Question

For a \(1^{\text {* }}\) order reaction \(\mathrm{A} \longrightarrow \mathrm{P}\), the temperature (T) dependent rate constant \((\mathrm{K})\) was found to follow the equation \(\log \mathrm{k}=-(2000) \frac{1}{\mathrm{~T}}+6\) The pre- exponential factor \(\mathrm{A}\) and activation energy Ea are respectively? (a) \(1 \times 10^{6} \mathrm{~S}^{-1}\) and \(9.2 \mathrm{~kJ} / \mathrm{M}\) (b) \(1 \times 10^{6} \mathrm{~S}^{-1}\) and \(38.3 \mathrm{~kJ} / \mathrm{M}\) (c) \(1 \times 10^{6} \mathrm{~S}^{-1}\) and \(16.6 \mathrm{~kJ} / \mathrm{M}\) (d) \(6 \mathrm{~S}^{-1}\) and \(16.6 \mathrm{~kJ} / \mathrm{M}\)

Short answer

(b) \(1 \times 10^{6} \text{ s}^{-1}\) and \(38.3 \text{ kJ/mol}\)

Step-by-step solution

Identify the Given Equation

The rate constant \(k\) follows the equation \(\log k = -\frac{2000}{T} + 6\). This is similar to the Arrhenius equation in logarithmic form: \(\log k = \log A - \frac{E_a}{2.303RT}\).

Determine the Pre-exponential Factor \(A\)

From the given equation, compare \(6\) with \(\log A\). Therefore, \(\log A = 6\). Solving for \(A\), we get \(A = 10^{6} \text{ s}^{-1}\).

Calculate the Activation Energy \(E_a\)

Comparing \(-\frac{2000}{T}\) with \(-\frac{E_a}{2.303RT}\), we identify that \(\frac{2000}{T} = \frac{E_a}{2.303RT}\). Therefore, \(E_a = 2000 \times 2.303 \times R\). Given that \(R = 8.314 \text{ J/mol K}\), calculate \(E_a = 2000 \times 2.303 \times 8.314 = 38281.62 \text{ J/mol}\) or \(38.3 \text{ kJ/mol}\).

Match the Correct Option

The calculated \(A\) is \(1 \times 10^{6} \text{ s}^{-1}\) and \(E_a\) is \(38.3 \text{ kJ/mol}\). Comparing these values with the options provided, the correct answer is option (b).

Key concepts

First Order Reaction

In a first order reaction, the rate at which the reaction occurs is directly proportional to the concentration of one reactant. This means if you double the concentration of that reactant, the rate of reaction also doubles. For such reactions, the rate can be expressed mathematically as:

• Rate = k[A]

where:

• A is the concentration of the reactant, and
• k is the rate constant.

First order reactions have a characteristic exponential decay in concentration over time. Such reactions are common in processes like radioactive decay or simple decomposition reactions.

Rate Constant

The rate constant, denoted by k, is a crucial factor in determining how fast a reaction proceeds. For first order reactions, k is measured in s^-1. It is important to note that the rate constant's value is affected by temperature and the presence of a catalyst.

For a given reaction, the Arrhenius equation provides insight into how k changes with temperature:

• \( k = A e^{-\frac{E_a}{RT}} \)

It can also be transformed into its logarithmic form for easier calculations:

• \( \log k = \log A - \frac{E_a}{2.303RT} \)

Understanding the rate constant helps in predicting how changes in reaction conditions affect the speed of the reaction.

Activation Energy

Activation energy is the minimum energy that reacting molecules need to successfully collide and start a reaction. Without sufficient activation energy, a reaction can hardly proceed even if reactants are present in ample amounts.

It is denoted as \( E_a \), and its value is given in energy units such as kJ/mol. Higher activation energy means that fewer molecules have sufficient energy to overcome the energy barrier, thus slowing the reaction down.

• \( E_a \) factors into the Arrhenius equation to show how temperature influences the reaction rate.
• In the solved problem, \( E_a \) was calculated as 38.3 kJ/mol.

Understanding activation energy is vital for controlling and optimizing reaction conditions in industrial processes.

Pre-exponential Factor

The pre-exponential factor, denoted as \( A \) in the Arrhenius equation, is a constant that provides insight into the frequency of collisions and the probability that collisions are successful. It is also sometimes referred to as the "frequency factor."

• \( A \) is independent of temperature changes.
• It represents the maximum rate of reaction activities when there are no energy barriers.

In the logarithmic form of the Arrhenius equation, \( \log A \) corresponds to the intercept when plotting \( \log k \) versus \( \frac{1}{T} \). For the problem solved, \( A \) was determined to be \( 1 \times 10^6 \; s^{-1} \). Understanding \( A \) is beneficial in demonstrating how variations in molecular orientation and motion influence reaction kinetics.