Question

Consider the following graph between rate constant log (k) and $\frac{1}{T}$

• A. Ea1 >Ea2 > Ea3
• B. Ea3 >Ea2 >Ea1
• C. Ea3 >Ea1 >Ea2
• D. Ea1 >Ea2 > Ea3

Answer 1

Answer: (B)

Ea3 >Ea2 >Ea1

Answer 2

The Arrhenius equation in logarithmic form is given by

$$\log k = -\frac{E_a}{2.303R} \cdot \left(\frac{1}{T}\right) + \log A.$$

This shows that the slope of $\log k$ versus $\frac{1}{T}$ is $-\frac{E_a}{2.303R}$. A steeper (more negative) slope corresponds to a higher activation energy. Given that Ea3 has the steepest slope, followed by Ea2 and then Ea1 (which has the shallowest slope), the correct order is

$$E_{a3} > E_{a2} > E_{a1}.$$

Core Explanation:

• Slope $\propto -E_a$.
• Steepest slope ⇒ highest $E_a$.
• Order: $E_{a3} > E_{a2} > E_{a1}$.