Consider the following plots of log of rate constant k (log k) vs $$\frac{1}{T}$$ for three different reactions. The correct order of activation energies of these reactions is :

To determine the correct order of activation energies, we use the relationship between the rate constant $$k$$, temperature $$T$$, and activation energy $$E_a$$ given by the Arrhenius equation.

The Arrhenius equation is

$$k=Ae^{-E_a/RT}$$

where

• $$k$$ is the rate constant,
• $$A$$ is the Arrhenius pre-exponential factor,
• $$E_a$$ is the activation energy,
• $$R$$ is the universal gas constant, and
• $$T$$ is the absolute temperature.

To relate this equation to the given graph of $$\log k$$ versus $$1/T$$, we take the logarithm of both sides.

This gives

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

This equation is of the form

$$y=mx+c,$$

where

• $$y=\log k,$$
• $$x=\dfrac{1}{T},$$
• the intercept is $$\log A,$$
• the slope is $$-\dfrac{E_a}{2.303R}.$$

The magnitude of the slope is directly proportional to the activation energy.

$$\left|\text{slope}\right|=\frac{E_a}{2.303R}.$$

Hence, a steeper line corresponds to a larger activation energy, while a flatter line corresponds to a smaller activation energy.

From the graph,

• Line 2 is the steepest, so it has the highest activation energy.
• Line 1 has intermediate steepness, so it has an intermediate activation energy.
• Line 3 is the least steep, so it has the lowest activation energy.

Therefore,

$$|slope_2|>|slope_1|>|slope_3|.$$

Hence, the correct order of activation energies is

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

Therefore, the correct answer is Option (A).