Problem 9 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The rate of a reaction quadruples when the temperature changes from 300 to \(310 \mathrm{~K}\). The activation energy of this reaction is : (Assume activation energy and pre-exponential factor are independent of temperature; \(\left.\ln 2=0.693 ; \mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) \(107.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(53.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(26.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(214.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The activation energy is (a) 107.2 kJ/mol.

Step 1: Understand the Arrhenius Equation

The rate of a chemical reaction can be represented by the Arrhenius equation: \( k = A e^{-\frac{E_a}{RT}} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

Step 2: Relate Rate Change to Temperature Change

From the problem, the rate quadruples when the temperature changes from 300 K to 310 K. This can be shown as: \( \frac{k_2}{k_1} = 4 \), where \( k_1 \) and \( k_2 \) are the rate constants at 300 K and 310 K respectively.

Step 3: Utilize the Arrhenius Equation Ratio

The ratio \( \frac{k_2}{k_1} = e^{-\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)} \) can be used, where \( T_1 = 300 \text{ K} \) and \( T_2 = 310 \text{ K} \). This gives \( 4 = e^{-\frac{E_a}{8.314}\left(\frac{1}{310} - \frac{1}{300}\right)} \).

Step 4: Solve for Activation Energy \( E_a \)

Taking natural logarithms on both sides: \( \ln 4 = -\frac{E_a}{8.314}\left(\frac{1}{310} - \frac{1}{300}\right) \). Since \( \ln 4 = 2 \times \ln 2 = 2 \times 0.693 = 1.386 \), equating gives \( 1.386 = -\frac{E_a}{8.314}\left(\frac{1}{310} - \frac{1}{300}\right) \).

Step 5: Calculate the Difference in Reciprocal Temperatures

Calculate \( \frac{1}{310} - \frac{1}{300} = \frac{300 - 310}{300 \times 310} = \frac{-10}{93000} = -0.0001075 \).

Step 6: Substitute and Solve for \( E_a \)

Substitute back: \( 1.386 = -\frac{E_a \times (-0.0001075)}{8.314} \). Solve for \( E_a \): \( E_a = \frac{1.386 \times 8.314}{0.0001075} = 107172.2 \text{ J/mol} \), which is \( 107.2 \text{ kJ/mol} \) when converted to kJ/mol.

Key Concepts

Arrhenius Equation

The Arrhenius equation is foundational to understanding the relationship between temperature and the rate of chemical reactions. This equation is expressed as:\[ k = A e^{-\frac{E_a}{RT}} \]where:

• \( k \) is the rate constant, indicating the speed at which a reaction occurs.
• \( A \) is the pre-exponential factor, also known as the frequency factor, which describes the number of collisions resulting in a reaction.
• \( E_a \) is the activation energy, the minimum energy needed to initiate a reaction.
• \( R \) is the universal gas constant, which is \( 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \).
• \( T \) is the absolute temperature in Kelvin.

These variables allow the Arrhenius equation to give a quantitative illustration of how reaction rates depend on temperature and activation energy.

Rate Constant

The rate constant \( k \) in the Arrhenius equation plays a crucial role in defining the reaction speed. It shows how fast or slow a reaction proceeds under certain conditions. Changes in temperature can affect this constant, altering the reaction rate.
In the exercise, you learned that at 300 K, the initial rate constant is \( k_1 \), and at 310 K, it's \( k_2 \). The problem stipulates that the reaction rate quadruples, implying:

• \( \frac{k_2}{k_1} = 4 \)

This illustrates how temperature influences the rate constant, making it vital to predict or control reaction conditions in practical applications.

Temperature Change Effect

Temperature plays a significant role in chemical reactions, often profoundly affecting the rate at which they happen. As temperature increases, the rate constant \( k \) typically increases too, leading to faster reaction rates.
In the provided exercise, a temperature change from 300 K to 310 K causes the reaction rate to quadruple. This illustrates the sensitivity of reaction rates to even small temperature changes. The impact is reflected in the modified Arrhenius equation:

• \( \frac{k_2}{k_1} = e^{-\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)} \)

The reciprocal difference \( \frac{1}{T_2} - \frac{1}{T_1} \) acts as a fine measure of how temperature shifts affect reactions, emphasizing the non-linear nature of this relationship.

Exponential Factor

The exponential factor \( e^{-\frac{E_a}{RT}} \) in the Arrhenius equation explains why even small changes in temperature can significantly impact reaction rates.
This factor reflects the fraction of molecules with enough energy to surpass the activation energy barrier \( E_a \). As temperature increases, this fraction rises exponentially, dramatically speeding up the reaction rate.

• A higher temperature means a smaller value for \( \frac{E_a}{RT} \), resulting in a larger exponential factor, thus a higher reaction rate.
• Conversely, a lower temperature results in a smaller exponential factor, decreasing the rate.

Understanding this principle helps chemists manipulate reaction conditions to optimize rates and yields, crucial in both research and industry.