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Make a graph of [A] versus time for zero-, first-, and second-order reactions. From these graphs, compare successive half-lives.

Short Answer

Expert verified
For zero-order reactions, the concentration of A decreases linearly with time, and successive half-lives have the same duration. The equation is [A] = -kt + [A]₀. For first-order reactions, the concentration of A decreases exponentially with time, and successive half-lives also have the same duration. The equation is [A] = [A]₀ * e^(-kt). For second-order reactions, the duration of successive half-lives increases as the concentration decreases. The equation is 1/[A] = kt + 1/[A]₀. In conclusion, zero and first-order reactions have constant half-lives, while second-order reactions have increasing half-lives.

Step by step solution

01

Understand the rate laws for different reaction orders

For a reaction A → products, the rate laws for different reaction orders are: 1. Zero-order: Rate = k[A]^0 = k 2. First-order: Rate = k[A]^1 = k[A] 3. Second-order: Rate = k[A]^2
02

Obtain the integrated rate laws for different reaction orders

For each reaction order, we can find the integrated rate law, which gives the relation between concentration [A] and time: 1. Zero-order: [A] = -kt + [A]₀ 2. First-order: ln[A] = -kt + ln[A]₀ or [A] = [A]₀ * e^(-kt) 3. Second-order: 1/[A] = kt + 1/[A]₀
03

Understand half-life for each reaction order

The half-life is the time required for the concentration of A to reduce to half of its initial value. For each reaction order, the half-life can be defined as: 1. Zero-order: t_(1/2) = ([A]₀ / 2) / k 2. First-order: t_(1/2) = ln(2) / k 3. Second-order: t_(1/2) = 1 / k[A]₀
04

Plot the graphs and compare successive half-lives for each order

1. Zero-order reaction: Plot [A] versus time, adjust the time scale to show each half-life. For zero-order reactions, the successive half-lives are the same duration because the amount of A decreases linearly. 2. First-order reaction: Plot [A] versus time, adjust the time scale to show each half-life. For first-order reactions, the successive half-lives also have the same duration, but the concentration of A decreases exponentially with time. 3. Second-order reaction: Plot [A] versus time, adjust the time scale to show each half-life. For second-order reactions, successive half-lives increase since the time required for the concentration of A to reduce to half of its value depends on the current concentration. To conclude, for zero and first-order reactions, successive half-lives have the same duration, while for second-order reactions, the duration of successive half-lives increases with time.

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Key Concepts

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

Rate Laws
Rate laws are fundamental to studying chemical kinetics, which is the branch of chemistry that deals with the speeds or rates of chemical reactions. These laws express the rate of a reaction as a function of the concentration of reactants. For instance, the general form of a rate law is expressed as \( Rate = k[A]^n \), where \( k \) is the rate constant, \( [A] \) represents the concentration of reactant \( A \) and \( n \) is the order of the reaction concerning \( A \) .

Zero-order reactions, where \( n = 0 \) , have a rate that is independent of the concentration of \( A \) . Thus the rate law simplifies to \( Rate = k \). In contrast, first-order reactions (\( n = 1 \) ) have a rate directly proportional to \( [A] \) , leading to \( Rate = k[A] \). For second-order reactions (\( n = 2 \) ), the rate is proportional to the square of the concentration of \( A \), which gives us \( Rate = k[A]^2 \). Understanding these distinctions is crucial because they influence how a reaction progresses with time and how to control or utilize these reactions effectively.
Integrated Rate Laws
Integrated rate laws are derived from the basic rate laws and provide a relationship between reactant concentrations and time. These equations are essential for predicting how concentrations will change over time and for calculating the half-life of a reaction.

For a zero-order reaction, the integrated rate law is \( [A] = -kt + [A]_0 \), where \( [A]_0 \) is the initial concentration. This equation predicts a linear decrease in concentration over time. In the case of a first-order reaction, the integrated rate law can take two forms. One is logarithmic: \( ln[A] = -kt + ln[A]_0 \), and the other is exponential: \( [A] = [A]_0e^{-kt} \). Both forms indicate an exponential decrease in concentration. Lastly, the integrated rate law for a second-order reaction is \( 1/[A] = kt + 1/[A]_0 \), showing that the inverse of the concentration increases linearly with time. Integrated rate laws are pivotal for plotting concentration vs. time graphs, which visually represent the progression of the reaction.
Half-life of Reactions
The half-life of a reaction, often denoted as \( t_{1/2} \), is the time it takes for the concentration of a reactant to decrease by half. It's a crucial concept in both chemical kinetics and practical applications like pharmacology.

Zero-order reactions have a half-life that is proportional to the initial concentration: \( t_{1/2} = [A]_0/2k \). On the other hand, the half-life of a first-order reaction is a constant that does not depend on the initial concentration, and it is given by \( t_{1/2} = ln(2)/k \). For second-order reactions, the half-life is inversely proportional to the initial concentration, described by \( t_{1/2} = 1/k[A]_0 \). The contrasting behaviors of half-lives in different reaction orders are critical for understanding how reactions behave over time and for designing processes that depend on certain reaction times.
Reaction Kinetics Graphs
Reaction kinetics graphs provide a visual representation of the concentration of reactants over time and can be used to quickly assess the order of a reaction. By plotting data from integrated rate laws, students and researchers can analyze how reactions progress.

For a zero-order reaction, the graph of \( [A] \) versus time will be a straight line with a negative slope, showing a constant rate of concentration decrease. In first-order reactions, the graph is a curve that depicts an exponential decay of concentration. Plotting the natural logarithm of \( [A] \) against time will yield a straight line, illustrating the linear relationship from the logarithmic form of the first-order integrated rate law.

A second-order reaction graph will show a curve where the concentration decreases more rapidly over time. By plotting \( 1/[A] \) versus time, the graph becomes a straight line, indicating a linear relationship similar to zero-order but with inversely related variables.

By comparing successive half-lives on these graphs, one can see that for zero-order and first-order reactions, the half-lives remain constant, whereas for second-order reactions, each half-life period increases as the reaction progresses due to the dependence on the changing concentration of \( A \).

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Most popular questions from this chapter

Consider the following statements: "In general, the rate of a chemical reaction increases a bit at first because it takes a while for the reaction to get 'warmed up.' After that, however, the rate of the reaction decreases because its rate is dependent on the concentrations of the reactants, and these are decreasing." Indicate everything that is correct in these statements, and indicate everything that is incorrect. Correct the incorrect statements and explain.

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 first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{~s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{~s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .{ }^{\circ} \mathrm{C}\), respectively. What is the value of the activation energy?

Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{~A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} M\), \([\mathrm{B}]_{0}=3.0 \mathrm{M}\), and \([\mathrm{C}]_{0}=2.0 \mathrm{M} .\) The reaction is started, and after \(8.0\) seconds, the concentration of \(\mathrm{A}\) is \(3.8 \times 10^{-3} \mathrm{M}\). a. Calculate the value of \(k\) for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after \(13.0\) seconds. d. Calculate the concentration of \(\mathrm{C}\) after \(13.0\) seconds.

One mechanism for the destruction of ozone in the upper atmosphere is $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & \text { Slov } \\ \mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \text { Fast } \\ \hline \end{array} $$ Overall reaction \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{~g})\) a. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{\mathrm{a}}\) for the uncatalyzed reaction $$ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2} $$ is \(14.0 \mathrm{~kJ} . E_{\mathrm{a}}\) for the same reaction when catalyzed is \(11.9 \mathrm{~kJ}\). What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C}\) ? Assume that the frequency factor \(A\) is the same for each reaction.

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