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 provided graph plots \( \log K \) against \( \frac{1}{T} \), a standard representation for the Arrhenius equation. The Arrhenius equation is given by \[K = A \exp\left(-\frac{E_a}{RT}\right)\] Taking the natural logarithm yields \[\log K = \log A - \frac{E_a}{2.303 R} \cdot \frac{1}{T}\] This linear form indicates that the slope of the graph is \( -\frac{E_a}{2.303 R} \). Observations from the graph: - The slope with the greatest magnitude (steepest negative slope) corresponds to the highest activation energy (\( E_{a1} \)), as the slope is inversely proportional to \( -E_a \). - The second steepest slope corresponds to \( E_{a2} \). - The least steep slope corresponds to \( E_{a3} \). Consequently, the graphical data implies the following order of activation energies: \[E_{a1}>E_{a2}>E_{a3}\] The correct sequence of activation energies is therefore \( E_{a1}>E_{a2}>E_{a3} \). The correct answer is (1) \( E_{a1}>E_{a2}>E_{a3} \).