Rate Constant Units: Videos & Practice Problems

The rate constant, denoted as k, is a crucial parameter in chemical kinetics, and its units vary based on the overall reaction order. Understanding these units is essential for analyzing reaction rates and mechanisms. The units of the rate constant for different reaction orders are as follows:

For zero-order reactions, the units of the rate constant k are expressed as molarity per second, which can also be written as molarity times inverse seconds: \[\text{Units of } k = \text{M} \cdot \text{s}^{-1}\]This indicates that the concentration of the reactant decreases at a constant rate over time.

In the case of first-order reactions, the units of the rate constant are simply inverse seconds:\[\text{Units of } k = \text{s}^{-1}\]This reflects that the rate of reaction is directly proportional to the concentration of a single reactant.

For second-order reactions, the units of the rate constant are given as inverse molarity times inverse seconds:\[\text{Units of } k = \text{M}^{-1} \cdot \text{s}^{-1}\]This indicates that the rate of reaction depends on the concentrations of two reactants or the square of the concentration of a single reactant.

One commonality among all these units is the presence of inverse seconds, which signifies the time dependency of the reaction rate. Additionally, the exponents of the units provide insight into the reaction order: for zero-order reactions, the sum of the exponents is zero; for first-order reactions, it is negative one; and for second-order reactions, it totals negative two. This relationship aids in memorizing the units associated with each reaction order.

By grasping these concepts, students can effectively apply their understanding of rate constants and reaction orders to solve related problems in chemical kinetics.

In this example problem, we explore the determination of the rate constant units for a given rate law. The rate law expresses the reaction velocity, denoted as \( v \), which is defined by the equation:

\[ v = k [A]^n \]

Here, \( k \) represents the rate constant, \( [A] \) is the concentration of the reactant, and \( n \) indicates the reaction order. In this case, since there is only one reactant and it is raised to the power of 2, we conclude that the reaction is second order.

To find the units of the rate constant \( k \), we need to consider the relationship between the units of the rate law and the overall reaction order. The common unit for reaction velocity is molarity per second, expressed as \( \text{M/s} \). For a second-order reaction, the sum of the exponents in the rate law must equal 2. This leads us to analyze the units of \( k \) in relation to the concentration units.

Since the overall reaction order is 2, we can express the units of \( k \) as follows:

\[ k = \frac{v}{[A]^2} \]

Substituting the units, we have:

\[ k = \frac{\text{M/s}}{(\text{M})^2} = \frac{\text{M/s}}{\text{M}^2} = \text{M}^{-1} \text{s}^{-1} \]

This indicates that the units of the rate constant \( k \) for a second-order reaction are \( \text{M}^{-1} \text{s}^{-1} \). Therefore, the correct answer aligns with option c, confirming that the rate constant units for this second-order reaction are indeed \( \text{M}^{-1} \text{s}^{-1} \).