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:
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:
Show Hint
- \( Ea_2>Ea_1>Ea_3 \)
- \( Ea_1>Ea_3>Ea_2 \)
- \( Ea_1>Ea_2>Ea_3 \)
- \( Ea_3>Ea_2>Ea_1 \)
The Correct Option is A
Approach Solution - 1
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: \(Ea_2 > Ea_1 > Ea_3\).
Approach Solution -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)
.
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