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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
In summary, the half-lives of zero-, first-, and second-order reactions are influenced by different factors: 1. Zero-order reaction: half-life depends directly on the initial concentration (C_0) and inversely on the rate constant (k), thus \(t_{1/2} = \frac{C_0}{2k}\). 2. First-order reaction: half-life depends only on the rate constant (k) and is independent of the initial concentration, thus \(t_{1/2} = \frac{ln2}{k}\). 3. Second-order reaction: half-life depends inversely on both the initial concentration (C_0) and the rate constant (k), thus \(t_{1/2} = \frac{1}{kC_0}\). These differences are due to the reaction orders' varying dependence on initial reactant concentration and rate constant.

Step by step solution

01

Define half-life

Half-life is the time required for the concentration of a reactant to decrease to half of its initial value. It depends on the reaction order and the rate constant (k) of the reaction. Zero-, first-, and second-order reactions have different expressions for their half-lives, so we'll discuss each of them to understand their differences conceptually.
02

Analyze the zero-order reaction half-life

A zero-order reaction is one in which the reaction rate is independent of the reactant concentration, so the rate equation is: \[Rate = k\] To find the half-life, we need to integrate the rate equation over time and solve for the time when the concentration has decreased to half of its initial value: \[C(t) = C_0 - kt\] Half-life (t1/2) can now be expressed as: \[t_{1/2} = \frac{C_0}{2k}\] Conceptually, the half-life of a zero-order reaction depends directly on the initial concentration (C_0) and inversely on the rate constant (k). So, the higher the initial concentration, the longer it will take for the reactant to decrease to half its value. On the other hand, with a higher rate constant, the reaction will happen faster, and thus, the half-life will be shorter.
03

Analyze the first-order reaction half-life

A first-order reaction is one in which the reaction rate is proportional to the reactant concentration, so the rate equation is: \[Rate = kC\] To find the half-life, we can integrate the rate equation over time and solve for the time when the concentration has decreased to half of its initial value: \[ln\frac{C}{C_0} = -kt\] Half-life (t1/2) can now be expressed as: \[t_{1/2} = \frac{ln2}{k}\] Conceptually, the half-life of a first-order reaction is independent of the initial concentration (C_0) and depends only on the rate constant (k). As the rate constant increases, the half-life will decrease, and the reaction will happen faster.
04

Analyze the second-order reaction half-life

A second-order reaction has a reaction rate proportional to the square of the reactant concentration, so the rate equation is: \[Rate = kC^2\] To find the half-life, we integrate the rate equation over time and solve for the time when the concentration has decreased to half of its initial value: \[\frac{1}{C} - \frac{1}{C_0} = kt\] Half-life (t1/2) can now be expressed as: \[t_{1/2} = \frac{1}{kC_0}\] Conceptually, the half-life of a second-order reaction depends inversely on both the initial concentration (C_0) and the rate constant (k). So, higher initial concentrations or higher rate constants will result in shorter half-lives.
05

Compare the half-lives of the three reaction orders

In summary, the half-lives of zero-, first-, and second-order reactions are influenced by different factors: 1. Zero-order reaction: half-life \(鈭� C_0\) and half-life \(鈭� \frac{1}{k}\) 2. First-order reaction: half-life \(鈭� \frac{1}{k}\) (independent of C_0) 3. Second-order reaction: half-life \(鈭� \frac{1}{kC_0}\) The main differences in the half-lives of these reactions can be attributed to their dependence on the initial concentration and the rate constant.

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Key Concepts

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

Zero-Order Reaction
A zero-order reaction is a chemical reaction where the rate is constant and independent of the reactant's concentration. In simple terms, no matter how much reactant you have, the reaction proceeds at the same pace. This might seem counterintuitive, as we often expect that having more reactants would lead to a faster reaction.

However, in a zero-order reaction, the reaction rate is determined by factors other than reactant concentration, such as surface area or catalyst availability. This is represented by the equation for the rate of a zero-order reaction: \[ Rate = k \]
The half-life of a zero-order reaction, the time it takes for half of the reactant to be consumed, can be expressed as:\[ t_{1/2} = \frac{C_0}{2k} \]
Here, we see that the half-life is proportional to the initial concentration \( C_0 \) and inversely proportional to the rate constant \( k \). This means that, for a zero-order reaction, doubling the initial concentration will double the half-life, while doubling the rate constant will halve the half-life.
First-Order Reaction
In contrast to zero-order reactions, first-order reactions have a rate that is directly proportional to the concentration of one reactant. This implies that the reaction speed will change according to how much reactant is present. A classic example of a first-order reaction is radioactive decay, where the rate at which a material decays is directly related to the amount of the substance remaining.

The mathematical relationship for the rate of a first-order reaction is:\[ Rate = kC \]
For first-order reactions, the half-life \( t_{1/2} \) is a constant that does not depend on the initial concentration of the reactant. The half-life equation is:\[ t_{1/2} = \frac{ln2}{k} \]
Here, \( ln2 \) (the natural logarithm of 2) is simply a constant that arises from the mathematical integration in deriving the half-life expression. This characteristic provides an important diagnostic tool in chemical kinetics; if the half-life remains the same regardless of the initial concentration, it suggests a first-order reaction.
Second-Order Reaction
Second-order reactions are a bit more complex, as the rate is proportional to the square of the concentration of one reactant or to the product of two reactant concentrations. This type of reaction usually occurs when two reactant molecules collide and react. An increase in reactant concentration has a more dramatic effect on the rate of the reaction compared to zero- and first-order reactions.

The rate equation for a second-order reaction is given by:\[ Rate = kC^2 \]
For second-order reactions, the half-life is inversely proportional to both the initial concentration and the rate constant, as shown in the equation:\[ t_{1/2} = \frac{1}{kC_0} \]
This infers that increasing the initial concentration, or the rate constant, results in a shorter half-life, which makes intuitive sense. If more reactant molecules are available, or if they're more likely to react (higher \( k \)), then it will take less time for half of the reactant to be used up.

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Most popular questions from this chapter

One reason suggested for the instability of long chains of silicon atoms is that the decomposition involves the transition state shown below: The activation energy for such a process is \(210 \mathrm{~kJ} / \mathrm{mol}\), which is less than either the \(\mathrm{Si}-\) Si or the \(\mathrm{Si}-\mathrm{H}\) bond energy. Why would a similar mechanism not be expected to play a very important role in the decomposition of long chains of carbon atoms as seen in organic compounds?

Upon dissolving \(\operatorname{InCl}(s)\) in \(\mathrm{HCl}, \operatorname{In}^{+}(a q)\) undergoes a disproportionation reaction according to the following unbalanced equation: $$ \operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ This disproportionation follows first-order kinetics with a half-life of \(667 \mathrm{~s}\). What is the concentration of \(\operatorname{In}^{+}(a q)\) after \(1.25 \mathrm{~h}\) if the initial solution of \(\mathrm{In}^{+}(a q)\) was prepared by dissolving \(2.38 \mathrm{~g} \operatorname{InCl}(s)\) in dilute \(\mathrm{HCl}\) to make \(5.00 \times 10^{2} \mathrm{~mL}\) of solution? What mass of \(\operatorname{In}(s)\) is formed after \(1.25 \mathrm{~h}\) ?

Write the rate laws for the following elementary reactions. a. \(\mathrm{CH}_{3} \mathrm{NC}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) b. \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)\) c. \(\mathrm{O}_{3}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) d. \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\)

A certain substance, initially present at \(0.0800 M\), decomposes by zero-order kinetics with a rate constant of \(2.50 \times 10^{-2}\) \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\). Calculate the time (in seconds) required for the system to reach a concentration of \(0.0210 \mathrm{M}\).

Consider the hypothetical reaction \(\mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(g) \longrightarrow\) \(2 \mathrm{AB}(g)\), where the rate law is: $$ -\frac{\Delta\left[\mathrm{A}_{2}\right]}{\Delta t}=k\left[\mathrm{~A}_{2}\right]\left[\mathrm{B}_{2}\right] $$ The value of the rate constant at \(302^{\circ} \mathrm{C}\) is \(2.45 \times 10^{-4} \mathrm{~L} / \mathrm{mol}\). \(\mathrm{s}\), and at \(508^{\circ} \mathrm{C}\) the rate constant is \(0.891 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\). What is the activation energy for this reaction? What is the value of the rate constant for this reaction at \(375^{\circ} \mathrm{C}\) ?
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