Which plot of ln k vs 1/T is consistent with Arrhenius equation?

The Arrhenius equation relates the rate constant, activation energy, and temperature of a chemical reaction, and is given by:

$k=A{e}^{-\frac{E_a}{RT}}$

Where:
- $k$ is the rate constant,
- $A$ is the pre-exponential factor (Arrhenius constant),
- $E_a$ is the activation energy,
- $R$ is the universal gas constant, and
- $T$ is the absolute temperature (in Kelvin).

Taking the natural logarithm ($\ln$) on both sides of the equation:

$\ln k=\ln \left(A{e}^{-\frac{E_a}{RT}}\right)$

Using the properties of logarithms, this simplifies to:

$\ln k=\ln A-\frac{E_a}{RT}$

Rearranging the terms to fit the slope-intercept form of a straight line ($y=mx+c$):

$\ln k=\left(-\frac{E_a}{R}\right)\left(\frac{1}{T}\right)+\ln A$

Comparing this with $y=mx+c$ where we plot $y=\ln k$ on the vertical axis and $x=\frac{1}{T}$ on the horizontal axis:
- The slope ($m$) is $-\frac{E_a}{R}$.
- The y-intercept ($c$) is $\ln A$.

Since activation energy ($E_a$) and the gas constant ($R$) are positive quantities, the slope of the line must be negative ($m<0$). Therefore, the plot of $\ln k$ versus $\frac{1}{T}$ is a straight line with a negative slope.

By analyzing the provided images:
- The first image shows a straight line with a positive slope starting below the origin.
- The second image shows a straight line with a positive slope starting above the origin.
- The third image shows a straight line with a positive slope passing through the origin.
- The fourth image shows a straight line with a negative slope and a positive y-intercept.

Thus, the fourth plot is the one consistent with the Arrhenius equation.