Question

For a first order reaction \( A \rightarrow P \), the temperature \( (T) \) dependent rate constant \( k \) was found to follow the equation \[ \log k = -\frac{2000}{T} + 6.0 \] The pre-exponential factor \( A \) and the activation energy \( E_a \) respectively, are

• A. \( 1.0 \times 10^5 \, \text{s}^{-1} \) and 9.2 kJ mol\(^{-1}\)
• B. \( 6.0 \times 10^5 \, \text{s}^{-1} \) and 16.6 kJ mol\(^{-1}\)
• C. \( 1.0 \times 10^6 \, \text{s}^{-1} \) and 16.6 kJ mol\(^{-1}\)
• D. \( 1.0 \times 10^6 \, \text{s}^{-1} \) and 38.3 kJ mol\(^{-1}\)

Hint:

To calculate the pre-exponential factor and activation energy, compare the given rate equation to the Arrhenius equation.

Answer 1

The Correct Option is C

Step 1: Use the Arrhenius equation.
The Arrhenius equation is given by: \[ k = A e^{-\frac{E_a}{RT}} \] where \( A \) is the pre-exponential factor and \( E_a \) is the activation energy. Comparing the given equation with the Arrhenius form, we can calculate \( A \) and \( E_a \).
Step 2: Conclusion.
Using the equation, we find that \( A = 1.0 \times 10^6 \, \text{s}^{-1} \) and \( E_a = 16.6 \, \text{kJ mol}^{-1} \).
Final Answer: \[ \boxed{1.0 \times 10^6 \, \text{s}^{-1} \text{ and } 16.6 \, \text{kJ mol}^{-1}} \]