Problem 11 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The rate of a reaction doubles when its temperature changes from \(300 \mathrm{~K}\) to \(310 \mathrm{~K}\). Activation energy of the reaction will be : \((\mathrm{R}=\) \(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) and \(\log 2=0.301\) ) (a) \(53.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(48.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(58.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(60.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The activation energy is approximately 53.6 kJ/mol, so the correct option is (a).

Step 1: Understand the Problem

We need to find the activation energy of a reaction given that its rate doubles when the temperature increases from 300 K to 310 K. Activation energy can be calculated using the Arrhenius equation and its temperature dependence.

Step 2: Use Arrhenius Equation

The rate of reaction is given by the Arrhenius equation: \( k = A e^{-E_a/RT} \), where \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

Step 3: Relation Between Temperature and Rate

Since the rate doubles as temperature changes from 300 K to 310 K, we use the relationship: \( \frac{k_2}{k_1} = 2 \). This gives us \( \frac{A e^{-E_a/(R \times 310)}}{A e^{-E_a/(R \times 300)}} = 2 \).

Step 4: Transform and Simplify the Equation

After canceling \( A \) and simplifying, this becomes: \( e^{E_a/(300 \times R) - E_a/(310 \times R)} = 2 \). Taking the natural logarithm on both sides: \( \frac{E_a}{R} \left( \frac{1}{300} - \frac{1}{310} \right) = \ln 2 \).

Step 5: Calculate the Ratio

Calculate the difference on the left side: \( \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = \frac{1}{9300} \).

Step 6: Solve for Activation Energy

Use the expression obtained: \( \frac{E_a}{8.314} \frac{1}{9300} = 0.693 \). Hence, \( E_a = 8.314 \times 0.693 \times 9300 \).

Step 7: Compute the Activation Energy

Finally, compute \( E_a \): \( E_a = 53558.7222 \mathrm{~J~mol}^{-1} = 53.6 \mathrm{~kJ~mol}^{-1} \).

Step 8: Choose the Right Option

Compare your result with the provided choices and confirm the correct answer. The calculated activation energy is closest to (a) 53.6 kJ/mol.

Key Concepts

Arrhenius Equation

The Arrhenius Equation is a fundamental formula in chemistry used to describe how the rate of a chemical reaction depends on temperature and activation energy. It's expressed as:\( k = A e^{-E_a/RT} \)where:

• \(k\) is the rate constant of the reaction,
• \(A\) is the pre-exponential factor, a constant for each chemical reaction,
• \(E_a\) is the activation energy, the minimum energy required for a reaction to occur,
• \(R\) is the universal gas constant (8.314 J/mol·K),
• \(T\) is the temperature in Kelvin.

The equation shows that even small changes in temperature can significantly affect the reaction rate. As temperature increases, molecules move faster, increasing the chance of successful collisions, and thus, the reaction rate increases. The exponential factor \(e^{-E_a/RT}\) accounts for the fraction of molecules that possess the energy needed to surpass the activation barrier.

Temperature Dependence of Reaction Rates

Temperature affects reaction rates because it influences the kinetic energy of reacting molecules. When temperature rises, more molecules have enough energy to overcome the activation energy barrier. This relationship is a core concept in studying reaction kinetics. To quantify this, the Arrhenius equation provides insight:

• For a 10 K increase in temperature, it's common to see the reaction rate double, which is the scenario posed by the original exercise.
• Higher temperatures lead to more collisions among molecules, increasing the reaction's chance of progressing.

These principles are critical in many fields to optimize reaction conditions, from industrial processes to biochemistry, where understanding how temperature changes alter reaction rates can have significant practical implications.

Natural Logarithm in Chemistry Calculations

The natural logarithm (\( \, \ln \, \)) often appears in chemistry calculations, particularly when dealing with exponential equations like the Arrhenius equation. It's a mathematical tool that helps linearize equations, making them easier to handle.In the context of the Arrhenius equation, taking the natural logarithm on both sides transforms the equation into a more usable form:\( \, \ln k = \ln A - \frac{E_a}{RT} \, \)This linear form is essential for analyzing data in experiments, as it allows chemists to determine activation energy graphically by plotting \( \, \ln k \, \) against \( \, 1/T \, \) and finding the slope.

• Using \( \, \ln \, \) simplifies complex multipliers in equations into additive terms.
• It provides a clearer insight into how sensitive reaction rates are to temperature changes.

Understanding the use of the natural logarithm in these calculations is vital for predicting how different experimental conditions will impact reaction kinetics.