Question

At a particular temperature, the magnitude of the rate constant of a reaction is $5 \times 10^{-5}$ and the unit of the pre-exponential factor of the Arrhenius equation for this reaction is $\text{mol L}^{-1}\text{ min}^{-1}$. Which of the following plots is correct for this reaction?
[Note: $[\text{R}]_0$ is the initial concentration and $\text{t}_{1/2}$ is the half-life of the reaction]}

• A. A
• B. B
• C. C
• D. D

Hint:

Always check units first in chemical kinetics!
$\text{mol L}^{-1}\text{ s}^{-1}$ (or $\text{min}^{-1}$) belongs exclusively to a zero-order reaction.
For zero order, $t_{1/2} \propto [\text{R}]_0$, which must be a straight line passing through the origin.

Answer 1

The Correct Option is A

Step 1 : Understanding the Question:
The question asks us to identify the correct plot of half-life ($t_{1/2}$) versus initial concentration ($[\text{R}]_0$) based on the unit of the pre-exponential factor ($A$) in the Arrhenius equation.
Step 2 : Key Formulas and Approach:
The Arrhenius equation is:
\[ k = A e^{-E_a/RT} \]
Since the exponential term $e^{-E_a/RT}$ is dimensionless, the unit of the pre-exponential factor $A$ is identical to the unit of the rate constant $k$.
We will determine the order of the reaction from the units of $k$ and then apply the half-life equation for that order.
Step 3 : Detailed Explanation:

Determining Reaction Order:
The unit of $A$ is given as $\text{mol L}^{-1}\text{ min}^{-1}$.
Thus, the unit of the rate constant $k$ is also $\text{mol L}^{-1}\text{ min}^{-1}$.
The general unit for a rate constant of an $n$-th order reaction is:
\[ (\text{mol L}^{-1})^{1-n} \text{ time}^{-1} \]
Comparing the units:
\[ 1 - n = 1 \implies n = 0 \]
This confirms the reaction is of zero order.

Half-Life of a Zero-Order Reaction:
For a zero-order reaction, the half-life ($t_{1/2}$) is given by the formula:
\[ t_{1/2} = \frac{[\text{R}]_0}{2k} \]
This represents a straight-line equation of the form $y = mx$, where:
\[ y = t_{1/2}, \quad x = [\text{R}]_0, \quad \text{and slope } m = \frac{1}{2k} \]
Since $k$ is positive, the slope is positive, meaning $t_{1/2}$ increases linearly with $[\text{R}]_0$ starting from the origin.

Step 4 : Final Answer:
This linear relationship starting from the origin is correctly depicted in Plot (a).
This matches Option (A).