JEE Chemistry 2025 Question with Solution

Given: The graph is a plot of $\log k$ versus $\frac{1}{T}$ for three reactions.

Find: The correct order of activation energies.

From the Arrhenius equation,

So, the slope of the straight line in a $\log k$ versus $\frac{1}{T}$ plot is

Thus, the magnitude of slope is directly proportional to the activation energy. A steeper line means a larger value of $E_a$.

From the given graph:

• Line $2$ has the steepest slope.
• Line $1$ has a moderate slope.
• Line $3$ has the least steep slope.

Therefore, the activation energies are in the order

Hence, the correct option is A.

Given: A plot of $\log k$ against $\frac{1}{T}$ is shown for three reactions.

Find: Compare $E_{a1}$, $E_{a2}$, and $E_{a3}$.

For an Arrhenius plot,

where $k$ is the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature in Kelvin.

Since the coefficient of $\frac{1}{T}$ is $-\frac{E_a}{2.303R}$, the line with greater steepness has greater magnitude of negative slope and therefore higher activation energy.

The graph indicates that reaction $2$ is steepest, reaction $1$ is next, and reaction $3$ is least steep. Therefore,

This matches option (1), so the correct option is A.