Question

The rate constant for a first order reaction is $0.58 \text{ s}^{-1}$ at $300 \text{ K}$ and $0.026 \text{ s}^{-1}$ at $290 \text{ K}$ . What is the energy of activation? $\left(\text{R} = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}\right)$}

• A. 124.48 kJ
• B. 224.55 kJ
• C. 348.18 kJ
• D. 513.21 kJ

Hint:

When temperature increases by 10K, the rate constant often increases significantly; use the log form of Arrhenius for $E_a$ calculations.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The dependence of a reaction's rate constant on temperature is mathematically described by the Arrhenius equation. By comparing the rate constants at two different temperatures, we can calculate the activation energy ($E_a$).
Step 2: Key Formula or Approach:
The two-point form of the Arrhenius equation is: \[ \log_{10} \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \] Step 3: Detailed Explanation:
Given values from the problem: $k_1 = 0.026 \text{ s}^{-1}$ at $T_1 = 290 \text{ K}$ $k_2 = 0.58 \text{ s}^{-1}$ at $T_2 = 300 \text{ K}$ $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$ Substitute these values into the Arrhenius equation: \[ \log_{10} \left( \frac{0.58}{0.026} \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{300 - 290}{300 \times 290} \right) \] \[ \log_{10} (22.307) = \frac{E_a}{19.147} \left( \frac{10}{87000} \right) \] The value of $\log_{10} (22.307)$ is approximately $1.3484$. \[ 1.3484 = \frac{E_a}{19.147} \times 1.1494 \times 10^{-4} \] Now, isolate and solve for $E_a$: \[ E_a = \frac{1.3484 \times 19.147}{1.1494 \times 10^{-4}} \] \[ E_a = \frac{25.818}{1.1494 \times 10^{-4}} \] \[ E_a \approx 224621 \text{ J mol}^{-1} \] Convert the activation energy from Joules to kilojoules: \[ E_a \approx 224.62 \text{ kJ mol}^{-1} \] Comparing this result with the given options, the closest matched value is $224.55 \text{ kJ}$.
Step 4: Final Answer:
The activation energy is approximately $224.55 \text{ kJ}$.