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: Set up the ratio:
Doubling of $k$ over a 30 K rise gives $\log(k_2/k_1) = \log 2 = 0.301$.

Step 2: Solve for Ea:
$E_a = \dfrac{2.303R \times 0.301\times T_1T_2}{T_2-T_1}$. Put in $2.303R = 19.15$ and $T_1T_2/(T_2-T_1) = 99000/30 = 3300$.

Step 3: Evaluate:
$E_a = 19.15\times0.301\times3300 = 19022$ J/mol, which is 19.02 kJ/mol, option (C).

Final Answer:
Activation energy is 19.02 kJ/mol. \[ \boxed{\text{(C) }19.02\ \text{kJ mol}^{-1}} \]