Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Chemical Kinetics

Question:

The activation energy of a reaction is 9 kcal/mole. The increase in the rate constant when its temperature is raised from 295 to 300 is:

Options:

14.9%

28.9%

78.9%

82.9%

Correct Answer:

28.9%

Explanation:

To determine the increase in the rate constant when the temperature is raised, we can use the Arrhenius equation:

\[ k = A \cdot e^{-\frac{E_a}{RT}} \]

Where:
\( k \) is the rate constant
\( A \) is the pre-exponential factor (frequency factor)
\( E_a \) is the activation energy
\( R \) is the gas constant
\( T \) is the temperature in Kelvin

To calculate the increase in the rate constant, we can compare the rate constants at two different temperatures.

Let's consider the temperatures 295 K and 300 K.

\[ \frac{k_2}{k_1} = \frac{A \cdot e^{-\frac{E_a}{RT_2}}}{A \cdot e^{-\frac{E_a}{RT_1}}} = e^{\frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)} \]

Substituting the given values:
\[ \frac{k_2}{k_1} = e^{\frac{9 \text{ kcal/mol}}{1.987 \, \text{ cal/(mol K)}} \left( \frac{1}{295 \, \text{ K}} - \frac{1}{300 \, \text{ K}} \right)} \]

Calculating the ratio:
\[ \frac{k_2}{k_1} \approx 1.289 \]

To find the increase in the rate constant, we subtract 1 from the ratio and multiply by 100%:

Increase in rate constant = (1.289 - 1) × 100% ≈ 28.9%

Therefore, the correct answer is (2) 28.9%.