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:
Answer & explanation
Correct answer: option 2
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%.