Plot Following Arrhenius Equation for Rate Constant Variation

According to the Arrhenius equation, the dependence of the rate constant $k$ on temperature $T$ (in Kelvin) is given by:
$k = A e^{-\frac{E_a}{RT}}$
where:

• $A$ is the pre-exponential factor (frequency factor),
• $E_a$ is the activation energy ($E_a > 0$ for typical reactions),
• $R$ is the universal gas constant,
• $T$ is the absolute temperature.

Let us analyze the behavior of $k$ as a function of $T$:

1. Boundary Limits:

   • As $T \to 0^+$, $-\frac{E_a}{RT} \to -\infty$, which gives:
     $\lim_{T \to 0^+} k = A e^{-\infty} = 0$
   • As $T \to \infty$, $-\frac{E_a}{RT} \to 0$, which gives:
     $\lim_{T \to \infty} k = A e^0 = A$

2. First Derivative (Slope):
   $\frac{dk}{dT} = A \cdot e^{-\frac{E_a}{RT}} \cdot \left( \frac{E_a}{RT^2} \right) = \frac{k E_a}{RT^2}$
   Since $k > 0$, $E_a > 0$, and $T > 0$, we have:
   $\frac{dk}{dT} > 0$
   This indicates that the rate constant $k$ increases monotonically with an increase in temperature $T$. This immediately rules out options (C) and (D), where $k$ decreases.

3. Second Derivative (Curvature):
   $\frac{d^2k}{dT^2} = \frac{d}{dT}\left( \frac{A E_a}{R T^2} e^{-\frac{E_a}{RT}} \right) = \frac{k E_a}{RT^3} \left( \frac{E_a}{RT} - 2 \right)$
   For conventional chemical reactions, $E_a$ is of the order of tens to hundreds of $\text{kJ}\cdot\text{mol}^{-1}$, making $\frac{E_a}{RT} \gg 2$ at practically accessible temperatures. Thus:
   $\frac{d^2k}{dT^2} > 0$
   This implies that the plot of $k$ versus $T$ is concave upward (exhibiting an exponential-like increase) and non-linear, which rules out the straight line in option (B).

Therefore, plot (A) accurately depicts the variation of $k$ with $T$ according to the Arrhenius equation.

Correct Answer: A