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Three different sets of data of \([\mathrm{A}]\) versus time are giv the following table for the reaction \(A \longrightarrow\) prod [Hint: There are several ways of arriving at answer each of the following six questions. $$\begin{array}{cccccc} \hline \text { I } & & \text { II } & & \text { III } & \\ \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } \\ \hline 0 & 1.00 & 0 & 1.00 & 0 & 1.00 \\ 25 & 0.78 & 25 & 0.75 & 25 & 0.80 \\ 50 & 0.61 & 50 & 0.50 & 50 & 0.67 \\ 75 & 0.47 & 75 & 0.25 & 75 & 0.57 \\ 100 & 0.37 & 100 & 0.00 & 100 & 0.50 \\ 150 & 0.22 & & & 150 & 0.40 \\ 200 & 0.14 & & & 200 & 0.33 \\ 250 & 0.08 & & & 250 & 0.29 \\ \hline \end{array}$$ Which of these sets of data corresponds to a (a) zero-order, (b) first-order, (c) second-order reaction?

Short Answer

Expert verified
Set I corresponds to a first-order reaction, Set II corresponds to a zero-order reaction, and Set III corresponds to a second-order reaction.

Step by step solution

01

Examine the Trends for Each Reaction Order

Start by understanding the trends for each reaction order. A zero-order reaction shows a linear decrease in concentration with time. A first-order reaction shows an exponential decay in concentration with time. A second-order reaction typically shows a rapid decrease in concentration at the beginning, which gradually slows down.
02

Identify Zero-Order Reaction

The zero-order reaction should have a linear decrease in the concentration of reactant A over time. When viewing the provided data sets, it can be seen that dataset II matches this description most closely. The concentration decreases from 1.00 M to 0.00 M over 100 seconds, indicating a linear decrease.
03

Identify First-Order Reaction

First-order reactions show exponential decay in the concentration of the reactant over time. In the given data sets, dataset I fits this description. The concentration decreases exponentially from 1.00 M to 0.08 M over 250 seconds.
04

Identify Second-Order Reaction

A second-order reaction shows a rapid decrease in concentration initially, which slows down with time. Dataset III depicts this trend. The concentration rapidly decreases from 1.00 M to 0.50 M in the first 100 seconds, then decreases slowly reaching 0.29 M at 250 seconds.

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Key Concepts

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

Zero-Order Reaction
In a zero-order reaction, the rate at which the reactant is consumed is constant over time. This means the reactant concentration decreases linearly with time. For example, when you have a zero-order reaction, you expect a linear plot with time on the x-axis and concentration on the y-axis. Alike a straight road that does not twist or turn!

One of the characteristics of zero-order reactions is that the rate of reaction doesn't depend on the concentration of the reactant. This is often seen in reactions where the catalyst or surface is saturated. Think of it like a bottleneck effect— regardless of how much reactant you have, the rate remains the same as capacity is maxed out.
  • The concentration changes linearly over time.
  • The rate is constant and independent of the concentration of reactants.
In terms of mathematical representation, the rate law for a zero-order reaction is given by: \[ [ ext{A}] = [ ext{A}_0] - kt \] where \(k\) is the rate constant. If you were following the data from the exercise, DataSet II beautifully follows this pattern.
First-Order Reaction
First-order reactions are the kind where the rate of reaction is directly proportional to the concentration of one reacting substance. This implies that the concentration of the reactant decreases exponentially over time. As the reaction goes on, the reactant concentration halves successively, demonstrating this exponential decay. Rather like watching a repeatedly halving pie getting smaller and smaller!

If you picture plotting time against the natural logarithm of concentration, you will see a stunning straight line, unlike the curvy profile seen with concentration only. The rule of thumb goes: higher concentration, faster reaction; lower concentration, slower reaction.
  • The concentration decreases exponentially over time.
  • The rate of reaction depends linearly on the concentration of the reactant.
For first-order reactions, \( ext{Rate} = k[ ext{A}] \), and the logarithmic form is: \[ ext{ln}[ ext{A}] = ext{ln}[ ext{A}_0] - kt \] where \(k\) is the rate constant. Dataset I from the exercise text book illustrates this concept perfectly with an elegant exponential decline.
Second-Order Reaction
Second-order reactions can be a bit more complex as they involve either two molecules of the same reactant or a combination of two different reactants. The rate of reaction is proportional to the square of the concentration of a single reactant or product of concentrations of two reactants. This results in an interesting curve as the concentration of reactant decreases with time—it starts off quickly and gradually slows down, like running in the beginning of a marathon and slowing down towards the end.

This reaction order gives you a nifty hyperbolic decline when plotting concentration against time. If you instead plot the reciprocal of concentration against time, a linear relationship would emerge, revealing its secret.
  • The rate is proportional to the square of the reactant's concentration.
  • The concentration change with time shows a rapid decrease initially, then slows.
The rate law for second-order reactions is expressed as:\[ rac{1}{[ ext{A}]} = rac{1}{[ ext{A}_0]} + kt \] where \(k\) is the rate constant. In your textbook exercise, Dataset III follows this trajectory, showing a rapid fall followed by a tapering off in concentration.

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Most popular questions from this chapter

The reaction \(A \longrightarrow\) products is first order in A. Initially, \([\mathrm{A}]=0.800 \mathrm{M}\) and after 54min, \([\mathrm{A}]=0.100 \mathrm{M}.\) (a) At what time is \([\mathrm{A}]=0.025 \mathrm{M} ?\) (b) What is the rate of reaction when \([\mathrm{A}]=0.025 \mathrm{M} ?\)

The following rates of reaction were obtained in three experiments with the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NOCl}(\mathrm{g}).\) $$\begin{array}{llll} \hline & \text { Initial } & \text { Initial } & \text { Initial Rate of } \\ \text { Expt } & \text { [NO], M } & \text { [Cl }_{2} \text { ], M } & \text { Reaction, } \mathrm{M} \mathrm{s}^{-1} \\ \hline 1 & 0.0125 & 0.0255 & 2.27 \times 10^{-5} \\ 2 & 0.0125 & 0.0510 & 4.55 \times 10^{-5} \\ 3 & 0.0250 & 0.0255 & 9.08 \times 10^{-5} \\ \hline \end{array}$$ What is the rate law for this reaction?

In the first-order decomposition of substance A the following concentrations are found at the indicated times: \(t=0 \mathrm{s},[\mathrm{A}]=0.88 \mathrm{M} ; t=50 \mathrm{s},[\mathrm{A}]=0.62 \mathrm{M} ; t=100 \mathrm{s},[\mathrm{A}]=0.44 \mathrm{M} ; t=150 \mathrm{s},[\mathrm{A}]=0.31 \mathrm{M}.\) Calculate the instantaneous rate of decomposition at \(t=100 \mathrm{s}.\)

The decomposition of \(\mathrm{HI}(\mathrm{g})\) at \(700 \mathrm{K}\) is followed for \(400 \mathrm{s},\) yielding the following data: at \(t=0,[\mathrm{HI}]=\) \(1.00 \mathrm{M} ;\) at \(t=100 \mathrm{s},[\mathrm{HI}]=0.90 \mathrm{M} ;\) at \(t=200 \mathrm{s}, [\mathrm{HI}]=0.81 \mathrm{M} ; t=300 \mathrm{s},[\mathrm{HI}]=0.74 \mathrm{M} ;\) at \(t=400 \mathrm{s}, [\mathrm{HI}]=0.68 \mathrm{M} .\) What are the reaction order and the rate constant for the reaction: $$\mathrm{HI}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{I}_{2}(\mathrm{g}) ?$$ Write the rate law for the reaction at 700 K.

The object is to study the kinetics of the reaction between peroxodisulfate and iodide ions. $$\begin{aligned} &\text { (a) } \mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq}) \end{aligned}$$ The \(I_{3}^{-}\) formed in reaction (a) is actually a complex of iodine, \(\mathrm{I}_{2},\) and iodide ion, \(\mathrm{I}^{-}\). Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) also present in the reaction mixture, reacts with \(\mathrm{I}_{3}^{-}\) just as fast as it is formed. $$\text { (b) } 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}+3 \mathrm{I}^{-}(\mathrm{aq})$$ When all of the thiosulfate ion present initially has been consumed by reaction (b), a third reaction occurs between \(\mathrm{I}_{3}^{-}(\mathrm{aq})\) and starch, which is also present in the reaction mixture. $$\text { (c) } \mathrm{I}_{3}^{-}(\mathrm{aq})+\operatorname{starch} \longrightarrow \text { blue complex }$$ The rate of reaction (a) is inversely related to the time required for the blue color of the starch-iodine complex to appear. That is, the faster reaction (a) proceeds, the more quickly the thiosulfate ion is consumed in reaction (b), and the sooner the blue color appears in reaction (c). One of the photographs shows the initial colorless solution and an electronic timer set at \(t=0 ;\) the other photograph shows the very first appearance of the blue complex (after 49.89 s). Tables I and II list some actual student data obtained in this study. $$\begin{array}{l} \hline\text { TABLE I } \\ \text { Reaction conditions at } 24^{\circ} \mathrm{C}: 25.0 \mathrm{mL} \text { of the } \\ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}_{2} \mathrm{O}_{8}(\text { aq) listed, } 25.0 \mathrm{mL} \text { of the } \mathrm{KI}(\mathrm{aq}) \\ \text { listed, } 10.0 \mathrm{mL} \text { of } 0.010 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}), \text { and } 5.0 \mathrm{mL} \\ \text { starch solution are mixed. The time is that of the } \\ \text { first appearance of the starch-iodine complex. } \\ \hline & \text { Initial Concentrations, } \mathrm{M} \\ \hline \text { Experiment } & \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}_{2} \mathrm{O}_{8} & \mathrm{KI} & \text { Time, s } \\ \hline 1 & 0.20 & 0.20 & 21 \\ 2 & 0.10 & 0.20 & 42 \\ 3 & 0.050 & 0.20 & 81 \\ 4 & 0.20 & 0.10 & 42 \\ 5 & 0.20 & 0.050 & 79 \\ \hline \end{array}$$ $$\begin{array}{l} \hline \text { TABLE II } \\ \text { Reaction conditions: those listed in Table I for } \\ \text { Experiment } 4, \text { but at the temperatures listed. } \\ \hline \text { Experiment } & \text { Temperature, }^{\circ} \mathrm{C} & \text { Time, } \mathrm{s} \\ \hline 6 & 3 & 189 \\ 7 & 13 & 88 \\ 8 & 24 & 42 \\ 9 & 33 & 21 \\ \hline \end{array}$$ (a) Use the data in Table I to establish the order of reaction (a) with respect to \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) and to I \(^{-}\). What is the overall reaction order? [Hint: How are the times required for the blue complex to appear related to the actual rates of reaction? (b) Calculate the initial rate of reaction in Experiment 1 expressed in \(\mathrm{M} \mathrm{s}^{-1} .\) [Hint: You must take into account the dilution that occurs when the various solutions are mixed, as well as the reaction stoichiometry indicated by equations \((a),(b), \text { and }(c) .]\) (c) Calculate the value of the rate constant, \(k,\) based on experiments 1 and 2 (d) Calculate the rate constant, \(k\), for the four different temperatures in Table II. (e) Determine the activation energy, \(E_{\mathrm{a}}\), of the peroxodisulfate- iodide ion reaction. (f) The following mechanism has been proposed for reaction (a). The first step is slow, and the others are fast. $$\begin{array}{c} \mathrm{I}^{-}+\mathrm{S}_{2} \mathrm{O}_{8}^{2-} \longrightarrow \mathrm{IS}_{2} \mathrm{O}_{8}^{3-} \\ \mathrm{IS}_{2} \mathrm{O}_{8}^{3-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}^{+} \\ \mathrm{I}^{+}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{2} \\ \mathrm{I}_{2}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}^{-} \end{array}$$ Show that this mechanism is consistent with both the stoichiometry and the rate law of reaction (a). Explain why it is reasonable to expect the first step in the mechanism to be slower than the others.

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