Question

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:

Choose the correct answer from the options given below:

• A. \( Ea_2>Ea_1>Ea_3 \)
• B. \( Ea_1>Ea_3>Ea_2 \)
• C. \( Ea_1>Ea_2>Ea_3 \)
• D. \( Ea_3>Ea_2>Ea_1 \)

Hint:

The slope of the Arrhenius plot is directly related to the activation energy. A steeper slope corresponds to a higher activation energy.

Answer 1

The Correct Option is A

The graph provided is a plot of the logarithm of the rate constant \((\log k)\) versus the reciprocal of temperature \((\frac{1}{T})\). According to the Arrhenius equation:

\(\log k = \log A - \frac{Ea}{2.303RT}\)

where:

• \(k\) is the rate constant.
• \(A\) is the pre-exponential factor.
• \(Ea\) is the activation energy.
• \(R\) is the universal gas constant.
• \(T\) is the temperature in Kelvin.

The slope of the plot is \(-\frac{Ea}{2.303R}\). Thus, a steeper slope indicates a larger value for the activation energy \((Ea)\).

In the provided graph, the slopes of the lines indicate the relative values of \(Ea\) for reactions 1, 2, and 3:

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

Therefore, the order of activation energies is \(Ea_2 > Ea_1 > Ea_3\).

The correct answer is: \(Ea_2 > Ea_1 > Ea_3\).

Answer 2

The activation energy \( E_a \) of a reaction is related to the slope of the plot of \( \log k \) vs \( \frac{1}{T} \) by the Arrhenius equation: \[ \log k = -\frac{E_a}{2.303R} \times \frac{1}{T} + \text{constant} \] where: -
\( \log k \) is the log of the rate constant, 
\( T \) is the temperature,
 \( R \) is the gas constant.

The steeper the slope, the higher the activation energy. In this plot: 
Reaction 1 (the line with the steepest slope) corresponds to the highest activation energy, i.e., \( E_{a1} \). 
Reaction 2 (the line with a less steep slope) corresponds to the middle activation energy, i.e., \( E_{a2} \).
Reaction 3 (the line with the least steep slope) corresponds to the lowest activation energy, i.e., \( E_{a3} \). 
Thus, the correct order of activation energies is \( E_{a2}>E_{a1}>E_{a3} \), which corresponds to option  (1)
.