What Is The Integrated Rate Law

The integrated rate law unveils the relationship between the concentration of reactants and time during a chemical reaction, providing a powerful tool for understanding and predicting reaction kinetics. Unlike the differential rate law, which focuses on the instantaneous rate of reaction, the integrated rate law allows us to determine reactant concentrations at specific points in time or to calculate the time required for a certain amount of reactant to be consumed.

Understanding Reaction Orders and Rate Laws

Before diving into the integrated rate law, let's briefly revisit reaction orders and the basic rate law. The rate of a chemical reaction is influenced by the concentration of reactants. The rate law expresses this relationship mathematically.

For a simple reaction:

aA → Products

The rate law can be written as:

Rate = k[A]^n

Where:

• Rate is the reaction rate (typically in units of M/s)
• k is the rate constant (units vary depending on the reaction order)
• [A] is the concentration of reactant A (typically in molarity, M)
• n is the order of the reaction with respect to reactant A

The order of a reaction (n) indicates how the concentration of a reactant affects the reaction rate. Common reaction orders include:

• Zero Order (n = 0): The rate is independent of the concentration of the reactant.
• First Order (n = 1): The rate is directly proportional to the concentration of the reactant.
• Second Order (n = 2): The rate is proportional to the square of the concentration of the reactant.

The Integrated Rate Laws: A Detailed Exploration

The integrated rate laws are derived from the differential rate laws using calculus. They provide equations that directly relate the concentration of reactants to time. Let's examine the integrated rate laws for zero-order, first-order, and second-order reactions.

1. Zero-Order Integrated Rate Law

For a zero-order reaction, the rate law is:

Rate = k[A]^0 = k

This means the reaction rate is constant and does not depend on the concentration of reactant A. To derive the integrated rate law, we start with the differential rate law:

-d[A]/dt = k

Integrating both sides with respect to time:

∫d[A] = -k ∫dt

This yields:

[A] = -kt + [A]₀

Where:

• [A] is the concentration of reactant A at time t
• [A]₀ is the initial concentration of reactant A at time t = 0
• k is the rate constant

The zero-order integrated rate law is:

[A] = -kt + [A]₀

Graphical Representation:

Plotting [A] versus time (t) for a zero-order reaction yields a straight line with a slope of -k and a y-intercept of [A]₀. This linear relationship is a key characteristic of zero-order reactions.

Half-Life:

The half-life (t₁/₂) of a reaction is the time required for the concentration of the reactant to decrease to half of its initial value. For a zero-order reaction, the half-life is:

t₁/₂ = [A]₀ / 2k

Notice that the half-life of a zero-order reaction is dependent on the initial concentration of the reactant.

Examples of Zero-Order Reactions:

• Decomposition of ammonia (NH₃) on a hot platinum wire. The rate is constant as long as there is sufficient ammonia to cover the surface of the catalyst.
• Enzyme-catalyzed reactions under saturated conditions. When the enzyme is saturated with substrate, adding more substrate does not increase the reaction rate.

2. First-Order Integrated Rate Law

For a first-order reaction, the rate law is:

Rate = k[A]

The rate is directly proportional to the concentration of reactant A. The differential rate law is:

-d[A]/dt = k[A]

To derive the integrated rate law, we separate variables and integrate:

∫d[A]/[A] = -k ∫dt

This yields:

ln[A] = -kt + ln[A]₀

Where:

• ln[A] is the natural logarithm of the concentration of reactant A at time t
• ln[A]₀ is the natural logarithm of the initial concentration of reactant A at time t = 0
• k is the rate constant

The first-order integrated rate law can be written in several forms:

• ln[A] = -kt + ln[A]₀
• ln([A]/[A]₀) = -kt
• [A] = [A]₀ * e^(-kt)

Graphical Representation:

Plotting ln[A] versus time (t) for a first-order reaction yields a straight line with a slope of -k and a y-intercept of ln[A]₀. Alternatively, plotting log[A] versus time will also yield a straight line, but the slope will be -k/2.303.

Half-Life:

For a first-order reaction, the half-life is:

t₁/₂ = ln(2) / k ≈ 0.693 / k

A crucial characteristic of first-order reactions is that the half-life is independent of the initial concentration of the reactant. This means that it takes the same amount of time for the concentration to halve, regardless of how much reactant you start with.

Examples of First-Order Reactions:

• Radioactive decay. The decay of radioactive isotopes follows first-order kinetics.
• Many unimolecular gas-phase reactions. For example, the decomposition of N₂O₅ into NO₂ and O₂.
• Hydrolysis of aspirin.

3. Second-Order Integrated Rate Law

For a second-order reaction where the rate law is Rate = k[A]², the differential rate law is:

-d[A]/dt = k[A]²

To derive the integrated rate law, we separate variables and integrate:

∫d[A]/[A]² = -k ∫dt

This yields:

-1/[A] = -kt - 1/[A]₀

Rearranging gives:

1/[A] = kt + 1/[A]₀

Where:

• [A] is the concentration of reactant A at time t
• [A]₀ is the initial concentration of reactant A at time t = 0
• k is the rate constant

Graphical Representation:

Plotting 1/[A] versus time (t) for a second-order reaction yields a straight line with a slope of k and a y-intercept of 1/[A]₀.

Half-Life:

For a second-order reaction (Rate = k[A]²), the half-life is:

t₁/₂ = 1 / (k[A]₀)

The half-life of a second-order reaction is inversely proportional to the initial concentration of the reactant.

Examples of Second-Order Reactions:

• Dimerization of butadiene.
• The reaction of NO₂ with CO.
• Reactions of hydroxide ions (OH⁻) with esters.

Summary of Integrated Rate Laws

Reaction Order	Rate Law	Integrated Rate Law	Linear Plot	Half-Life
Zero	Rate = k	[A] = -kt + [A]₀	[A] vs. t	t₁/₂ = [A]₀ / 2k
First	Rate = k[A]	ln[A] = -kt + ln[A]₀	ln[A] vs. t	t₁/₂ = ln(2) / k
Second	Rate = k[A]²	1/[A] = kt + 1/[A]₀	1/[A] vs. t	t₁/₂ = 1 / (k[A]₀)

Determining Reaction Order Using Integrated Rate Laws

One of the key applications of integrated rate laws is to determine the order of a reaction experimentally. This can be done by:

1. Collecting Experimental Data: Measure the concentration of a reactant at different time intervals during the reaction.
2. Plotting the Data: Plot the data in different ways, corresponding to the integrated rate laws for zero-order, first-order, and second-order reactions.
   • Plot [A] vs. t (for zero order)
   • Plot ln[A] vs. t (for first order)
   • Plot 1/[A] vs. t (for second order)
3. Identifying the Linear Plot: The plot that yields a straight line indicates the order of the reaction. For example, if the plot of ln[A] vs. t is linear, the reaction is first order.
4. Calculating the Rate Constant: The slope of the linear plot provides the rate constant (k). Remember to account for the sign in the slope.

Applications of Integrated Rate Laws

Integrated rate laws have numerous applications in chemistry, including:

• Predicting Reactant Concentrations: Calculating the concentration of reactants at a specific time during a reaction.
• Determining Reaction Time: Calculating the time required for a certain amount of reactant to be consumed.
• Calculating Half-Lives: Determining the half-life of a reaction.
• Understanding Reaction Mechanisms: Providing insights into the steps involved in a chemical reaction.
• Pharmacokinetics: Predicting drug concentrations in the body over time.
• Environmental Science: Modeling the degradation of pollutants in the environment.
• Industrial Chemistry: Optimizing reaction conditions for industrial processes.

Beyond Simple Rate Laws: Complex Reactions

While the integrated rate laws discussed above apply to relatively simple reactions, many reactions involve more complex rate laws. These can include:

• Reactions with Multiple Reactants: The rate law may depend on the concentrations of multiple reactants.
• Reversible Reactions: Reactions that proceed in both the forward and reverse directions.
• Consecutive Reactions: Reactions that involve multiple steps, where the product of one step becomes the reactant of the next.
• Parallel Reactions: Reactions that proceed simultaneously through different pathways.

Analyzing these more complex reactions often requires more sophisticated mathematical techniques and may involve numerical methods to solve the rate equations. However, understanding the basic principles of integrated rate laws provides a solid foundation for tackling these more challenging scenarios.

Factors Affecting Reaction Rates

Several factors influence reaction rates, and understanding these factors is crucial for applying integrated rate laws effectively. These include:

• Temperature: Increasing temperature generally increases reaction rates. This is because higher temperatures provide more energy for molecules to overcome the activation energy barrier. The Arrhenius equation quantifies the relationship between temperature and the rate constant.
• Catalysts: Catalysts are substances that speed up a reaction without being consumed in the process. Catalysts lower the activation energy, making it easier for the reaction to occur.
• Concentration: As we've seen, the concentration of reactants directly affects the reaction rate (except for zero-order reactions).
• Surface Area: For reactions involving solids, increasing the surface area can increase the reaction rate. This is because more reactant molecules are exposed to the reaction environment.
• Pressure: For gas-phase reactions, increasing the pressure can increase the reaction rate by increasing the concentration of reactants.

Practical Examples and Calculations

Let's illustrate the use of integrated rate laws with a few practical examples:

Example 1: First-Order Decomposition

The decomposition of dinitrogen pentoxide (N₂O₅) into nitrogen dioxide (NO₂) and oxygen (O₂) is a first-order reaction:

2N₂O₅(g) → 4NO₂(g) + O₂(g)

The rate constant for this reaction at a certain temperature is k = 5.0 x 10⁻⁴ s⁻¹. If the initial concentration of N₂O₅ is 0.10 M, what is the concentration of N₂O₅ after 1000 seconds?

Solution:

Using the first-order integrated rate law:

ln[N₂O₅] = -kt + ln[N₂O₅]₀

ln[N₂O₅] = -(5.0 x 10⁻⁴ s⁻¹)(1000 s) + ln(0.10 M)

ln[N₂O₅] = -0.5 - 2.303

ln[N₂O₅] = -2.803

[N₂O₅] = e⁻².⁸⁰³ ≈ 0.0607 M

Therefore, the concentration of N₂O₅ after 1000 seconds is approximately 0.0607 M.

Example 2: Determining Reaction Order

A reaction A → Products is studied, and the following data is obtained:

Time (s)	[A] (M)
0	1.00
10	0.67
20	0.50
30	0.40

Determine the order of the reaction.

Solution:

We need to plot the data in different ways and see which plot yields a straight line:

• Zero Order: Plot [A] vs. t.
• First Order: Plot ln[A] vs. t.
• Second Order: Plot 1/[A] vs. t.

Calculating the necessary values:

Time (s)	[A] (M)	ln[A]	1/[A]
0	1.00	0.00	1.00
10	0.67	-0.40	1.49
20	0.50	-0.69	2.00
30	0.40	-0.92	2.50

After plotting the data, we observe that the plot of 1/[A] vs. t yields a straight line. Therefore, the reaction is second order.

Conclusion

The integrated rate law is a fundamental concept in chemical kinetics that provides a powerful tool for understanding and predicting the behavior of chemical reactions. By understanding the relationship between reactant concentrations and time, we can gain valuable insights into reaction mechanisms, optimize reaction conditions, and make predictions about the progress of chemical reactions. Mastery of integrated rate laws is essential for students and professionals in chemistry, chemical engineering, and related fields.