Integrated Rate Laws

Introduction

Key Concepts

Understanding Rate Laws

A rate law expresses the relationship between the rate of a chemical reaction and the concentrations of its reactants. It is generally represented as: $$ rate = k [A]^m [B]^n $$ where:

• rate is the reaction rate.
• k is the rate constant.
• [A] and [B] are the concentrations of reactants A and B.
• m and n are the orders of the reaction with respect to A and B.

Order of Reaction

The order of a reaction indicates how the rate is affected by the concentration of reactants. It can be determined experimentally and is not necessarily the same as the stoichiometric coefficients in the balanced equation.

• Zero-Order Reactions: The rate is independent of the concentration of reactant(s). Rate law: \( rate = k \).
• First-Order Reactions: The rate is directly proportional to the concentration of one reactant. Rate law: \( rate = k [A] \).
• Second-Order Reactions: The rate depends on the concentration of either two reactants or the square of one reactant. Rate law: \( rate = k [A]^2 \) or \( rate = k [A][B] \).

Integrated Rate Laws

Integrated Rate Laws link the concentration of reactants to time, allowing the determination of concentration at any given moment or the calculation of rate constants from experimental data.

• Zero-Order Integrated Rate Law: $$ [A] = [A]_0 - kt $$ where \( [A]_0 \) is the initial concentration, and \( t \) is time.
• First-Order Integrated Rate Law: $$ \ln[A] = \ln[A]_0 - kt $$ or $$ [A] = [A]_0 e^{-kt} $$
• Second-Order Integrated Rate Law: $$ \frac{1}{[A]} = \frac{1}{[A]_0} + kt $$

Graphical Representation

Each order of reaction exhibits a characteristic plot when concentration is graphed against time:

• Zero-Order: Linear decrease in concentration with time. Plot of [A] vs. \( t \) yields a straight line with slope \( -k \).
• First-Order: Semi-logarithmic plot. Plot of \( \ln[A] \) vs. \( t \) yields a straight line with slope \( -k \).
• Second-Order: Reciprocal plot. Plot of \( \frac{1}{[A]} \) vs. \( t \) yields a straight line with slope \( k \).

Determining Reaction Order

To determine the order of a reaction, experimental data is analyzed using the integrated rate laws. By plotting data according to each order and identifying which plot yields a straight line, the reaction order can be established.

• If [A] vs. \( t \) is linear, the reaction is zero-order.
• If \( \ln[A] \) vs. \( t \) is linear, the reaction is first-order.
• If \( \frac{1}{[A]} \) vs. \( t \) is linear, the reaction is second-order.

Half-Life (\( t_{1/2} \))

The half-life of a reaction is the time required for the concentration of a reactant to decrease by half.

• Zero-Order: $$ t_{1/2} = \frac{[A]_0}{2k} $$ The half-life decreases as the reaction proceeds.
• First-Order: $$ t_{1/2} = \frac{0.693}{k} $$ The half-life is constant and independent of the initial concentration.
• Second-Order: $$ t_{1/2} = \frac{1}{k [A]_0} $$ The half-life increases as the reaction proceeds.

Integrated Rate Laws for Reversible Reactions

In reversible reactions, both the forward and reverse reactions contribute to the rate. The integrated rate laws become more complex, often requiring advanced techniques such as the steady-state approximation or equilibrium assumptions to solve. For a reversible reaction: $$ aA + bB \leftrightarrow cC + dD $$ The rate of the forward reaction (\( r_f \)) and the reverse reaction (\( r_r \)) are: $$ r_f = k_f [A]^a [B]^b $$ $$ r_r = k_r [C]^c [D]^d $$ At equilibrium, \( r_f = r_r \), and the integrated rate laws must account for these opposing rates.

Complex Reactions and Integrated Rate Laws

Many real-world reactions involve multiple steps and complex mechanisms. In such cases, the overall rate law may not correspond directly to the stoichiometry of the overall reaction but rather to the rate-determining step. For example, consider the reaction: $$ A + B \rightarrow C $$ with a rate-determining step: $$ A + B \rightarrow C \quad (rate = k[A][B]) $$ The integrated rate laws must consider the mechanism to accurately describe concentration changes over time.

Applications of Integrated Rate Laws

Integrated Rate Laws are crucial in various applications, including:

• Drug Metabolism: Understanding the elimination rate of drugs from the body.
• Environmental Chemistry: Modeling the degradation of pollutants.
• Industrial Chemistry: Designing reactors and optimizing reaction conditions.
• Forensic Science: Determining the time of death based on chemical changes.

Limitations and Challenges

While Integrated Rate Laws are powerful tools, they have limitations:

• Complex Mechanisms: Multistep reactions may not fit simple rate laws.
• Temperature Dependence: Rate constants vary with temperature, affecting predictions.
• Experimental Errors: Inaccurate concentration measurements can lead to incorrect rate laws.
• Assumptions: Assumptions like constant temperature and pressure may not hold in all scenarios.

Mathematical Derivation of Integrated Rate Laws

Deriving Integrated Rate Laws involves solving differential equations based on the rate laws.

• Zero-Order: $$ \frac{d[A]}{dt} = -k $$ Integrating: $$ [A] = [A]_0 - kt $$
• First-Order: $$ \frac{d[A]}{dt} = -k[A] $$ Separating variables and integrating: $$ \ln[A] = \ln[A]_0 - kt $$ or $$ [A] = [A]_0 e^{-kt} $$
• Second-Order: $$ \frac{d[A]}{dt} = -k[A]^2 $$ Separating variables and integrating: $$ \frac{1}{[A]} = \frac{1}{[A]_0} + kt $$

Determining Rate Constants

Once the reaction order is known, the rate constant \( k \) can be determined using the slope of the appropriate integrated rate law plot.

• Zero-Order: From the slope of [A] vs. \( t \), \( k = -slope \).
• First-Order: From the slope of \( \ln[A] \) vs. \( t \), \( k = -slope \).
• Second-Order: From the slope of \( \frac{1}{[A]} \) vs. \( t \), \( k = slope \).

Example Problem

Problem: Given the concentration of reactant A decreases from 0.10 M to 0.050 M in 25 seconds, determine the order of the reaction and the rate constant. Solution:

• Zero-Order: $$ [A] = [A]_0 - kt \\ 0.050 = 0.10 - k(25) \\ k = \frac{0.10 - 0.050}{25} = 0.002 \, \text{M/s} $$
• First-Order: $$ \ln[A] = \ln[A]_0 - kt \\ \ln(0.050) = \ln(0.10) - k(25) \\ k = \frac{\ln(0.10) - \ln(0.050)}{25} \approx 0.0277 \, \text{s}^{-1} $$
• Second-Order: $$ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \\ \frac{1}{0.050} = \frac{1}{0.10} + k(25) \\ 20 = 10 + 25k \\ k = \frac{10}{25} = 0.4 \, \text{L/mol.s} $$

Comparison Table

Summary and Key Takeaways