Question

The rate constant for the first order reaction is given by \(\ln k=15-(1.5\times 10^{4})/T\). Calculate the activation energy.

• A. \(125\ \text{kJ mol}^{-1}\)
• B. \(120\ \text{kJ mol}^{-1}\)
• C. \(130\ \text{kJ mol}^{-1}\)
• D. \(140\ \text{kJ mol}^{-1}\)

Hint:

Compare the given equation with ln k = ln A - Ea/RT and take the slope as Ea/R.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The Arrhenius equation in log form is \(\ln k=\ln A-\dfrac{E_a}{RT}\).

Step 2: Compare
Given \(\ln k=15-\dfrac{1.5\times10^{4}}{T}\), so \(\dfrac{E_a}{R}=1.5\times10^{4}\) K.

Step 3: Calculation
\(E_a=1.5\times10^{4}\times8.314\approx1.247\times10^{5}\) J/mol \(\approx125\) kJ/mol.

Final Answer:
The activation energy is about \(125\ \text{kJ mol}^{-1}\). \[ \boxed{\text{(A) }125\ \text{kJ mol}^{-1}} \]