According to Arrhenius equation \(k = Ae^{\frac{E_a}{RT}}\). If \(ln k\) vs \(\frac{1}{T}\) graph plotted, what will be the intercept of this plot?

\(ln A\)

The correct answer is option 3. \(ln A\).

When we plot \(ln k\) vs \(\frac{1}{T}\) for the Arrhenius equation \(k = Ae^{\frac{-E_a}{RT}}\), we are essentially graphing the natural logarithm of the rate constant (\(ln k\)) against the reciprocal of the temperature (\(\frac{1}{T}\)). This plot is typically linear, as the Arrhenius equation can be rearranged to fit the form of a linear equation:

\[ln k = \frac{-E_a}{R} \cdot \frac{1}{T} + ln A\]

Here, \(ln k\) is the dependent variable, and \(\frac{1}{T}\) is the independent variable. The slope of this line is \(\frac{-E_a}{R}\), and the intercept is \(ln A\).

The intercept of this plot, \(ln A\), represents the natural logarithm of the pre-exponential factor (\(A\)) in the Arrhenius equation. The pre-exponential factor accounts for the frequency of collisions between reactant molecules and determines the rate of the reaction at the molecular level. It reflects the probability that the reactants will have sufficient energy to overcome the activation energy barrier and react.

Therefore, when we plot \(ln k\) vs \(\frac{1}{T}\), the intercept of the plot is \(ln A\), which represents the natural logarithm of the pre-exponential factor in the Arrhenius equation.