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Chapter 12: Problem 8

Provide a conceptual rationale for the differences in the halflives of zero-, first-, and second-order reactions.

Short Answer

Expert verified
The differences in half-lives of zero-, first-, and second-order reactions can be attributed to their distinct rate laws and how they depend on the concentration of reactants. Zero-order reactions have a half-life that is directly proportional to the initial concentration and inversely proportional to the rate constant (\(t_{1/2} = \frac{[A]_0}{2k}\)). First-order reactions have a half-life that is independent of the initial concentration and only depends on the rate constant (\(t_{1/2} = \frac{\ln{2}}{k}\)). Second-order reactions have a half-life that is inversely proportional to both the rate constant and the initial concentration (\(t_{1/2} = \frac{1}{k[A]_0}\)). These differences arise due to the characteristics of each type of reaction and the role concentration plays in their reaction rates.

Step by step solution

01

Understand zero-order reactions

A zero-order reaction is a reaction where the rate is independent of the concentration of the reactant. In other words, the rate remains constant during the reaction. The rate law for a zero-order reaction is given by: \[r = k\] where \(r\) is the rate, and \(k\) is the zero-order rate constant.
02

Half-life for zero-order reactions

The half-life formula for a zero-order reaction can be derived from their rate law, and is given by: \[t_{1/2} = \frac{[A]_0}{2k}\] where \(t_{1/2}\) is the half-life, [\(A\)] is the initial concentration of the reactant, and \(k\) is the rate constant. Notice that the half-life of a zero-order reaction is inversely related to the rate constant and directly proportional to the initial concentration.
03

Understand first-order reactions

A first-order reaction is a reaction where the rate is directly proportional to the concentration of one reactant. The rate law for a first-order reaction is given by: \[r = k[A]\] where \(r\) is the rate, [\(A\)] is the concentration, and \(k\) is the first-order rate constant.
04

Half-life for first-order reactions

The half-life formula for a first-order reaction can be derived from their rate law, and it is given by: \[t_{1/2} = \frac{\ln{2}}{k}\] where \(t_{1/2}\) is the half-life, and \(k\) is the rate constant. Notice that the half-life for a first-order reaction is independent of the initial concentration and only depends on the rate constant.
05

Understand second-order reactions

A second-order reaction is a reaction where the rate is directly proportional to the square of the concentrations of one reactant or the product of concentrations of two different reactants. The rate law for a second-order reaction is given by: \[r = k[A]^2\] or \[r = k[A][B]\] depending on whether it is the square of one reactant or the product of two reactants. In this case, we will focus on the single reactant case.
06

Half-life for second-order reactions

The half-life formula for a second-order reaction can be derived by integrating the rate law, and it is given by: \[t_{1/2} = \frac{1}{k[A]_0}\] where \(t_{1/2}\) is the half-life, [\(A\)] is the initial concentration of the reactant, and \(k\) is the rate constant. Notice that the half-life for a second-order reaction depends on both the rate constant and the initial concentration, inversely.
07

Conclusion

In conclusion, the differences in half-lives of zero-, first-, and second-order reactions can be summarized as follows: 1. Zero-order: Half-life is directly proportional to the initial concentration and inversely proportional to the rate constant. 2. First-order: Half-life is independent of the initial concentration and only depends on the rate constant. 3. Second-order: Half-life is inversely proportional to both the rate constant and the initial concentration. These differences arise due to the way in which the reactions occur and how their rates depend on the concentration of the reactants involved. The half-life formulas also show how the values of half-life for each reaction differ conceptually.

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