For the reaction \(\mathrm{A}+\mathrm{B} \longrightarrow 2 \mathrm{C},\) which proceeds by a single-step bimolecular elementary process, (a) \(t_{1 / 2}=0.693 / k ;\) (b) rate of appearance of C= - rate of disappearance of \(\mathrm{A} ;\) (c) rate of reaction = \(k[\mathrm{A}][\mathrm{B}] ;\) (d) \(\ln [A]_{t}=-k t+\ln [A]_{0}.\)

Short Answer

Expert verified
Statements (b) and (c) are correct, while statements (a) and (d) are incorrect.

Step by step solution

01

Verifying Statement (a)

The half-life equation \(t_{1 / 2}=0.693 / k\) is applicable for first-order reactions. But in this case, the reaction is a bimolecular (second-order) process. So, statement (a) is incorrect.
02

Verifying Statement (b)

In any reaction where new products are formed, the rate of appearance of the product is indeed equivalent to the rate of disappearance of the reactants. So, statement (b) holds true in this context and is correct.
03

Verifying Statement (c)

The rate of reaction for a bimolecular process indeed follows the form \(rate = k[\mathrm{A}][\mathrm{B}]\), where [A] and [B] are the molar concentrations of reactants A and B, respectively, and k is the rate constant. So, statement (c) is correct.
04

Verifying Statement (d)

The equation \(\ln [A]_{t}=-k t+\ln [A]_{0}\) is valid for first-order reactions and it relates the concentration of reactant A at any time t to the initial concentration of A. In this case, as the reaction is a bimolecular process (second-order), this equation is not applicable. Hence, statement (d) is incorrect.

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Key Concepts

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

Reaction Kinetics
Reaction kinetics is the branch of physical chemistry that studies the rates of chemical processes. It examines how various factors such as concentration, temperature, and catalysts affect the speed of a chemical reaction. Understanding kinetics helps predict how a reaction proceeds over time and the time it takes for reactants to transform into products.

In the context of a bimolecular elementary process, reaction kinetics focuses on how two molecules come together to react. Bimolecular reactions are one subset of elementary reactions, indicating that the reaction occurs in one step and involves the simultaneous collision of two reactant molecules.
Rate of Reaction
The rate of a reaction measures the speed at which reactants are converted into products over time. It is typically expressed in terms of concentration change per unit time, such as moles per liter per second. Factors like temperature, the presence of catalysts, and the concentrations of reactants can influence this rate.

For a bimolecular reaction such as \(\mathrm{A} + \mathrm{B} \longrightarrow 2\mathrm{C}\), the rate is proportional to the product of the concentrations of the reactants, \(k[\mathrm{A}][\mathrm{B}]\), where \(k\) is the rate constant. The rate of appearance of product \(\mathrm{C}\) is directly related to the rate of disappearance of reactants \(\mathrm{A}\) and \(\mathrm{B}\).
Half-life of Reaction
The half-life of a reaction, often represented by \(t_{1/2}\), is the time required for half the quantity of a reactant to be consumed or for the concentration of the reactant to fall to half its initial value. For first-order reactions, \(t_{1/2}\) is a constant that is independent of the initial concentration and can be calculated using the formula \(t_{1/2} = 0.693 / k\).

However, for second-order reactions, the half-life depends on the initial concentration of the reactants, and the formula for the first-order reactions does not apply. In a half-life context, this highlights the importance of knowing the reaction order when conducting kinetic analysis.
First-Order Reactions
First-order reactions are characterized by a rate that is directly proportional to the concentration of one reactant. The integrated rate law for a first-order reaction is given by \(\ln[A]_t = -kt + \ln[A]_0\), where \(\ln[A]_t\) is the natural logarithm of the concentration of reactant \(A\) at time \(t\), \(\ln[A]_0\) is the natural logarithm of the initial concentration of \(A\), and \(k\) is the first-order rate constant. In these reactions, the half-life is constant and does not depend on the initial concentration of reactants.
Second-Order Reactions
Second-order reactions are defined by a rate proportional to the square of the concentration of one reactant or to the product of the concentrations of two reactants. The rate expression for a second-order reaction with two reactants \(A\) and \(B\) would be \(rate = k[A][B]\).

Unlike first-order reactions, the half-life of a second-order reaction depends on the initial concentrations of the reactants, and therefore, it changes over time. The integrated rate law and half-life expressions for second-order reactions differ significantly from those of first-order reactions, reflecting the dependency on initial reactant concentrations.

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Most popular questions from this chapter

You want to test the following proposed mechanism for the oxidation of HBr. $$\begin{array}{c} \mathrm{HBr}+\mathrm{O}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{HOOBr} \\\ \mathrm{HOOBr}+\mathrm{HBr} \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{HOBr} \\\ \mathrm{HOBr}+\mathrm{HBr} \stackrel{k_{3}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}+\mathrm{Br}_{2} \end{array}$$ You find that the rate is first order with respect to HBr and to \(\mathrm{O}_{2}\). You cannot detect HOBr among the products. (a) If the proposed mechanism is correct, which must be the rate-determining step? (b) Can you prove the mechanism from these observations? (c) Can you disprove the mechanism from these observations?

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.

We have used the terms order of a reaction and molecularity of an elementary process (that is, unimolecular, bimolecular). What is the relationship, if any, between these two terms?

A kinetic study of the reaction \(A \longrightarrow\) products yields the data: \(t=0 \mathrm{s},[\mathrm{A}]=2.00 \mathrm{M} ; 500 \mathrm{s}, 1.00 \mathrm{M}; 1500 \mathrm{s}, 0.50 \mathrm{M} ; 3500 \mathrm{s}, 0.25 \mathrm{M} .\) Without performing detailed calculations, determine the order of this reaction and indicate your method of reasoning.

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.

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