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The half-lives of both zero-order and second-order reactions depend on the initial concentration, as well as on the rate constant. In one case, the half- life gets longer as the initial concentration increases, and in the other it gets shorter. Which is which, and why isn't the situation the same for both?

Short Answer

Expert verified
In the case of zero-order reactions, the half-life increases as the initial concentration increases. For second-order reactions, the half-life decreases as the initial concentration increases. This is because the rate of a zero-order reaction is independent of the concentration of reactant, while the rate of a second-order reaction is highly dependent on it.

Step by step solution

01

Understand zero-order reactions

A zero order reaction has the rate law: rate = k[A]^0 = k. Its half-life equation is \(t_{1/2} = \frac{[A]_0}{2k}\), where [A]_0 is the initial concentration, and k is the rate constant. As we can see, as the initial concentration [A]_0 increases, the half-life also increases, meaning the reaction takes longer.
02

Understand second-order reactions

A second order reaction has the rate law: rate = k[A]^2. Its half-life equation is \(t_{1/2} = \frac{1}{2k[A]_0}\). In this case, as the initial concentration [A]_0 increases, the half-life decreases, meaning the reaction takes less time.
03

Explain why they differ

The reason why these two reactions have opposite dependencies on the initial concentration relates to their rate laws. For a zero-order reaction, the rate is independent of the concentration of reactant, while for a second-order reaction, the rate is highly dependent on the concentration of the reactant. Therefore, having a large initial concentration for a second-order reaction means there are more reactant molecules available to react, thus the reaction completes faster.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Zero-Order Reactions
In reaction kinetics, a zero-order reaction is one where the rate of reaction is constant and independent of the concentration of reactants. This can be expressed with the rate law: \( \text{rate} = k[A]^0 = k \), which implies that the rate is equal to the rate constant, \( k \). These reactions can occur when a catalyst is saturated by the reactant, making the reaction speed constant.

An important aspect of zero-order reactions is how the half-life changes with concentration. The half-life (the time it takes for half of the reactant to be consumed) for zero-order reactions is given by: \( t_{1/2} = \frac{[A]_0}{2k} \). Here, \( [A]_0 \) represents the initial concentration of the reactant, and this equation shows that the half-life increases with increasing initial concentration.

Thus, in a zero-order reaction, the more reactant you start with, the longer it will take for half of it to be used because the reaction rate doesn't change."
Second-Order Reactions
Second-order reactions exhibit a different behavior compared to zero-order reactions because their rate depends on the concentration of the reactants. The rate law for a second-order reaction is: \( \text{rate} = k[A]^2 \). This equation shows that the rate is proportional to the square of the concentration of the reactant.

Now, let’s discuss the half-life for second-order reactions. The half-life is determined by the equation: \( t_{1/2} = \frac{1}{2k[A]_0} \). In this case, \( [A]_0 \) is still the initial concentration, and you can observe that the half-life decreases as the initial concentration increases.

This means, for second-order reactions, a higher initial concentration results in a shorter half-life. Why? Because the reaction rate increases with more reactant available. More molecules mean more frequent collisions and reactions, leading to quicker depletion of reactants."
Half-Life Dependency
The concept of half-life in chemical kinetics gives us insight into how long it takes for half of a given amount of reactant to be consumed in a reaction.

For zero-order and second-order reactions, the dependency of half-life on initial concentration varies significantly because of the different ways the reaction rates depend on concentration.

  • Zero-order reactions: The half-life increases as the initial concentration increases, due to the constant reaction rate.
  • Second-order reactions: The half-life decreases with higher initial concentrations, due to the increasing rate as more reactants are present.
In summary, these differences arise from the core nature of their rate laws. While zero-order reactions are independent of concentration, second-order reactions speed up as the reactant concentration increases. Understanding this distinction helps in predicting how long it will take for half of a reactant to be used in each type of reaction.

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Most popular questions from this chapter

The following three-step mechanism has been proposed for the reaction of chlorine and chloroform. $$\begin{aligned} & \text { (1) } \quad \mathrm{Cl}_{2}(\mathrm{g}) \stackrel{k_{1}}{\rightleftharpoons_{k-1}} 2 \mathrm{Cl}(\mathrm{g})\\\ & \text { (2) } \quad \mathrm{Cl}(\mathrm{g})+\mathrm{CHCl}_{3}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{HCl}(\mathrm{g})+\mathrm{CCl}_{3}(\mathrm{g})\\\ &\text { (3) } \quad \mathrm{CCl}_{3}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{CCl}_{4}(\mathrm{g}) \end{aligned}$$ The numerical values of the rate constants for these steps are \(k_{1}=4.8 \times 10^{3} ; \quad k_{-1}=3.6 \times 10^{3} ; \quad k_{2}=1.3 \times 10^{-2} ; k_{3}=2.7 \times 10^{2} .\) Derive the rate law and the magnitude of \(k\) for the overall reaction.

Explain the important distinctions between each pair of terms: (a) first-order and second-order reactions; (b) rate law and integrated rate law; (c) activation energy and enthalpy of reaction; (d) elementary process and overall reaction; (e) enzyme and substrate.

For the disproportionation of \(p\)-toluenesulfinic acid, $$3 \mathrm{ArSO}_{2} \mathrm{H} \longrightarrow \mathrm{ArSO}_{2} \mathrm{SAr}+\mathrm{ArSO}_{3} \mathrm{H}+\mathrm{H}_{2} \mathrm{O}$$ (where \(\mathrm{Ar}=p-\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4}-\) ), the following data were obtained: \(t=0 \min ,[\mathrm{ArSO}_{2} \mathrm{H}]=0.100 \mathrm{M} ; 15 \mathrm{min}, 0.0863 \mathrm{M} ; 30 \mathrm{min}, 0.0752 \mathrm{M} ; 45 \mathrm{min}, 0.0640 \mathrm{M} ; 60 \mathrm{min}, 0.0568 \mathrm{M} ; 120 \mathrm{min}, 0.0387 \mathrm{M} ; 180 \mathrm{min}, 0.0297 \mathrm{M}; 300 \mathrm{min}, 0.0196 \mathrm{M}.\) (a) Show that this reaction is second order. (b) What is the value of the rate constant, \(k ?\) (c) At what time would \(\left[\mathrm{ArSO}_{2} \mathrm{H}\right]=0.0500 \mathrm{M} ?\) (d) At what time would \(\left(\mathrm{ArSO}_{2} \mathrm{H}\right)=0.0250 \mathrm{M} ?\) (e) At what time would \(\left[\mathrm{ArSO}_{2} \mathrm{H}\right]=0.0350 \mathrm{M} ?\)

For the reaction \(A \longrightarrow\) products the following data are obtained. $$\begin{array}{cll} \hline {\text { Experiment 1 }} & &{\text { Experiment 2 }} \\ \hline [\mathrm{A}]=1.204 \mathrm{M} & t=0 \mathrm{min} & {[\mathrm{A}]=2.408 \mathrm{M}} & t=0 \mathrm{min}\\\ {[\mathrm{A}]=1.180 \mathrm{M}} & t=1.0 \mathrm{min} & {[\mathrm{A}]=?} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=0.602 \mathrm{M}} & t=35 \mathrm{min} & {[\mathrm{A}]=?} & t=30 \mathrm{min} \\ \hline \end{array}$$ (a) Determine the initial rate of reaction in Experiment 1. (b) If the reaction is second order, what will be \([\mathrm{A}]\) at \(t=1.0\) min in Experiment 2? (c) If the reaction is first order, what will be \([\mathrm{A}]\) at 30 min in Experiment 2?

In the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+3 \mathrm{D},\) reactant \(\mathrm{A}\) is found to disappear at the rate of \(6.2 \times 10^{-4} \mathrm{M} \mathrm{s}^{-1}.\) (a) What is the rate of reaction at this point? (b) What is the rate of disappearance of \(\mathrm{B}\) ? (c) What is the rate of formation of D?

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