Chapter 15

Data in the table were collected at \(540 \mathrm{K}\) for the following reaction: $$\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g})$$ (a) Derive the rate equation. (b) Determine the reaction order with respect to each reactant. (c) Calculate the rate constant, giving the correct units for \(k.\) $$\begin{array}{lll}\hline \text { Initial Concentration }(\mathrm{mol} / \mathrm{L}) & {\text { Initial Rate }} \\\\\hline[\mathrm{C} 0] & {\left[\mathrm{NO}_{2}\right]} & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\\\\hline 5.1 \times 10^{-4} & 0.35 \times 10^{-4} & 3.4 \times 10^{-8} \\\5.1 \times 10^{-4} & 0.70 \times 10^{-4} & 6.8 \times 10^{-8} \\\5.1 \times 10^{-4} & 0.18 \times 10^{-4} & 1.7 \times 10^{-8} \\\1.0 \times 10^{-3} & 0.35 \times 10^{-4} & 6.8 \times 10^{-8} \\\1.5 \times 10^{-3} & 0.35 \times 10^{-4} & 10.2 \times 10^{-8} \\\\\hline\end{array}$$

Ammonium cyanate, NH_NCO, rearranges in water to give urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}:\) $$\mathrm{NH}_{4} \mathrm{NCO}(\mathrm{aq}) \longrightarrow\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}(\mathrm{aq})$$ $$\begin{array}{ll}\hline \begin{array}{l}\text { Time } \\\\(\min )\end{array} & \begin{array}{l}{\left[\mathrm{NH}_{4} \mathrm{NCO}\right]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} \\\\\hline 0 & 0.458 \\\4.50 \times 10^{1} & 0.370 \\\1.07 \times 10^{2} & 0.292 \\\2.30 \times 10^{2} & 0.212 \\\6.00 \times 10^{2} & 0.114 \\\\\hline\end{array}$$ Using the data in the table: (a) Decide whether the reaction is first order or second order. (b) Calculate \(k\) for this reaction. (c) Calculate the half-life of ammonium cyanate under these conditions. (d) Calculate the concentration of \(\mathrm{NH}_{4} \mathrm{NCO}\) after \(12.0 \mathrm{h}\).

Nitrogen oxides, \(\mathrm{NO}_{x}\) (a mixture of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) collectively designated as \(\mathrm{NO}_{x}\) ), play an essential role in the production of pollutants found in photochemical smog. The \(\mathrm{NO}_{x}\) in the atmosphere is slowly broken down to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) in a first-order reaction. The average half-life of NO in the smokestack emissions in a large city during daylight is \(3.9 \mathrm{h}\) (a) Starting with \(1.50 \mathrm{mg}\) in an experiment, what quantity of \(\mathrm{NO}_{x}\) remains after \(5.25 \mathrm{h} ?\) (b) How many hours of daylight must have elapsed to decrease \(1.50 \mathrm{mg}\) of \(\mathrm{NO}_{x}\) to \(2.50 \times 10^{-6} \mathrm{mg} ?\)

At temperatures below \(500 \mathrm{K},\) the reaction between carbon monoxide and nitrogen dioxide $$ \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) $$ has the following rate equation: Rate \(=k\left[\mathrm{NO}_{2}\right]^{2} .\) Which of the three mechanisms suggested here best agrees with the experimentally observed rate equation? Mechanism 1 \(\quad\) single, elementary step $$\mathrm{NO}_{2}+\mathrm{CO} \longrightarrow \mathrm{CO}_{2}+\mathrm{NO}$$ Mechanism \(2 \quad\) Two steps $$\begin{aligned}&\text { Slow } \quad \mathrm{NO}_{2}+\mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\mathrm{NO}\\\&\text { Fast } \quad \mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}\end{aligned}$$ Mechanism 3 \(\quad\) Two steps $$\begin{aligned}&\text { Slow } \quad \mathrm{NO}_{2} \longrightarrow \mathrm{NO}+\mathrm{O}\\\&\text { Fast } \quad \mathrm{CO}+\mathrm{O} \longrightarrow \mathrm{CO}_{2}\end{aligned}$$

The decomposition of dinitrogen pentaoxide $$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ has the following rate equation: \(-\Delta\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] / \Delta t=k\left[\mathrm{N}_{2} \mathrm{O}_{5}\right].\) It has been found experimentally that the decomposition is \(20 \%\) complete in \(6.0 \mathrm{h}\) at \(300 \mathrm{K}\). Calculate the rate constant and the half-life at \(300 \mathrm{K}\)

The data in the table give the temperature dependence of the rate constant for the reaction \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow\) \(2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) .\) Plot these data in the appropriate way to derive the activation energy for the reaction. $$\begin{aligned}&\\\&\begin{array}{ll}\hline T(\mathrm{K}) & k\left(\mathrm{s}^{-1}\right) \\ \hline 338 & 4.87 \times 10^{-3} \\\328 & 1.50 \times 10^{-3} \\\318 & 4.98 \times 10^{-4} \\\308 & 1.35 \times 10^{-4} \\\298 & 3.46 \times 10^{-5} \\\273 & 7.87 \times 10^{-7} \\\\\hline\end{array}\end{aligned}$$

The decomposition of gaseous dimethyl ether at ordinary pressures is first order. Its half-life is 25.0 min at \(500^{\circ} \mathrm{C}\) $$\mathrm{CH}_{3} \mathrm{OCH}_{3}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ (a) Starting with \(8.00 \mathrm{g}\) of dimethyl ether, what mass remains (in grams) after 125 min and after 145 min? (b) Calculate the time in minutes required to decrease \(7.60 \mathrm{ng}\) (nanograms) to 2.25 ng. (c) What fraction of the original dimethyl ether remains after 150 min?

Radioactive iodine-131, which has a half-life of 8.04 days, is used in the form of sodium iodide to treat cancer of the thyroid. If you begin with \(25.0 \mathrm{mg}\) of \(\mathrm{Na}^{131} \mathrm{I},\) what quantity of the material remains after 31 days?

The ozone in the earth's ozone layer decomposes according to the equation $$2 \mathrm{O}_{3}(\mathrm{g}) \longrightarrow 3 \mathrm{O}_{2}(\mathrm{g})$$ The mechanism of the reaction is thought to proceed through an initial fast equilibrium and a slow step: Step 1 \(\quad\) Fast, Reversible \(\quad \mathrm{O}_{3}(\mathrm{g}) \rightleftarrows \mathrm{O}_{2}(\mathrm{g})+\mathrm{O}(\mathrm{g})\) Step 2 Slow \(\quad \mathrm{O}_{3}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{g})\) Show that the mechanism agrees with this experimental rate law: \(-\Delta\left[\mathrm{O}_{3}\right] / \Delta t=k\left[\mathrm{O}_{3}\right]^{2} /\left[\mathrm{O}_{2}\right]\)

Hundreds of different reactions can occur in the stratosphere, among them reactions that destroy the earth's ozone layer. The table below lists several (second-order) reactions of Cl atoms with ozone and organic compounds; each is given with its rate constant. $$\begin{array}{ll}\hline & \text { Rate Constant } \\\\\text { Reaction } & \left(298 \mathrm{K}, \mathrm{cm}^{3} / \mathrm{molecule} \cdot \mathrm{s}\right) \\\\\hline \text { (a) } \mathrm{Cl}+0_{3} \longrightarrow \mathrm{Cl} 0+0_{2} & 1.2 \times 10^{-11} \\\\\text {(b) } \mathrm{Cl}+\mathrm{CH}_{4} \longrightarrow\mathrm{HCl}+\mathrm{CH}_{3} & 1.0 \times 10^{-13} \\\\\text {(c) } \mathrm{Cl}+\mathrm{C}_{3} \mathrm{H}_{8} \longrightarrow\mathrm{HCl}+\mathrm{C}_{3} \mathrm{H}_{7} & 1.4 \times 10^{-10} \\\\\text {(d) } \mathrm{Cl}+\mathrm{CH}_{2} \mathrm{FCl} \longrightarrow\mathrm{HCl}+\mathrm{CHFCl} & 3.0 \times 10^{-18} \\\\\hline\end{array}$$ For equal concentrations of Cl and the other reactant, which is the slowest reaction? Which is the fastest reaction?

Data for the reaction $$\begin{aligned}\left[\mathrm{Mn}(\mathrm{CO})_{5}\left(\mathrm{CH}_{3}\mathrm{CN}\right)\right]^{+}+\mathrm{NC}_{5}\mathrm{H}_{5} \longrightarrow & \\\&\left[\mathrm{Mn}(\mathrm{CO})_{5}\left(\mathrm{NC}_{5}\mathrm{H}_{5}\right)\right]^{+}+\mathrm{CH}_{3} \mathrm{CN}\end{aligned}$$ are given in the table. Calculate \(E_{\mathrm{a}}\) from a plot of \(\ln k\) versus \(1 / T\) $$\begin{array}{ll}\hline T(\mathrm{K}) & k\left(\min ^{-1}\right) \\\\\hline 298 & 0.0409 \\\308 & 0.0818 \\\318 & 0.157 \\\\\hline\end{array}$$

The gas-phase reaction $$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ has an activation energy of \(103 \mathrm{kJ},\) and the rate constant is \(0.0900 \min ^{-1}\) at \(328.0 \mathrm{K}\). Find the rate constant at \(318.0 \mathrm{K}\)

A Two molecules of the unsaturated hydrocarbon 1,3-butadiene \(\left(\mathrm{C}_{4} \mathrm{H}_{6}\right)\) form the "dimer" \(\mathrm{C}_{8} \mathrm{H}_{12}\) at higher temperatures. $$2 \mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g}) \longrightarrow \mathrm{C}_{8} \mathrm{H}_{12}(\mathrm{g})$$ Use the following data to determine the order of the reaction and the rate constant, \(k\). (Note that the total pressure is the pressure of the unreacted \(\mathrm{C}_{4} \mathrm{H}_{6}\) at any time and the pressure of the \(\mathrm{C}_{8} \mathrm{H}_{12} .\)) $$\begin{array}{cl}\hline \text { Time (min) } & \text { Total Pressure (mm Hg) } \\\\\hline 0 & 436 \\\3.5 & 428 \\\11.5 & 413 \\\18.3 & 401 \\\25.0 & 391 \\\32.0 & 382 \\\41.2 & 371 \\\\\hline\end{array}$$

Hypofluorous acid, HOF, is very unstable, decomposing in a first-order reaction to give HF and \(\mathrm{O}_{2}\), with a halflife of only 30 min at room temperature: $$\mathrm{HOF}(\mathrm{g}) \longrightarrow \mathrm{HF}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ If the partial pressure of HOF in a \(1.00-\mathrm{L}\). flask is initially \(1.00 \times 10^{2} \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C},\) what is the total pressure in the flask and the partial pressure of HOF after exactly 30 min? After 45 min?

We know that the decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}.\) $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ with a half-life of 245 min at \(600 \mathrm{K}\). If you begin with a partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) of \(25 \mathrm{mm}\) Hg in a 1.0 -L. flask, what is the partial pressure of each reactant and product after 245 min? What is the partial pressure of each reactant after \(12 \mathrm{h} ?\)

The substitution of \(\mathrm{CO}\) in \(\mathrm{Ni}(\mathrm{CO})_{4}\) by another group \(\mathrm{L}\) [where L is an electron-pair donor such as \(\left.\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\right]\) was studied some years ago and led to an understanding of some of the general principles that govern the chemistry of compounds having metal-CO bonds. (See J. P. Day, F. Basolo, and R. G. Pearson: Journal of the American Chemical Society, Vol. \(90, \text { p. } 6927,1968 .)\) A detailed study of the kinetics of the reaction led to the following mechanism: Slow \(\quad \mathrm{Ni}(\mathrm{CO})_{4} \longrightarrow \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{CO}\) Fast \(\quad \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{L} \longrightarrow \mathrm{Ni}(\mathrm{CO})_{3} \mathrm{L}\) (a) What is the molecularity of each of the elementary reactions? (b) Doubling the concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) increased the reaction rate by a factor of \(2 .\) Doubling the concentration of L. had no effect on the reaction rate. Based on this information, write the rate equation for the reaction. Does this agree with the mechanism described? (c) The experimental rate constant for the reaction, when \(\mathrm{L}=\mathrm{P}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3},\) is \(9.3 \times 10^{-3} \mathrm{s}^{-1}\) at \(20^{\circ} \mathrm{C} .\) If the initial concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(0.025 \mathrm{M},\) what is the concentration of the product after 5.0 min?

Hydrogenation reactions, processes wherein \(\mathrm{H}_{2}\) 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.

The following statements relate to the reaction with the following rate law: Rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right].\) $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})$$ Determine which of the following 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.

Chlorine atoms contribute to the destruction of the earth's ozone layer by the following sequence of reactions: $$\begin{aligned}\mathrm{Cl}+\mathrm{O}_{3} \longrightarrow & \mathrm{ClO}+\mathrm{O}_{2} \\ \mathrm{ClO}+\mathrm{O} \longrightarrow & \mathrm{Cl}+\mathrm{O}_{2}\end{aligned}$$ where the O atoms in the second step come from the decomposition of ozone by sunlight: $$\mathrm{O}_{3}(\mathrm{g}) \longrightarrow \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ What is the net equation on summing these three equations? Why does this lead to ozone loss in the stratosphere? What is the role played by Cl in this sequence of reactions? What name is given to species such as ClO?

Describe each of the following statements as true or false. If false, rewrite the sentence to make it correct. (a) The rate-determining elementary step in a reaction is the slowest step in a mechanism. (b) It is possible to change the rate constant by changing the temperature. (c) As a reaction proceeds at constant temperature, the rate remains constant. (d) A reaction that is third order overall must involve more than one step.

Identify which of the following statements are incorrect. If the statement is incorrect, rewrite it to be correct. (a) Reactions are faster at a higher temperature because activation energies are lower. (b) Rates increase with increasing concentration of reactants because there are more collisions between reactant molecules. (c) At higher temperatures a larger fraction of molecules have enough energy to get over the activation energy barrier. (d) Catalyzed and uncatalyzed reactions have identical mechanisms.

The reaction cyclopropane \(\longrightarrow\) propene occurs on a platinum metal surface at \(200^{\circ} \mathrm{C}\). (The platinum is a catalyst.) The reaction is first order in cyclopropane. Indicate how the following quantities change (increase, decrease, or no change) as this reaction progresses, assuming constant temperature. (a) [ cyclopropane] (b) [propene] (c) [catalyst] (d) the rate constant, \(k\) (e) the order of the reaction (f) the half-life of cyclopropane

Isotopes are often used as "tracers" to follow an atom through a chemical reaction, and the following is an example. Acetic acid reacts with methanol by eliminating a molecule of water and forming methyl acetate (See Chapter 11 ). $$\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{CH}_{3} \mathrm{OH} \quad \longrightarrow \quad \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3}+\mathrm{H}_{2} \mathrm{O}$$ Explain how you could use the isotope \(^{18} \mathrm{O}\) to show whether the oxygen atom in the water comes from the - OH of the acid or the - OH of the alcohol.

Draw a reaction coordinate diagram for an exothermic reaction that occurs in a single step. Mark the activation energy, and identify the net energy change for the reaction on this diagram. Draw a second diagram that represents the same reaction in the presence of a catalyst. Identify the activation energy of this reaction and the energy change. Is the activation energy in the two drawings different? Does the energy evolved in the two reactions differ?

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 Fast, reversible \(\quad \mathrm{HA} \rightleftarrows \mathrm{H}^{+}+\mathrm{A}^{-}\) Step 2 Fast, reversible \(\quad \mathrm{X}+\mathrm{H}^{+} \rightleftarrows \mathrm{XH}^{+}\) Step 3 Slow \(\mathrm{XH}^{+} \longrightarrow\) products What rate law is derived from this mechanism? What is the order of the reaction with respect to HA? How would doubling the concentration of HA affect the reaction?