mathrm{J/mol}}{8.314 \mathrm{J/mol·K}} (\frac{1}{333.15 \mathrm{K}} - \frac{1}{293.15 \mathrm{K}})}\) After calculating, we get: \(k_2 ≈ 1.37 \times 10^{-2}~\mathrm{s}^{-1}\) So the value of k at \(60^{\circ} \mathrm{C}\) for part (b) is approximately \(1.37 \times 10^{-2}~\mathrm{s}^{-1}\).
04

Discuss assumptions for part (c)

In order to calculate the answers for parts (a) and (b), we need to make several assumptions: 1. The Arrhenius equation is valid for the given temperature range (i.e., the relationship between rate constant, activation energy, and temperature follows the equation for both reactions). 2. The activation energy and frequency factor A remain constant over the temperature range of interest. 3. The reaction is indeed a first-order reaction. By making these assumptions, we were able to use the Arrhenius equation to calculate the rate constants for the reactions at \(60^{\circ} \mathrm{C}\).

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