Question

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 (a) \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ;\) (b) \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol}\) ?

Short answer

For a first-order reaction with a rate constant of $2.75 \times 10^{-2} \mathrm{~s}^{-1}$ at $20^{\circ} \mathrm{C}$, the rate constant at $60^{\circ} \mathrm{C}$ is: (a) \(1.32 \times 10^{-1} \, \mathrm{s^{-1}}\) when \(E_{a} = 75.5 \, \mathrm{kJ/mol}\) (b) \(9.08 \times 10^{-1} \, \mathrm{s^{-1}}\) when \(E_{a} = 125 \, \mathrm{kJ/mol}\)

Step-by-step solution

Find the Arrhenius equation in its linear form

The Arrhenius equation in its linear form is: \(\ln{k} = \ln{A} - \frac{E_\text{a}}{RT}\) Here, k is the rate constant, A is the pre-exponential factor, Eₐ is the activation energy, R is the gas constant (8.314 J/molK,) and T is the temperature in Kelvin. We will be using this equation to find the rate constant at the two temperatures given for each of the activation energy values.

Convert the temperatures to Kelvin

The given temperatures are in Celsius, and we need to convert them to Kelvin for use in the Arrhenius equation. To do this, add 273.15 to the Celsius temperature. T₁ = 20°C + 273.15 = 293.15 K T₂ = 60°C + 273.15 = 333.15 K

Set up the equation for each temperature

We can now set up the Arrhenius equation for each of the temperatures: \(\ln{k_1} = \ln{A} - \frac{E_\text{a}}{RT_1}\) \(\ln{k_2} = \ln{A} - \frac{E_\text{a}}{RT_2}\)

Subtract the two equations

Subtract the second equation from the first to eliminate the pre-exponential factor A: \(\ln{k_1} - \ln{k_2} = - \frac{E_\text{a}}{R} (\frac{1}{T_1} - \frac{1}{T_2})\)

Rearrange the equation for k₂

Rearrange the equation to solve for k₂: \(\ln{k_2} = \ln{k_1} + \frac{E_\text{a}}{R} (\frac{1}{T_1} - \frac{1}{T_2})\)

Calculate k for each activation energy

Now, we will use the equation above to find k₂ for each activation energy: (a) Eₐ = 75.5 kJ/mol (convert to J/mol: 75500 J/mol) \(\ln{k_2} = \ln{2.75 \times 10^{-2}} + \frac{75500}{8.314} (\frac{1}{293.15} - \frac{1}{333.15})\) Use a calculator to find the natural logarithm of k₂ and then exponentiate to get k₂: \(k_2 = 1.32 \times 10^{-1} \, \mathrm{s^{-1}}\) (b) Eₐ = 125 kJ/mol (convert to J/mol: 125000 J/mol) \(\ln{k_2} = \ln{2.75 \times 10^{-2}} + \frac{125000}{8.314} (\frac{1}{293.15} - \frac{1}{333.15})\) Use a calculator to find the natural logarithm of k₂ and then exponentiate to get k₂: \(k_2 = 9.08 \times 10^{-1} \, \mathrm{s^{-1}}\) So, the rate constant k at 60°C is: (a) 1.32 × 10⁻¹ s⁻¹ when Eₐ = 75.5 kJ/mol and (b) 9.08 × 10⁻¹ s⁻¹ when Eₐ = 125 kJ/mol.

Key concepts

Reaction Rate Constant

The reaction rate constant, denoted as 'k', is a pivotal parameter in chemical kinetics. It quantifies the speed of a chemical reaction under specific conditions. For example, a high value of 'k' implies that the reaction occurs rapidly, while a low value indicates a slower process.

The value of 'k' is influenced by several factors such as temperature, presence of catalysts, and the physical state of the reactants. Temperature is especially crucial because, as temperature increases, molecules move faster and collide more frequently with the necessary energy to initiate a reaction. The Arrhenius equation is key to understanding how temperature affects the reaction rate constant. In the given exercise, we are asked to determine the value of 'k' at a higher temperature, starting from its known value at a lower temperature, which directly applies this concept.

Activation Energy

Activation energy, denoted as 'Eₐ', is the minimum amount of energy that reacting particles need to collide with each other and form products. Think of it like the energy barrier that must be overcome for a reaction to proceed. The higher the activation energy, the slower the reaction rate, since fewer particles will possess sufficient energy to surpass the barrier at a given temperature.

In the exercise, two different activation energies are presented, indicating how varying 'Eₐ' can impact the calculation of the new rate constant 'k' at a different temperature. The case with the higher 'Eₐ' would intuitively result in a greater change in 'k' with temperature, precisely because it's harder for reactants to overcome the energy barrier.

Temperature Conversion

Temperature conversion is essential when working with the Arrhenius equation because it requires the temperature to be in Kelvin (K). To convert from Celsius to Kelvin, one simply adds 273.15 to the Celsius temperature. This is important as the Kelvin scale starts at absolute zero, the point at which particles theoretically stop moving.

For the formulas we're using, it is vital not to forget this step, as using Celsius directly would result in incorrect values for 'k'. This is one of those 'simple, yet easy to miss' steps that could lead to an incorrect solution, as the student needs to be careful with units of measurement in chemistry.

Natural Logarithm

The natural logarithm is the logarithm to the base of the mathematical constant e, approximately equivalent to 2.71828. It is denoted by 'ln'. In kinetic equations, natural logarithms are often used because they can simplify the mathematics involved in exponential processes, which are common in chemical kinetics.

In our exercise, natural logarithms help us linearize the Arrhenius equation, making it possible to isolate and solve for the unknown 'k'. Understanding how to work with 'ln' is crucial, as taking the logarithm of both sides of an equation is a standard technique for handling equations that involve exponentiation. Students should get comfortable with properties of logarithms, such as how \(\ln{a} - \ln{b}\) simplifies to \(\ln{\frac{a}{b}}\), which help in manipulating the Arrhenius equation.