Consider the following plots of rate constant versus $$\frac{1}{T}$$ for four different reactions. Which of the following orders is correct for the activation energies of these reactions?

The logarithmic form of the Arrhenius equation is expressed as:

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

When plotting $$\log k$$ on the y-axis against $$\frac{1}{T}$$ on the x-axis, the graph yields a straight line ($$y = mx + c$$) where:

• The intercept ($$c$$) equals $$\log A$$.
• The slope ($$m$$) of the line equals $$-\frac{E_a}{2.303 R}$$.

Relating Slope to Activation Energy

Because the slope is negative ($$m = -\frac{E_a}{2.303 R}$$), the steepness of the downward incline tells us the magnitude of the activation energy:

$$\text{Magnitude of Slope } |m| = \frac{E_a}{2.303 R} \implies E_a \propto |m|$$

A steeper downward line (a larger negative slope value in terms of absolute magnitude) corresponds directly to a **higher activation energy ($$E_a$$)**.

Comparing the Plots:

By analyzing the steepness of the four lines in the graph from the steepest incline to the flattest:

1. Line $$c$$: Has the steepest downward slope, meaning it has the highest activation energy ($$E_c$$).
2. Line $$a$$: Is the next steepest, so it has the second-highest activation energy ($$E_a$$).
3. Line $$d$$: Has a gentler slope, meaning intermediate activation energy ($$E_d$$).
4. Line $$b$$: Is nearly horizontal (the flattest line), meaning its rate constant is minimally affected by temperature, giving it the lowest activation energy ($$E_b$$).

Conclusion:

Ordering the activation energies based on the absolute values of their slopes gives:

$$\mathbf{E_c > E_a > E_d > E_b}$$

Answer: Option C — $$E_c > E_a > E_d > E_b$$