Question

The rate constant of reaction is \(1.5\times 10^7 \text{s}^{-1}\) at 300 K and \(3.0\times 10^7 \text{s}^{-1}\) at 330 K. What is the activation energy for the reaction? [ \(R\times 2.303 = 19.15 \text{JK}^{-1}\text{mol}^{-1}\)]

• A. \(18.02 \text{kJ mol}^{-1}\)
• B. \(20.1 \text{kJ mol}^{-1}\)
• C. \(19.02 \text{kJ mol}^{-1}\)
• D. \(21.5 \text{kJ mol}^{-1}\)

Hint:

Use the two-temperature Arrhenius equation with 2.303R given.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The Arrhenius equation compares rate constants at two temperatures and links them to the activation energy \(E_a\).

Step 2: Key Formula or Approach:
\[ \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1 T_2}\right) \]

Step 3: Detailed Explanation:
\(k_2/k_1 = 3.0\times10^7 / 1.5\times10^7 = 2\), so \(\log 2 = 0.301\).
\(\dfrac{T_2-T_1}{T_1T_2} = \dfrac{30}{300\times330} = \dfrac{1}{3300}\).
\[ E_a = \frac{0.301 \times 19.15 \times 3300}{1} = 19022\ \text{J mol}^{-1} \approx 19.02\ \text{kJ mol}^{-1} \]
Options (A), (B) and (D) differ from this result in the third digit or more.

Step 4: Final Answer:
\(E_a = 19.02\) kJ mol\(^{-1}\), option (C).

Final Answer:
Rate doubles from 300 K to 330 K, so Ea is 19.02 kJ/mol. \[ \boxed{\text{(C) }19.02\ \text{kJ mol}^{-1}} \]