Question

The rate constant of a first order reaction was doubled when the temperature was increased from 300 to 310 K. What is its approximate activation energy (in kJ mol\(^{-1}\))?
(R = 8.3 J mol\(^{-1}\) K\(^{-1}\), \(\log 2 = 0.3\))

• A. 5.33
• B. 53.3
• C. 5333
• D. 53.33

Hint:

For first-order reactions, if the rate constant doubles with a small temperature increase, use the logarithmic Arrhenius equation to estimate activation energy.

Answer 1

The Correct Option is D

We use the Arrhenius equation in logarithmic form: \[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \] Step 1: Given data
- \( T_1 = 300 \) K, \( T_2 = 310 \) K
- \( k_2 = 2k_1 \) (since the rate constant is doubled)
- \( R = 8.3 \) J mol\(^{-1}\)K\(^{-1}\)
- \( \log 2 = 0.3 \)
Step 2: Substituting the values
\[ \log 2 = \frac{E_a}{2.303 \times 8.3} \times \left( \frac{10}{300 \times 310} \right) \] \[ 0.3 = \frac{E_a}{19.1} \times \left( \frac{10}{93000} \right) \] Step 3: Solving for \( E_a \)
\[ E_a = \frac{0.3 \times 19.1 \times 93000}{10} \] \[ E_a = 53.33 \text{ kJ mol}^{-1} \] Step 4: Identifying the correct answer
Thus, the activation energy is approximately 53.33 kJ mol\(^{-1}\), which corresponds to option (D).