Chapter 14

A simplified rate law for the reaction \(2 \mathrm{O}_{3}(\mathrm{g}) \longrightarrow\) \(3 \mathrm{O}_{2}(\mathrm{g})\) is $$\text { rate }=k=\frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]}$$ For this reaction, propose a two-step mechanism that consists of a fast, reversible first step, followed by a slow second step.

One proposed mechanism for the formation of a double helix in DNA is given by $$\left(S_{1}+S_{2}\right)=\left(S_{1}: S_{2}\right)^{*} \quad \text { (fast) }$$ $$\left(S_{1}: S_{2}\right)^{*} \longrightarrow S_{1}: S_{2} \quad \text { (slow) }$$ where \(S_{1}\) and \(S_{2}\) represent strand 1 and \(2,\) and \(\left(S_{1}: S_{2}\right)^{*}\) represents an unstable helix. Write the rate of reaction expression for the formation of the double helix.

Suppose that the reaction in Example 14-8 is first order with a rate constant of \(0.12 \mathrm{min}^{-1}\). Starting with \([\mathrm{A}]_{0}=1.00 \mathrm{M},\) will the curve for \([\mathrm{A}]\) versus \(t\) for the first-order reaction cross the curve for the second-order reaction at some time after \(t=0 ?\) Will the two curves cross if \([\mathrm{A}]_{0}=2.00 \mathrm{M} ?\) In each case, if the curves are found to cross, at what time will this happen?

We have seen that the unit of \(k\) depends on the overall order of a reaction. Derive a general expression for the units of \(k\) for a reaction of any overall order, based on the order of the reaction (o) and the units of concentration (M) and time (s).

Hydroxide ion is involved in the mechanism of the following reaction but is not consumed in the overall reaction. $$\mathrm{OCI}^{-}+\mathrm{I}^{-} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{OI}^{-}+\mathrm{Cl}^{-}$$ (a) From the data given, determine the order of the reaction with respect to \(\mathrm{OCl}^{-}, \mathrm{I}^{-},\) and \(\mathrm{OH}^{-}\) (b) What is the overall reaction order? (c) Write the rate equation, and determine the value of the rate constant, \(k.\) $$\begin{array}{lccl} \hline & & & \text { Rate Formation } \\ {\left[\mathrm{OC}^{-}\right], \mathrm{M}} & {\left[\mathrm{l}^{-}\right], \mathrm{M}} & {\left[\mathrm{OH}^{-}\right], \mathrm{M}} & \mathrm{O}^{-}, \mathrm{M} \mathrm{s}^{-1} \\ \hline 0.0040 & 0.0020 & 1.00 & 4.8 \times 10^{-4} \\ 0.0020 & 0.0040 & 1.00 & 5.0 \times 10^{-4} \\ 0.0020 & 0.0020 & 1.00 & 2.4 \times 10^{-4} \\ 0.0020 & 0.0020 & 0.50 & 4.6 \times 10^{-4} \\ 0.0020 & 0.0020 & 0.25 & 9.4 \times 10^{-4} \\ \hline \end{array}$$

The half-life for the first-order decomposition of nitramide, \(\mathrm{NH}_{2} \mathrm{NO}_{2}(\mathrm{aq}) \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1),\) is \(123 \min\) at \(15^{\circ} \mathrm{C} .\) If \(165 \mathrm{mL}\) of a \(0.105 \mathrm{M} \mathrm{NH}_{2} \mathrm{NO}_{2}\) solution is allowed to decompose, how long must the reaction proceed to yield \(50.0 \mathrm{mL}\) of \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) collected over water at \(15^{\circ} \mathrm{C}\) and a barometric pressure of \(756 \mathrm{mm} \mathrm{Hg} ?\) (The vapor pressure of water at \(15^{\circ} \mathrm{C}\) is \(12.8 \mathrm{mmHg} .)\)

The decomposition of ethylene oxide at \(690 \mathrm{K}\) is monitored by measuring the total gas pressure as a function of time. The data obtained are \(t=10 \mathrm{min}, P_{\text {tot }}= 139.14 \mathrm{mmHg} ; 20 \mathrm{min}, 151.67 \mathrm{mmHg} ; 40 \mathrm{min}, 172.65 \mathrm{mmHg} ; 60 \mathrm{min}, 189.15 \mathrm{mmHg} ;\) \(100 \mathrm{min}, 212.34\) \(\mathrm{mmHg} ; 200 \mathrm{min}, 238.66 \mathrm{mmHg} ; \infty, 249.88 \mathrm{mmHg}\) What is the order of the reaction \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) ?\)

The following data are for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow\) products. Establish the order of this reaction with respect to A and to B. $$\begin{array}{cccc} \hline \text { Expt 1, }[\mathrm{B}]=1.00 \mathrm{M} & & {\text { Expt 2, }[\mathrm{B}]=0.50 \mathrm{M}} \\ \hline \begin{array}{cccc} \text { Time, } \\ \text { min } \end{array} & \begin{array}{c} \text { [A], M } \\ \end{array} & \text { Time, } \text { min } &\text { [A], M } \\ \hline 0 & 1.000 \times 10^{-3} & 0 & 1.000 \times 10^{-3} \\ 1 & 0.951 \times 10^{-3} & 1 & 0.975 \times 10^{-3} \\ 5 & 0.779 \times 10^{-3} & 5 & 0.883 \times 10^{-3} \\ 10 & 0.607 \times 10^{-3} & 10 & 0.779 \times 10^{-3} \\ 20 & 0.368 \times 10^{-3} & 20 & 0.607 \times 10^{-3} \\ \hline \end{array}$$

Derive a plausible mechanism for the following reaction in aqueous solution, \(\mathrm{Hg}_{2}^{2+}+\mathrm{Tl}^{3+} \longrightarrow 2 \mathrm{Hg}^{2+}+\mathrm{Tl}^{+}\) for which the observed rate law is: rate \(=k\left[\mathrm{Hg}_{2}^{2+1}\right]\) \(\left.[\mathrm{T}]^{3+}\right] /\left[\mathrm{Hg}^{2+}\right].\)

The following three-step mechanism has been proposed for the reaction of chlorine and chloroform. $$\begin{aligned} & \text { (1) } \quad \mathrm{Cl}_{2}(\mathrm{g}) \stackrel{k_{1}}{\rightleftharpoons_{k-1}} 2 \mathrm{Cl}(\mathrm{g})\\\ & \text { (2) } \quad \mathrm{Cl}(\mathrm{g})+\mathrm{CHCl}_{3}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{HCl}(\mathrm{g})+\mathrm{CCl}_{3}(\mathrm{g})\\\ &\text { (3) } \quad \mathrm{CCl}_{3}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{CCl}_{4}(\mathrm{g}) \end{aligned}$$ The numerical values of the rate constants for these steps are \(k_{1}=4.8 \times 10^{3} ; \quad k_{-1}=3.6 \times 10^{3} ; \quad k_{2}=1.3 \times 10^{-2} ; k_{3}=2.7 \times 10^{2} .\) Derive the rate law and the magnitude of \(k\) for the overall reaction.

For the reaction \(A \longrightarrow\) products, derive the integrated rate law and an expression for the half-life if the reaction is third order.

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. $$\ln \frac{[\mathrm{A}]_{0} \times[\mathrm{B}]_{t}}{[\mathrm{B}]_{0} \times[\mathrm{A}]_{t}}=\left([\mathrm{B}]_{0}-[\mathrm{A}]_{0}\right) \times k t$$

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 decomposition of nitric oxide occurs through two parallel reactions: $$\mathrm{NO}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad k_{1}=25.7 \mathrm{s}^{-1}$$ $$\mathrm{NO}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\frac{1}{4} \mathrm{O}_{2}(\mathrm{g}) \quad k_{2}=18.2 \mathrm{s}^{-1}$$ (a) What is the reaction order for these reactions? (b) Which reaction is the slow reaction? (c) If the initial concentration of \(\mathrm{NO}(\mathrm{g})\) is \(2.0 \mathrm{M},\) what is the concentration of \(\mathrm{N}_{2}(\mathrm{g})\) after 0.1 seconds? (d) If the initial concentration of \(\mathrm{NO}(\mathrm{g})\) is \(4.0 \mathrm{M},\) what is the concentration of \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) after 0.025 seconds?

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.

In your own words, define or explain the following terms or symbols: (a) \([\mathrm{A}]_{0} ;\) (b) \(\dot{k} ;\) (c) \(t_{1 / 2} ;\) (d) zeroorder reaction; (e) catalyst.

Briefly describe each of the following ideas, phenomena, or methods: (a) the method of initial rates; (b) activated complex; (c) reaction mechanism; (d) heterogeneous catalysis; (e) rate-determining step.

Explain the important distinctions between each pair of terms: (a) first-order and second-order reactions; (b) rate law and integrated rate law; (c) activation energy and enthalpy of reaction; (d) elementary process and overall reaction; (e) enzyme and substrate.

The rate equation for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) is found to be rate \(=k[\mathrm{A}][\mathrm{B}] .\) For this reaction, we can conclude that (a) the unit of \(k=s^{-1} ;\) (b) \(t_{1 / 2}\) is constant; (c) the value of \(k\) is independent of the values of \([\mathrm{A}]\) and \([\mathrm{B}] ;\) (d) the rate of formation of \(\mathrm{C}\) is twice the rate of disappearance of A.

A first-order reaction, \(\mathrm{A} \longrightarrow\) products, has a halflife of \(75 \mathrm{s},\) from which we can draw two conclusions. Which of the following are those two (a) the reaction goes to completion in 150 s; (b) the quantity of \(A\) remaining after 150 s is half of what remains after 75 s; (c) the same quantity of A is consumed for every 75 s of the reaction; (d) one- quarter of the original quantity of A is consumed in the first 37.5 s of the reaction; (e) twice as much A is consumed in 75 s when the initial amount of \(\mathrm{A}\) is doubled; (f) the amount of \(\mathrm{A}\) consumed in 150 s is twice as much as is consumed in 75 s.

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 \([\mathrm{A}]_{0}=[\mathrm{A}]_{0}=0.25 \mathrm{M}.\)

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 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) \(\ln [A]_{t}=-k t+\ln [A]_{0}.\)

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 \(t=100 \mathrm{s}.\)

A reaction is \(50 \%\) complete in 30.0 min. How long after its start will the reaction be \(75 \%\) complete if it is (a) first order; (b) zero order?

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.

For the reaction \(A \longrightarrow\) products the following data are obtained. $$\begin{array}{cll} \hline {\text { Experiment 1 }} & &{\text { Experiment 2 }} \\ \hline [\mathrm{A}]=1.204 \mathrm{M} & t=0 \mathrm{min} & {[\mathrm{A}]=2.408 \mathrm{M}} & t=0 \mathrm{min}\\\ {[\mathrm{A}]=1.180 \mathrm{M}} & t=1.0 \mathrm{min} & {[\mathrm{A}]=?} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=0.602 \mathrm{M}} & t=35 \mathrm{min} & {[\mathrm{A}]=?} & t=30 \mathrm{min} \\ \hline \end{array}$$ (a) Determine the initial rate of reaction in Experiment 1. (b) If the reaction is second order, what will be \([\mathrm{A}]\) at \(t=1.0\) min in Experiment 2? (c) If the reaction is first order, what will be \([\mathrm{A}]\) at 30 min in Experiment 2?

For the reaction \(\mathrm{A}+2 \mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D},\) the rate law is rate of reaction \(=k[\mathrm{A}][\mathrm{B}]\) (a) Show that the following mechanism is consistent with the stoichiometry of the overall reaction and with the rate law. $$\begin{array}{l} \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{I} \quad(\text { slow }) \\ \mathrm{I}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D} \quad(\text { fast }) \end{array}$$ (b) Show that the following mechanism is consistent with the stoichiometry of the overall reaction, but not with the rate law. $$\begin{array}{c} 2 \mathrm{B} \stackrel{k_{1}}{\mathrm{k}_{1}} \mathrm{B}_{2} \text { (fast) } \\\ \mathrm{A}+\mathrm{B}_{2} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}+\mathrm{D} \text { (slow) } \end{array}$$

If the plot of the reactant concentration versus time is nonlinear, but the concentration drops by \(50 \%\) every 10 seconds, then the order of the reaction is (a) zero order; (b) first order; (c) second order; (d) third order.

If the plot of the reactant concentration versus time is linear, then the order of the reaction is (a) zero order; (b) first order; (c) second order; (d) third order.

One example of a zero-order reaction is the decomposition of ammonia on a hot platinum wire, \(2 \mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If the concentration of ammonia is doubled, the rate of the reaction will (a) be zero; (b) double; (c) remain the same; (d) exponentially increase.