Which of the following plot represents the variation of ln k versus \( \frac{1}{T} \) in accordance with the Arrhenius equation?
The Correct Option is C
Solution and Explanation
To determine which plot best represents the variation of \( \ln k \) versus \( \frac{1}{T} \) in accordance with the Arrhenius equation, we need to understand the Arrhenius equation itself.
The Arrhenius equation is given by:
\(k = Ae^{-\frac{E_a}{RT}}\)
Where:
- \(k\) is the rate constant
- \(A\) is the pre-exponential factor
- \(E_a\) is the activation energy
- \(R\) is the universal gas constant
- \(T\) is the temperature in Kelvin
Taking the natural logarithm of both sides, we have:
\(\ln k = \ln A - \frac{E_a}{RT}\)
This equation is of the form \(y = mx + c\), where:
- \(y = \ln k\)
- \(m = -\frac{E_a}{R}\) (the slope)
- \(x = \frac{1}{T}\)
- \(c = \ln A\) (the intercept)
Thus, the plot of \(\ln k\) versus \(\frac{1}{T}\) will be a straight line with a negative slope.
Now, we need to determine which of the provided plots represents this relationship. We are given the following options:
The correct answer is the plot which shows a straight line with a negative slope.
The correct plot is:
Therefore, the plot that represents the variation of \(\ln k\) versus \(\frac{1}{T}\) in accordance with the Arrhenius equation is the third plot.
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