Question

A certain reaction rate increases \(1000\) folds in the presence of a catalyst at \(27^\circ\mathrm{C}\). The activation energy of the original pathway is \(98\mathrm{\ kJ/mol}\). What is the activation energy of the new pathway?

• A. \(80.77\mathrm{\ kJ}\)
• B. \(56.38\mathrm{\ kJ}\)
• C. \(24.67\mathrm{\ kJ}\)
• D. \(90.43\mathrm{\ kJ}\)

Hint:

Use Arrhenius equation: \(\ln(k_2/k_1) = (E_{a1} - E_{a2})/RT\).

Answer 1

The Correct Option is A

Step 1: Formula / Definition}
\[ \frac{k_2}{k_1} = e^{(E_{a1} - E_{a2})/RT} \]
Step 2: Calculation / Simplification}
\(\ln(1000) = \frac{E_{a1} - E_{a2}}{RT}\)
\(6.9078 = \frac{98000 - E_{a2}}{8.314 \times 300}\)
\(6.9078 \times 2494.2 = 98000 - E_{a2}\)
\(17228 = 98000 - E_{a2} \Rightarrow E_{a2} = 98000 - 17228 = 80772\mathrm{\ J/mol} = 80.77\mathrm{\ kJ/mol}\)
Step 3: Final Answer
\[ 80.77\mathrm{\ kJ} \]