CHAPTER 13 | Chemical Kinetics: Clearing the Airpostonp/ch223/pdf/Chemat_ISM_Ch13.pdf39 CHAPTER 13 |...

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CHAPTER 13 | Chemical Kinetics: Clearing the Air

13.1. Collect and Organize For the plot of Figure P13.1, we are to identify which curves represent [N2O] and [O2] over time for the conversion of N2O to N2 and O2 according to the equation

2 N2O(g) → 2 N2(g) + O2(g) Analyze As the reaction proceeds, the concentration of the reactant, N2O, decreases and the concentration of the product, O2, increases. The rate at which N2O is used up in the reaction is twice the rate at which O2 is produced. Solve The green line represents [N2O], and the red line represents [O2]. Think about It [N2], represented by the blue line, increases twice as fast as [O2] because two N2 molecules are produced for every one O2 molecule in the reaction.

13.2. Collect and Organize For the plot of Figure P13.2, we are to identify which curves represent [SO2] and [O2] over time for the reaction of SO2 with O2 to form SO3 according to the equation

12 2 32SO ( ) O ( ) SO ( )+ →g g g

Analyze As the reaction proceeds, the concentrations of the reactants, SO2 and O2, decrease over time. The rate at which SO2 decreases in the reaction is twice the rate at which O2 is used up according to the balanced equation. Solve The blue line represents [SO2], and the green line represents [O2]. Think about It Because one molecule of SO3 is produced for every molecule of SO2 reacted, [SO3] (red line) increases at the same rate at which [SO2] decreases.

13.3. Collect and Organize For three different initial concentrations of reactant A shown in Figure P13.3, we are to choose which would have the fastest rate for the conversion 2 A→B. Analyze We are given that the reaction is second order in A. The rate law is written as follows:

Rate = k[A]2

As the concentration of A increases, the rate increases. Solve Figure P13.3(b) has the fastest reaction rate because it has the highest concentration of A. Think about It The higher the concentration of reactant molecules, the more often they collide, which increases the rate of the reaction.

40 | Chapter 13

13.4. Collect and Organize Knowing that the rate of the reaction

A + B → C is first order in both A and B, we are to choose the sample in Figure P13.4 that produces the fastest initial rate. Analyze The initial rate of a reaction depends on the concentration of the reactants involved in the rate law expression. For the reaction described in this problem, the rate law would be

Rate = k[A][B] Therefore, the higher the concentrations of both A and B, the faster the rate of the reaction. Solve Figure P13.4(c) has the highest concentration of A (seven red spheres) and B (six blue spheres) compared to sample (a), with 4 A + 5 B, and sample (b), with 6 A + 5 B, so sample (c) has the fastest initial rate. Think about It The lower the concentration of the reactants, the slower the initial rate of the reaction. In this problem, sample (a) would give the slowest initial rate.

13.5. Collect and Organize For the reaction profile in Figure P13.5, we are to identify the parts of the energy diagram. Analyze The energy of the reactants is indicated at the beginning of the reaction; the energy of the products is indicated at the end of the reaction. The activation energy is the energy that the reaction must attain to form products; for the forward reaction it will be the energy change from the reactants to the energy of the transition state, and for the reverse reaction it will be the energy change from the products to the energy of the transition state. The energy change of the reaction is the energy difference between that of the products and that of the reactants. Solve (a) The energy of the reactants is shown as (1). (b) The energy of the products is shown as (5). (c) The activation energy of the forward reaction is shown as (2). (d) The activation energy for the reverse reaction is shown as (4). (e) The energy change of the reaction is shown as (3).

Think about It Be careful not to assume that reaction (b) is slow because it is nonspontaneous. The rate of a reaction does not depend on the thermodynamics of the reactants and the products.

13.6. Collect and Organize From the five reaction profiles in Figure P13.6, we are to choose the ones that corresponds in overall reaction energy and magnitude of activation energy and stability of an intermediate.

Chemical Kinetics | 41

Analyze A reaction profile for an endothermic reaction will show the energy of the products higher than the energy of the reactants; a reaction profile for an exothermic reaction will show the energy of the products lower than the energy of the reactants. The Ea is the energy difference between the reactants (on the left-hand side of the graph) and the transition state (the highest point on the reaction profile curve). Large activation energies are shown by a large difference between the transition-state energy and the reactant energy; small activation energies are shown by a small difference between the transition-state energy and the reactant energy. A reaction with a stable intermediate will show a low-energy dip in the middle of the reaction profile. Solve (a) Reaction profile 3 shows a highly exothermic reaction with a large activation energy.

(b) Reaction profile 2 shows a highly endothermic reaction with a moderate activation energy.

(c) Reaction profile 5 shows a reaction that forms a stable intermediate.

Think about It Reaction profile 1 shows an exothermic reaction with a small activation energy, and reaction profile 4 shows a reaction profile that has little or no enthalpy change but also shows a barrier over which the reactants must proceed to form products.

13.7. Collect and Organize

For the three reaction profiles in Figure P13.7, we are to choose the one that has the smallest rate constant.

42 | Chapter 13

Analyze The rate of a reaction (and therefore the rate constant) is determined by the activation energy (Ea) of the slowest step. All the reactions shown consist of a single step. The Ea is the energy difference between the reactants (on the left-hand side of the graph) and the transition state (the highest point on the reaction profile curve). Solve Reaction profile (b) has the largest Ea and therefore is the slowest reaction. Think about It Be careful not to assume that reaction (b) is slow because it is nonspontaneous. The rate of a reaction does not depend on the thermodynamics of the reactants and the products.

13.8. Collect and Organize We are given three reaction profiles in Figure P13.8 and, from the relative activation energies (Ea), we can determine which reaction has the largest rate constant. Analyze The activation energy is the energy barrier that the reactants must overcome to form products; the lower the activation energy barrier, the faster the reaction and the greater the rate constant for the reaction. Solve Reaction profile (a) has the lowest Ea (energy difference from reactants to the transition state at the highest point in energy in the reaction profile) and therefore has the fastest reaction. Think about It Profile (c) represents the reaction with the slowest rate. The ΔE for this reaction is zero.

13.9. Collect and Organize We are to match the reaction profile in Figure P13.9 with the possible reactions given. Analyze The reaction profile shows a two-step reaction that has a slightly larger activation energy for its second step than for its first step. Solve Reaction (c) is correct because it is a two-step reaction with the first step faster than the second.

Think about It Reaction (b), which occurs in a single step, would show only one transition state and one activation energy in its reaction profile.


We are to choose the mechanism consistent with the reaction profile in Figure P13.10 showing two activation energies.


Analyze The presence of two activation energies in the reaction profile means that two steps are in the mechanism. The relative activation energies of the two steps tell us the relative rates of the two steps, with the step having the smaller Ea being the fastest. Solve Reaction mechanism (a) is consistent with the reaction profile because two transition states are shown and the activation energy of the second step is lower than that of the first step, so the second step proceeds at a faster rate than the first step.

Think about It A single-step reaction, as in (b), would give a reaction profile showing only a single transition state and one activation energy barrier.

13.11. Collect and Organize Given the reaction profile of an uncatalyzed reaction (Figure P13.11), we are to choose the reaction profile corresponding to the catalyzed reaction. Analyze A catalyst increases the rate of reaction by decreasing the activation energy of the reaction through an alternate pathway to the products. This alternate pathway usually involves more steps. Solve Reaction profile (b) correctly shows the catalyzed reaction. Think about It Reaction profiles (a) and (c) cannot be correct because the initial uncatalyzed nonspontaneous reaction is represented as spontaneous. A catalyst cannot change a nonspontaneous reaction into a spontaneous reaction.

13.12. Collect and Organize Given the reaction profile of an uncatalyzed reaction (Figure P13.12), we are asked to choose the reaction profile for the catalyzed reaction. Analyze A catalyst increases the rate of a reaction by lowering the activation energy barrier by providing a different, lower-energy path from reactants to products. Solve Figure P13.12(c) best represents the reaction profile of the catalyzed reaction because the largest activation energy barrier of the catalyzed reaction is lower than the energy barrier of the uncatalyzed reaction. The difference in pathways is shown by the two transition states present in the catalyzed reaction profile compared to the one transition state shown in the uncatalyzed reaction profile.

44 | Chapter 13

Think about It Reaction profiles (a) and (b) cannot be correct because in both the relative energy of the products and reactants is different from the uncatalyzed reaction; the energy of the reactants and the energy of the products do not change in a catalyzed reaction.

13.13. Collect and Organize Of the highlighted elements in Figure P13.13, we are to choose which forms gaseous oxides associated with photochemical smog. Analyze Photochemical smog is the result of sunlight interacting with NOx produced by automobile emissions and volatile organic compounds (VOCs) released into the atmosphere. Solve Nitrogen (light blue) forms the volatile oxides that are components of photochemical smog. Think about It Sunlight causes a reaction of NOx and VOCs to produce peroxyacyl nitrates that are very irritating to the lungs.

13.14. Collect and Organize Of the highlighted elements in Figure P13.14, we are to choose which forms noxious oxides that are removed from automobile exhaust by catalytic converters. Analyze We read in Chapter 13 that catalytic converters remove hydrocarbons, CO, and NOx gases. Solve Carbon (green) and nitrogen (light blue) form oxide gases in automobile engines that are removed by a catalytic converter. Think about It In the catalytic converter, hydrocarbons and CO are converted into CO2 and H2O, whereas NOx is converted into N2 and O2.

13.15. Collect and Organize Of the highlighted elements in Figure P13.15, we are to identify which are widely used as heterogeneous catalysts. Analyze Heterogeneous catalysts have a different phase from the reactants. We read in Section 13.6 about the specific metals used in catalytic converters. Solve Both the transition metals, palladium (blue) and platinum (orange), can serve as heterogeneous catalysts and were identified in the chapter as catalysts. Think about It Because catalytic converters contain precious metals such as rhodium, platinum, and palladium, interest in recycling the metals from catalytic converters is great.

13.16. Collect and Organize Of the highlighted elements in Figure P13.16, we are to identify which forms oxide radicals (odd-electron species) that catalyze the destruction of the ozone layer. Analyze The destruction of ozone in the upper atmosphere is described in Section 13.6.


Solve Free chlorine atoms (yellow) that are generated by UV light interacting with chlorofluorocarbons in the upper atmosphere react with O3 to form ClO. Think about It The average amount of ozone depletion is currently stabilized and the ozone layer is expected to recover by 2050, due mostly to the worldwide agreement in the Montreal Protocol to limit the production of chlorofluorocarbons.

13.17. Collect and Organize By considering the levels of O3 during the day as seen in Figure 13.2, we are to explain why [O3]max occurs later in the day than [NO]max and [NO2]max. Analyze Ozone in the troposphere (the lowest portion of Earth’s atmosphere) is due to the reaction of O2 with O generated from the interaction of UV light with NO2. Solve The presence of NO2 in the atmosphere and ample sunlight allows the O atoms to react with O2 to generate O3. The reactant NO2 is present in the atmosphere due to automobile exhausts, which build up during the day. The buildup of O3 lags behind until later in the day, until [NO2] increases and the sunlight becomes stronger as midday approaches. Think about It Ozone is a very reactive gas and is irritating to lung tissues.

13.18. Collect and Organize For a plot of the concentration of reactants and products for

A → B → C we are to assess whether [C]max always appears after [B]max. Analyze From the equation for the reaction, we know that A must react to first form B. Then B can react to form C. Solve Yes, the maximum concentration of C appears after the maximum concentration of B because the formation of C relies on the buildup in concentration of B. Think about It Our plot would not necessarily show that all of A has reacted when product C is being formed.

13.19. Collect and Organize We are to explain why NO concentration does not increase after the evening rush hour. Analyze The reaction in the troposphere (lower atmosphere) that produces NO is

sunlight2NO ( ) NO( ) + O( )⎯⎯⎯→g g g

Solve In the evening the sunlight (and UV radiation) is less intense, so the photochemical breakdown of NO2 does not occur to as great an extent as after the morning rush hour. Think about It The use of catalytic converters to reduce the NOx to N2 and O2 in automobile exhaust has greatly helped to reduce photochemical smog in large urban and suburban centers.

46 | Chapter 13

13.20. Collect and Organize We consider why [O3] reaches a peak in the early afternoon even though O3 may react with NO to form NO2 and O2. Analyze The generation of ozone is due to the photochemical reaction

2NO ( ) NO( ) + O( )hg g gν⎯⎯→ so that O reacts with O2 to give ozone:

2 3O( ) + O ( ) O ( )ν⎯⎯→hg g g Solve The production of ozone is greatest when the sunlight is intense during the midday. This causes the production of O3 to be faster than its reaction with NO. Think about It The concentration of NO in the atmosphere is expected to be small, so we might expect the rate of its reaction with O3 to be slow.

13.21. Collect and Organize Using ΔHf

for NO, O2, and NO2 in Appendix 4, we can calculate ∆Hrxn° for

2 NO(g) + O2(g) → 2 NO2(g) Analyze The ΔHrxn

can be calculated by using

ΔHrxn

= n∑ productsΔHf,products – nproducts∑ ΔHf,reactants

Solve

ΔHrxn = 2 mol NO2 × 33.2 kJ/mol( )− 2 mol NO × 90.3 kJ/mol( ) + 1 mol O2 × 0.0 kJ/mol( )⎡⎣ ⎤⎦

= −114.2 kJ

Think about It This reaction is exothermic and, therefore, favored by enthalpy.

13.22. Collect and Organize Using the enthalpies of formation for O3, NO, NO2, and O2 gases in Appendix 4, we can calculate the enthalpy of the reaction in which O3 reacts with NO to produce O2 and NO2. Analyze The enthalpy of reaction may be calculated from

ΔHrxn

= n∑ productsΔHf,products – nproducts∑ ΔHf,reactants

Solve

ΔHrxn = 1 mol O2 × 0.0 kJ/mol( ) + 1 mol NO2 × 33.2 kJ/mol( )⎡⎣ ⎤⎦

− 1 mol O3 ×142.7 kJ/mol( ) + 1 mol NO × 90.3 kJ/mol( )⎡⎣ ⎤⎦= −199.8 kJ

Think about It This reaction is favored by enthalpy and is exothermic.


13.23. Collect and Organize / Analyze For the reaction of N2 with O2 to produce N2O and N2O5, we are to write balanced chemical equations. Solve (a) 1

2 2 22

2 2 2

N ( ) O ( ) N O( )or 2 N ( ) O ( ) 2 N O( )

g g gg g g+ →+ →

(b) 52 2 2 52

2 2 2 5

N ( ) O ( ) N O ( )or 2 N ( ) 5 O ( ) 2 N O ( )

g g gg g g+ →+ →

Think about It Balanced chemical equations usually are written with whole-number coefficients.

13.24. Collect and Organize We are to write balanced chemical equations for the reaction of N2O with O2 to give NO2 and for the decomposition of N2O5 to NO2 and O2. Analyze We can balance these reactions by inspection. Solve (a) 3

2 2 22

2 2 2

N O( ) O ( ) 2 NO ( )or 2 N O( ) 3 O ( ) 4 NO ( )

g g gg g g+ →

+ →

(b) 12 5 2 22

2 5 2 2

N O ( ) 2 NO ( ) O ( )or 2 N O ( ) 4 NO ( ) O ( )

g g gg g g→ +

→ +

Think about It NO2 itself is reactive because it is a radical species with an odd number of electrons.

13.25. Collect and Organize We are to explain the difference between the average rate and the instantaneous rate of a reaction. Analyze The rate of reaction is measured from the change in concentration of a reactant or product over time. The difference between the average and instantaneous rates is the length of time over which the change in concentration is measured. Solve The average rate is the rate averaged over a fairly long time, whereas the instantaneous rate is the rate at a specific moment in time (or over a very, very short time).

48 | Chapter 13

Think about It The rate of a reaction is always positive.

13.26. Collect and Organize We consider whether the average and the instantaneous rate may be equal. Analyze The average rate is the rate averaged over a fairly long period, whereas the instantaneous rate is the rate at a specific moment in time (or over a very, very short period). Solve Yes, the average rate and the instantaneous rate may be the same, as long as the points in time over which the average rate is measured are chosen so that the slope of the average rate line is equal to the slope of the tangent to the line for the instantaneous rate.

Think about It Also, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.

13.27. Collect and Organize We are to explain why the average rates of most reactions change over time. Analyze The forward rate of a reaction as measured from the average rate depends on the concentration of reactants. Solve As the reaction proceeds, the concentrations of the reactants decrease. Because most reactions depend on the availability (that is, concentration) of reactants to proceed, the decrease in reactant concentrations lowers the reaction rate.


Think about It Reactions that have no dependence on the concentrations of the reactants, although rare, do not show a change in rate as the reaction proceeds.

13.28. Collect and Organize We consider whether the instantaneous rate of a reaction changes with time. Analyze The instantaneous rate is the rate of the reaction measured at a specific moment in time. It is the absolute value of the slope of the line that is tangent to the curve of reactant or product concentration versus time. Solve Yes, for most reactions, the instantaneous rate changes (decreases) with time until the reaction reaches equilibrium.

Think about It When we study chemical equilibria in Chapter 16, we will learn that at equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction.


Given the balanced equation for the reaction of ammonia with oxygen to produce H+, NO2–, and H2O, we are

asked to relate the formation of products and consumption of reactants. Analyze The coefficients in the balanced chemical equation tell us how the rate of formation of products and consumption of reactants are related to each other. For the generic chemical equation where A, B, C, and D are the reactants and products and a, b, c, and d are their stoichiometric coefficients,

a A + b B → c C + d D the relationship of the rates is given by

Rate = − 1

aΔ A⎡⎣ ⎤⎦Δt

= − 1bΔ B⎡⎣ ⎤⎦Δt

= 1cΔ C⎡⎣ ⎤⎦Δt

= 1dΔ D⎡⎣ ⎤⎦Δt

Solve For the reaction given

Rate = – 1

2Δ NH3⎡⎣ ⎤⎦

Δt= – 1

3Δ O2⎡⎣ ⎤⎦Δt

= 12Δ H+⎡⎣ ⎤⎦Δt

= 12Δ NO2

–⎡⎣ ⎤⎦Δt

= 12Δ H2O⎡⎣ ⎤⎦

Δt

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(a) [ ] + –23 H NONH

t t t

⎡ ⎤ ⎡ ⎤Δ ΔΔ ⎣ ⎦ ⎣ ⎦− = =Δ Δ Δ

, the rate of consumption of NH3, is equal to the rate of formation of H+

and NO2–.

(b) [ ]–2 2NO O2

3t t

⎡ ⎤Δ Δ⎣ ⎦ = −Δ Δ

, the rate of formation of NO2–, is two-thirds the rate of consumption of O2.

(c) [ ] [ ]3 2NH O23t t

Δ Δ=

Δ Δ, the rate of consumption of NH3, is two-thirds the rate of consumption of O2.

Think about It The negative sign is used in front of the expressions involving the consumption of a reactant to give a positive reaction rate because the change in concentration, [X]f – [X]i, or Δ[X], is negative for reactants because [X]f < [X]i.

13.30. Collect and Organize Using the balanced equation describing the reaction of CO with NO, we are to write expressions to compare the rates of formation of products and the rates of consumption of reactants. Analyze From the balanced equation, we see that the reaction may be expressed as

Rate =

Δ N2⎡⎣ ⎤⎦Δt

= 12Δ CO2⎡⎣ ⎤⎦

Δt = – 1

2Δ CO⎡⎣ ⎤⎦

Δt = – 1

2Δ NO⎡⎣ ⎤⎦

Δt

Solve

(a) Because [ ] [ ]2N CO1 –2t t

Δ Δ=

Δ Δ, the rate of formation of N2 is one-half the rate of consumption of CO.

(b) Because [ ] [ ]2CO NO1 1–2 2t tΔ Δ

=Δ Δ

, the rate of formation of CO2 is the equal to the rate of consumption

of NO.

(c) Because [ ] [ ]CO NO1 1 – –2 2t tΔ Δ

=Δ Δ

, the rate of consumption of the two reactants is equal.

Think about It These relative rates make sense according to the stoichiometry of the reaction. For every 2 mol of CO and NO used in the reaction, 2 mol of CO2 and 1 mol of N2 are produced. So, for example, the concentration of CO decreases at twice the rate that the concentration of N2 increases.


For each of three reactions, we are to write expressions for the rate of formation of products and consumption of reactants. Analyze The coefficients in the balanced chemical equation tell us how the rate of formation of products and consumption of reactants are related to each other. For the generic chemical equation where A, B, C, and D are the reactants and products and a, b, c, and d are their stoichiometric coefficients,

a A + b B → c C + d D the relationship of the rates is given by

Rate = − 1

aΔ A⎡⎣ ⎤⎦Δt

= − 1bΔ B⎡⎣ ⎤⎦Δt

= 1cΔ C⎡⎣ ⎤⎦Δt

= 1dΔ D⎡⎣ ⎤⎦Δt


Solve

(a) [ ] [ ]2 2H O OH1Rate2t t

Δ Δ= − =

Δ Δ

(b) [ ] [ ] [ ]2 3O ClOClORate

t t tΔ ΔΔ

= − = − =Δ Δ Δ

(c) [ ] [ ] [ ]2 5 2 3N O H O HNO1Rate2t t t

Δ Δ Δ= − = − =

Δ Δ Δ


13.32. Collect and Organize For each reaction we are to write an expression relating the rate of formation of products to the rate of consumption of reactants. Analyze In the expression relating the rate of a reaction in terms of the formation of products and consumption of reactants, the denominator of the fraction for each product’s or reactant’s change of concentration versus time is the coefficient of that product or reactant in the overall balanced equation:

a A + b B = c C+ d D

Rate = − 1a

ΔA⎡⎣ ⎤⎦Δt

= − 1b

ΔB⎡⎣ ⎤⎦Δt

= 1c

ΔC⎡⎣ ⎤⎦Δt

= 1d

ΔD⎡⎣ ⎤⎦Δt

Solve

(a) Rate = −

Δ Cl2O2⎡⎣ ⎤⎦Δt

= 12Δ ClO⎡⎣ ⎤⎦

Δt

(b) Rate = −

Δ N2O5⎡⎣ ⎤⎦Δt

=Δ NO2⎡⎣ ⎤⎦

Δt=Δ NO3⎡⎣ ⎤⎦

Δt

(c) Rate = − 1

2Δ INO⎡⎣ ⎤⎦

Δt=Δ I2⎡⎣ ⎤⎦Δt

= 12Δ NO⎡⎣ ⎤⎦

Δt


13.33. Collect and Organize Using the balanced equation describing the reaction of SO2 with CO, we are to write expressions to compare the rates of formation of products and the rates of consumption of reactants. Analyze From the balanced equation, we see that the reaction may be expressed as

Rate = −

Δ SO2⎡⎣ ⎤⎦Δt

= − 13Δ CO⎡⎣ ⎤⎦

Δt= 1

2Δ CO2⎡⎣ ⎤⎦

Δt=Δ COS⎡⎣ ⎤⎦

Δt

Solve

(a) [ ] [ ]2CO CO2Rate3t t

Δ Δ= = −

Δ Δ

52 | Chapter 13

(b) [ ] [ ]2SOCOSRate

t tΔΔ

= = −Δ Δ

(c) [ ] [ ]2SOCORate 3

t tΔΔ

= =Δ Δ

Think about It These relative rates make sense according to the stoichiometry of the reaction. For every 1 mol of SO2 used in the reaction, 2 mol of CO2 and 1 mol of COS are produced. So, for example, the concentration of CO2 will increase twice as fast as the concentration of SO2 decreases.

13.34. Collect and Organize Using the balanced equation describing the reaction of CH4 with NO, we are to write expressions to compare the rates of formation of products and the rates of consumption of reactants. Analyze From the balanced equation, we see that the reaction may be expressed as

Rate = −

Δ CH4⎡⎣ ⎤⎦Δt

= − 14Δ NO⎡⎣ ⎤⎦

Δt= 1

2Δ N2⎡⎣ ⎤⎦Δt

=Δ CO2⎡⎣ ⎤⎦

Δt= 1

2Δ H2O⎡⎣ ⎤⎦

Δt

Solve

(a) [ ] [ ]2 2N CORate 2

t tΔ Δ

= =Δ Δ

(b) [ ] [ ]2CO NO1Rate4t t

Δ Δ= = −

Δ Δ

(c) [ ] [ ]4 2CH H O1Rate2t t

Δ Δ= − =

Δ Δ



Using the relative rate expressions and the rate of the consumption of ClO in two reactions, we are to calculate the rate of change in the formation of the products of the two reactions. Analyze (a) For this reaction,

Rate = − 1

2Δ ClO⎡⎣ ⎤⎦

Δt=Δ Cl2⎡⎣ ⎤⎦

Δt=Δ O2⎡⎣ ⎤⎦Δt

(b) For this reaction,

Rate = −

Δ ClO⎡⎣ ⎤⎦Δt

= –Δ O3⎡⎣ ⎤⎦Δt

=Δ O2⎡⎣ ⎤⎦Δt

=Δ ClO2⎡⎣ ⎤⎦

Δt

For each reaction we are given ∆[ClO]/∆t. We can use this value in the relationships above to calculate the rate of change in (a) the concentration of Cl2 and O2 and (b) the concentration of O2 and ClO2. Solve

(a)

Rate =Δ Cl2⎡⎣ ⎤⎦


= − 12Δ ClO⎡⎣ ⎤⎦

Δt= – 1

2× –2.3×107 M/s

= 1.2×107 M/s


(b)

Rate =Δ O2⎡⎣ ⎤⎦Δt

=Δ ClO2⎡⎣ ⎤⎦

Δt= −

Δ ClO⎡⎣ ⎤⎦Δt

= – –2.9×104 M/s( )= 2.9×104 M/s

Think about It The rates of the formation of products are positive because [X]f > [X]i, so [X]f – [X]i = Δ[X] is positive.

13.36. Collect and Organize Using the relative rate expressions and the given rates of the consumption of NO3 in two reactions, we can calculate the rates of change in the formation of the products of the two reactions. Analyze

(a) For this reaction, Rate = −

Δ NO3⎡⎣ ⎤⎦Δt

= –Δ NO⎡⎣ ⎤⎦

Δt= 1

2Δ NO2⎡⎣ ⎤⎦

Δt

(b) For this reaction, Rate = − 1

2Δ NO3⎡⎣ ⎤⎦

Δt= 1

2Δ NO2⎡⎣ ⎤⎦


For each reaction we are given ∆[NO3]/∆t. We can use this value in the rate relationships to calculate the rates of change in the concentration of NO2. Solve

(a)

12Δ NO2⎡⎣ ⎤⎦

Δt= −


orΔ NO2⎡⎣ ⎤⎦

Δt= −2


= –2× –2.2×105 mM/min( )= 4.4×105 mM/min

(b)

12Δ NO2⎡⎣ ⎤⎦

Δt= − 1

2Δ NO3⎡⎣ ⎤⎦

Δtor


= −Δ NO3⎡⎣ ⎤⎦

Δt= – –2.3 mM/min( )= 2.3 mM/min

Think about It Because we always express the rate of reaction as a positive value, we assign the rate of consumption of a reactant a negative sign because [X]f – [X]i < 0, since [X]f < [X]i.

13.37. Collect and Organize Given the [O3] over time when it reacts with NO2

–, we are to calculate the average reaction rate for two intervals. Analyze The average rate of reaction can be found according to

Δ O3⎡⎣ ⎤⎦Δt

=O3⎡⎣ ⎤⎦f

– O3⎡⎣ ⎤⎦i

tf – ti

Solve Between 0 and 100 µs:

–Δ O3⎡⎣ ⎤⎦Δt

=9.93×10–3– 1.13×10–2( ) M

100 – 0( ) µs= 1.4×10–5 M/µs

54 | Chapter 13

Between 200 and 300 µs:

–Δ O3⎡⎣ ⎤⎦Δt

=8.15×10–3 – 8.70×10–3( ) M

300 – 200( ) µs= 5.50×10–6 M/µs

Think about It As the reaction proceeds, the rate of consumption of ozone decreases. This is due to the decreasing reactant concentrations.

13.38. Collect and Organize Given the [N2O5] over time when it decomposes, we are to calculate the average rate of the reaction between all the consecutive times for the data presented. Analyze The average rate of the reaction can be found according to

[ ] [ ] [ ]2 5 2 52 5 f i

f i

N O – N O N O–t t t

Δ=

Δ

Solve Between 0 and 1.45 s:

[ ] ( )( )

12 12 32 5 10 3

1.357 10 –1.500 10 molecules/cmN O– 9.86 10 molecules (cm s)

1.45 – 0 s× ×Δ

= = × ⋅Δt

Between 1.45 s and 2.90 s:

[ ] ( )( )

12 12 32 5 10 3


2.90 –1.45 s× ×Δ

= = × ⋅Δt

Between 2.90 and 4.35 s:

[ ] ( )( )

12 12 32 5 10 3


4.35 – 2.90 s× ×Δ

= = × ⋅Δt

Between 4.35 and 5.80 s:

[ ] ( )( )

12 12 32 5 10 3


5.80 – 4.35 s× ×Δ

= = × ⋅Δt

Think about It As the reaction proceeds, the rate of disappearance of dinitrogen pentoxide decreases. This is due, in part, to the decreasing reactant concentrations.

13.39. Collect and Organize After we plot [ClO] versus time and [Cl2O2] versus time, we can determine the instantaneous rate of change at 1 s for each compound. Analyze The instantaneous rate is the slope of the line that is tangent to the curve at the time we are interested in. We can estimate fairly well the instantaneous rate at 1 s from the plots by choosing two points that are close to 1 s to calculate the slope. Since we are given only values for [ClO], we need to calculate [Cl2O2] for each time.

[Cl2O2]t =ClO⎡⎣ ⎤⎦0 – ClO⎡⎣ ⎤⎦t( )

2where [ClO]0 = 2.60×1011M


Time, s [ClO]t, molecules/cm3 [Cl2O2]t, molecules/cm3

0 2.60 × 1011 0.00 1 1.08 × 1011 7.60 × 1010 2 6.83 × 1010 9.59 × 1010 3 4.99 × 1010 1.05 × 1011 4 3.93 × 1010 1.10 × 1011 5 3.24 × 1010 1.14 × 1011 6 2.76 × 1010 1.16 × 1011

Initially, no Cl2O2 is present, so [Cl2O2]0 = 0 molecules/cm3. Solve For the change in concentration of ClO versus time, we obtain the following plot:

To get a fairly good estimate from the data given for the instantaneous rate, we can choose two points from the data set that surrounds the data point of interest and calculate ∆[ClO]/∆t. For t = 1 s, we can use the points t = 0 s and t = 2 s:

10 11 310 –3 –1[ClO] (6.83 10 2.60 10 ) molecules/cm– 9.59 10 molecules cm s

(2 – 0) t sΔ × − ×

= = × ⋅ ⋅Δ

Using a graphing program that calculates the slope of the tangent to the line at t = 1 s, we get an instantaneous rate of 8.28 × 1010 molecules ⋅ cm–3 ⋅ s.

For the change in concentration of Cl2O2 versus time, we obtain the following plot.

56 | Chapter 13

To get a fairly good estimate from the data given for the instantaneous rate, we can choose two points from the data set that surrounds the data point of interest and calculate ∆[Cl2O2]/∆t. For t = 1 s, we can use the points t = 0 s and t = 2 s:

10 310 –3 –12 2[Cl O ] (9.59 10 0) molecules/cm 4.80 10 molecules cm s

(2 – 0) t sΔ × −

= = × ⋅ ⋅Δ

Using a graphing program that calculates the slope of the tangent to the line at t = 1 s, we get an instantaneous rate of 4.13 × 1010 molecules ⋅ cm–3 ⋅ s. Think about It Because we expect the rate of disappearance of ClO to be twice the rate of appearance of Cl2O2 from the balanced equation

2 ClO(g) → Cl2O2(g) our answers make sense:

10 –3 1

10 –3 12 2

[ClO]9.59 10 molecules cm s 2.0[Cl O ] 4.80 10 molecules cm s

−

−

Δ× ⋅ ⋅Δ = =Δ × ⋅ ⋅

Δ

t

t


To determine the instantaneous rate of the reaction of O3 with NO at 0.000 s and 0.052 s, we have to plot [NO] versus time. Analyze On the plot, we will draw the tangent to the line where t = 0.00 s and t = 0.052 s. The slopes of these lines are the instantaneous rates of the reaction. We can estimate fairly well the instantaneous rate at the two times during the reaction from the plot by choosing two points from the data set given that surrounds those times of interest to calculate the slope. Solve

To get a fairly good estimate from the data given for the instantaneous rate, we can choose two points from the data set that surrounds the data point of interest and calculate ∆[NO]/∆t. For t = 0.00, we can use the points t = 0.00 and t = 0.011:

–8 –8–7[NO] (1.8 10 2.0 10 ) – – 1.82 10 /s

(0.011– 0.000) sΔ × − ×

= = ×Δ

M Mt

Using a graphing program that calculates the slope of the tangent to the line at t = 0.00 s, we get an instantaneous rate of 1.92 × 10–7 M/s.


For t = 0.052, we can use the points where t = 0.027 s and t = 0.102 s: –8 –8

–8[NO] (1.2 10 1.6 10 ) – – 5.33 10 /st (0.102 – 0.027) s

M MΔ × − ×= = ×

Δ

Using a graphing program that calculates the slope of the tangent to the line at t = 1 s, we get an instantaneous rate of 6.27 × 10–8 M/s. Think about It As the reaction progresses, the rate of the reaction decreases.

13.41. Collect and Organize We consider whether two different chemical reactions can have the same rate law expression. Analyze The rate law expression is of the form

Rate = k[A]x where k is the rate constant, [A] is the concentration of the reactant(s) on which the rate depends, and x is the order of the reaction for that concentration of reactant. Solve Yes, two different reactions can have the same form of the rate law. They both may have the same dependence on the concentration of reactants, yet yield different products. For example, A may decompose by two routes:

A → B + C A → D + E

If both reactions are first order (or even second order) in [A], they would have the same rate law: Rate = k[A]

Think about It The value of the rate constant, however, is expected to be different for the two reactions.

13.42. Collect and Organize We are to explain why the units of the rate constant change with the overall order of the reaction. Analyze The form of the rate law is

Rate = k[A]x Rearranging and solving for k:

k = Rate

A⎡⎣ ⎤⎦x

Solve The rate of a reaction is expressed as concentration per time (often in moles per liter per second) and the concentration of A is usually expressed in moles per liter (M ). The order of the reaction may be any number (fractional, whole numbers, or even negative). The order of the reaction changes the units of k. Some examples are as follows:

For a first-order reaction, 1s sMkM

−= =

For a third-order reaction, 2 13

s sMk MM

− −= =

For a one-half-order reaction, k = M s

M 1/2 = M 1/2s−1

Think about It Once the form of the rate law is known, we can determine the units of the rate constant.

58 | Chapter 13

13.43. Collect and Organize We are asked whether the units of the half-life for a second-order reaction are the same as those of the half-life for a first-order reaction. Analyze The half-life is the time for the amount of reactant originally present to decrease by one-half. Solve Yes. Because the half-life is a time measurement, the units, no matter what the order of the reaction, are always in units of time (s, min, hr, yr, etc.). Think about It The half-life of a reaction depends on the value of the reaction’s rate constant and, except for first-order reactions, on the initial concentration. The larger the rate constant, the faster the reaction and the shorter the half-life of the reaction.

13.44. Collect and Organize We consider whether the half-life of a first-order reaction depends on the concentration of the reactants. Analyze For a first-order reaction,

1/ 20.693

=tk

Solve No, there is no dependence of 1/ 2t for a first-order reaction on the concentration of reactants because [A]0 does not appear in the expression for the half-life. Think about It The half-life for a first-order reaction is inversely related to the rate constant, k.

13.45. Collect and Organize For a second-order reaction, we are to predict the effect of doubling [A]0 on the half-life. Analyze For a second-order reaction,

[ ]1/ 20

1A

=tk

Solve From the equation for the half-life of a second-order reaction, we see that doubling [A]0 halves the half-life. Think about It The half-life of a second-order reaction, like that of a first-order reaction, is inversely related to the rate constant.

13.46. Collect and Organize For two decomposition reactions that have the same rate constant, k, we are to determine whether these reactions will also have the same 1/ 2.t Analyze The half-life of a first-order reaction is

1/ 20.693

=tk


Solve Because t1/2 for a first-order reaction is dependent only on the value of k, yes, two reactions with the same rate constant have the same half-life. Think about It The reactions may not necessarily have the same value of k at a different temperature, though.

13.47. Collect and Organize For each rate law expression, we are to determine the order of the reaction with respect to each reactant and the overall reaction order. Analyze The order of a reaction is the experimentally determined dependence of the rate of a reaction on the concentration of the reactants involved in the reaction. In the rate law expression, the order is shown as the power to which the concentration of a particular reactant is raised. The overall reaction order is the sum of the powers of the reactants in the rate law expression. Solve (a) For the rate law expression Rate = k[A][B], the reaction is first order in both A and B and second order overall. (b) For Rate = k[A]2[B], the reaction is second order in A, first order in B, and third order overall. (c) For Rate = k[A][B]3, the reaction is first order in A, third order in B, and fourth order overall. Think about It The higher the order of the reaction for a particular reactant, the greater the effect of a change in concentration of that reactant on the reaction rate.

13.48. Collect and Organize For each rate law expression, we are to determine the order of the reaction with respect to each reactant and the overall reaction order. Analyze The order of a reaction is the experimentally determined dependence of the rate of a reaction on the concentration of the reactants involved in the reaction. In the rate law expression, the order is shown as the power to which the concentration of a particular reactant is raised. The overall reaction order is the sum of the powers of the reactants in the rate law expression. Solve (a) For the rate law expression Rate = k[A]2[B]1/2, the reaction is second order in A, one-half order in B, and two-an-one-half order overall. (b) For Rate = k[A]2[B][C], the reaction is second order in A, first order in both B and C, and fourth order overall. (c) For Rate = k[A][B]3[C]1/2, the reaction is first order in A, third order in B, one-half order in C, and four-and-one-half order overall. Think about It The higher the order of the reaction for a particular reactant, the greater the effect of a change in concentration of that reactant on the reaction rate.

13.49. Collect and Organize For each reaction described, we are to write the rate law and determine the units for k, using the units M for concentration and s for time. Analyze The general form of the rate law is

Rate = k[A]x[B]y

60 | Chapter 13

where k is the rate constant, A and B are the reactants, and x and y are the orders of the reaction with respect to each reactant as determined by experiment. Solve (a) Rate = k[O][NO2] Because rate has units of M/s and each concentration has units of M,

1 12

s sMk MM

− −= =

(b) Rate = k[NO]2[Cl2] Because rate has units of M/s and each concentration has units of M,

2 13

s sMk MM

− −= =

(c) Rate = k [CHCl3][Cl2]12

Because rate has units of M/s and each concentration has units of M, 1

2 13/ 2

s s− −= =Mk MM

(d) Rate = k[O3]2[O]–1 Because rate has units of M/s and each concentration has units of M,

12

s sMkM M

−= =

Think about It The units of the rate constant clearly depend on the overall order of the reaction.


Given the order of the reaction for each of the reactants in the chemical reaction A + B → C

we are to write the rate law expression and determine the units of the rate constant k, using the units M for concentration and s for time. Analyze In the rate law, the order of the reaction for each reactant is placed as an exponent to the concentration:

Rate = k[A]x[B]y where x and y are the orders of the reaction with respect to A and B. The rate of the reaction is expressed as M/s, so the rate constant units will have to be 1/Mz–1 ⋅ s, where z is the overall order of the reaction. Solve (a) Rate = k[A][B]2, k in units of 1/M

2 ⋅ s (b) Rate = k[A][B], k in units of 1/M ⋅ s (c) Rate = k[B]2, k in units of 1/M ⋅ s (d) Rate = k[A]2[B]2, k in units of 1/M

3 ⋅ s Think about It A reaction rate that is independent of the concentration of a reactant is zero order in that reactant. So for part c,

Rate = k[A]0[B]2 where [A]0 = 1.

13.51. Collect and Organize Given the changes in rate of the decomposition of BrO to Br2 and O2 when [BrO] is changed, we are to predict the rate law in each case.


Analyze The general form of the rate law for the reaction is

Rate = k[BrO]x where x is an experimentally determined exponent. Solve (a) If the rate doubles when [BrO] doubles, then x = 1 and the rate law is

Rate = k[BrO] (b) If the rate quadruples when [BrO] doubles, then x = 2 and the rate law is

Rate = k[BrO]2 (c) If the rate is halved when [BrO] is halved, then x = 1 and the rate law is

Rate = k[BrO] (c) If the rate is unchanged when [BrO] is doubled, then x = 0 and the rate law is

Rate = k[BrO]0 = k Think about It For this reaction, the relationship is straightforward between the change in rate when the concentration of the reactant was changed to determine x. To determine x for more complicated reactions, use

rate2

rate1

=A⎡⎣ ⎤⎦2

A⎡⎣ ⎤⎦1

⎛

⎝⎜

⎞

⎠⎟

x

lnrate2

rate1

⎛

⎝⎜⎞

⎠⎟= x ln

A⎡⎣ ⎤⎦2

A⎡⎣ ⎤⎦1

⎛

⎝⎜

⎞

⎠⎟

x =ln

rate2

rate1

⎛⎝⎜

⎞⎠⎟

lnA⎡⎣ ⎤⎦2

A⎡⎣ ⎤⎦1

⎛

⎝⎜

⎞

⎠⎟


Using the information of how the rate of the reaction changes when the concentration of a reactant is changed, we can predict the rate law for the reaction of NO with Br2 to give NOBr2. Analyze (a) If the rate doubles when [NO] doubles while [Br2] remains constant, then the reaction is first order in NO. (b) If the rate doubles when [Br2] doubles while [NO] remains constant, then the reaction is first order in Br2. (c) If the rate of the reaction increases by 1.56 times when [NO] increases by 1.25 times while [Br2] remains constant, then the order of the reaction for NO is

1.25x = 1.56 x ln(1.25) = ln(1.56)

x = 1.993 = 2 (d) If the rate is halved when [NO] is doubled and [Br2] remains constant, then the order of the reaction of NO is

2x = 0.5 x ln(2) = ln(0.5)

x = –1 Solve (a) Rate = k[NO][Br2]y, where y is undefined (b) Rate = k[NO]y[Br2], where y is undefined

62 | Chapter 13

(c) Rate = k[NO]2[Br2]y, where y is undefined (d) Rate = k[Br2]y/[NO], where y is undefined Think about It The order for a particular reactant in a rate law expression may be any number (positive, negative, zero, fraction, or whole) and can be determined only by experiment, not by looking at the overall balanced chemical equation.

13.53. Collect and Organize Given that the rate of the reaction quadruples when both [NO] and [ClO] are doubled, we are to identify what additional information we would need to write the rate law for the reaction

NO(g) + ClO(g) → NO2(g) + Cl(g) Analyze The general form of the rate law for this reaction is

Rate = k[NO]x[ClO]y Solve If the rate quadruples when both [NO] and [ClO] are doubled, the rate law could be any of the following:

Rate = k[NO][ClO] Rate = k[NO]2

Rate = k[ClO]2

To differentiate among these, we need to determine the change in the rate when only [NO] or [ClO] is changed. Think about It If the rate is only doubled when [NO] and [ClO] are independently changed, then the rate law is

Rate = k[NO][ClO] If the rate is quadrupled when [NO] is doubled but remains constant if [ClO] is doubled, then the rate law is

Rate = k[NO]2 If the rate is quadrupled when [ClO] is doubled but remains constant if [NO] is doubled, then the rate law is

Rate = k[ClO]2

13.54. Collect and Organize Given the information that the rate for the reaction between M (a molecule that is unchanged in the reaction) with ClO and NO2 is first order in NO2 and in ClO, we are to write the rate law and determine the order of the reaction for M. Analyze Knowing that the reaction is first order in both NO and ClO means that the exponent for the concentrations in the rate law for both of these reactants is 1. Because no dependence on the rate was given in the problem for M, we can assume that changing [M] does not affect the rate of the reaction. Solve (a) Rate = k[ClO][NO] (b) For the rate law to show no dependence on [M], the order of the reaction with respect to M must be zero. Think about It The molecule M may function in this reaction to transfer energy from one of the reactants to another to enable the reaction to take place or to bring the reactants closer together to facilitate the reaction between them.

13.55. Collect and Organize For the reaction of NO2 with O3 to produce NO3 and O2, we are to write the rate law given that the reaction is first order in both NO2 and O3. From the rate law and given the rate constant, we can calculate the rate of the


reaction for a given [NO2] and [O3]. From this we can calculate the rate of appearance of NO3 and the rate of the reaction when [O3] is doubled. Analyze The general form of the rate law for this reaction is

Rate = k[NO2]x[O3]y The rate of consumption of reactants and formation of products is

Rate = −


= –Δ O3⎡⎣ ⎤⎦Δt

=Δ NO3⎡⎣ ⎤⎦


Solve (a) Rate = k[NO2][O3]

(b) 4

–8 –7 –111.93 10Rate 1.8 10 1.4 10 4.9 10 /ss

M M MM×

= × × × × = ×⋅

(c) [ ]3 –11NORate 4.9 10 /sM

tΔ

= = ×Δ

(d) When [O3] is doubled, the rate of the reaction doubles. Think about It When [O3] = 2.8 × 10–7 M (double that in part b) the rate of reaction is 9.73 × 10–11 M/s, which is twice that calculated in part b, so our prediction in part d is correct.

13.56. Collect and Organize We are to write the rate law for the reaction between N2O5 and H2O, and from that rate law we are to calculate the rate constant for the reaction for a specified reaction rate and concentrations of N2O5 and H2O. Analyze (a) We are given that the reaction is first order in both reactants, so the exponents for the concentrations of both N2O5 and H2O are 1. (b) To calculate k we rearrange the rate law to solve for k and use the given rate of the reaction and the concentrations of the reactants in the equation. Solve (a) Rate = k[N2O5][H2O]

(b) –4

–5 –1 –1

2 5 2

rate 4.55 10 m /min= 1.50 10 m min[N O ][H O] 0.132 m 230 m

Mk MM M

×= = ×

×

Think about It The rate constant does not change (except with temperature), although changing the initial concentrations of the reactants would change the rate of the reaction.

13.57. Collect and Organize By comparing the rate constants for four reactions that are all second order, we can determine which reaction is the fastest if all the initial concentrations are the same. Analyze The reaction with the largest rate constant has the fastest reaction rate. Solve Reaction (c) has the largest value of k, so it proceeds the fastest. Think about It The slowest reaction is a reaction with the smallest value of k.

64 | Chapter 13

13.58. Collect and Organize We are asked to compare a first-order reaction with a second-order reaction, each with the same magnitude of rate constant, k. Analyze The first-order reaction rate law is

RateA = k[A] The second-order reaction rate law is

RateB = k[B]2 Solve (a) Neither. If the initial concentrations of A and B are both 1.0 mM, the reaction rates are identical because

RateA = k[1.0 mM] RateB = k[1.0 mM]2

RateA = RateB (b) If the initial concentrations of A and B are both 2.0 M, the reaction rates differ:

RateA = k[2.0 M] RateB = k[2.0 M]2 = k × 4.0 M 2

For these concentrations of reactants, the second-order reaction proceeds faster. Think about It For these two reactions, any [A] = [B] less than 1.0 mM will have a faster rate for the first-order reaction, but for [A] = [B] greater than 1.0 mM the second order-reaction will be faster.

13.59. Collect and Organize Given the information that the rate of the reaction between NO and NO2 with water doubles when either [NO] or [NO2] doubles and the rate does not depend on [H2O], we can write the rate law for the reaction. Analyze The general form of the rate law for this reaction is

Rate = k[NO]x[NO2]y[H2O]z Doubling of the reaction rate with doubling of [NO] or [NO2] means the reaction is first order in those reactants. Because the rate does not depend on [H2O], the reaction is zero order in that reactant. Solve The rate law for this reaction is

Rate = k[NO]1[NO2]1[H2O]0 = k[NO][NO2] Think about It If [NO] and [NO2] are doubled simultaneously, the rate of reaction quadruples.

13.60. Collect and Organize Given the effect on the rate of the reaction between HO2 and SO3 when their concentrations are doubled, we can write the rate law for the reaction. Analyze If the rate of the reaction doubles when either reactant concentration is doubled, then the exponent for each reactant in the rate law expression is 1. Solve

Rate = k[HO2][SO3] Think about It Although this reaction is first order in each reactant, it is second order overall.


13.61. Collect and Organize In the reaction of ClO2 with OH– the rate of the reaction was measured for various concentrations of both reactants. From the data we are to determine the rate law and calculate the rate constant, k. Analyze To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two experiments in which the concentration of that reactant changes but the concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law expression. To calculate the rate constant for the reaction, we can rearrange the rate law to solve for k and use the data from any of the experiments. Solve Using experiments 1 and 2, we find that the order of the reaction with respect to ClO2 is 1:

2 0,11

2 2 0,2

[ClO ]raterate [ClO ]

0.0248 /s 0.060 0.00827 /s 0.020

3.00 3.011

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠==

x

x

x

M MM M

x

Using experiments 2 and 3, we find that the order of the reaction with respect to [OH–] is also 1: –

0,33–

2 0,2

[OH ]raterate [OH ]

0.0247 /s 0.090 0.00827 /s 0.030

2.99 3.01

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠==

x

x

x

M MM M

x

The rate law for this reaction is Rate = k[ClO2][OH–]. Rearranging the rate law expression to solve for k and using the data from experiment 1 gives

–1 –1–

2

rate 0.0248 /s 14 s0.060 0.030 [ClO ][OH ]

Mk MM M

= = =×

Think about It We may use any of the experiments in the table to calculate k. Each experiment’s data give the same value of k as long as the experiments were all run at the same temperature.

13.62. Collect and Organize In the reaction of NO2

– with O3, the rate of the reaction was measured for various concentrations of both reactants. From the data we are to determine the rate law and calculate the rate constant, k. Analyze To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two experiments in which the concentration of that reactant changes but the concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law expression. To calculate the rate constant for the reaction, we can rearrange the rate law to solve for k and use the data from any of the experiments.

66 | Chapter 13

Solve Using experiments 1 and 2, we find that the order of the reaction with respect to NO2

– is 1: –

2 0,22–

1 2 0,1

[NO ]raterate [NO ]

37.5 /s 0.0150 25 /s 0.0100

1.5 1.501

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠==

x

x

x

M MM M

x

Using experiments 3 and 4, we find that the order of the reaction with respect to O3 is also 1:

3 0,44

3 3 0,3

[O ]raterate [O ]

200.0 /s 0.0200 50.0 /s 0.0050

4.00 4.01

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠==

x

x

x

M MM M

x

The rate law for this reaction is Rate = k[NO2–][O3].

Rearranging the rate law expression to solve for k and using the data from experiment 1 gives 5 –1 –1

–2 3

rate 25 /s 5.0 10 s0.0100 0.0050 [NO ][O ]

Mk MM M

= = = ××


13.63. Collect and Organize In the reaction of H2 with NO, the rate of the reaction was measured for various concentrations of both reactants. From the data we are to determine the rate law and calculate the rate constant, k. Analyze To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two experiments in which the concentration of that reactant changes, but the concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law expression. To calculate the rate constant for the reaction, we can rearrange the rate law to solve for k and use the data from any of the experiments. Solve Using experiments 1 and 2, we find that the order of the reaction with respect to NO is 2:

0,22–

1 2 0,1

[NO]raterate [NO ]

0.0991 /s 0.272 0.0248 /s 0.136

4.00 2.002

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠==

x

x

x

M MM M

x


Using experiments 3 and 4, we find that the order of the reaction with respect to H2 is 1:

2 0,44

3 2 0,3

[H ]raterate [H ]

1.59 /s 0.848 0.793 /s 0.424

2.01 2.001

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠==

x

x

x

M MM M

x

The rate law for this reaction is Rate = k[NO]2[H2]. Rearranging the rate law expression to solve for k and using the data from experiment 1 gives

( )–2 –1

2

0.0248 /s 6.32 s0.136 0.212

Mk MM M

= =×


13.64. Collect and Organize In the reaction of NO2 with CO, the rate of the reaction was measured for various concentrations of both reactants. From the data we are to determine the rate law and calculate the rate constant, k. We are also to calculate the rate of appearance of CO2 when [NO2] = [CO] = 0.500 M. Analyze To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two experiments in which the concentration of that reactant changes but the concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law expression. To calculate the rate constant for the reaction, we can rearrange the rate law to solve for k and use the data from any of the experiments. Finally we can calculate the rate of the formation of a product from the concentrations of the reactants by substituting those concentrations and the calculated value of k into the rate law expression, making sure the result agrees with the stoichiometry of the reaction. Solve (a) Using experiments 1 and 2, we find that the order of the reaction with respect to CO is 0:

0,22

1 0,1

–5

–5

[CO]raterate [CO]

1.44 10 /s 0.413 0.826 1.44 10 /s

1.00 0.5000

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

× ⎛ ⎞= ⎜ ⎟× ⎝ ⎠==

x

x

x

M MMM

x

Using experiments 2 and 3, we find that the order of the reaction with respect to NO2 is 2:

2 0,33

2 2 0,2

–5

–5

[NO ]raterate [NO ]

5.76 10 /s 0.526 0.263 1.44 10 /s

4.00 2.002

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

× ⎛ ⎞= ⎜ ⎟× ⎝ ⎠==

x

x

x

M MMM

x

The rate law for this reaction is Rate = k[CO]0[NO2]2 = k[NO2]2.

68 | Chapter 13

(b) Rearranging the rate law expression to solve for k and using the data from experiment 1 gives –5

–4 –1 –12 2

2

rate 1.44 10 /s 2.08 10 s[NO ] (0.263 )

Mk MM

×= = = ×

(c) The rate of formation of CO2 when [NO2] = [CO] = 0.500 M is

Rate = 2.08×10–4

M ⋅s× (0.500 M )2 = 5.20×10–5 M /s

Think about It Because the concentration of CO has no effect on the rate of the reaction, [CO] has an exponent of zero and the carbon monoxide concentration does not appear in the rate law expression.

13.65. Collect and Organize From the data given for the concentration of NO3 over time as it decomposes to NO2 and O2, we are to calculate the value of k for this reaction.

Analyze We are given a single data set and the fact that the reaction is second order. For this second-order reaction the plot of 1/[NO3] versus time gives a straight line with a slope equal to k, the reaction rate constant.

Solve

The slope = k = 0.32 µM–1 ⋅ min–1.

Think about It The half-life of this second-order reaction is

t1/2 =1

k[NO3]0

= 10.3202 µM –1 ⋅min–1( )× 1.470×10–3µM( )

= 2124 min, or 35.4 h

13.66. Collect and Organize From the data given for the concentration of ClOO over time as it decomposes to Cl2 and O2, we are to determine the rate law and calculate the value of k for this reaction.

Analyze We are given a single data set. If a plot of [ClOO] versus time yields a straight line, the reaction is zero order. If the plot of ln[ClOO] versus time yields a straight line, the reaction is first order. If a plot of 1/[ClOO] versus time yields a straight line, the reaction is second order. The slope of the line on the appropriate graph gives the rate constant k for the reaction (slope = –k for a zero- or first-order reaction and slope = +k for a second-order reaction).

y = 0.3202x + 680.24 R² = 1

6.00E+02

6.50E+02

7.00E+02

7.50E+02

8.00E+02

8.50E+02

9.00E+02

0 100 200 300 400 500 600

1/[N

O3]

(µM

–1)

Time (min)


Solve

The first-order plot is linear with the slope = –k = –3.0, so k = 3.0 µs–1. The rate law is Rate = k[ClOO].

Think about It The half-life of this first-order reaction is

t1/2 =

0.693k

= 0.6933.0 µs–1 = 0.23 µs

13.67. Collect and Organize From the data given for concentration of NH3 over time as it decomposes to N2 and H2, we are to determine the rate law and calculate the value for k for this reaction.

Analyze We are given a single data set. If a plot of [NH3] versus time yields a straight line, the reaction is zero order. If a plot of ln[NH3] versus time yields a straight line, the reaction is first order. If a plot of 1/[NH3] versus time yields a straight line, the reaction is second order. The slope of the line on the appropriate graph gives the rate constant k for the reaction (slope = –k for a zero- or first-order reaction and slope = +k for a second-order reaction).

y = –5E–07x + 1E–06 R² = 0.58722

-5.00E-07

0.00E+00

5.00E-07

1.00E-06

1.50E-06

2.00E-06

0 0.5 1 1.5 2 2.5 3

[ClO

O] (

M)

Time (µs)

Zero-Order Plot

y = –3.0125x – 13.215 R² = 0.99987

-30 -28 -26 -24 -22 -20 -18 -16 -14 -12 -10

0 0.5 1 1.5 2 2.5 3

ln[C

lOO

] (M

)

Time (µs)

First-Order Plot

y = 8E+08x – 5E+08 R² = 0.61021

-1.00E+09

-5.00E+08

0.00E+00

5.00E+08

1.00E+09

1.50E+09

2.00E+09

2.50E+09

3.00E+09

0 0.5 1 1.5 2 2.5 3

1/[C

lOO

] (1/M

)

Time (µs)

Second-Order Plot

70 | Chapter 13

Solve

The first-order plot is linear, so that the rate law is Rate = k[NH3]. The slope = –k = –0.003 s–1 and thus k = 0.003 s–1.


t1 2 =0.693k

= 0.6930.0030 s–1 = 231 s, or 3.9 min

13.68. Collect and Organize From the data given for the concentration of HO2 over time as it decomposes to H2O2 and O2, we are to determine the rate law and calculate the value of k for this reaction.

Analyze We are given a single data set. If a plot of [HO2] versus time yields a straight line, the reaction is zero order. If a plot of ln[HO2] versus time yields a straight line, the reaction is first order. If a plot of 1/[HO2] versus time yields a straight line, the reaction is second order. The slope of the line on the appropriate graph gives the rate constant k for the reaction (slope = –k for a zero- or first-order reaction and slope = +k for a second-order reaction).

y = –3.12E–05x + 2.36E–02 R² = 9.45E–01

0.00E+00

5.00E-03

1.00E-02

1.50E-02

2.00E-02

2.50E-02

3.00E-02

0 200 400 600 800

[NH

3] (M

)

Time (s)

Zero-Order Plot

y = –3.00E–03x – 3.67E+00 R² = 1.00E+00

-6.5

-6

-5.5

-5

-4.5

-4

-3.5

-3

0 200 400 600 800

ln[N

H3]

Time (s)

First-Order Plot

y = 0.408x + 20.126 R² = 0.935

0.00 50.00

100.00 150.00 200.00 250.00 300.00 350.00 400.00

0 200 400 600 800

1/[N

H3]

Time (s)

Second-Order Plot


Solve

The first-order plot is linear with the slope = –k = –0.86, so k = 0.86 µs–1. The rate law is Rate = k[HO2].


t1/2 =

0.693k

= 0.6930.86 µs–1 = 0.81 µs

13.69. Collect and Organize For the decomposition reaction of N2O5 to NO2 and O2, we are given that the reaction is first order with a rate constant, k, equal to 6.32 ×10–4 s–1. We are to calculate the amount of N2O5 that remains after 1 hr of reaction time when the initial concentration of N2O5 is 0.50 mol/L and determine the percentage of N2O5 that has reacted.

Analyze The integrated rate law for a first-order reaction is

ln[A] = –kt + ln[A]0

Solve The concentration of N2O5 remaining after 1 hr is

ln[N2O5 ]= −kt + ln[N2O5 ]0

ln[N2O5 ]= − 6.32 ×10−4

s⎛⎝⎜

⎞⎠⎟× 1 h × 60 min

1 h× 60 s

1 min⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ + ln 0.50 mol/L( ) = −2.9683

[N2O5 ]= e−2.9683 = 0.051 mol/L

The percentage of N2O5 reacted is 0.50 mol/L − 0.0514 mol/L

0.50 mol/L×100 = 90%

y = –3.0054x + 7.3899 R² = 0.90948

0

2

4

6

8

10

0 0.5 1 1.5 2 2.5 3

[HO

2] (µM

)

Time (µs)

Zero-Order Plot

y = –0.8543x + 2.1399 R² = 0.99985

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5 3

ln[H

O2]

Time (µs)

First-Order Plot

y = 0.3225x + 0.0198 R² = 0.91219

0.00

0.20

0.40

0.60

0.80

1.00

0 0.5 1 1.5 2 2.5 3

1/[H

O2]

Time (µs)

Second-Order Plot

72 | Chapter 13

Think about It The number of half-lives needed to reduce the concentration of N2O5 to 10% can be calculated:

10100

= 0.50n

0.10 = 0.50n

ln0.10 = n ln0.50n = 3.32


For the decomposition reaction of PH3 to P4 and H2, we are given that the reaction is first order with a rate constant, k, equal to 0.023 s–1 3 that reacts in 1 min if the initial partial pressure of PH3 is 375 torr. Analyze The integrated rate law for a first-order reaction is

ln[A] = –kt + ln[A]0 Solve The partial pressure of PH3 remaining after 1 min is

ln[PH3]= −kt + ln[PH3]0

ln[PH3]= − 0.023s

⎛⎝⎜

⎞⎠⎟ × 60 s⎡

⎣⎢⎤⎦⎥+ ln 375 torr( ) = −4.547

[PH3]= e−4.547 = 94.3 torr

The percentage of PH3 that reacts in 1 min is 375 torr − 94.3 torr

375 torr×100 = 75%

Think about It The number of half-lives needed to reduce the concentration of N2O5 to 10% can be calculated:

25100

= 0.50n

0.25 = 0.50n

ln0.25 = n ln0.50n = 2.0


For the decomposition reaction of N2O to N2 and O2, we are given that the plot of ln[N2O] versus time is linear. We are to write the rate law and then determine the number of half-lives it would take for the concentration of N2O to become 6.25% of its original concentration. Analyze The integrated rate laws for zero-, first-, and second-order reactions with their half-life equations are as follows:

[A] = –kt + [A]0 zero order t1/2 = [A]0/2k ln[A] = –kt + ln[A]0 first order t1/2 = 0.693/k 1/[A] = kt + 1/[A]0 second order t1/2 = 1/k[A]0

Solve (a) Since we are given that the plot of ln[N2O] versus time is linear, the reaction is first order and the rate law is Rate = k[N2O].


(b) The number of half-lives needed to reduce the concentration to 6.25% would be, where n = number of half-lives,

6.25 0.50100

0.0625 = 0.50ln 0.0625 ln 0.50

4

n

n

nn

=

==

Think about It The number of half-lives needed to reduce the concentration of a reactant to products is not dependent on the order of the reaction or on the magnitude of the rate constant.

13.72. Collect and Organize From the information given about the linearity of the plots of [C4H6], ln[C4H6], and 1/[C4H6] versus time, we can determine the rate law for the dimerization of C4H6 and then calculate the number of half-lives required for the concentration of C4H6 to be reduced to 3.1% of its original concentration. Analyze The integrated rate laws for zero-, first-, and second-order reactions with their half-life equations are as follows:

[A] = –kt + [A]0 zero order t1/2 = [A]0/2k ln[A] = –kt + ln[A]0 first order t1/2 = 0.693/k 1/[A] = kt + 1/[A]0 second order t1/2 = 1/k[A]0

Solve (a) Since we are given that the plot of 1/[C4H6] versus time is linear, the reaction is second order and the rate law is Rate = k[C4H6]2. (b) The number of half-lives needed to reduce the concentration of C4H6 to 3.1% would be, where n = number of half-lives,

3.1 0.50100

0.031 = 0.50ln 0.031 ln 0.50

5.0

n

n

nn

=

==

Think about It The number of half-lives needed to reduce the concentration of a reactant to a certain amount of product is not dependent on the order of the reaction or on the magnitude of the rate constant.

13.73. Collect and Organize From the data given for the concentration of 32P over time, we are to determine the rate law and calculate the value for k for this radioactive decay. Analyze Radioactive decay follows first-order kinetics. This is confirmed in part b, where we are told to determine the first-order rate constant. A plot of ln[32P] versus time for this decay gives a straight line with slope = –k. The half-life of a first-order decay is given by

1/ 20.693tk

=

74 | Chapter 13

Solve (a)

The rate law for this radioactive decay is Rate = k[32P].(b) k = –slope = 0.0485 day–1

(c) 1/ 2 –1

0.693 14.3 days0.0485 day

t = =

Think about It The time needed for [32P] to reduce to 1.00% of its original concentration, where n = number of half-lives, is

n

1.00 0.50100

0.0100 0.50ln 0.0100 ln 0.50

6.64 half-lives

n

nn

=

===

The number of days is 1/ 2 6.64 14.3 days 95 days.n t× = × =

13.74. Collect and Organize From the data given for the second-order reaction that follows the concentration of HNO2 over time as it decomposes to NO, NO2, and H2O, we are to calculate the value of k and the half-life for this reaction.

Analyze Because this reaction is second order in HNO2, when we plot the concentration of HNO2 versus time according to the second-order integrated rate equation,

1/[A] = kt + 1/[A]0 the slope of the line is equal to k and the half-life will be t1/2 = 1/k[A]0.

Solve (a) The rate law for this second-order reaction is Rate = k[HNO2]2. A plot of the reciprocal of the concentration of HNO2 versus time for this second-order reaction gives the slope = k = 4.097 × 10–4 µM–1 min–1.

y = –0.0485x + 2.303!R² = 1!

1 1.2 1.4 1.6 1.8

2 2.2 2.4

0 5 10 15 20 R

adio

activ

ity (l

n pC

i) Time (d)

y = 4.097E–04x + 6.410E+00 R² = 1.000E+00

6

6.5

7

7.5

8

0 1000 2000 3000 4000

1/[H

NO

2] (µ

M–1

)

Time (min)


(b) The half-life of this reaction is

t1/2 =

1k[A]0

= 1(4.097 ×10–4 µM –1 min–1)× (0.1560 µM )

= 1.565×104 min, or 10.87 d

Think about It The half-life of a second-order reaction decreases as the initial concentration of the reactant increases. For an initial concentration of 0.312 µM for this particular reaction, the half-life is

t1/2 =

1k[A]0

= 1(4.097 ×10–4 µM –1 min–1)× (0.3120 µM )

= 7.823×103 min, or 5.433d

13.75. Collect and OrganizeGiven that the dimerization of ClO to Cl2O2 is second order, we are to determine the value of the rate constant and to calculate the half-life of this reaction.

Analyze We are given a single data set and knowledge that the reaction is second order. For this second-order reaction a plot of 1/[ClO] versus time gives a straight line with a slope equal to k, the reaction rate constant. The half-life of this second-order reaction is

1/ 20

1[ClO]

tk

=

where [ClO]0 is the initial concentration of ClO used in the reaction.

Solve

The value of k for this reaction is k = slope = 5.40 × 10–12 cm3 molecules–1 s–1, and the half-life of the reaction is as follows:

t1/2 =1

k[ClO]0

= 15.40×10−12cm3

molecules ⋅s× 2.60×1011 molecules

cm3

= 0.712 s

Think about It For a second-order reaction, as we increase the initial concentration of the reactant the half-life gets shorter.

13.76. Collect and Organize From the data given for the concentration of Cl2O2 over time as it decomposes to ClO and the fact that the reaction is first order, we can calculate the rate constant and the half-life for the reaction.

y = 5.398E–12x + 3.851E–12 R² = 1.000E+00

0.00E+00

5.00E-12

1.00E-11

1.50E-11

2.00E-11

2.50E-11

3.00E-11

0 1 2 3 4 5

1/[C

lO],

(mol

ecul

es/c

m3 )

–1

Time (s)

76 | Chapter 13

Analyze For a first-order reaction, the slope of the line of the linear plot of ln[A] versus time equals –k. Once we have the value of k, the half-life may be calculated from

1/ 20.693tk

=

Solve

Because the slope = –8.72 × 10–4, k = 8.72 × 10–4 µs–1. The half-life of the reaction is

t1/2 =

0.6938.72×10–4 µs–1 = 795 µs

Think about It From this information we could calculate how long it might take, for example, for the concentration of Cl2O2 to decrease to 15% of its original concentration. The integrated rate law for first-order reactions is

ln[A] = –kt + ln[A]0 where, for this example, [A] = 15, [A]0 = 100, and k = 8.715 × 10–4 µs–1:

ln(15) = –(8.715 × 10–4 µs–1 × t) + ln(100) t = 2.18 × 103 µs

13.77. Collect and Organize From the data for the pseudo-first-order hydrolysis of sucrose provided, we are to write the rate law and determine the value of the pseudo-first-order rate constant, k′.

Analyze To obtain the value of k′, we plot ln[sucrose] over time. The slope of the line is equal to –k.

Solve The rate law is Rate = k[C12H22O11][H2O] = k′[C12H22O11].

The pseudo-first-order plot gives k′ = –slope = 5.19 × 10–5 s–1.

y = –8.715E–04x – 1.653E+01 R² = 1.00E+00

-17.4

-17.2

-17

-16.8

-16.6

-16.4

0 200 400 600 800 1000

ln[C

l 2O2]

(ln M

)

Time (µs)

y = –5.190E–05x – 5.878E–01 R² = 9.726E-01

-0.9

-0.8

-0.7

-0.6

-0.5

0 1000 2000 3000 4000 5000 6000

ln[s

ucro

se] (

ln M

)

Time (s)


Think about It If we knew the concentration of water in the hydrolysis reaction, we could calculate the value of k by using

2 0[H O]k

k′

=

13.78. Collect and Organize For the reaction of HO2 with O3, we are asked to determine the pseudo-first-order rate constant given the experimental data for [HO2] and [O3] over time where the [O3] is large and therefore remains constant throughout the reaction. Because we are given the actual [O3], we can also calculate the second-order rate constant for the reaction.

Analyze To determine the pseudo-first-order rate constant, we treat the data for [HO2] versus time as a first-order reaction (plot ln[HO2] versus t), which gives a straight line with slope = k' (the pseudo-first-order rate constant). The second-order rate constant can be calculated from the known experimental concentration of ozone and the value of the pseudo-first-order rate constant:

k' = k[O3], or k = k'/[O3]

Solve The first-order plot gives k' = –slope = 1.03 × 10–2 ms–1.

The second-order rate constant, k, is

k = 1.03×10–2 ms–1

1.0×10–3 M=10.3 M –1 ms–1, or 1.0×104 M –1 s–1

Think about It When a second-order reaction is run so that one reactant is in large excess, the reaction appears to obey first-order kinetics for the other reactant. By this method, the rate constants of reactions involving more than one reactant can be experimentally determined.

13.79. Collect and Organize We are asked to explain how the magnitude of the activation energy for a reaction is related to the rate of the reaction.

Analyze The activation energy is the minimum amount of energy required for colliding molecules to react.

Solve The greater the activation energy, the more energy required for the reaction to occur and the slower the reaction. This is because the higher the activation energy, fewer molecules will possess the minimum energy required for reaction. With fewer molecules reacting, the slower the reaction.

y = –1.034E–02x – 1.265E+01 R² = 9.991E–01

-13.6

-13.4

-13.2

-13

-12.8

-12.6

0 20 40 60 80 100

ln[H

O2]

(ln M

)

Time (ms)

78 | Chapter 13

Think about It A reaction with a lower activation energy is faster than one that is higher at the same temperature.

13.80. Collect and Organize We are to explain whether all spontaneous reactions happen instantly at room temperature. Analyze In a spontaneous reaction, the free energy of the products is lower than the free energy of the reactants. In an instantaneous reaction the activation energy is near zero. Solve No, just because a reaction is spontaneous (negative free energy), it does not mean that the reaction is instantaneous (or even fast). A spontaneous reaction can be either fast (low activation energy) or quite slow (large activation energy).

Think about It A very fast, spontaneous reaction can be an explosion.

13.81. Collect and Organize We are asked to describe the circumstances for a reaction in which the activation energy of the forward reaction is less than that of the reverse reaction. Analyze For the activation energy of the forward reaction to be less than that of the reverse reaction, the energy of the reactants must be closer to the energy of the transition state than the energy of the products. Solve For the activation energy of the forward reaction to be less than the activation energy of the reverse reaction, the energy of the products must be lower than the energy of the reactants and therefore the reaction is spontaneous (–ΔG).


Think about It Just because a reaction is favored thermodynamically does not necessarily mean that it is fast. This reaction could have a quite high activation energy in the forward direction and therefore be quite slow.

13.82. Collect and Organize We are asked to describe the circumstances for a reaction in which the activation energy of the forward reaction is greater than that of the reverse reaction. Analyze For the activation energy of the forward reaction to be greater than that of the reverse reaction, the energy of the products must be closer to the energy of the transition state than the energy of the reactants. Solve For the activation energy of the forward reaction to be greater than the activation energy of the reverse reaction, the energy of the reactants must be lower than the energy of the products and therefore the reaction is nonspontaneous (+ΔG).

Think about It Just because a reaction is not favored thermodynamically under standard conditions does not necessarily mean that it does not proceed at all under nonstandard conditions. We will look at this more carefully when studying equilibrium and thermodynamics (Section 14.9).

13.83. Collect and Organize We are to explain why the order of a reaction is independent of temperature, yet the value of k changes with temperature. Analyze We need to consider how temperature affects the motion and collision of the reactants. Solve An increase in temperature increases the frequency and the kinetic energy at which the reactants collide. This speeds up the reaction, changing the value of k. The activation energy of the slowest step in the reaction, however, is not affected by a change in temperature and, therefore, the order of the reaction is unaffected. Think about It As a general

13.84. Collect and Organize For a reduction in the activation energy by ½, we are to assess whether that will increase the value of the rate constant by a factor of 2. Analyze The rate constant is related to the activation energy by the equation

k = Ae−Ea /RT

80 | Chapter 13

Solve In the equation

k = Ae−Ea /RT if A and T are held constant (R is constant as well), k will change by a factor of e–1/2 when the activation energy is halved. The value of e–1/2 is 0.6065. For the value of k to double, e–1/2 would have to equal 2. Therefore, halving the activation energy would not double the value of k. Think about It The rate of the reaction is increased, as the value of the rate constant is reduced by a factor of approximately 0.61. In this problem we would not expect the value of the rate constant to double (meaning that the reaction is slower) if we reduce the activation energy.

13.85. Collect and Organize In comparing two first-order reactions with different activation energies, we are to decide which would show a larger increase in its rate as the reaction temperature is increased. Analyze We can use the Arrhenius equation to mathematically determine which reaction would be most accelerated by an increase in temperature:

aln ln –E

k ART

=

Solve Let’s assume that T2 = 2T1. For either reaction the difference in the rate constants is as follows:

1 2

2 1

a a

1 2

a a a

2 1 1 2

ln ln – ln ln –

– 1 1ln – ln

= =

⎛ ⎞= + = −⎜ ⎟

⎝ ⎠

T T

T T

E Ek A k ART RT

E E Ek kRT RT R T T

But because T2 = 2T1,

2 1

a a

1 1 1

1 1ln – ln2 2

⎛ ⎞= − =⎜ ⎟

⎝ ⎠T T

E Ek kR T T RT

For Ea = 150 kJ/mol,

2 11

150 kJ/molln – ln2T Tk kRT

=

For Ea = 15 kJ/mol,

2 11

15 kJ/molln – ln2T Tk kRT

=

Comparing these as a ratio,

2 1

2 1

a 1

a

1

150 kJ/molln – ln for 150 kJ/mol 2

1015 kJ/molln – ln for 15 kJ/mol2

T T

T T

k k E RTk k E

RT

== =

=

Therefore, the reaction with the larger activation energy (150 kJ/mol) would be accelerated more than the reaction with the lower activation energy (15 kJ/mol) when heated. Think about It Our derivation demonstrates that different reactions with different activation energies will not accelerate in the same way when they are heated.


13.86. Collect and Organize By looking at the Arrhenius equation, we can determine whether Ea depends on temperature. Analyze The Arrhenius equation is

aln ln –E

k ART

=

Solve No. From the form of the equation, we see a dependence of k on temperature because the equation is in the form of y = mx + b, where y = ln k and x = 1/T. The rate constant, not Ea, changes with temperature. Think about It The Arrhenius equation was actually first proposed by Jacobus van ’t Hoff, but Svante Arrhenius interpreted it and introduced the idea of an activation energy.

13.87. Collect and Organize We can use an Arrhenius plot of rate constant versus temperature for the reaction of O and O3 to determine the activation energy (Ea) and the value of the frequency factor (A). Analyze The Arrhenius equation is

a– 1ln lnE

k AR T

⎛ ⎞= +⎜ ⎟⎝ ⎠

If we plot ln k (y-axis) versus 1/T (x-axis), we obtain a straight line with slope m = –Ea/R. The activation energy is therefore calculated from

Ea = –slope × R where R = 8.314 J/mol ⋅ K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or eb = A. Solve The Arrhenius plot gives a slope of –2060 and a y-intercept of 0.00160.

Ea = –(–2060 K × 8.314 J/mol ⋅ K) = 1.713 × 104 J/mol, or 17.1 kJ/mol

b = 0.00160 = ln A A = e0.00160 = 1.002

Think about It Once we have the values of Ea and A from the plot, we can calculate the value of k at any temperature.

13.88. Collect and Organize We can use an Arrhenius plot of rate constant versus temperature for the reaction of NO2 and O3 to determine the activation energy (Ea) and to find the rate constant for the reaction at 300 K.

82 | Chapter 13

Analyze The Arrhenius equation is

a– 1ln lnE

k AR T

⎛ ⎞= +⎜ ⎟⎝ ⎠


Ea = –slope × R where R = 8.314 J/mol ⋅ K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or eb = A. Once Ea and A are known, we may use another form of the Arrhenius equation to calculate k at any temperature:

aexp –⎛ ⎞= ⎜ ⎟⎝ ⎠Ek ART

Solve The Arrhenius plot gives a slope of –2450.6 and a y-intercept of 25.005.

(a) Ea = –(–2450.6 K × 8.314 J/mol ⋅ K) = 2.04 × 104 J/mol, or 20.4 kJ/mol (b) To calculate k at 300 K we first need the value of A:

A = e25.005 = 7.237 × 1010 The value of k at 300 K is

410 7 –1 –12.04 10 J/mol7.237 10 exp – 2.03 10 s

8.314 J/mol K 300 K⎛ ⎞×= × = ×⎜ ⎟⋅ ×⎝ ⎠

k M

Think about It Alternatively, k can be calculated from the original Arrhenius equation:

4a

7 –1 –1

– –2.04 10 J/molln ln 25.005 16.838.314 J/mol K 300 K

2.04 10 s

Ek A

RTk M

×= + = + =

⋅ ×= ×


We can use an Arrhenius plot of rate constant versus temperature for the reaction of N2 with O2 to form NO to determine the activation energy (Ea), the frequency factor (A), and the rate constant for the reaction at 300 K. Analyze The Arrhenius equation is

a– 1ln lnE

k AR T

⎛ ⎞= +⎜ ⎟⎝ ⎠


Ea = –slope × R


where R = 8.314 J/mol ⋅ K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or eb = A. Once Ea and A are known, we may use another form of the Arrhenius equation to calculate k at any temperature:

aexp –E

k ART

⎛ ⎞= ⎜ ⎟⎝ ⎠

Solve The Arrhenius plot gives a slope of –37758 and a y-intercept of 24.641.

(a) Ea = –(–37758 K × 8.314 J/mol ⋅ K) = 3.14 × 105 J/mol, or 314 kJ/mol (b) A = e24.641 = 5.03 × 1010

510 –44 –1/2 –13.14 10 J/mol(c) 5.03 10 exp 1.06 10 s

8.314 J/mol K 300 K⎛ ⎞×= × − = ×⎜ ⎟⋅ ×⎝ ⎠

k M

Think about It Alternatively, k can be calculated from the original Arrhenius equation:

5a

101.25 44 –1/ 2 –1

– –3.14 10 J/molln ln 24.6418.314 J/mol K 300 K

101.251.07 10 s

Ek A

RT

k e M− −

×= + = +

⋅ ×= −= = ×


We can use an Arrhenius plot of rate constant versus temperature for the decomposition of N2O5 to determine the activation energy (Ea) and the rate constant for the reaction at 300 K. Analyze The Arrhenius equation is

a– 1ln lnE

k AR T

⎛ ⎞= +⎜ ⎟⎝ ⎠


Ea = –slope × R where R = 8.314 J/mol ⋅ K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or eb = A. Once Ea and A are known, we may use another form of the Arrhenius equation to calculate k at any temperature:

aexp –E

k ART

⎛ ⎞= ⎜ ⎟⎝ ⎠

Solve The Arrhenius plot gives a slope of –11781 and a y-intercept of 30.164.

84 | Chapter 13

(a) Ea = –(–11781 K × 8.314 J/mol ⋅ K) = 9.79 × 104 J/mol, or 97.9 kJ/mol (b) To calculate k at 300 K we first need the value of A

A = e30.164 = 1.26 × 1013 The value of k at 300 K is

413 –4 –19.79 10 J/mol1.26 10 exp – 1.13 10 s

8.314 J/mol K 300 K⎛ ⎞×= × = ×⎜ ⎟⋅ ×⎝ ⎠

k

Think about It This reaction will be relatively slow, as k < 1.

13.91. Collect and Organize We can use an Arrhenius plot of rate constant versus temperature for the reaction of ClO2 and O3 to determine the activation energy (Ea) and the value of the frequency factor (A). Analyze The Arrhenius equation is

a– 1ln lnE

k AR T

⎛ ⎞= +⎜ ⎟⎝ ⎠


Ea = –slope × R where R = 8.314 J/mol ⋅ K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or eb = A. Solve The Arrhenius plot gives a slope of –4698.7 and a y-intercept of 27.872.

Ea = –(– 4698.7 K × 8.314 J/mol ⋅ K) = 3.91 × 104 J/mol, or 39.1 kJ/mol b = 27.872 = ln A

A = e27.872 = 1.27 × 1012


Think about It Once we have the values of Ea and A from the plot, we can calculate the value of k at any temperature.

13.92. Collect and Organize Given the value of the rate constant for the reaction of Cl with CH4 to form HCl and CH3 at 298 K and at three additional temperatures, we can use an Arrhenius plot to calculate the activation energy (Ea) and the frequency factor (A) for the reaction. Analyze The Arrhenius equation is

a– 1ln lnE

k AR T

⎛ ⎞= +⎜ ⎟⎝ ⎠


Ea = –slope × R where R = 8.314 J/mol ⋅ K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or eb = A. Solve The Arrhenius plot gives a slope of –1387.5 and a y-intercept of 22.567.

Ea = –(–1387.5 K × 8.314 J/mol ⋅ K) = 1.15 × 104 J/mol, or 11.5 kJ/mol A = e22.567 = 6.32 × 109

Think about It The rate constant for this reaction is fairly large, so we expect that the reaction will proceed quickly to products.

13.93. Collect and Organize , we are to

mperature, the rate of a reaction doubles. Analyze We can use the form of the Arrhenius equation relating the rate constants for a reaction at different temperatures:

ln

k1

k2

=Ea

R1T2

− 1T1

⎛

⎝⎜⎞

⎠⎟

Solve For the equation we use k2 = 2.00 × 108 s–1, T2 T1 Ea = 2.14 × 104 J/mol with R = 8.314 J/mol ⋅ K.

86 | Chapter 13

lnk1

2.00×108 s−1 =2.14×104 J/mol8.314 J/mol ⋅K

1873 K

− 1883 K

⎛⎝⎜

⎞⎠⎟= 0.03391

k1

2.00×108 s−1 = e0.03391 = 1.034

k1 = 1.034× 2.00×108 s−1( ) = 2.07 ×108 s−1

The ratio of these rate constants is k610 Ck600 C

= 2.07 ×108s−1

2.00 ×108s−1= 1.04

does not apply to this temperature domain. This rule applies best when the reaction temperatures are near 298K. Think about It This problem illustrates well how general rules do not apply to every situation, so be cautious in using general rules to make predictions.

13.94. Collect and Organize For the decomposition of CH3CHF2 to CH2CHF and HF we are to determine to what temperature we would

k is 5.8 × 10–6 s–1 and the activation energy is 265 kJ/mol. Analyze We can use the form of the Arrhenius equation relating the rate constants for a reaction at different temperatures to solve this problem.

ln

k1

k2

=Ea

R1T2

− 1T1

⎛

⎝⎜⎞

⎠⎟

Solve For the equation we can use the ratio of k1/k2 = 4 with T2 Ea = 2.65 × 105 J/mol with R = 8.314 J/mol ⋅ K.

ln41= 2.65×105 J/mol

8.314 J/mol ⋅K1

733 K− 1

T1

⎛

⎝⎜⎞

⎠⎟

ln4× 8.314 J/mol ⋅K2.65×105 J/mol

= 1733 K

− 1T1

⎛

⎝⎜⎞

⎠⎟= 4.3493×10−5 K–1

4.3493×10−5 K–1 − 1733 K

= − 1T1

= −1.3207 ×10−3 K–1

T1 = 757 K

Think about It This is only a increase in temperature to make the reaction go four times as fast.

13.95. Collect and Organize By comparing the rate laws for two reactions, we can determine whether their mechanisms are similar. Analyze For the reaction between NO and H2,

Rate = k[NO]2[H2] For the reaction between NO and Cl2,

Rate = k[NO][Cl2]


Solve No. The different rate laws for the two reactions indicate different mechanisms. Think about It However, even if two rate laws were similar, this would not necessarily mean that their reaction mechanisms are similar.

13.96. Collect and Organize For two reactions, one of NO with Cl2 and the other of NO2 with F2, we consider whether their identical rate laws mean that they might have similar mechanisms. Analyze No difference exists in the forms of the two rate laws given. Both are first order in the two reactants, NO or NO2 and X2, and both are second order overall. Solve Yes, when the rate laws are the same, the reactions can have similar mechanisms. Think about It Be careful here. Reactions with the same form of the rate laws may proceed by very different mechanisms.

13.97. Collect and Organize / Analyze We are to identify the conditions under which a bimolecular reaction shows pseudo-first-order behavior. Solve Pseudo-first-order kinetics occurs when one reactant is in sufficiently high concentration that it does not change appreciably during the reaction. Think about It We solved problems relating to pseudo-first-order reactions earlier in this chapter (Problems 13.77 and 13.78).

13.98. Collect and Organize For a reaction that is zero order in a reactant, we are to determine whether that means the reactant is not involved in any collisions with other reactants during the reaction. Analyze When a reactant is zero order it does not appear in the overall rate equation. Solve No. The reactant may be used and therefore must collide with other reactants, but it is not involved in the rate-determining step of the reaction. Think about It The rate law expresses only the concentration dependence of those reactants (and intermediates) involved in the slowest (rate determining) step of the mechanism.

13.99. Collect and Organize We are asked to draw reaction profiles that fit the reaction A → B (which has Ea = 50.0 kJ/mol) for three different mechanisms: (a) a single elementary step, (b) a two-step reaction with Ea,step 2 = 15 kJ/mol, and (c) a two-step reaction in which the second step is rate determining. Analyze The reaction profile plots energy versus reaction progress and shows the presence of an activated complex at highest energy, the reaction intermediates in the “valleys” between reactant and products, and the relative activation energies in multistep reactions. The rate-determining step is the slowest step in the mechanism and has the largest activation energy in the reaction profile.

88 | Chapter 13

Solve

Think about It These reaction profiles could also be drawn for a nonspontaneous reaction. Reactant A would then be lower in energy than product B.

13.100. Collect and Organize We are asked to draw reaction profiles that fit the reaction A + B → C → D + E for three different mechanisms. Analyze The reaction profile plots energy versus reaction progress and shows the presence of an activated complex at highest energy, the reaction intermediates in the “valleys” between reactant and products, and the relative activation energies in multistep reactions. The rate-determining step is the slowest step in the mechanism and has the largest activation energy in the reaction profile. Solve (a) When C is an activated complex:

(b) When the first step is rate determining and C is an intermediate:


(c) When the second step is rate determining and C is an intermediate:

Think about It These reaction profiles could also be drawn for a nonspontaneous reaction. For example, when the first step of a two-step reaction is rate determining with C as the intermediate, the reaction profile would look like this:

13.101. Collect and Organize For each elementary step given, we are to write the rate law and determine whether the step is uni-, bi-, or termolecular. Analyze The rate law for an elementary step in a mechanism is written in the form

Rate = k[A]x[B]y[C]z

where A, B, and C are the reactants involved in the elementary reaction and x, y, and z are the stoichiometric coefficients for the respective reactants in the elementary reaction. Solve (a) Rate = k[SO2Cl2]. Because this elementary step involves only a molecule of SO2Cl2, it is unimolecular. (b) Rate = k[NO2][CO]. Because this elementary step involves a molecule of NO2 and a molecule of CO, it is bimolecular. (c) Rate = k[NO2]2. Because this elementary step involves two molecules of NO2, it is bimolecular. Think about It Termolecular elementary reactions are rare.

13.102. Collect and Organize For each elementary step given, we are to write the rate law and determine whether the step is uni-, bi-, or termolecular. Analyze The rate law for an elementary step in a mechanism is written in the form

Rate = k[A]x[B]y[C]z

where A, B, and C are the reactants involved in the elementary reaction and x, y, and z are the stoichiometric coefficients for the respective reactants in the elementary reaction.

90 | Chapter 13

Solve (a) Rate = k[Cl][O3]. Because this elementary step involves a Cl atom and an O3 molecule, it is bimolecular. (b) Rate = k[NO2]2. Because this elementary step involves two NO2 molecules, it is bimolecular. (c) Rate = k[14C]. Because this elementary step involves only a 14C atom, it is unimolecular. Think about It Termolecular elementary reactions are rare.

13.103. Collect and Organize From three elementary steps that describe a reaction mechanism, we are to write the overall chemical equation. Analyze To write the overall chemical reaction we need to add the elementary steps, being sure to cancel the intermediates in the reaction. Solve

2 5 3 N O ( ) NO ( )→g g 2

3

NO ( ) NO ( )

+ gg 2NO ( ) O( )→ +g g

2 2

2 5 2 2

O( ) O ( )

N O ( ) O( ) 2 NO ( ) O ( )

→

+ → +

g g

g g g g

Think about It In this reaction, NO3 is a reaction intermediate. It is generated in the reaction but consumed in a subsequent step in the mechanism.

13.104. Collect and Organize From three elementary steps that describe a reaction mechanism, we are to write the overall chemical equation. Analyze To write the overall chemical reaction we need to add the elementary steps, being sure to cancel the intermediates in the reaction. Solve

ClO– (aq) + H2O() → HClO(aq) + OH– (aq)

I– (aq) + HClO(aq) → HIO(aq) +Cl– (aq)OH– (aq) + HIO(aq) → H2O() + IO– (aq)

ClO– (aq) + I– (aq)→ Cl– (aq) + IO– (aq)

Think about It In this reaction HClO, OH–, and HIO are reaction intermediates. They are generated in the reaction but consumed in subsequent steps in the mechanism.

13.105. Collect and Organize We are given the mechanism by which N2 reacts with O2 to form NO. For a given rate law of

Rate = k[N2][O2]1/2 we are to determine which step in the mechanism is the rate-determining step. Analyze To determine which step in the proposed mechanism might be the slowest, we can write the rate law for the mechanism when the first, second, or third step is slow and then match the theoretical rate law to the experimental rate law.


Solve If the first step is slow, the rate law is

Rate = k1[O2] This does not match the experimental rate law, so the first step is not the slowest step in the mechanism. If the second step is slow, the rate law is

Rate = k2[O][N2] Because O is an intermediate, we use the first step to express its concentration in terms of concentrations of the reactants. For a fast step occurring before a slow step in a mechanism,

Rateforward = Ratereverse k1[O2] = k–1[O]2

Rearranging to solve for [O],

[ ]1/ 2

12

–1

[O] O⎛ ⎞

= ⎜ ⎟⎝ ⎠

kk

Substituting this into the rate law from the second step gives

Rate = k2

k1

k–1

O2⎡⎣ ⎤⎦⎛

⎝⎜⎞

⎠⎟

1/2

N2⎡⎣ ⎤⎦ = k O2⎡⎣ ⎤⎦1/2

N2⎡⎣ ⎤⎦

This rate law matches the experimental rate law, so the rate-determining step is the second step. We should check to see whether the mechanism of the third step is the slow step and might also give the experimental rate law. From the logic above,

Rate = k3[N][O] From the second fast step in the mechanism,

k2[O][N2] = k–2[NO][N] Solving for [N], an intermediate, gives

[ ][ ][ ]

2 2

–2

O N[N]

NOkk

=

From the first fast step in the mechanism,

k1[O2] = k–1[O]2 solving for [O]2 gives

[ ]2 12

–1

[O] Okk

=

Substituting these expressions into the rate law from the third step in the mechanism gives

Rate = k3

k2

k–2

k1

k–1

[O2][N2]

[NO]

=k[N2][O2]

[NO]

This rate law does not match the experimental rate law. Think about It Just because the rate law for a mechanism matches the experimental rate law does not mean that the mechanism is the correct mechanism. Another mechanism might also give the same experimental rate law.

13.106. Collect and Organize Given a proposed mechanism for the decomposition of H2O2, we are to determine which step is the rate-determining (slow) step if the reaction is known by experiment to be first order in H2O2. Analyze The slowest step in a mechanism determines the rate law. From each elementary reaction we can write a rate law expression.

92 | Chapter 13

Solve If the first step is the rate-determining step, the rate law is

Rate = k1[H2O2] This is consistent with the reaction being first order in the concentration of H2O2, so the first step is likely to be the rate-determining step. Think about It If the second step were rate determining, the rate law would be

Rate = k2[H2O2][OH] This is also first order in H2O2 but is also first order in OH concentration. OH in this reaction is a reaction intermediate. Because the first step is fast and reversible, we can write

kf[H2O2] = kr[OH]2 Rearranging this to solve for the concentration of OH, the reaction intermediate, gives

1 2

f2 2

r

[OH] [H O ]⎛ ⎞

= ⎜ ⎟⎝ ⎠

kk

Substituting this concentration of OH into the rate law expression for the rate-determining step gives

Rate = k2[H2O2]×

kf

kr

[H2O2]⎛

⎝⎜⎞

⎠⎟

1 2

= k2

kf

kr

⎛

⎝⎜⎞

⎠⎟

1 2

[H2O2]3/2

Thus, the rate law derived from the mechanism where the second step is rate determining is inconsistent with the experimentally observed rate law.

13.107. Collect and Organize We are given the mechanism by which NO reacts with Cl2 to produce NOCl2. For a given rate law of

Rate = k[NO][Cl2] we are to determine which step in the mechanism is the rate-determining step. Analyze To determine which step in the proposed mechanism might be the slowest, we can write the rate law for the mechanism when the first, second, or third step is slow and then match the rate law to the experimental rate law. Solve If the first step is slow, the rate law is

Rate = k1[NO][Cl2] This matches the experimental rate law, so the first step is the rate-determining step. We should check to see whether the rate law for the mechanism with the second step as the slow step,

Rate = k2[NOCl2][NO] might also give the experimental rate law. If the second step is slow, then

Rate1 = Rate–1 k1[NO][Cl2] = k–1[NOCl2]

[ ][ ]12 2

–1

[NOCl ] NO Clkk

=

Substituting this into the rate law expression gives

Rate = k2

k1

k–1

NO⎡⎣ ⎤⎦ Cl2⎡⎣ ⎤⎦⎛

⎝⎜⎞

⎠⎟NO⎡⎣ ⎤⎦ = k NO⎡⎣ ⎤⎦

2Cl2⎡⎣ ⎤⎦

This does not match the experimental rate law, so the second step in the mechanism is not the rate-determining step. Think about It Just because the rate law for a mechanism matches the experimental rate law does not mean the mechanism is the correct mechanism. Another possible mechanism might also give the same experimental rate law.


13.108. Collect and Organize Given a proposed mechanism for the thermal destruction of ozone, we are to consider the relative rates and reversibility of the elementary steps to determine what properties are consistent with the experimental observation that the reaction is second order in O3. Analyze The rate law is determined by the slowest step in the mechanism. We can write the rate expressions for the two possibilities—the first step as slow or the second step as slow—and then consider the consistency of those two possible rate laws with the observation that the reaction is second order in ozone. Solve If the first step is slow,

Rate = k[O3] This is only first order in O3, so the first step as the rate-determining step in the mechanism is inconsistent with the experimental observation for this reaction. If the second step is slow, then the first step is fast and reversible and the rate of the forward reaction is equal to the rate of the reverse reaction:

kf[O3] = kr[O][O2] When the second step is rate determining, the rate law expression will be

Rate = k2[O][O3] Combining these equations by replacing [O] in the rate law expression for the second slow step with

f 3

r 2

[O ][O]

[O ]kk

=

obtained from the reversible first step of the mechanism gives

Rate =

kf k2[O3]2

kr[O2]

This rate law is consistent with the observation that the reaction is second order in O3. Think about It The rate law for this reaction also shows an inverse dependence of the reaction rate on the concentration of O2. This means that if the concentration of O2 increases, the reaction slows down.

13.109. Collect and Organize From the mechanisms given, we are to determine which are possible for the thermal decomposition and which are possible for the photochemical decomposition of NO2. We are given the rate laws: for the thermal decomposition reaction, Rate = k[NO2]2; for the photochemical decomposition, Rate = k[NO2]. Analyze Using the slowest elementary step in the mechanism, we can write the rate law expression for each mechanism and then determine which is consistent with the order of the reaction given for each process. Solve For mechanism a, the first step in the mechanism is slow, so the rate law is

Rate = k[NO2] For mechanism b, the second step in the mechanism is slow, so the rate law is

Rate = k2[N2O4] Using the first step to express [N2O4] in terms of the concentrations of the reactant NO2 gives

k1[NO2]2 = k–1[N2O4]

[ ]212 4 2

–1

[N O ] NOkk

=

94 | Chapter 13

Substituting into the rate expression from the second step,

Rate =

k2k1

k–1

NO2⎡⎣ ⎤⎦2= k NO2⎡⎣ ⎤⎦

2

For mechanism c, the first step in the mechanism is slow, so the rate law is Rate = k[NO2]2

Therefore, mechanisms b and c are consistent with the thermal decomposition of NO2 and mechanism a is consistent with the photochemical decomposition of NO2. Think about It To distinguish between the two possible mechanisms for thermal decomposition, we might try to detect the different intermediates formed in each. Detection of the formation of N2O4 would support mechanism b over mechanism c.

13.110. Collect and Organize From the possible mechanisms, we are to determine which occurs for the thermal decomposition and which for the photochemical decomposition of NO2. We are given that the thermal decomposition reaction is second order in NO2 and that the photochemical decomposition is first order in NO2. Analyze Using the slowest elementary step in the mechanism, we can write the rate law expression for each mechanism and then determine which is consistent with the order of the reaction given for each process. Solve Because the first step in each mechanism is the rate-determining step, the rate laws are as follows: (a) Rate = k[NO2]2 (b) Rate = k[NO2]2

(c) Rate = k[NO2] Therefore, mechanisms a and b are consistent with the thermal decomposition process and mechanism c is consistent with the photochemical decomposition process. Think about It To distinguish between the two possible mechanisms for thermal decomposition, we might try to detect the different intermediates formed in each. The detection of N2O4, N2O3, or N2O2 would support mechanism a, whereas the detection of NO3 would support mechanism b.

13.111. Collect and Organize We are asked whether a catalyst affects both the rate and the rate constant of a reaction. Analyze A catalyst speeds up a reaction by providing an alternate pathway (mechanism) to the products; this alternate pathway has a lower activation energy. Solve Yes. Because the reaction is faster (affecting the rate) and the activation energy is lowered (affecting the value of k), a catalyst affects both the rate of the reaction and the value of the rate constant. Think about It A “negative” catalyst that slows down a reaction would increase Ea and decrease k for a reaction. We call these “negative catalysts” inhibitors.

13.112. Collect and Organize We are asked whether the rate law for a catalyzed reaction is the same as that for the uncatalyzed reaction. Analyze A catalyst is used in the reaction but later is regenerated.


Solve The rate law for a catalyzed reaction, like any other reaction, depends on the species involved in the rate-limiting step. The rate law for a catalyzed reaction may be the same as the rate law for an uncatalyzed reaction, but if, for example, the catalyst is involved in the rate-limiting step, its concentration appears in the rate law. This pathway, then, would have a different rate law from that of the uncatalyzed reaction. Think about It Remember that catalysts enhance the rate of a reaction by providing a different, lower-activation-energy pathway to the products.

13.113. Collect and Organize We are asked whether a substance (catalyst) that increases the rate of a reaction also increases the rate of the reverse reaction. Analyze A catalyst speeds up a reaction by providing an alternate, lower-activation-energy pathway (mechanism) to the products. Solve Yes, both the reverse and forward reaction rates are increased when a catalyst is added to a reaction. The activation energies of both processes are lowered by the different pathway that the catalyst provides for the reaction.

Think about It Likewise, an inhibitor would decrease the rates of both forward and reverse reactions.


We are asked whether CO is a catalyst in the reaction between NO2 and CO because it does not appear in the rate law. Analyze A catalyst is used in the reaction but later is regenerated, and a catalyst speeds up a reaction by providing an alternate, lower-activation-energy pathway (mechanism) to the products. Solve No. To function as a catalyst, the CO would have to speed up the reaction and not be consumed. If it does not lower the Ea for the reaction and later is generated in the reaction, it is not a catalyst. Think about It The CO may not appear in the rate law because it may not be involved in the rate-limiting step of the mechanism.

96 | Chapter 13

13.115. Collect and Organize We are to explain why the concentration of a homogeneous catalyst does not appear in the rate law. Analyze A catalyst is used in a reaction and later regenerated. Solve The concentration of a homogeneous catalyst may not appear in the rate law because the catalyst itself is not involved in the rate-limiting step. Think about It If the catalyst is involved in the slowest step of the mechanism, however, it is involved in the rate law.

13.116. Collect and Organize We are to identify the better way to determine the rate constant for a slow reaction: adding a catalyst or increasing the temperature. Analyze A catalyst provides a lower activation energy to the reaction through a different mechanism to speed up a reaction. Heating a reaction causes the reactants to collide more frequently and more effectively without changing the mechanism. Solve Raising the temperature to determine the rate constant for a slow reaction would be better because the mechanism for the reaction is not changed. Think about It Using the form of Arrhenius’s equation in which we compare two rate constants at two temperatures,

a1

1

a2

2

a22 1

1 1 2

1ln ln

1ln ln

1 1ln – ln ln

⎛ ⎞−= +⎜ ⎟

⎝ ⎠⎛ ⎞−

= +⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Ek AR T

Ek AR T

Ekk kk R T T

we could calculate k1 at the lower temperature T1, having measured k2 at the higher temperature T2, and knowing Ea.

13.117. Collect and Organize Given the mechanism for the reaction of NO and N2O to form N2 and O2, we are to determine whether NO or N2O is used in the reaction as the catalyst. Analyze A catalyst is used in a reaction and later regenerated and provides a lower-energy pathway to the products by lowering the activation energy of the reaction, thereby speeding up the reaction. Solve We can assume that the presence of either NO or N2O, if either is a catalyst, increases the rate of the reaction. In examining the mechanism we see that N2O is a reactant, not a catalyst, but that NO is a catalyst because it is used in the reaction and then regenerated. Thus, NO is a catalyst for the decomposition of N2O. Think about It If the slow step of this mechanism were the first step, the rate law would be

Rate = k[NO][N2O]


If the second step were slow, the rate law would be

Rate = kNO⎡⎣ ⎤⎦

2N2O⎡⎣ ⎤⎦

2

N2⎡⎣ ⎤⎦2


Given the mechanism for the destruction of O3 in the presence of NO, we are to explain why NO is functioning as a catalyst. Analyze A catalyst is used in a reaction and later regenerated. It provides a lower-energy pathway to the products by lowering the activation energy of the reaction, thereby speeding up the reaction. Solve NO is used in the reaction in step 1 and regenerated in step 2. Thus, as long as this mechanism has a lower energy pathway to form O2 from O and O3, NO is functioning as a catalyst. Think about It If, however, the presence of NO slows down the reaction, it would be an inhibitor.

13.119. Collect and Organize Using the Arrhenius equation, we can compute and compare the rate constants for the reaction of O3 with O versus the reaction of O3 with Cl. Analyze We are given values A and Ea for each reaction at 298 K. The Arrhenius equation is

a–ln ln

Ek A

RT= +

Solve For the reaction of O3 with O,

ln k = –17.1×103 J/mol8.314 J/mol ⋅K × 298 K

+ ln 8.0×10–12 cm3 /molecules ⋅s( )ln k = –32.45

k = 8.05×10–15cm3 /molecules ⋅s

For the reaction of O3 with Cl,

ln k = –2.16×103 J/mol8.314 J/mol ⋅K × 298 K

+ ln 2.9×10–11 cm3/molecules ⋅s( )ln k = –25.14

k =1.21×10–11cm3/molecules ⋅s

Therefore, the reaction of O3 with Cl has the larger rate constant. Think about It Our answer is consistent with a qualitative look at the activation energies and frequency factors for the two reactions. The higher activation energy and lower frequency factor for the reaction of O3 with O give a smaller reaction rate constant.

13.120. Collect and Organize Using the Arrhenius equation, we can compute and compare the rate constants for the reaction of O3 with Cl versus the reaction of O3 with NO.

98 | Chapter 13

Analyze We are given values of A and Ea for each reaction at 298 K. The Arrhenius equation is

a–ln ln

Ek A

RT= +

Solve For the reaction of O3 with Cl,

ln k = –2.16×103 J/mol8.314 J/mol ⋅K × 298 K

+ ln(2.9×10–11 cm3/molecules ⋅s)

ln k = –25.14

k =1.21×10–11 cm3/molecules ⋅s

For the reaction of O3 with NO,

ln k = –11.6×103 J/mol8.314 J/mol ⋅K × 298 K

+ ln(2.0×10–12 cm3/molecules ⋅s)

ln k = –31.62

k =1.85×10–14 cm3/molecules ⋅s

The reaction with the larger rate constant is the reaction of O3 with Cl. Think about It Our answer is consistent with a qualitative look at the activation energies and frequency factors for the two reactions. The higher activation energy and lower frequency factor for the reaction of O3 with NO give a smaller reaction rate constant.

13.121. Collect and Organize We are to explain why a glowing wood splint burns faster in a test tube filled with O2 than in air. Analyze Air is composed of about 21% O2. Solve When the concentration of a reactant (O2 for the combustion reaction) increases, the rate of reaction also increases. As we place the glowing wood in pure O2, the rate of combustion increases. Think about It If the wood splint were placed in a test tube filled with argon, the combustion reaction would stop.

13.122. Collect and Organize We are to explain why a spark is required to ignite the spontaneous combustion of propane in a barbecue grill. Analyze The spark adds energy to the propane–air (oxygen) mixture. Solve The spark provides the energy to the propane–air mixture needed to overcome the activation energy barrier to start the combustion reaction. Think about It Once started, the combustion reaction itself is self-sustaining.

13.123. Collect and Organize We are to explain why a person submerged in cold water is less likely to have a lack of oxygen for a given period than a person submerged in a warm pool.


Analyze Chemical reactions are slower at colder temperatures than at warmer temperatures. Solve The bodily reactions that use O2 are slower at colder temperatures, so the person submerged in an ice-covered lake uses less of the already dissolved oxygen in his or her system than the person in a warm pool. Think about It Rapid-cooling technology is being investigated at Argonne National Laboratory for use in surgery patients and heart attack victims to reduce the damage done to cells by lack of oxygen in the blood.

13.124. Collect and Organize We are asked to explain why when we have a quadrupling of a reaction rate that it may not correspond to a reaction rate order of 4. Analyze Rate expressions show the relationship between the rate of a reaction and the concentration of the reactants raised to an exponent.

Rate ∝ A⎡⎣ ⎤⎦

x

Solve To quadruple the rate of a reaction means that rate2 = 4 × rate1. To have a reactant be fourth order means that the concentration term in the rate equation will be raised to the fourth power. For example, if

Rate1 ∝ NO⎡⎣ ⎤⎦4, where NO⎡⎣ ⎤⎦ =1 M

Rate2 ∝16× rate1 if NO⎡⎣ ⎤⎦ is doubled to 2 M

Think about It If a reaction is second order, as in the case of

Rate1 ∝ NO⎡⎣ ⎤⎦

2

then if [NO] is doubled from 1 M to 2 M Rate2 = 4×Rate1


In the case where ratereverse << rateforward, we are to consider whether the method to determine the rate law (initial concentrations and initial rates) would work at other times, not just at the start of the reaction. If so, we are to specify which concentrations might be used to determine the rate law. Analyze The method that uses initial rates and concentrations to determine the rate law is under the condition in which no reverse reaction is occurring. Solve Yes, we could use this method at other times, not just t = 0, to determine the rate law if the rate of the reverse reaction is much slower than the forward reaction as long as [products] << [reactants] at the time so that no appreciable reverse reaction is occurring. Think about It We will see in Chapter 15 that when the rate of the reverse reaction equals the rate of the forward reaction, the reaction is at equilibrium.

100 | Chapter 13

13.126. Collect and Organize Given a statement relating reaction rate and the rate constant directly to the number of collisions occurring in a reaction and the concentrations of the reactants, we are to determine what is incorrect about the statement. Analyze The rate law expression for a reaction describes the relationship between the concentration of reactants in the reaction and the rate of the reaction. The rate constant in the rate law (k) relates the rate of the reaction to the concentration of the reactants in a chemical reaction. Solve In this statement, it is true that the reaction rate depends on the concentration of the reactants as stated explicitly by the rate law expression. The reaction rate is also dependent on the number of collisions of the reactants, as more frequent collisions (which can be accomplished by increasing the reactant concentrations) will increase the reaction rate. However, the rate constant is not dependent on either the concentrations or the number of collisions of the reactants. The value of k is unique to a reaction and changes only with temperature or the use of a catalyst. Think about It The Arrhenius equation

k = Ae−Ea RT shows that k is dependent on the frequency factory, A; the activation energy, Ea; and the temperature, T.

13.127. Collect and Organize In the plot of 1/[X] – 1/[X]0 as a function of time, t, we are asked how the rate constant, k, can be determined. Analyze The plot of 1/[X] – 1/[X]0 as a function of time, t, is the plot for a second-order rate equation.

Rate = k X⎡⎣ ⎤⎦2

1X⎡⎣ ⎤⎦

= kt + 1X⎡⎣ ⎤⎦0

Solve In this plot 1/[X] – 1/[X]0 divided by t – t0 is the slope of the line that corresponds to k, the reaction rate constant. All we need to do to determine k from this plot is to determine the slope of the line. Think about It The integrated form of the rate law allows us to obtain the value of k from the concentration-versus-time data from a single experiment.

13.128. Collect and Organize We are asked to explain why first-order reactions are common, second-order reactions are less common, and third-order reactions are rare. Analyze First-order reactions involve a single reactant molecule only. Second-order reactions involve two molecules colliding. Third-order reactions involve three molecules colliding simultaneously. Solve For a first-order reaction to occur, the single reactant molecule needs to have only sufficient energy to react to become products. It does not depend directly on collisions with other molecules, so first-order reactions are quite common. For a second-order reaction to occur, two reactant molecules must collide with sufficient energy and in the correct orientation to produce products. This is a higher requirement for a reaction, so second-order reactions are less common than first-order reactions. Third-order reactions are very rare because they require


that three molecules collide simultaneously with sufficient energy and in the correct orientation for the reaction to occur. Think about It Termolecular reactions are often explained by two consecutive reactions, such as where once two species have reacted to form a highly energetic intermediate (AB*), an inert chemical species (M) collides with the intermediate. Excess energy from the intermediate is then transferred to M.

A + B → AB* AB* + M → C + M


We are asked to explain why an elementary step may not have a rate law that is zero order. Analyze An elementary step in a reaction mechanism describes the collisions of molecular or atomic species taking place in a reaction. Solve For an elementary step to take place, some involvement from a molecular or atomic species must occur. Therefore, there cannot be no dependence (or zero order) of the reactant in an elementary step of a reaction mechanism. Think about It Because most elementary steps are either unimolecular or bimolecular, the rate expressions of elementary steps are usually first or second order.

13.130. Collect and Organize Given the balanced equation for the decomposition of N2O5, we are asked to relate the rate of change in [N2O5] to that of [NO2] and [O2]. Analyze From the balanced equation, the rate of formation of products and consumption of reactants is

Rate = − 1

2Δ N2O5⎡⎣ ⎤⎦

Δt= 1

4Δ NO2⎡⎣ ⎤⎦


Solve The rate of consumption of N2O5 is one-half the rate of formation of NO2 and twice the rate of formation of O2. Think about It Generally, in measuring reaction kinetics we monitor the rate of consumption of a reactant to obtain the initial rate of the reaction.

13.131. Collect and Organize Given the balanced equation for the reaction between NO2 and O3 to produce N2O5 and O2, we are asked to relate the rates of change in [NO2], [O3], [N2O5], and [O2]. Analyze From the balanced equation, the rate of formation of products and consumption of reactants is

Rate = − 1

2Δ NO2⎡⎣ ⎤⎦

Δt= −

Δ O3⎡⎣ ⎤⎦Δt

=Δ N2O5⎡⎣ ⎤⎦


Solve The rate of consumption of O3 is the same as the rate of formation of N2O5 and O2 and one-half the rate of consumption of NO2.

102 | Chapter 13

Think about It The rate of consumption of N2O5 is half the rate of formation of NO2 and twice the rate of formation of O2.

13.132. Collect and Organize Given data for two experiments in which the initial rate of the decomposition of N2O5 was measured at different concentrations of N2O5, we are to determine the order of the reaction. Analyze We can determine the order of the reaction with respect to N2O5 by comparing the rate of the two reactions on changing the concentration:

2 5 0,22

1 2 5 0,1

[N O ]raterate [N O ]

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

x

Solve

3.6×10–5 M /s1.8×10–5 M /s

= 0.100 M0.050 M

⎛⎝⎜

⎞⎠⎟

x

2.0 = 2.0x

x =1

This reaction is first order. Think about It Not all decomposition reactions are first order. The decomposing species might have to collide with another molecule of itself in the rate-determining step of the reaction, in which case the reaction is second order in the decomposing reactant.

13.133. Collect and Organize We can write the rate law from the order of the decomposition reaction determined in Problem 13.132. From that we are to calculate the value of the rate constant at the experimental temperature and write the complete rate law expression. Analyze From Problem 13.132, we know that the reaction is first order in [N2O5]. Solve Because this reaction is a first-order decomposition reaction, the rate law expression is

Rate = k[N2O5] Using the data in experiment 1 in Problem 13.132,

1.8 × 10–5 M/s = k × 0.050 M k = 3.6 × 10–4 s–1

The complete rate law expression is then Rate = (3.6 × 10–4 s–1) × [N2O5]

Think about It The data from experiment 2 in Problem 13.132 would give the same value of k.

13.134. Collect and Organize In the reaction of NO with Cl2, the rate of reaction was measured for various concentrations of both reactants. From the data we can determine the rate law and calculate the rate constant k, which will allow us to determine the initial rate of reaction in experiment 4. Analyze To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two experiments in which the concentration of that reactant changes but the


concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law expression. To calculate the rate constant for the reaction, we rearrange the rate law to solve for k and use the data from any of the experiments. Finally, we can calculate the rate of the reaction for two different concentrations of the reactants by substituting those concentrations and the calculated value of k into the rate law expression. Solve Using experiments 1 and 2, we find that the order of the reaction with respect to Cl2 is 2:

2 0,22

1 2 0,1

[Cl ]rate rate [Cl ]

5.70 /s 0.30 0.63 /s 0.10

9.0 3.02

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠==

x

x

x

M MM M

x

Using experiments 1 and 3, we find that the order of the reaction with respect to NO is 1:

0,33

1 0,1

[NO]rate rate [NO]

2.58 /s 0.80 0.63 /s 0.20

4.1 4.01

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠==

x

x

x

M MM M

x

This gives a rate law for the reaction of Rate = k[NO][Cl2]2

Using experiment 1, we can calculate k:

( ) ( )2

–2 –1

0.63 0.20 0.10 s

315 s

= × ×

=

M k M M

k M

Using the calculated value of k and the [NO]0 and [Cl2]0, we calculate the rate for experiment 4: Rate = 315 M–2 s–1 × 0.40 M × (0.20 M)2

= 5.0 M/s Think about It We may use any of the experiments in the table to calculate k. Each experiment’s data give the same value of k as long as the experiments were all run at the same temperature.

13.135. Collect and Organize For the reaction of NO with O3 to produce NO2 and O2 we can use the information that the reaction is first order in both NO and O3 along with the values of the rate constants at two different temperatures to determine whether the reaction occurs in a single or many steps and to calculate the activation energy, the rate of the reaction at another concentration of the reactants, and the rate constants at two other temperatures. Analyze To answer the questions we need to use the Arrhenius equation,

a–ln ln

Ek A

RT= +

and the rate law expression, which states that the reaction is first order in both NO and O3: Rate = k[NO][O3]

104 | Chapter 13

Solve (a) Because the rate law in which the reaction is first order in both NO and O3 is consistent with that in which the reaction would occur in a single step, this reaction might indeed occur in a single step. (b) We can calculate the activation energy for the reaction by comparing the rate constant at the two temperatures:

a25 C

a75 C

– 1ln ln298 K

– 1ln ln348 K

⎛ ⎞= +⎜ ⎟⎝ ⎠⎛ ⎞= +⎜ ⎟⎝ ⎠

o

o

Ek AREk AR

Subtracting ln k k

ln k75C

− ln k25C

=–Ea

R1

348 K− 1

298 K⎛⎝⎜

⎞⎠⎟= ln

k75C

k25C

⎛

⎝⎜

⎞

⎠⎟

ln3000 M –1 s–1

80 M –1 s–1

⎛⎝⎜

⎞⎠⎟=

–Ea

8.314 J/mol ⋅K1

348 K− 1

298 K⎛⎝⎜

⎞⎠⎟

Ea = 6.25×104 J/mol, or 62.5 kJ/mol

(c) We can use the rate × 10–6 M and [O3] = 5 × 10–9 M. We are given in the statement of the problem that k M–1 s–1.

Rate = 80 M–1 s–1 × (3 × 10–6 M) × (5 × 10–9 M) = 1.2 × 10–12 M/s (d) To use the Arrhenius equation to calculate the values of k , we have to first determine the value of the frequency factor A. To do this we can use the value for Ea and k

( )4

–1 –1

12

–6.25 10 J/mol 1ln 80 s ln8.314 J/mol K 298 K

ln 29.6087.22 10

× ⎛ ⎞= +⎜ ⎟⎝ ⎠⋅== ×

M A

AA

So the value of k 4

12

–1 –1

–6.25 10 J/mol 1ln ln(7.22 10 )8.314 J/mol K 283 K

ln 3.04521 s

× ⎛ ⎞= + ×⎜ ⎟⎝ ⎠⋅==

k

kk M

The value of k 4

12

2 –1 –1

–6.25 10 J/mol 1ln ln(7.22 10 )8.314 J/mol K 308 K

ln 5.201.8 10 s

× ⎛ ⎞= + ×⎜ ⎟⎝ ⎠⋅== ×

k

kk M

Think about It The values of k respectively, than the value of k k be used to determine the activation energy and the frequency factor values in this problem.

13.136. Collect and Organize Given the balanced reaction between NH3 and HNO2, we are asked to write the rate law and determine the units of the rate constant, k; compare the rate law to an alternate expression that includes the species NH4

+ and NO2–;

calculate the enthalpy of the overall reaction; and draw the reaction energy profile for the reaction occurring by a two-step mechanism.


Analyze This problem puts together many of the concepts in this chapter (writing rate laws from experimental results, drawing a reaction profile for a multistep reaction) along with calculation of the enthalpy of a reaction from Chapter 5. Solve (a) Because the reaction is first order in [NH3] and second order in [HNO2], the rate law is Rate = k[NH3][HNO2]2. If the rate is in units of molar per second, then the units of k are

–2 –12

/s s( )( )M MM M

=

(b) Yes, the expression is equivalent to the rate law in part a. If the formation of NH4

+ and NO2– is an equilibrium step, then

NH3 + HNO2 = NH4+ + NO2

– k1[NH3][HNO2] = k–1[NH4

+][NO2–]

Solving for [NH3][HNO2] gives

[ ][ ] –11 3 2 4 2

1

NH HNO = NH NOk

kk

+− ⎡ ⎤ ⎡ ⎤= ⎣ ⎦ ⎣ ⎦

Substituting into the rate equation in part a gives

Rate = kk−1

k1

NH4+⎡⎣ ⎤⎦ NO2

–⎡⎣ ⎤⎦ HNO2⎡⎣ ⎤⎦

= k NH4+⎡⎣ ⎤⎦ NO2

–⎡⎣ ⎤⎦ HNO2⎡⎣ ⎤⎦

(c) ΔH rxn = 1 mol N2 × 0.0 kJ/mol( )+ 2 mol H2O × –285.8 kJ/mol( )⎡⎣ ⎤⎦

− 1 mol NH3 × – 46.1 kJ/mol( )+ 1 mol HNO2 × – 43.1 kJ/mol( )⎡⎣ ⎤⎦= −482.4 kJ

(d)

Think about It From the reaction energy diagram, we see that the second step is the rate-limiting step.

13.137. Collect and Organize We consider the mechanism for the exchange of H2

16O around a Na+ cation for H218O to write the rate law for

the rate-determining step. We also need to think about the relative energies of the reactants and products if we are to sketch the reaction energy profile. Analyze (a) We are given that the first step of the reaction is rate determining, so this is the step from which we write the rate law expression. (b) In deciding which has the higher energy, the reactants or products, for the reaction profile, we need to consider the relative strength of the Na+–16OH2 interaction versus that of the Na+–18OH2 interaction.

106 | Chapter 13

Solve (a) Rate = k[Na(H2O)6

+] (b) Neither. The ion–dipole interaction should not be significantly different for H2

18O versus H216O, so the

energy of the reactants and the products in the reaction profile will be about the same. Think about It In reality, an isotope effect exists, in which the H2

18O–Na+ interaction is slightly stronger than the H216O–

Na+ interaction, so the energy of the product in this reaction is slightly lower than the energy of the reactants.

13.138. Collect and Organize For the second-order decomposition of PAN, we are to calculate k for the reaction given t1/2, determine the rate of the reaction when the partial pressure of PAN is 10.5 torr, and draw a plot of PPAN versus time from 0 to 200 hr. Analyze This problem puts together many of the concepts in this chapter (using t1/2 to calculate k, calculating the rate of a reaction from the rate law, and plotting change in the concentration, here partial pressure, of a reactant over time). Solve (a) From t1/2 we calculate k for this second-order reaction, where PPAN = 10.5 torr:

t1/2 =1

k A⎡⎣ ⎤⎦0

, or k = 1t1/2 A⎡⎣ ⎤⎦0

= 1100 hr ×10.5 torr

= 9.52×10–4 torr–1 hr–1

(b) For a second-order reaction, the rate law is

Rate = k A⎡⎣ ⎤⎦0

2= k PPAN( )2

= 9.52×10−4

torr ⋅hr× 10.5 torr( )2

= 0.105 torr/hr

(c)

Think about It From the plot in part c, we may determine the instantaneous rate of the reaction at any time by determining the slope of the tangent line at that point.

13.139. Collect and Organize For the reaction of NO with ONOO–, we are to use the provided data to determine the rate law for the reaction. We are also to draw the Lewis structure with resonance forms to describe the bonding in the ONOO– anions and then use bond energies to estimate ∆Hrxn

° .. Analyze (a) To determine the rate law, we can compare the effects of changing the concentrations of NO and ONOO– on the rate of the reaction.


(b) After drawing the resonance forms for ONOO– we can determine which is the preferred structure by assigning formal charges to the atoms in each resonance form. (c) The ∆Hrxn

° may be estimated from bond energies by using

ΔHrxn

= ΔHbond breaking +∑ ΔHbond forming∑ Solve (a) For the order of reaction with respect to ONOO–, we can compare experiments 1 and 2:

–0,22

–1 0,1

–11 –4

–11 –4

[ONOO ]raterate [ONOO ]

1.02 10 /s 0.625 102.03 10 /s 1.25 10

0.502 0.5001

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞× ×= ⎜ ⎟× ×⎝ ⎠==

x

x

x

M MM M

x

The data table does not have two experiments for which the concentration of ONOO– stays the same. For the order of reaction with respect to NO, we can compare experiments 2 and 3 as long as we take into account the knowledge that the reaction is first order in ONOO–. Between these two experiments we see that as the ONOO– concentration is quadrupled, we would expect that the rate would be quadrupled. The rate of the reaction when the NO concentration is halved simultaneously with the quadrupling of the rate on changing the ONOO–, however, doubles:

–113

–112

rate 2.03 10 /s 1.99rate 1.02 10 /s

⎛ ⎞×= =⎜ ⎟×⎝ ⎠

MM

Because the ∆[ONOO–] = 4, which would quadruple the rate of the reaction, the ∆[NO] must affect the rate by 1.99/4 = 0.498. Therefore,

–4

–4

0.625 100.4981.25 10

0.498 0.5001

⎛ ⎞×= ⎜ ⎟×⎝ ⎠==

x

x

MM

x

The rate law for this reaction is Rate = k[NO][ONOO–]. We can use any experiment to calculate the value of k. Using the data from experiment 1:

2.03 × 10–11 M/s = k × (1.25 × 10–4 M) × (1.25 × 10–4 M) k = 1.30 × 10–3 M–1 s–1

(b)

(c)

Bonds broken = O—O (146 kJ/mol) Bonds formed = N—O (201 kJ/mol)

∆Hrxn° = [(1 mol × 146 kJ/mol) + (1 mol × –201 kJ/mol)] = –55 kJ

Think about It To solve this problem you had to draw on several concepts you have learned so far in this course. You had to review not only how to write a rate law given kinetic data but also how to draw resonance structures and how bond energies might be used to estimate the enthalpy of a reaction.

108 | Chapter 13

13.140. Collect and Organize For the first-order decomposition of C6H13SNO, we can use a plot of ln[C6H13SNO] versus time to calculate the value of k. We are also asked which amino acids might be a source of S-nitrosothiols.

Analyze The slope of the line of the plot of ln[C6H13SNO] versus time is equal to –k. The amino acids that might be precursors to S-nitrosothiols must contain sulfur.

Solve (a)

k = –slope = 6.50 × 10–3 min–1

Think about It S-Nitrosothiol drugs that generate NO in the body are being developed as treatments for respiratory and pulmonary diseases.

13.141. Collect and Organize Using data for [HNO2] over time, we can determine the order of the isotopic exchange reaction with respect to [HNO2]. We are also asked whether the rate of the reaction will depend on [H2

18O].

Analyze We are given a single data set. If the plot of [HNO2] versus time yields a straight line, the reaction is zero order. If the plot of ln[HNO2] versus time yields a straight line, the reaction is first order. If the plot of 1/[HNO2] versus time yields a straight line, the reaction is second order.

Solve (a)

y = –6.50E–03x – 6.86E+00 R² = 1.00E+00

-7.3

-7.2

-7.1

-7

-6.9

-6.8

0 10 20 30 40 50 60 70

ln[C

6H13

SNO

] (ln

M)

Time (min)

y = –0.0008x + 0.0384 R² = 0.61471

-0.02

0.00

0.02

0.04

0.06

0 10 20 30 40 50 60 70

[HN

18O

2]

Time (min)

Zero-Order Plot

y = –0.073x – 3.8486 R² = 0.78631

-9 -8 -7 -6 -5 -4 -3 -2 -1 0

0 10 20 30 40 50 60 70

ln [H

NO

18O

2]

Time (min)

First-Order Plot


This reaction is second order in [HNO2]. (b) Because the reaction mixture has a very large [H2

18O], we cannot observe a change in [H218O] over time.

The rate might be dependent on [H218O], but we cannot tell from the information given.

Think about It If it were possible, we could place the reaction in a nonreactive solvent and then vary the [H2

18O] over time to determine the rate’s dependence on its concentration.

13.142. Collect and Organize In the reaction of C2H4 with O3, the rate of the reaction was measured for various concentrations of both reactants. From the data we are to determine the rate law and calculate the rate constant, k. Also, given the dependence of k on temperature, we are to calculate Ea for this reaction.

Analyze (a) To determine the dependence of the rate on a change in the concentration of a particular reactant, we can compare the reaction rates for two experiments in which the concentration of that reactant changes but the concentrations of the other reactants remain constant. Once we have the order of the reaction for each reactant, we can write the rate law expression. To calculate the rate constant for the reaction, we can rearrange the rate law to solve for k and use the data from any of the experiments. Finally, we can calculate the rate of the reaction for two different concentrations of the reactants by substituting those concentrations and the calculated value of k into the rate law expression. (b) The Arrhenius equation is

a– 1ln lnE

k AR T

⎛ ⎞= +⎜ ⎟⎝ ⎠If we plot ln k (y-axis) versus 1/T (x-axis), we can obtain a straight line with slope m = –Ea/R. The activation energy is therefore calculated from

Ea = –slope × R where R = 8.314 J/mol ⋅ K. The y-intercept (b) is equal to ln A, so we can calculate the frequency factor from b = ln A or eb = A. Once Ea and A are known, we may use another form of the Arrhenius equation to calculate kat any temperature:

aexp –E

k ART

⎛ ⎞= ⎜ ⎟⎝ ⎠

Solve (a) Experiments 1 and 2 show that the rate dependence on [O3] is 1:

rate1

rate2

=[O3]1

[O3]2

⎛

⎝⎜⎞

⎠⎟

x

y = 31.729x + 24.884 R² = 0.99993

0.00

500.00

1000.00

1500.00

2000.00

2500.00

0 10 20 30 40 50 60 70 1/

[HN

18O

2]

Time (min)

Second-Order Plot

110 | Chapter 13

0.0877 M /s0.0439 M /s

= 0.86×10–2 M0.43×10–2 M

⎛⎝⎜

⎞⎠⎟

x

2.00 = 2.0x

x = 1

Using this result, we compare experiments 2 and 3 to obtain the order of the reaction with respect to C2H4: 1–2 –2

–2 –2

1

0.43 10 1.00 10 0.0439 /s0.0110 /s0.22 10 0.50 10

1.95 2 3.991

⎛ ⎞ ⎛ ⎞× ×× =⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠× ==

x

x

M M MMM M

x

The rate law for this reaction is Rate = k[O3][C2H4]. Using experiment 1 to calculate the value of k,

k = 0.0877 M /s0.0086 M( )× 0.0100 M( ) =1020 M –1 s–1

(b)

Ea = –slope × R = 2629.8 × 8.314 J/mol ⋅ K = 2.19 × 104 J/mol, or 21.9 kJ/mol Think about It Knowing the value of the activation energy, we can use the value of k at any of the temperatures given in the data table to calculate the value of the frequency factor, A.

13.143. Collect and Organize Using the raw data obtained for four experiments in which the concentrations of NH2 and NO were varied, we are to write the rate law and determine the value of k for the reaction between NH2 and NO at 1200 K. Analyze To determine the rate law, we can compare the effects of changing the concentrations of NH2 and NO on the rate of the reaction. Once we have determined the order of the reaction with respect to each reactant, we can write the rate law expression and use any of the experiments to calculate the value of the rate constant, k. Solve (a) For the order of the reaction with respect to NH2, we can use the data from experiments 1 and 2:

rate2

rate1

=[NH2]0,2

[NH2]0,1

⎛

⎝⎜

⎞

⎠⎟

x


0.24 M/s0.12 M/s

= 2.00×10–5 M1.00×10–5 M

⎛⎝⎜

⎞⎠⎟

x

2.0 = 2.00x

x = 1

For the order of the reaction with respect to NO, we can use the data from experiments 2 and 3:

0,33

2 0,2

–5

–5

[NO]raterate [NO]

0.36 /s 1.50 100.24 /s 1.00 10

1.5 1.501

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞×= ⎜ ⎟×⎝ ⎠==

x

x

x

M MM M

x

The rate law expression is Rate = k[NH2][NO]. (b) Using experiment 1 to calculate the value of k,

0.12 M/s = k × (1.00 × 10–5 M) × (1.00 × 10–5 M) k = 1.2 × 109 M–1 s–1 Think about It Any of the experiments listed in the data table would give us the same value of k in the calculation in part b.

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