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The reaction \(A \longrightarrow\) products is second order. The initial rate of decomposition of \(A\) when \([\mathrm{A}]_{0}=0.50 \mathrm{M}\) is \((\mathrm{a})\) the same as the initial rate for any other value of \([\mathrm{A}]_{0} ;\) (b) half as great as when \([\mathrm{A}]_{0}=1.00 \mathrm{M} ;(\mathrm{c})\) five times as great as when \([\mathrm{A}]_{0}=[\mathrm{A}]_{0}=0.25 \mathrm{M}.\)

Short Answer

Expert verified
The initial rate of reaction for a second order reaction is not the same for all initial concentrations of the reactant. When the initial concentration is halved from 1.00 M to 0.50 M, the initial rate is halved. When the initial concentration is decreased to a quarter (0.25 M), the initial rate becomes five times greater.

Step by step solution

01

Identify the Rate Law

For a second-order reaction, the rate equation is given by \(rate = k[A]^2\). This means that the rate of reaction is directly proportional to the square of the concentration of the reactant A.
02

Calculate the Rate for Different Concentrations

(a) The rate of reaction will not be the same for any other value of [A], as it is dependent on the concentration of A. (b) When [A] = 1.00 M, the rate will be (1.00 M)^2 = 1, which is twice as large as when [A] = 0.50 M (rate = (0.50 M)^2 = 0.25), thus the rate is half when [A] = 0.50 M. (c) When [A] = 0.25 M, the rate is (0.25 M)^2 = 0.0625, which is five times less than when [A] = 0.50 M.
03

Summarize the Results

In conclusion, the initial rate of reaction is dependent on the square of the initial concentration of the reactant for a second-order reaction. The rate when [A] = 0.50 M is twice as large as when [A] = 1.00 M and five times as large as when [A] = 0.25 M.

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Key Concepts

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

Rate Law
Understanding the rate law is essential in the field of chemical kinetics. It's a mathematical equation that relates the reaction rate to the concentration of the reactants. In the context of a second-order reaction, the rate law expression takes the form \( rate = k[A]^2 \), where \(k\) is the rate constant, and \[A\] represents the concentration of reactant A. This specific form implies that the rate of the reaction is directly proportional to the square of the concentration of the reactant. This quadratic relationship indicates a stark increase in reaction rate with increasing concentration, making precise concentration measurements crucial for predicting reaction behavior.
Reaction Rates
The concept of reaction rates dives into how quickly reactants are transformed into products in a chemical reaction. It's a dynamic aspect of chemistry that can be influenced by multiple factors including temperature, catalysts, and reactant concentration. For second-order reactions, as the concentration of reactant A doubles, the reaction rate quadruples, reflecting the squared concentration dependency. This means that minor changes in the concentration can lead to significant differences in how fast the reaction proceeds, which is a fundamental concept when predicting how a chemical process will unfold over time.
Chemical Kinetics
Chemical kinetics is the study of the speeds or rates at which chemical reactions occur. It not only considers the reaction rates but also the steps that comprise the reaction mechanism and the factors that affect these rates. It's a significant branch of physical chemistry owing to its role in industrial applications, environmental science, and even in biological processes. By comprehending kinetics, chemists can design experiments to measure and control the rates of chemical reactions, ensuring the efficiency and safety of chemical processes.
Concentration Dependence
Concentration dependence is a pivotal concept in chemical kinetics that describes how the change in concentration of the reactants affects the reaction rate. For a second-order reaction, this dependence is more pronounced as the rate is proportional to the square of the concentration. This squared term implies that if you were to cut the concentration of reactant A in half, the rate of the reaction would decrease to a quarter of its original rate. This steep dependence on concentration is key to predicting how a reaction will propagate and is particularly important in scenarios where precise control of reaction speed is desired.

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Most popular questions from this chapter

For the reaction \(A \longrightarrow\) products, the following data were obtained: \(t=0 \mathrm{s},[\mathrm{A}]=0.715 \mathrm{M} ; 22 \mathrm{s}, 0.605 \mathrm{M}\) 74 s, 0.345 M; 132 s, 0.055 M. (a) What is the order of this reaction? (b) What is the half-life of the reaction?

The reaction \(A+B \longrightarrow\) products is first order in \(A\) first order in \(\mathrm{B},\) and second order overall. Consider that the starting concentrations of the reactants are \([\mathrm{A}]_{0}\) and [ \(\mathrm{B}]_{0},\) and that \(x\) represents the decrease in these concentrations at the time \(t .\) That is, \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-x\) and \([\mathrm{B}]_{t}=[\mathrm{B}]_{0}-x .\) Show that the integrated rate law for this reaction can be expressed as shown below. $$\ln \frac{[\mathrm{A}]_{0} \times[\mathrm{B}]_{t}}{[\mathrm{B}]_{0} \times[\mathrm{A}]_{t}}=\left([\mathrm{B}]_{0}-[\mathrm{A}]_{0}\right) \times k t$$

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.

The rate of a chemical reaction generally increases rapidly, even for small increases in temperature, because of a rapid increase in (a) collision frequency; (b) fraction of reactant molecules with very high kinetic energies; (c) activation energy; (d) average kinetic energy of the reactant molecules.

For the first-order reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ \(t_{1 / 2}=22.5 \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) and \(1.5 \mathrm{h}\) at \(40^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) If the Arrhenius constant \(A=2.05 \times 10^{13} \mathrm{s}^{-1}\) determine the value of \(k\) at \(30^{\circ} \mathrm{C}\).

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