Chapter 11

The reaction of \(\mathrm{NO}_{2}(\mathrm{~g})\) and \(\mathrm{CO}(\mathrm{g})\) to form \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{NO}(\mathrm{g})\) is thought to occur in two steps: Step 1: \(\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{~g}) \quad\) slow Step \(2: \mathrm{NO}_{3}(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) \quad\) fast (a) Show that the elementary steps add up to give the overall, stoichiometric equation. (b) Determine the molecularity of each step. (c) For this mechanism to be consistent with kinetic data, what must be the experimental rate equation? (d) Identify any intermediates in this reaction.

For the reaction $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NOCl}(\mathrm{g}) $$ the currently accepted mechanism is $$ \mathrm{NO}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{NOCl}_{2} $$ fast $$ \mathrm{NOCl}_{2}+\mathrm{NO} \longrightarrow 2 \mathrm{NOCl} $$ slow (a) Determine the rate law for this mechanism. (Be sure to express it in terms of concentrations of reactants or products of the overall reaction, not in terms of intermediates.) (b) Suggest another mechanism that agrees with the same rate law. (c) Suggest another mechanism that does not agree with the same rate law.

The mechanism for the reaction of \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{HBr}\) is believed to involve two steps. The overall reaction is exothermic. $$ \text { Step 1: } \mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \text {fast, endothermic } $$ $$ \text { Step } 2: \mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O} $$ slow (a) Write an equation for the overall reaction. (b) Draw a reaction energy diagram for this reaction. (c) Show that the rate law for this reaction is $$ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right] $$

Which of these statements is (are) true? If a statement is false, reword it so that it becomes true. (a) The concentration of a homogeneous catalyst may appear in the rate law. (b) A catalyst is always consumed in the reaction. (c) A catalyst must always be in the same phase as the reactants. (d) A catalyst can change the course of a reaction and allow different products to be produced.

Hydrogenation reactions-processes in which \(\mathrm{H}_{2}\) is added to a molecule-are usually catalyzed. An excellent catalyst is a very finely divided metal suspended in the reaction solvent. Tell why finely divided rhodium, for example, is a much more efficient catalyst than a small block of the metal that has the same mass.

Which of these reactions appears to involve a catalyst? In those cases where a catalyst is present, tell whether it is homogeneous or heterogeneous. (a) \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow\) $$ \begin{array}{c} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{OH}(\mathrm{aq}) \\ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]\left[\mathrm{H}^{+}\right] \end{array} $$ (b) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})\) $$ \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] $$ (c) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{O}_{2}\right]\) (area of \(\mathrm{Pt}\) surface) $$ \begin{array}{l} \text { (d) } \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{CO}(\mathrm{g}) \\ \text { Rate }=k\left[\mathrm{H}_{2}\right]^{1 / 2}[\mathrm{CO}] \end{array} $$

In acid solution, methyl formate forms methyl alcohol and formic acid. \(\mathrm{HCO}_{2} \mathrm{CH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{HCOOH}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{OH}(\mathrm{aq})\) methyl formate \(\begin{array}{ll}\text { formic acid } & \text { methyl alcohol }\end{array}\) The rate law is Rate \(=k\left[\mathrm{HCO}_{2} \mathrm{CH}_{3}\right]\left[\mathrm{H}^{+}\right] .\) Why does \(\mathrm{H}^{+}\) appear in the rate law but not in the overall equation for the reaction?

A biological catalyst lowers the activation energy of a reaction from \(215 \mathrm{~kJ} / \mathrm{mol}\) to \(206 \mathrm{~kJ} / \mathrm{mol}\). Calculate by what factor the rate constant, \(k,\) would increase at \(25^{\circ} \mathrm{C}\). Assume that the frequency factors \((A)\) are the same for the uncatalyzed and catalyzed reactions.

When enzymes are present at very low concentration, their effect on reaction rate can be described by firstorder kinetics. Calculate by what factor the rate of an enzyme-catalyzed reaction changes when the enzyme concentration is changed from \(1.5 \times 10^{-7} \mathrm{M}\) to \(4.5 \times 10^{-6} \mathrm{M}\)

When substrates are present at relatively high concentration and are catalyzed by enzymes, the effect on reaction rate of changing substrate concentration can be described by zeroth-order kinetics. Calculate by what factor the rate of an enzyme-catalyzed reaction changes when the substrate concentration is changed from \(1.5 \times 10^{-2} \mathrm{M}\) to \(4.5 \times 10^{-2} \mathrm{M}\)

Many biochemical reactions are catalyzed by acids. A typical mechanism consistent with the experimental results (in which HA is the acid and \(X\) is the reactant) is Step 1: reversible HA(aq) \(\rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq}) \quad\) fast Step 2: reversible \(\mathrm{X}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{XH}^{+}(\mathrm{aq}) \quad\) fast Step 3: \(\mathrm{XH}^{+}(\mathrm{aq}) \longrightarrow\) products slow (a) Derive the rate law from this mechanism. (b) Determine the order of reaction with respect to HA. (c) Determine how doubling the concentration of HA would affect the rate of the reaction.

Why are homogeneous catalysts harder to separate from products and leftover reactants than are heterogeneous reactants?

In an automobile catalytic converter the catalysis is accomplished on a surface consisting of platinum and other precious metals. The metals are deposited as a thin layer on a ceramic support that is a fine grid (see the photo). (a) Why is the ceramic support arranged in the grid-like geometry? (b) Why are the metals deposited on the ceramic surface instead of being used as strips or rods?

Draw a reaction energy diagram for an exothermic process. Mark the positions of reactants, products, and activated complex. Indicate the activation energies of the forward and reverse processes and explain how \(\Delta_{\mathrm{r}} E\) for the reaction can be calculated from the diagram.

Draw a reaction energy diagram for an endothermic process. Mark the positions of reactants, products, and activated complex. Indicate the activation energies of the forward and reverse processes and explain how \(\Delta_{r} E\) for the reaction can be calculated from the diagram.

Under certain conditions, biphenyl, \(\mathrm{C}_{12} \mathrm{H}_{10},\) can be produced by the decomposition of cyclohexane, \(\mathrm{C}_{6} \mathrm{H}_{12}\) \(2 \mathrm{C}_{6} \mathrm{H}_{12} \longrightarrow \mathrm{C}_{12} \mathrm{H}_{10}+7 \mathrm{H}_{2}\) This table represents part of the data obtained in the kinetics experiment. \begin{tabular}{lccc} \hline Time (s) & {\(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right](\mathrm{M})\)} & {\(\left[\mathrm{C}_{12} \mathrm{H}_{10}\right](\mathrm{M})\)} & {\(\left[\mathrm{H}_{2}\right](\mathrm{M})\)} \\ \hline 0.0 & 0.200 & 0.000 & 0.000 \\ 1.00 & 0.159 & 0.021 & \(-\) \\ 2.00 & 0.132 & \(-\) & \(-\) \\ 3.00 & \(-\) & 0.044 & \(-\) \\ \hline \end{tabular} (a) Fill in the missing concentrations. (b) Calculate the rate of reaction at \(1.5 \mathrm{~s}\).

The statements below relate to this reaction: $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g}) \quad \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] $$ Determine which statements are true. If a statement is false, indicate why it is incorrect. (a) The reaction must occur in a single step. (b) This is a second-order reaction overall. (c) Raising the temperature will cause the value of \(k\) to decrease. (d) Raising the temperature lowers the activation energy for this reaction. (e) If the concentrations of both reactants are doubled, the rate will double. (f) Adding a catalyst in the reaction will cause the initial rate to increase.

Indicate whether each of these statements is true or false. Change the wording of each false statement to make it true. (a) It is possible to change the rate constant for a reaction by changing the temperature. (b) The reaction rate remains constant as a first-order reaction proceeds at a constant temperature. (c) The rate constant for a reaction is independent of reactant concentrations. (d) As a second-order reaction proceeds at a constant temperature, the rate constant changes.

For the reaction of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) at \(660 \mathrm{~K}\), \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})\) \begin{tabular}{lcc} \hline \multicolumn{2}{c} { Concentration (mol/L) } & \\ \hline [NO] & {\(\left[\mathrm{O}_{2}\right]\)} & Rate of Disappearance of \(\mathrm{NO}\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 0.010 & 0.010 & \(2.5 \times 10^{-5}\) \\ 0.020 & 0.010 & \(1.0 \times 10^{-4}\) \\ 0.010 & 0.020 & \(5.0 \times 10^{-5}\) \\ \hline \end{tabular} (a) Determine the order with respect to each reactant. (b) Write the rate equation for the reaction. (c) Calculate the rate constant. (d) Calculate the rate when [NO] \(=0.025 \mathrm{~mol} / \mathrm{L}\) and \(\left[\mathrm{O}_{2}\right]=0.050 \mathrm{~mol} / \mathrm{L}\) (e) If \(\mathrm{O}_{2}\) disappears at a rate of \(1.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}, \mathrm{cal}-\) culate the rate at which NO is consumed. Calculate the rate at which \(\mathrm{NO}_{2}\) is formed.

Nitryl fluoride is an explosive compound that can be made by oxidizing nitrogen dioxide with fluorine: \(2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2} \mathrm{~F}(\mathrm{~g})\) Several kinetics experiments, all done at the same temperature and involving formation of nitryl fluoride, are summarized in this table: \begin{tabular}{cccccc} \hline & \multicolumn{2}{c} { Initial Concentration (mol/L) } & & \\ \cline { 2 - 4 } \cline { 6 } Experiment & {\(\left[\mathrm{NO}_{2}\right]\)} & {\(\left[\mathrm{F}_{2}\right]\)} & {\(\left[\mathrm{NO}_{2} F\right]\)} & & Initial Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 1 & 0.0010 & 0.0050 & 0.0020 & & \(2.0 \times 10^{-4}\) \\ 2 & 0.0020 & 0.0050 & 0.0020 & & \(4.0 \times 10^{-4}\) \\ 3 & 0.0020 & 0.0020 & 0.0020 & & \(1.6 \times 10^{-4}\) \\ 4 & 0.0020 & 0.0020 & 0.0010 & & \(1.6 \times 10^{-4}\) \\ \hline \end{tabular} (a) Write the rate law for the reaction. (b) Determine what the order of the reaction is with respect to each reactant and each product. (c) Calculate the rate constant \(k\) and express it in appropriate units.

The deep blue compound \(\mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}\) can be made from the chromate ion by using hydrogen peroxide in an acidic solution. \(\mathrm{HCrO}_{4}^{-}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\ell) $$ The kinetics of this reaction have been studied, and the rate equation is Rate of disappearance of \(\mathrm{HCrO}_{4}^{-}=k\left[\mathrm{HCrO}_{4}^{-}\right]\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\left[\mathrm{H}^{+}\right]\) One of the mechanisms suggested for the reaction is $$ \begin{array}{c} \mathrm{HCrO}_{4}^{-}+\mathrm{H}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{CrO}_{4} \\ \mathrm{H}_{2} \mathrm{CrO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}+\mathrm{H}_{2} \mathrm{O} \end{array} $$ \(\mathrm{H}_{2} \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{CrO}\left(\mathrm{O}_{2}\right)_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (a) Give the order of the reaction with respect to each reactant. (b) Show that the steps of the mechanism agree with the overall equation for the reaction. (c) Which step in the mechanism is rate-limiting? Explain your answer.

Two mechanisms are proposed for the reaction $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g}) $$ Mechanism $$ \text { 1: } \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3} $$ fast $$ \mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2} $$ slow $$ \text { Mechanism } 2: \mathrm{NO}+\mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} $$ fast Show that each mechanism is consistent with the observed rate law: Rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\).

For a reaction involving the decomposition of a hypothetical substance \(\mathrm{Y}\), these data are obtained: Rate \([\mathrm{Y}]\) \(\begin{array}{lllll}\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1}\right) & 0.288 & 0.245 & 0.202 & 0.158 \\ & 0.200 & 0.170 & 0.140 & 0.110\end{array}\) (a) Determine the order of the reaction. (b) Write the rate law for the decomposition of \(Y\). (c) Calculate \(k\) for the experiment above.

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is first-order with a rate constant of \(2.5 \times 10^{-4} \mathrm{~s}^{-1}\) (a) Calculate the half-life for decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (b) Calculate how long it takes for the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(\frac{1}{32}\) of its original value. (c) Calculate how long it takes for the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop from \(3.4 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\) to \(2.3 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\).

The rate constant for decomposition of azomethane at $$ \begin{aligned} 425^{\circ} \mathrm{C} \text { is } 0.68 \mathrm{~s}^{-1} & \\ \mathrm{CH}_{3} \mathrm{~N}=\mathrm{NCH}_{3}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) \end{aligned} $$ (a) Based on the units of the rate constant, determine if this reaction is zeroth-, first-, or second-order. (b) If \(2.0 \mathrm{~g}\) azomethane is placed in a \(2.0-\mathrm{L}\) flask and heated to \(425^{\circ} \mathrm{C},\) calculate the mass of azomethane that remains after \(5.0 \mathrm{~s}\). (c) Calculate how long it takes for the mass of azomethane to drop from \(2.0 \mathrm{~g}\) to \(0.24 \mathrm{~g}\). (d) Calculate the mass of nitrogen that would be found in the flask after \(0.50 \mathrm{~s}\) of reaction.

Cyclopropane isomerizes to propene when heated. Rate constants for the reaction cyclopropane \(\longrightarrow\) propene are \(1.10 \times 10^{-4} \mathrm{~s}^{-1}\) at \(470.0^{\circ} \mathrm{C}\) and \(1.02 \times 10^{-3} \mathrm{~s}^{-1}\) at \(510.0^{\circ} \mathrm{C}\) (a) Calculate the activation energy, \(E_{\mathrm{a}},\) for this reaction. (b) Calculate how long it takes at \(500 .{ }^{\circ} \mathrm{C}\) for the concentration of cyclopropane to drop from \(0.10 \mathrm{M}\) to \(0.023 \mathrm{M}\).

The rate of decay of a radioactive solid is independent of the temperature of that solid-at least for temperatures easily obtained in the laboratory. What does this observation imply about the activation energy for this process?

Calculate the half-life of a first-order reaction if \(30.5 \mathrm{~s}\) after the reaction starts the concentration of the reactant is \(0.0451 \mathrm{M}\) and \(45.0 \mathrm{~s}\) after the reaction starts it is 0.0321 M. Calculate how many seconds after the start of the reaction it takes for the reactant concentration to decrease to \(0.0100 \mathrm{M}\).

Which of these mechanisms (more than one may be chosen) is compatible with the rate law? Rate \(=k\left[\mathrm{Cl}_{2}\right]^{3 / 2}[\mathrm{CO}]\) $$ \text { (a) } \frac{1}{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl} $$ \(\mathrm{Cl}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl}_{3}\) fast \(\mathrm{Cl}_{3}+\mathrm{CO} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl} \quad\) slow $$ \mathrm{Cl} \rightleftharpoons \frac{1}{2} \mathrm{Cl}_{2} $$ (b) \(\mathrm{Cl}_{2}+\mathrm{CO} \longrightarrow \mathrm{CCl}_{2}+\mathrm{O}\) slow \(\mathrm{O}+\mathrm{Cl}_{2} \longrightarrow \mathrm{Cl}_{2} \mathrm{O}\) fast $$ \mathrm{Cl}_{2} \mathrm{O}+\mathrm{CCl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl}_{2} $$ fast s (c) \(\frac{1}{2} \mathrm{Cl}_{2} \rightleftharpoons \mathrm{Cl}\) $$ \mathrm{Cl}+\mathrm{CO} \rightleftharpoons \mathrm{COCl} $$ fast $$ \begin{array}{l} \mathrm{COCl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl} \\\ \mathrm{Cl} \rightleftharpoons \frac{1}{2} \mathrm{Cl}_{2} \end{array} $$ slow (d) \(\mathrm{Cl}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{COCl}+\mathrm{Cl}\) fast \(\mathrm{COCl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}+\mathrm{Cl}\) slow \(\mathrm{Cl}+\mathrm{Cl} \longrightarrow \mathrm{Cl}_{2}\) fast

Let \(x\) represent the number of half-lives that have elapsed during the course of a first-order reaction. That is, the elapsed time \(t\) is \(x\) times the half-life \(t_{1 / 2}: t=x t_{1 / 2}\). (a) Show that at time \(t\) the concentration \([\mathrm{A}]_{t}\) of reactant is related to the initial concentration \([\mathrm{A}]_{0}\) by $$ [\mathrm{A}]_{t}=[\mathrm{A}]_{0}\left(\frac{1}{2}\right)^{x} $$ (b) Use your result in part (a) to show that $$ \log \left(\frac{[\mathrm{A}]_{t}}{[\mathrm{~A}]_{0}}\right)=\frac{t}{t_{1 / 2}}(-\log 2)=\frac{t}{t_{1 / 2}}(-0.301) $$ (c) Use your result in part (b) to show that for any firstorder reaction a plot of \(\log [\mathrm{A}]_{t},\) versus \(t\) will be a straight line and that half-life can be obtained from the slope of the line.

Measurements of the initial rate of hydrolysis of benzenesulfonyl chloride in aqueous solution at \(15^{\circ} \mathrm{C}\) in the presence of fluoride ion yielded the results in the table for a fixed concentration of benzenesulfonyl chloride of \(2 \times\) \(10^{-4} \mathrm{M}\). The reaction rate is known to be proportional to the concentration of benzenesulfonyl chloride. \begin{tabular}{cc} \hline & Initial Rate \(\times 10^{7}\) \\ {\(\left[\mathrm{~F}^{-}\right] \times 10^{2}(\mathrm{~mol} / \mathrm{L})\)} & \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 0 & 2.4 \\ 0.5 & 5.4 \\ 1.0 & 7.9 \\ 2.0 & 13.9 \\ 3.0 & 20.2 \\ 4.0 & 25.2 \\ 5.0 & 32.0 \\ \hline \end{tabular} Note that some reaction must be occurring in the absence of any fluoride ion, because at zero concentration of fluoride the rate is not zero. This residual rate should be subtracted from each observed rate to give the rate of the reaction being studied. (a) Derive the complete rate law for the reaction. (b) Calculate the rate constant \(k\) and express it in appropriate units.

When formic acid is heated, it decomposes to hydrogen and carbon dioxide in a first-order decay. \(\mathrm{HCOOH}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) The rate of reaction is monitored by measuring the total pressure in the reaction container. \begin{tabular}{rc} \hline Time (s) & \(P\) (torr) \\ \hline 0 & 220 \\ 50 & 324 \\ 100 & 379 \\ 150 & 408 \\ 200 & 423 \\ 250 & 431 \\ 300 & 435 \\ \hline \end{tabular} Calculate the rate constant and half-life, in seconds, for the reaction. At the start of the reaction (time \(=0\) ), only formic acid is present. (Hint: Find the partial pressure of formic acid [use Dalton's law of partial pressure ( \(\in\) Sec. \(8-6\) ) and the reaction stoichiometry to find \(P_{\mathrm{HCOOH}}\) at each time]).

When heated, cyclobutane, \(\mathrm{C}_{4} \mathrm{H}_{8},\) decomposes to ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}\) \(\mathrm{C}_{4} \mathrm{H}_{8}(\mathrm{~g}) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) The activation energy, \(E_{a},\) for this reaction is \(262 \mathrm{~kJ} / \mathrm{mol}\). (a) If the rate constant \(k=0.032 \mathrm{~s}^{-1}\) at \(800 . \mathrm{K},\) calculate the value of \(k\) at \(900 . \mathrm{K}\). (b) Calculate the cyclobutane concentration after \(2 \mathrm{~h}\) at \(850 . \mathrm{K}\) if the initial cyclobutane concentration was \(0.0427 \mathrm{M}\)

For each reaction listed with its rate law, propose a reasonable mechanism. (a) \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow\) $$ \begin{array}{l} \text { CH }_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{OH} \text { (aq) } \\ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]\left[\mathrm{H}^{+}\right] \\ \text {(b) } \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g}) \\ \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] \end{array} $$ For each reaction listed with its rate law, propose a reasonable mechanism. (a) \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow\) $$ \begin{array}{l} \text { CH }_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{OH}(\mathrm{aq}) \\ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]\left[\mathrm{H}^{+}\right] \\ \text {(b) } \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g}) \\ \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] \end{array} $$

In a time-resolved picosecond spectroscopy experiment, Sheps, Crowther, Carrier, and Crim (Journal of Physical Chemistry \(A,\) Vol. 110,\(2006 ;\) pp. \(3087-3092\) ) generated chlorine atoms in the presence of pentane. The pentane was dissolved in dichloromethane, \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\). The chlorine atoms are free radicals and are very reactive. After a nanosecond the chlorine atoms have reacted with pentane molecules, removing a hydrogen atom to form \(\mathrm{HCl}\) and leaving behind a pentane radical with a single unpaired electron. The equation is \(\mathrm{Cl} \cdot(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{12}(\mathrm{dcm}) \longrightarrow \mathrm{HCl}(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{11} \cdot(\mathrm{dcm})\) where \((\mathrm{dcm})\) indicates that a substance is dissolved in dichloromethane. Measurements of the concentration of chlorine atoms were made as a function of time at three different concentrations of pentane in the dichloromethane. These results are shown in the table. of Chlorine Atoms (M) \begin{tabular}{clcc} & \multicolumn{3}{c} { Concentration of Chlorine Atoms (m) } \\ Time (ps) & \multicolumn{4}{c} { for Different \(\mathrm{C}_{5} \mathrm{H}_{12}\) Concentrations } \\ \cline { 3 - 5 } & \(0.15-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.30-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.60-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) \\ \hline 100.0 & \(4.42 \times 10^{-5}\) & \(3.11 \times 10^{-5}\) & \(1.77 \times 10^{-5}\) \\ 140.0 & \(3.823 \times 10^{-5}\) & \(2.39 \times 10^{-5}\) & \(1.15 \times 10^{-5}\) \\\ 180.0 & \(3.38 \times 10^{-5}\) & \(1.94 \times 10^{-5}\) & \(8.48 \times 10^{-6}\) \\\ 220.0 & \(3.03 \times 10^{-5}\) & \(1.49 \times 10^{-5}\) & \(6.04 \times 10^{-6}\) \\\ 260.0 & \(2.68 \times 10^{-5}\) & \(1.19 \times 10^{-5}\) & \(4.12 \times 10^{-6}\) \\\ 300.0 & \(2.42 \times 10^{-5}\) & \(9.45 \times 10^{-6}\) & \(3.14 \times 10^{-6}\) \\\ 340.0 & \(2.08 \times 10^{-5}\) & \(7.75 \times 10^{-6}\) & \(2.38 \times 10^{-6}\) \\\ 380.0 & \(1.91 \times 10^{-5}\) & \(6.35 \times 10^{-6}\) & \(1.75 \times 10^{-6}\) \\\ 420.0 & \(1.71 \times 10^{-5}\) & \(4.58 \times 10^{-6}\) & \(1.61 \times 10^{-6}\) \\\ 460.0 & \(1.53 \times 10^{-5}\) & \(3.77 \times 10^{-6}\) & \(9.98 \times 10^{-7}\) \\\ \hline \end{tabular} (a) Determine the order of the reaction with respect to chlorine. (b) Determine whether the reaction rate depends on the concentration of pentane in dichloromethane. If so, determine the order of the reaction with respect to pentane. (c) Explain why the concentration of pentane in dichloromethane does not affect the data analysis that you performed in part (a). (d) Write the rate law for the reaction and calculate the rate of reaction for a concentration of chlorine atoms equal to \(1.0 \mu \mathrm{M}\) and a pentane concentration of \(0.23 \mathrm{M}\). (e) Sheps, Crowther, Carrier, and Crim found that the rate of formation of \(\mathrm{HCl}\) matched the rate of disappearance of Cl. From this they concluded that there were no intermediates and side reactions were not important. Explain the ic