Question

The rate constants of a first-order reaction at 300 K is $ k_1 $ and 400 K is $ k_2 $. What is the value of $ \frac{k_2}{k_1} $ if activation energy of reaction is $ 41.5 \, \text{kJ mol}^{-1} $?
($R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1}$

• A. 1.809
• B. 4.166
• C. 2.083
• D. 3.618

Hint:

For calculating rate constants, remember the Arrhenius equation and how to convert the activation energy units to match the gas constant.

Answer 1

The Correct Option is A

We use the Arrhenius equation, which relates the rate constant to temperature:
\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Where:
- \( E_a = 41.5 \, \text{kJ/mol} = 41500 \, \text{J/mol} \)
- \( T_1 = 300 \, \text{K}, T_2 = 400 \, \text{K} \)
- \( R = 8.314 \, \text{J/mol K} \)
Substituting the values into the equation:
\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{41500}{8.314} \left( \frac{1}{300} - \frac{1}{400} \right) \] \[ \ln \left( \frac{k_2}{k_1} \right) = 0.595 \] \[ \frac{k_2}{k_1} = e^{0.595} \approx 1.809 \]
Thus, the correct answer is \( 1.809 \).