Question

The rate constant of a first order reaction at 400 K and 500 K is respectively \( 2 \times 10^{-5} \, {s}^{-1} \) and \( 4 \times 10^{-3} \, {s}^{-1} \). What is the approximate activation energy (in kJ/mol\(^{-1}\))? (R = 8.3 J mol\(^{-1}\) K\(^{-1}\), log 2 = 0.3)

• A. \( 880 \)
• B. \( 88 \)
• C. \( 38.2 \)
• D. \( 8.8 \)

Hint:

The Arrhenius equation is key to calculating activation energy using temperature and rate constants. Remember to convert all units properly and use logarithmic relations for accuracy.

Answer 1

The Correct Option is B

Step 1: Apply the Arrhenius equation in logarithmic form. \[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Substitute the known values: \[ \log \left( \frac{4 \times 10^{-3}}{2 \times 10^{-5}} \right) = \frac{E_a}{2.303 \times 8.3} \left( \frac{1}{400} - \frac{1}{500} \right) \] Step 2: Simplify and solve for \( E_a \). \[ \log(200) = 2.3010, \quad \left( \frac{1}{400} - \frac{1}{500} \right) = 2.5 \times 10^{-5} \] \[ 2.3010 = \frac{E_a}{19.1159} \times 2.5 \times 10^{-5} \] \[ E_a = \frac{2.3010 \times 19.1159}{2.5 \times 10^{-5}} = 88 \, {kJ/mol}^{-1} \]