Question

What are the units of the rate constant for a third-order reaction?

Short answer

The units are \( \text{L}^{2} \text{mol}^{-2} \text{s}^{-1} \).

Step-by-step solution

Understanding Rate Law

For a third-order reaction, the rate law is expressed as: \[ \text{Rate} = k[A]^x[B]^y[C]^z \]where x + y + z = 3 (because it's a third-order reaction), and k is the rate constant. Our goal is to find the units for this rate constant.

Formula for Units of Rate

The units of reaction rate are typically given as concentration change per unit time. For example, it could be \( \text{mol L}^{-1} \text{s}^{-1} \). We use this understanding to deduce the units for the rate constant.

Calculating Units for Third-Order Reaction

A third-order reaction's rate expression is:\[ \text{Rate} = k[A][B][C] \]In terms of units, it's:\[ \text{mol L}^{-1} \text{s}^{-1} = k (\text{mol L}^{-1})^3 \]Simplifying gives:\[ k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^3 \text{L}^{-3}} \]

Unit Cancellation and Rearrangement

From the above expression:\[ k = \frac{1}{\text{mol}^2 \text{L}^{-2} \text{s}} \]So, the units of k for a third-order reaction are:\[ \text{L}^{2} \text{mol}^{-2} \text{s}^{-1} \]

Key concepts

Rate Constant

In chemical kinetics, the rate constant is a key factor that helps determine how quickly a reaction proceeds. When you think of a reaction, imagine it as a race to the finish line; the rate constant is like the speed of the runners. For third-order reactions, which involve three reactants, the rate constant must account for all their concentrations.Understand that the rate constant, often denoted as \( k \), is not just a simple number. It's a proportionality factor in the rate equation. This means it ties together the concentration of reactants to the reaction rate. The value of \( k \) can show how temperature and other factors affect the reaction rate.For a third-order reaction, the rate equation looks like this:- \( \text{Rate} = k[A][B][C] \)Here, \([A]\), \([B]\), and \([C]\) are concentrations of the reactants.Remember, the units of the rate constant vary depending on the reaction order. For third-order reactions, we calculated the units as \( \text{L}^2 \text{mol}^{-2} \text{s}^{-1} \). This is derived by balancing the units in the equation so that the rate units \( \text{mol L}^{-1} \text{s}^{-1} \) remain consistent throughout.

Reaction Rate

Reaction rate is an important concept because it provides a measure of how fast a reaction goes. It tells us how quickly reactants are turned into products over time. Typically, the rate is expressed in terms of concentration change per unit time.Think of reaction rate as the speedometer of a reaction; it shows how fast or slow the reaction 'vehicle' is moving. High reaction rates mean rapid product formation, while low rates indicate slower conversions.For a simple example, consider how hydrogen and oxygen form water. In a lab setting, measuring how fast hydrogen gets used up or water is formed can give insights into the reaction rate. Usually expressed in terms like \( \text{mol L}^{-1} \text{s}^{-1} \), this measurement allows chemists to compare the speed of different reactions.When dealing with a third-order reaction, remember to consider all reactants because the rate depends on each concentration. Calculating these rates accurately helps us understand the efficiency and dynamics of the reaction process.

Rate Law Units

Rate law units are vital as they reveal the relationship between reactant concentrations and the rate constant within the rate equation. Each order of reaction possesses unique units for its rate constant, essential for balance and consistency in calculations.For a third-order reaction, combining three reactants, the unit calculation follows a distinct path. The rate is expressed in units of \( \text{mol L}^{-1} \text{s}^{-1} \).Here's a brief breakdown:- - Rate = \( \text{mol L}^{-1} \text{s}^{-1} \) - For third-order: \( [A]^x[B]^y[C]^z = \text{(mol L}^{-1})^3 \)This leads us to the rate constant being:- - \( k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{ ext{mol}^3 \text{L}^{-3}} \) - Resulting in: \( \text{L}^2 \text{mol}^{-2} \text{s}^{-1} \)Understanding these units ensures you avoid calculation errors and maintain consistency in your work. It serves as a powerful tool to relate reactant behavior in different reaction scenarios, offering deeper insights into the chemistry at play.