Question

Consider the following graph between Rate Constant (K) and \( \frac{1}{T} \): Based on the graph, determine the correct order of activation energies \( E_{a1}, E_{a2}, \) and \( E_{a3} \).

• A. \( E_{a1}>E_{a2}>E_{a3} \)
• B. \( E_{a3}>E_{a2}>E_{a1} \)
• C. \( E_{a1}>E_{a3}>E_{a2} \)
• D. \( E_{a1}>E_{a2}>E_{a4} \)

Hint:

In the Arrhenius equation, a steeper slope of the \( \log K \) vs. \( \frac{1}{T} \) plot indicates a higher activation energy.

Answer 1

The Correct Option is A

The graph provided is a plot of \( \log K \) versus \( \frac{1}{T} \), which is typically used to study the Arrhenius equation. According to the Arrhenius equation: \[ K = A \exp\left(-\frac{E_a}{RT}\right) \] Taking the natural logarithm of both sides, we get: \[ \log K = \log A - \frac{E_a}{2.303 R} \cdot \frac{1}{T} \] This equation represents a straight line, where the slope is \( -\frac{E_a}{2.303 R} \). From the graph: - The steepest slope corresponds to the highest activation energy \( E_{a1} \), as the slope is proportional to \( -E_a \). - The second steepest slope corresponds to \( E_{a2} \), and - The least steep slope corresponds to \( E_{a3} \). Therefore, based on the graph: \[ E_{a1}>E_{a2}>E_{a3} \] Thus, the correct order of activation energies is \( E_{a1}>E_{a2}>E_{a3} \). Therefore, the correct answer is (1) \( E_{a1}>E_{a2}>E_{a3} \).