Chapter 14
Chemistry: The Central Science · 79 exercises
Problem 88
Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a \(1 \times 10^{5}\)-fold increase in the reaction rate?
6 step solution
Problem 89
Consider the reaction \(A+B \longrightarrow C+D\). Is each of the following statements true or false? (a) The rate law for the reaction must be Rate \(=k[\mathrm{~A}][\mathrm{B}]\). (b) If the reaction is an elementary reaction, the rate law is second order. (c) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be greater than that for the forward reaction.
4 step solution
Problem 90
Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at \(28{ }^{\circ} \mathrm{C}\) is \(3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(2.0 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2} \mathrm{~s} \mathrm{} 0.025 \mathrm{M}\), what is the rate of formation of \(\mathrm{Cl}^{-}\)?
5 step solution
Problem 91
The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\). When \([\mathrm{NO}]=0.040 \mathrm{M}\), and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M}\), the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8\) ?
5 step solution
Problem 92
You perform a series of experiments for the reaction \(A \longrightarrow B+C\) and find that the rate law has the form rate \(=k[\mathrm{~A}]^{x}\). Determine the value of \(x\) in each of the following cases: (a) There is no rate change when \([\mathrm{A}]_{0}\) is tripled. (b) The rate increases by a factor of 9 when \([\mathrm{A}]_{0}\) is tripled. (c) When \([\mathrm{A}]_{0}\) is doubled, the rate increases by a factor of 8 .
3 step solution
Problem 94
The following kinetic data are collected for the initial rates of a reaction \(2 \mathrm{X}+\mathrm{Z} \longrightarrow\) products: $$ \begin{array}{llll} \hline \text { Experiment } & {[\mathrm{X}]_{0}(M)} & {[\mathrm{Z}]_{0}(M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1 & 0.25 & 0.25 & 4.0 \times 10^{1} \\ 2 & 0.50 & 0.50 & 3.2 \times 10^{2} \\ 3 & 0.50 & 0.75 & 7.2 \times 10^{2} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(X\) is \(0.75 M\) and that of \(Z\) is \(1.25 M ?\)
4 step solution
Problem 95
The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Based on the units for \(k\), is the reaction first or second order in \(\mathrm{NO}_{2}\) ? (b) If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M}\), how would you determine how long it would take for the concentration to decrease to \(0.025 \mathrm{M}\) ?
4 step solution
Problem 96
Consider two reactions. Reaction (1) has a constant half-life, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?
4 step solution
Problem 97
A first-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) has the rate constant \(k=3.2 \times 10^{-3} \mathrm{~s}^{-1}\). If the initial concentration of \(\mathrm{A}\) is \(2.5 \times 10^{-2} M\), what is the rate of the reaction at \(t=660 \mathrm{~s}\) ?
4 step solution
Problem 98
(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\) is first order. Near room temperature, the rate constant equals \(7.0 \times 10^{-4} \mathrm{~s}^{-1}\). Calculate the half-life at this temperature. (b) At \(415^{\circ} \mathrm{C},\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}\) decomposes in the gas phase, \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g)\). If the reaction is first order with a half-life of \(56.3 \mathrm{~min}\) at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).
7 step solution
Problem 99
Americium-241 is used in smoke detectors. It has a first-order rate constant for radioactive decay of \(k=1.6 \times 10^{-3} \mathrm{yr}^{-1}\). By contrast, iodine-125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of \(k=0.011\) day \(^{-1}\). (a) What are the half-lives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a \(1.00\)-mg sample of each isotope remains after 3 half-lives? (d) How much of a 1.00-mg sample of each isotope remains after 4 days?
5 step solution
Problem 100
Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$ \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M}\), the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M}\) ? (c) What is the halflife for this reaction at \(61.05^{\circ} \mathrm{C}\) ?
3 step solution
Problem 101
The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{~nm}\). The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{~nm}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is \(0.605\) at the beginning of the reaction. (b) The absorbance falls to \(0.250\) at \(30.0 \mathrm{~min}\). Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100\) ?
8 step solution
Problem 102
A colored dye compound decomposes to give a colorless product. The original dye absorbs at \(608 \mathrm{~nm}\) and has an extinction coefficient of \(4.7 \times 10^{4} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at that wavelength. You perform the decomposition reaction in a 1 -cm cuvette in a spectrometer and obtain the following data: From these data, determine the rate law for the reaction "dye \(\longrightarrow\) product" and determine the rate constant.
4 step solution
Problem 103
Cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) reacts with itself to form dicyclopentadiene $\left(\mathrm{C}_{10} \mathrm{H}_{12}\right)\(. A \)0.0400 \mathrm{M}\( solution of \)\mathrm{C}_{5} \mathrm{H}_{6}$ was monitored as a function of time as the reaction $2 \mathrm{C}_{5} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12}$ proceeded. The following data were collected: $$ \begin{array}{cc} \hline \text { Time (s) } & {\left[\mathrm{C}_{5} \mathrm{H}_{6}\right](M)} \\\ \hline 0.0 & 0.0400 \\ 50.0 & 0.0300 \\ 100.0 & 0.0240 \\ 150.0 & 0.0200 \\ 200.0 & 0.0174 \\ \hline \end{array} $$ Plot \(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, $\ln \left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\( versus time, and \)1 /\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]$ versus time. (a) What is the order of the reaction? (b) What is the value of the rate constant?
3 step solution
Problem 106
The following is a quote from an article in the August 18, 1998 , issue of The New York Times about the breakdown of cellulose and starch: "A drop of 18 degrees Fahrenheit [from \(77^{\circ} \mathrm{F}\) to \(59^{\circ} \mathrm{F}\) ] lowers the reaction rate six times; a 36-degree drop [from \(77^{\circ} \mathrm{F}\) to \(41^{\circ} \mathrm{F}\) ] produces a fortyfold decrease in the rate." (a) Calculate activation energies for the breakdown process based on the two estimates of the effect of temperature on rate. Are the values consistent? (b) Assuming the value of \(E_{a}\) calculated from the \(36^{\circ}\) drop and that the rate of breakdown is first order with a half-life at \(25^{\circ} \mathrm{C}\) of \(2.7 \mathrm{yr}\), calculate the half-life for breakdown at a temperature of \(-15^{\circ} \mathrm{C}\).
9 step solution
Problem 107
The following mechanism has been proposed for the reaction of NO with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and $\mathrm{H}_{2} \mathrm{O} :$ $$\mathrm{NO}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)$ $\mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the mechanism. (c) Identify any intermediates in the mechanism. (d) The observed rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\) . If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?
7 step solution
Problem 108
Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{aligned} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) & \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) & \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? (c) What is the intermediate in the reaction?
4 step solution
Problem 110
The following mechanism has been proposed for the gasphase reaction of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) and chlorine: Step 1: \(\quad \mathrm{Cl}_{2}(g) \underset{k_{1}}{\stackrel{k_{-1}}{\rightleftharpoons}} 2 \mathrm{Cl}(g) \quad\) (fast) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g) \quad\) (slow) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{CCl}_{4}\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the ratedetermining step? (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)
5 step solution
Problem 111
Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\). The following two-step mechanism is proposed for the reaction: Step 1: \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X}\) Step 2: A \(+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D}\) \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).
3 step solution
Problem 112
In a hydrocarbon solution, the gold compound \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) decomposes into ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and a different gold compound, \(\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\). The following mechanism has been proposed for the decomposition of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) : Step 1: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3}\) (fast) Step 2: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{\mathrm{k}_{1}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au} \quad\) (slow) Step 3: \(\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{k_{3}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary steps? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (f) What would be the effect on the reaction rate of adding \(\mathrm{PH}_{3}\) to the solution of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) ?
3 step solution
Problem 113
Platinum nanoparticles of diameter \(\sim 2 \mathrm{~nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(3.924 \AA\). (a) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3}\). Recall that \(1 \dot{A}=1 \times 10^{-10} \mathrm{~m}\) and \(1 \mathrm{~nm}=1 \times 10^{-9} \mathrm{~m}\). (b) Estimate how many platinum atoms are on the surface of a \(2.0\)-nm Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of \(2.8 \mathrm{~A}\). (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a \(5.0\)-nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?
5 step solution
Problem 114
One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the \(\mathrm{CO}_{2}\) accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?
3 step solution
Problem 116
Enzymes are often described as following the two-step mechanism: $$ \begin{aligned} &\mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES} \quad \text { (fast) } \\\ &\mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \quad \text { (slow) } \end{aligned} $$ where \(\mathrm{E}=\) enzyme, \(\mathrm{S}=\) substrate, \(\mathrm{ES}=\) enzyme-substrate complex, and \(\mathrm{P}=\) product. (a) If an enzyme follows this mechanism, what rate law is expected for the reaction? (b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of \(E\) with \(I\), an inhibitor.
6 step solution
Problem 117
Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\). Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from \(1.00 \mathrm{~L}\) of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{~h}\) if the gas is collected in a \(10.0\)-L container. (Assume that the products do not dissolve in chloroform.)
4 step solution
Problem 118
The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}(l)+\mathrm{I}^{-}(\mathrm{alc})\), has an activation energy of \(86.8 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C}\). (b) \(\mathrm{A}\) solution of \(\mathrm{KOH}\) in ethanol is made up by dissolving \(0.335 \mathrm{~g} \mathrm{KOH}\) in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Similarly, \(1.453 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(35^{\circ} \mathrm{C}\) ? (c) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? (d) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(50^{\circ} \mathrm{C}\).
4 step solution
Problem 120
The gas-phase reaction of \(\mathrm{NO}\) with \(\mathrm{F}_{2}\) to form \(\mathrm{NOF}\) and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{~kJ} / \mathrm{mol}\). and a frequency factor of \(A=6.0 \times 10^{8} M^{-1} \mathrm{~s}^{-1}\). The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g) $$ (a) Calculate the rate constant at \(100{ }^{\circ} \mathrm{C}\). (b) Draw the Lewis structures for the \(\mathrm{NO}\) and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the shape for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.
5 step solution
Problem 123
Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carbon- containing fragment such as \(\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2}\), and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a \(\mathrm{C}\) atom for " \(\mathrm{R}^{w}\) ). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?
2 step solution
Problem 124
The \(\mathrm{NO}_{x}\) waste stream from automobile exhaust includes species such as \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\). Catalysts that convert these species to \(\mathrm{N}_{2}\). are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of \(\mathrm{NO}, \mathrm{NO}_{2}\), and \(\mathrm{N}_{2}\) - (b) Using a resource such as Table \(8.4\), look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of \(\mathrm{NO}_{x}\) into \(\mathrm{N}_{2}\), describing what wavelengths of light need to be monitored as a function of time.
3 step solution