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

According to the Arrhenius equation, the rate constant $k$ of a chemical reaction depends on the absolute temperature $T$ as follows:
$k=A{e}^{-\frac{E_a}{RT}}$
where:
• $k$ is the rate constant,
• $A$ is the pre-exponential factor (Arrhenius frequency factor),
• $E_a$ is the activation energy of the reaction,
• $R$ is the universal gas constant, and
• $T$ is the absolute temperature in Kelvin.

Taking the natural logarithm (ln) on both sides of the Arrhenius equation:
$\ln k=\ln \left(A{e}^{-\frac{E_a}{RT}}\right)$
$\ln k=\ln A+\ln \left(e^{-\frac{E_a}{RT}}\right)$
$\ln k=\ln A-\frac{E_a}{R}\left(\frac{1}{T}\right)$

We can compare this result with the standard equation of a straight line:
$y=mx+c$
where:
• $y=\ln k$ (plotted on the vertical axis),
• $x=\frac{1}{T}$ (plotted on the horizontal axis),
• $m=-\frac{E_a}{R}$ represents the slope of the line, and
• $c=\ln A$ represents the y-axis intercept.

Since the activation energy $E_a$ and the gas constant $R$ are both positive quantities, the slope of the line $m=-\frac{E_a}{R}$ must be negative. Thus, the plot of $\ln k$ versus $\frac{1}{T}$ is a straight line sloping downwards from left to right, with a positive vertical intercept of $\ln A$.

Analyzing the provided option images:
• The first graph (Image 1) shows a straight line with a positive slope and a negative y-intercept.
• The second graph (Image 2) shows a straight line with a positive slope and a positive y-intercept.
• The third graph (Image 3) shows a straight line with a positive slope passing through the origin.
• The fourth graph (Image 4) shows a straight line with a negative slope, which is consistent with the negative slope of $-\frac{E_a}{R}$.

Therefore, the fourth plot is the correct graph representing the Arrhenius equation.