Integrated Rate Laws: 3 Practical Examples

Understanding Integrated Rate Laws

Integrated rate laws are essential for understanding how the concentration of reactants changes over time in chemical reactions. They provide a mathematical relationship between concentration and time, which is vital for predicting reaction behavior. Here, we explore three diverse, practical examples of integrated rate laws that illustrate their applications in real-world scenarios.

Example 1: The Decomposition of Hydrogen Peroxide

In this example, we will explore the decomposition of hydrogen peroxide (H₂O₂) into water and oxygen gas. This reaction is a first-order reaction, making it a perfect candidate for integrated rate law applications.

Hydrogen peroxide decomposes according to the following equation:

2 H₂O₂ → 2 H₂O + O₂

In a laboratory setting, scientists can measure the concentration of hydrogen peroxide over time to determine the rate of the reaction. By plotting the natural logarithm of the concentration of H₂O₂ against time, they can derive the integrated rate law:

ln[H₂O₂] = -kt + ln[H₂O₂]₀

Where:

• [H₂O₂] is the concentration of hydrogen peroxide at time t.
• k is the rate constant.
• [H₂O₂]₀ is the initial concentration of hydrogen peroxide.

This equation allows chemists to calculate the concentration of hydrogen peroxide at any given time, which is crucial for applications in industries like cosmetics and food preservation, where H₂O₂ is commonly used as a disinfectant.

Notes:

• The rate constant k can be determined experimentally, and it varies with temperature.
• This first-order reaction indicates that the rate of reaction depends on the concentration of H₂O₂.

Example 2: The Saponification of Esters

Saponification is a classic reaction in organic chemistry where an ester reacts with a base to produce an alcohol and a soap. This reaction can be investigated using second-order integrated rate laws. A common example is the reaction between ethyl acetate and sodium hydroxide (NaOH).

The reaction can be represented as:

C₂H₅OCOCH₃ + NaOH → C₂H₅OH + CH₃COONa

For a second-order reaction, the integrated rate law is expressed as:

rac{1}{[A]} - rac{1}{[A]₀} = kt

Where:

• [A] is the concentration of the reactant (ethyl acetate) at time t.
• [A]₀ is the initial concentration of ethyl acetate.
• k is the rate constant.

In practical applications, understanding the rate of saponification is essential for soap production, allowing manufacturers to optimize the reaction conditions to achieve desired soap properties.

Notes:

• The reaction is second-order because it depends on the concentration of both the ester and the base.
• Temperature and concentration can significantly affect the rate constant k in this reaction.

Example 3: The Reaction Between Acetic Acid and Sodium Bicarbonate

In this example, we will examine the reaction between acetic acid (CH₃COOH) and sodium bicarbonate (NaHCO₃), which is a classic acid-base reaction often used in baking and cooking.

The overall reaction can be summarized as:

CH₃COOH + NaHCO₃ → CH₃COONa + CO₂ + H₂O

This reaction can be represented as a pseudo-first-order reaction because one reactant is in large excess compared to the other. Thus, we can simplify our analysis by focusing on the acetic acid concentration.

The integrated rate law can be expressed as:

ln[CH₃COOH] = -kt + ln[CH₃COOH]₀

Where:

• [CH₃COOH] is the concentration of acetic acid at time t.
• k is the effective rate constant for the reaction.

This example is particularly relevant in culinary applications where the reaction produces carbon dioxide gas, contributing to the leavening of baked goods.

Notes:

• This reaction is notable for its practical implications in cooking and food science.
• The rate constant k can be influenced by factors such as temperature and concentration of the reactants.