Question

(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

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

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. \(T_1 = 20^{\circ} \mathrm{C} + 273.15 = 293.15~\mathrm{K}\) \(T_2 = 60^{\circ} \mathrm{C} + 273.15 = 333.15~\mathrm{K}\)

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 \(5.35 \times 10^{-2}~\mathrm{s}^{-1}\).

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 \(1.37 \times 10^{-2}~\mathrm{s}^{-1}\).

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 \(60^{\circ} \mathrm{C}\).

Key concepts

First-Order Reaction

A first-order reaction is a type of chemical reaction where the rate of reaction is directly proportional to the concentration of a single reactant. This means if you double the concentration of the reactant, the rate of reaction also doubles. In mathematical terms, the rate law for a first-order reaction is written as: \[ \text{Rate} = k[A] \] where \(k\) is the rate constant, and \([A]\) is the concentration of the reactant. Understanding this can help in predicting how fast a reaction progresses as the concentration of reactants changes.

• First-order reactions are common in processes like radioactive decay.
• The half-life of a first-order reaction is constant, meaning it doesn't depend on the initial concentration.
• The units of the rate constant \(k\) in a first-order reaction are per time, such as \(\mathrm{s}^{-1}\).

Temperature Conversion

When dealing with reaction rates and the Arrhenius equation, converting temperatures from Celsius to Kelvin is crucial. This conversion is necessary because scientific calculations often require absolute temperature, which is measured in Kelvin. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example:

• \(20^{\circ}\mathrm{C} + 273.15 = 293.15~\mathrm{K}\)
• \(60^{\circ}\mathrm{C} + 273.15 = 333.15~\mathrm{K}\)

Using Kelvin ensures that there are no issues with negative temperatures in calculations, which would not be meaningful in the context of chemical reactions.

Activation Energy

Activation energy \(E_a\) is the minimum amount of energy required for a chemical reaction to occur. It acts as an energy barrier that reactants must overcome for transformation into products. This energy is a crucial factor in determining the rate of a reaction. A higher activation energy means that fewer molecules will have sufficient energy to react at a given temperature, leading to a slower reaction.

• An increase in temperature often helps more molecules overcome this barrier, speeding up the reaction.
• The Arrhenius equation makes it possible to calculate the rate constant of a reaction at different temperatures by incorporating the activation energy.
• For the exercise, activation energies of \(75.5~\mathrm{kJ/mol}\) and \(125~\mathrm{kJ/mol}\) were used to calculate rate constants at different temperatures.

Rate Constant

The rate constant \(k\) is a crucial factor in a reaction rate equation, indicating how quickly a reaction proceeds. It varies with temperature and is specific to each reaction. The Arrhenius equation is used to understand the relationship between the rate constant and temperature: \[ k = Ae^{-\frac{E_a}{RT}} \] where:

• \(A\) is the frequency factor, relating to the number of times that reactants collide with the correct orientation.
• \(E_a\) is the activation energy.
• \(R\) is the universal gas constant, \(8.314~\mathrm{J/mol \cdot K}\).
• \(T\) is the temperature in Kelvin.

This formula allows us to compare how the rate constant changes with different temperatures. Higher temperatures often result in increased rate constants as molecules move faster and collide more energetically, overcoming activation energy barriers more easily.