chapter twelve chemical kinetics - … · chapter twelve chemical kinetics questions 9. ... 13. a...

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CHAPTER TWELVE

CHEMICAL KINETICS

Questions

9. The rate of a chemical reaction varies with time. Consider the general reaction:

A ÷ Products where rate =

If we graph [A] vs. t, it would roughly look like the dark line in the following plot.

An instantaneous rate is the slope of a tangent line to the graph of [A] vs. t. We can determine the

1instantaneous rate at any time during the reaction. On the plot, tangent lines at t = 0 and t = t are

1drawn. The slope of these tangent lines would be the instantaneous rates at t . 0 and t = t . We callthe instantaneous rate at t . 0 the initial rate. The average rate is measured over a period of time. Forexample, the slope of the line connecting points a and c is the average rate of the reaction over the

2 entire length of time 0 to t (average rate = )[A]/)t). An average rate is determined over some timeperiod while an instantaneous rate is determined at one specific time. The rate which is largest isgenerally the initial rate. At t . 0, the slope of the tangent line is greatest, which means the rate islargest at t . 0.

10. a. An elementary step (reaction) is one in which the rate law can be written from the molecularity,i.e., from the coefficients in the balanced equation.

CHAPTER 12 CHEMICAL KINETICS292

b. The mechanism of a reaction is the series of proposed elementary reactions that may occur to givethe overall reaction. The sum of all the steps in the mechanism gives the balanced chemicalreaction.

c. The rate-determining step is the slowest elementary reaction in any given mechanism.

11. a. The greater the frequency of collisions, the greater the opportunities for molecules to react, and,hence, the greater the rate.

b. Chemical reactions involve the making and breaking of chemical bonds. The kinetic energy of thecollisions can be used to break bonds. So, as the kinetic energy of the collisions increases, the rateincreases.

c. For a reaction to occur, it is the reactive portion of each molecule that must be involved in acollision. Only some of all the possible collisions have the correct orientation to convert reactantsto products.

12. In a unimolecular reaction, a single reactant molecule decomposes to products. In a bimolecularreaction, two molecules collide to give products. The probability of the simultaneous collision of threemolecules with enough energy and orientation is very small, making termolecular steps very unlikely.

13. A catalyst increases the rate of a reaction by providing reactants with an alternate pathway(mechanism) to convert to products. This alternate pathway has a lower activation energy, thusincreasing the rate of the reaction.

A heterogeneous catalyst is in a different phase than the reactants. The catalyst is usually a solid,although a catalyst in a liquid phase can act as a heterogeneous catalyst for some gas phase reactions.Since the catalyzed reaction has a different mechanism than the uncatalyzed reaction, the catalyzedreaction most likely will have a different rate law.

14. Some energy must be added to get the reaction started, that is, to overcome the activation energybarrier. Chemically what happens is:

2Energy + H ÷ 2 H

2 2The hydrogen atoms initiate a chain reaction that proceeds very rapidly. Collisions of H and Omolecules at room temperature do not have sufficient kinetic energy to form hydrogen atoms andinitiate the reaction.

15. a. Activation energy and )E are independent of each other. Activation energy depends on the pathreactants to take to convert to products. The overall energy change, )E, depends only on theinitial and final energy states of the reactants and products. )E is path independent.

b. The rate law can be determined only from experiment, not from the overall balanced reaction.

c. Most reactions occur by a series of steps. The rate of the reaction is determined by the rate of theslowest step in the mechanism.

CHAPTER 12 CHEMICAL KINETICS 293

16. All of these choices would affect the rate of the reaction, but only b and c affect the rate by affectingthe value of the rate constant k. The value of the rate constant is dependent on temperature. The valueof the rate constant also depends on the activation energy. A catalyst will change the value of k

2because the activation energy changes. Increasing the concentration (partial pressure) of either H orNO does not affect the value of k, but it does increase the rate of the reaction because bothconcentrations appear in the rate law.

Exercises

Reaction Rates

17. The coefficients in the balanced reaction relate the rate of disappearance of reactants to the rate of

4production of products. From the balanced reaction, the rate of production of P will be 1/4 the rate

3 2 3of disappearance of PH , and the rate of production of H will be 6/4 the rate of disappearance of PH .By convention, all rates are given as positive values.

Rate = = = 2.4 × 10 mol/LCs-3

= 2.4 × 10 /4 = 6.0 × 10 mol/LCs-3 -4

= 6(2.4 × 10 )/4 = 3.6 × 10 mol/LCs-3 -3

18. ; So,

or

Ammonia is produced at a rate equal to 2/3 of the rate of consumption of hydrogen.

19. a. The units for rate are always mol/LCs. b. Rate = k; k must have units of mol/LCs.

c. Rate = k[A], d. Rate = k[A] , 2

k must have units of s . k must have units of L/molCs.-1

e. L /mol Cs2 2

320. Rate = k[Cl] [CHCl ], , k must have units of L /mol Cs.1/2 1/2 1/2

Rate Laws from Experimental Data: Initial Rates Method


221. a. In the first two experiments, [NO] is held constant and [Cl ] is doubled. The rate also doubled.

2 2Thus, the reaction is first order with respect to Cl . Or mathematically: Rate = k[NO] [Cl ]x y

, 2.0 = 2.0 , y = 1y

We can get the dependence on NO from the second and third experiments. Here, as the NO

2concentration doubles (Cl concentration is constant), the rate increases by a factor of four. Thus,the reaction is second order with respect to NO. Or mathematically:

2, 4.0 = 2.0 , x = 2; So, Rate = k[NO] [Cl ]x 2

Try to examine experiments where only one concentration changes at a time. The more variablesthat change, the harder it is to determine the orders. Also, these types of problems can usually besolved by inspection. In general, we will solve using a mathematical approach, but keep in mindyou probably can solve for the orders by simple inspection of the data.

b. The rate constant k can be determined from the experiments. From experiment 1:

, k = 180 L /mol Cmin2 2

From the other experiments:

k = 180 L /mol Cmin (2nd exp.); k = 180 L /mol Cmin (3rd exp.)2 2 2 2

meanThe average rate constant is k = 1.8 × 10 L /mol Cmin.2 2 2

2 822. a. Rate = k[I ] [S O ] ; , 2.00 = 2.0 , x = 1- x 2- y x

2 8, 2.00 = 2.0 , y = 1; Rate = k[I ][S O ]y - 2-

b. For the first experiment:

, k = 3.9 × 10 L/molCs -3

meanEach of the other experiments also gives k = 3.9 × 10 L/molCs, so k = 3.9 × 10 L/molCs.-3 -3

23. a. Rate = k[NOCl] ; Using experiments two and three:n


, 4.01 = 2.0 , n = 2; Rate = k[NOCl]n 2

b. , k = 6.6 × 10 cm /moleculesCs-29 3

The other three experiments give (6.7, 6.6 and 6.6) × 10 cm /moleculesCs, respectively.-29 3

The mean value for k is 6.6 × 10 cm /moleculesCs.-29 3

c.

2 524. Rate = k[N O ] ; The rate laws for the first two experiments are:x

2.26 × 10 = k(0.190) and 8.90 × 10 = k(0.0750)-3 x -4 x

2 5Dividing: 2.54 = = (2.53) , x = 1; Rate = k[N O ]x

meank = = 1.19 × 10 s ; k = 1.19 × 10 s-2 -1 -2 -1

25. a. Rate = k[Hb] [CO] ; Comparing the first two experiments, [CO] is unchanged, [Hb] doubles, andx y

the rate doubles. Therefore, the reaction is first order in Hb. Comparing the second and thirdexperiments, [Hb] is unchanged, [CO] triples. and the rate triples. Therefore, y = 1 and thereaction is first order in CO.

b. Rate = k[Hb][CO]

c. From the first experiment:

0.619 µmol/LCs = k (2.21 µmol/L)(1.00 µmol/L), k = 0.280 L/µmolCs

meanThe second and third experiments give similar k values, so k = 0.280 L/µmolCs.

d. Rate = k[Hb][CO] = = 2.26 µmol/LCs

226. a. Rate = k[ClO ] [OH ] ; From the first two experiments:x - y

2.30 × 10 = k(0.100) (0.100) and 5.75 × 10 = k(0.0500) (0.100)-1 x y -2 x y

Dividing the two rate laws: 4.00 = = 2.00 , x = 2x

Comparing the second and third experiments:


2.30 ×10 = k(0.100)(0.100) and 1.15 × 10 = k(0.100)(0.0500)-1 y -1 y

Dividing: 2.00 = = 2.00 , y = 1y

2The rate law is: Rate = k[ClO ] [OH ]2 -

mean2.30 × 10 mol/LCs = k(0.100 mol/L) (0.100 mol/L), k = 2.30 × 10 L /mol Cs = k -1 2 2 2 2

2b. Rate = k[ClO ] [OH ] = = 0.594 mol/LCs2 -

Integrated Rate Laws

27. The first assumption to make is that the reaction is first order. For a first-order reaction, a graph of

2 2ln [H O ] vs time will yield a straight line. If this plot is not linear, then the reaction is not first orderand we make another assumption. The data and plot for the first-order assumption is below.

2 2 2 2Time [H O ] ln [H O ] (s) (mol/L)

0 1.00 0.000120. 0.91 -0.094300. 0.78 -0.25600. 0.59 -0.53

1200. 0.37 -0.991800. 0.22 -1.512400. 0.13 -2.043000. 0.082 -2.503600. 0.050 -3.00

Note: We carried extra significant figures in some of the ln values in order to reduce round-off error. For the plots, we will do this most of the time when the ln function is involved.

2 2The plot of ln [H O ] vs. time is linear. Thus, the reaction is first order. The rate law and

2 2 2 2 2 2 ointegrated rate law are: Rate = k[H O ] and ln [H O ] = -kt + ln [H O ] .

2 2We determine the rate constant k by determining the slope of the ln [H O ] vs time plot (slope = -k). Using two points on the curve gives:

slope = -k = = -8.3 × 10 s , k = 8.3 × 10 s-4 -1 -4 -1

2 2 2 2 oTo determine [H O ] at 4000. s, use the integrated rate law where at t = 0, [H O ] = 1.00 M.


2 2 2 2 oln [H O ] = -kt + ln [H O ] or = -kt

2 2 2 2 = -8.3 × 10 s × 4000. s, ln [H O ] = -3.3, [H O ] = e = 0.037 M-4 -1 -3.3

28. a. Since the ln[A] vs time plot was linear, the reaction is first order in A. The slope of the ln[A] vs.time plot equals -k. Therefore, the rate law, the integrated rate law and the rate constant value are:

oRate = k[A]; ln[A] = -kt + ln[A] ; k = 2.97 × 10 min-2 -1

b. The half-life expression for a first-order rate law is:

1/2 1/2t = , t = = 23.3 min

c. 2.50 × 10 M is 1/8 of the original amount of A present initially, so the reaction is 87.5%-3

complete. When a first-order reaction is 87.5% complete (or 12.5% remains), the reaction hasgone through 3 half-lives:

1/2100% ÷ 50.0% ÷ 25% ÷ 12.5%; t = 3 × t = 3 × 23.3 min = 69.9 min

1/2 1/2 1/2t t t

Or we can use the integrated rate law:

= -kt, = - (2.97 × 10 min ) t, t = -2 -1

= 70.0 min

229. Assume the reaction is first order and see if the plot of ln [NO ] vs. time is linear. If this isn’t linear,

2try the second-order plot of 1/[NO ] vs. time. The data and plots follow.

2 2 2Time (s) [NO ] (M) ln [NO ] 1/[NO ] (M )-1

0 0.500 -0.693 2.00

1.20 × 10 0.444 -0.812 2.253

3.00 × 10 0.381 -0.965 2.623

0.3404.50 × 10 -1.079 2.943

9.00 × 10 0.250 -1.386 4.003

0.1741.80 × 10 -1.749 5.754


2 2The plot of 1/[NO ] vs. time is linear. The reaction is second order in NO . The rate law and

2integrated rate law are: Rate = k[NO ] and .2

2The slope of the plot 1/[NO ] vs. t gives the value of k. Using a couple of points on the plot:

slope = k = = 2.08 × 10 L/molCs-4

2 2 oTo determine [NO ] at 2.70 × 10 s, use the integrated rate law where 1/[NO ] = 1/0.500 M 4

= 2.00 M .-1

× 2.70 × 10 s + 2.00 M 4 -1

2 = 7.62, [NO ] = 0.131 M

30. a. Since the 1/[A] vs. time plot was linear, the reaction is second order in A. The slope of the 1/[A]vs. time plot equals the rate constant k. Therefore, the rate law, the integrated rate law and therate constant value are:

Rate = k[A] ; = kt + ; k = 3.60 × 10 L/molCs2 -2

1/2b. The half-life expression for a second-order reaction is: t =

1/2For this reaction: t = = 9.92 × 10 s3

1/2Note: We could have used the integrated rate law to solve for t where [A] = (2.80 × 10 /2) mol/L.-3

c. Since the half-life for a second-order reaction depends on concentration, we must use the


integrated rate law to solve.

1.43 × 10 - 357 = 3.60 × 10 t, t = 2.98 × 10 s3 -2 4

2 5 2 531. a. Since the [C H OH] vs. time plot was linear, the reaction is zero order in C H OH. The

2 5slope of the [C H OH] vs. time plot equals -k. Therefore, the rate law, the integrated rate law and

2 5 2 5 2 5 othe rate constant value are: Rate = k[C H OH] = k; [C H OH] = -kt + [C H OH] ; 0

k = 4.00 × 10 mol/LCs.-5

1/2 ob. The half-life expression for a zero-order reaction is: t = [A] /2k.

1/2t = = 156 s

1/2Note: we could have used the integrated rate law to solve for t where

2 5[C H OH] = (1.25 × 10 /2) mol/L.-2

2 5 2 5 oc. [C H OH] = -kt + [C H OH] , 0 mol/L = -(4.00 × 10 mol/LCs) t + 1.25 × 10 mol/L-5 -2

t = = 313 s

2 532. From the data, the pressure of C H OH decreases at a constant rate of 13 torr for every 100. s. Since

2 5the rate of disappearance of C H OH is not dependent on concentration, the reaction is zero order in

2 5C H OH.

k = = 1.7 × 10 atm/s-4

The rate law and integrated rate law are:

Rate = k = 1.7 × 10 atm/s; = -kt + 250. torr = -kt + 0.329 atm-4

At 900. s: = -1.7 × 10 atm/s × 900. s + 0.329 atm = 0.176 atm = 0.18 atm = 130 torr-4

33. The first assumption to make is that the reaction is first order. For a first-order reaction, a graph of

4 6 4 6ln [C H ] vs. t should yield a straight line. If this isn't linear, then try the second-order plot of 1/[C H ]vs. t. The data and the plots follow.


Time 195 604 1246 2180 6210 s

4 6[C H ] 1.6 × 10 1.5 × 10 1.3 × 10 1.1 × 10 0.68 × 10 M-2 -2 -2 -2 -2

4 6ln [C H ] -4.14 -4.20 -4.34 -4.51 -4.99

4 61/[C H ] 62.5 66.7 76.9 90.9 147 M -1

Note: To reduce round-off error, we carried extra sig. figs. in the data points.

The natural log plot is not linear, so the reaction is not first order. Since the second-order plot of

4 61/[C H ] vs. t is linear, we can conclude that the reaction is second order in butadiene. The rate lawis:

4 6Rate = k[C H ]2

For a second order reaction, the integrated rate law is:

The slope of the straight line equals the value of the rate constant. Using the points on the line at 1000.and 6000. s:

k = slope = = 1.4 × 10 L/molCs-2

34. a. First, assume the reaction to be first order with respect to O. A graph of ln [O] vs. t should belinear if the reaction is first order.

t(s) [O] (atoms/cm ) ln[O]3

0 5.0 × 10 22.339

10. × 10 1.9 × 10 21.37-3 9

20. × 10 6.8 × 10 20.34-3 8

30. × 10 2.5 × 10 19.34-3 8


Since the graph is linear, we can conclude that the reaction is first order with respect to O.

2b. The overall rate law is: Rate = k[NO ][O]

2Since NO was in excess, its concentration is constant. So for this experiment, the rate law is:

2Rate = kN[O] where kN = k[NO ] . In a typical first-order plot, the slope equals -k. For this

2experiment, the slope equals -kN = -k[NO ]. From the graph:

slope = = -1.0 × 10 s , kN = -slope = 1.0 × 10 s2 -1 2 -1

To determine k, the actual rate constant:

2kN = k[NO ], 1.0 × 10 s = k(1.0 × 10 molecules/cm ), k = 1.0 × 10 cm /moleculesCs2 -1 13 3 -11 3

35. Since the 1/[A] vs. time plot is linear with a positive slope, the reaction is second order with respect

oto A. The y-intercept in the plot will equal 1/[A] . Extending the plot, the y-intercept will be about

o10, so 1/10 = 0.1 M = [A] .

36. The slope of the 1/[A] vs time plot in Exercise 12.35 with equal k.

slope = k = = 10 L/molCs

a. = = 100, [A] = 0.01 M

1/2b. For a second-order reaction, the half-life does depend on concentration: t = .

1/2First half-life: t = = 1 s

o 1/2Second half-life ([A] is now 0.05 M): t = 1/(10 × 0.05) = 2 s

o 1/2Third half-life ([A] is now 0.025 M): t = 1/(10 × 0.025) = 4 s


o37. a. [A] = - kt + [A] , [A] = -(5.0 × 10 mol/LCs) t + 1.0 × 10 mol/L -2 -3

1/2b. The half-life expression for a zero-order reaction is: t =

1/2t = = 1.0 × 10 s-2

c. [A] = -5.0 × 10 mol/LCs × 5.0 × 10 s + 1.0 × 10 mol/L = 7.5 × 10 mol/L-2 -3 -3 -4

Since 7.5 × 10 M A remains, 2.5 × 10 M A reacted, which means that 2.5 × 10 M B has been-4 -4 -4

produced.

38. = -kt; k = = 4.85 × 10 d-2 -1

oIf [A] = 100.0, then after 95.0% completion, [A] = 5.0.

ln = -4.85 × 10 d × t, t = 62 days-2 -1

39. a. If the reaction is 38.5% complete, then 38.5% of the original concentration is consumed, leaving61.5%.

o o[A] = 61.5% of [A] or [A] = 0.615 [A] ; = -kt, ln = -k(480. s)

ln(0.615) = -k(480. s), -0.486 = -k(480. s), k = 1.01 × 10 s-3 -1

1/2b. t = (ln 2)/k = 0.6931/1.01 × 10 s = 686 s-3 -1

oc. 25% complete: [A] = 0.75 [A] ; ln(0.75) = -1.01 × 10 (t), t = 280 s-3

o75% complete: [A] = 0.25 [A] ; ln(0.25) = -1.01 × 10 (t), t = 1.4 × 10 s-3 3

1/2Or, we know it takes 2 × t for reaction to be 75% complete. t = 2 × 686 s = 1370 s

o95% complete: [A] = 0.05 [A] ; ln(0.05) = -1.01 × 10 (t), t = 3 × 10 s-3 3

o40. For a first-order reaction, the integrated rate law is: ln([A]/[A] ) = -kt. Solving for k:

ln = -k × 120. s, k = 0.0116 s-1

ln = - 0.0116 s × t, t = 150. s-1

1/241. For a second-order reaction: t = or k =


k = = 0.12 L/molCs

o42. a. The integrated rate law for a second-order reaction is: 1/[A] = kt + 1/[A] , and the half-life

1/2 o 1/2 expression is: t = 1/k[A] . We could use either to solve for t . Using the integrated rate law:

, k = = 0.555 L/molCs

b. = 0.555 L/molCs × t + , t = = 16 s

43. Successive half-lives increase in time for a second-order reaction. Therefore, assume the reaction issecond order in A.

1/2t = , k = = 1.0 L/molCmin

a. = kt + × 80.0 min + = 90. M , [A] = 1.1 × 10 M-1 -2

b. 30.0 min = 2 half-lives, so 25% of original A is remaining.

[A] = 0.25(0.10 M) = 0.025 M

o o44. Since [B] >>[A] , the B concentration is essentially constant during this experiment, so rate = k[A]!

where k = k[B] . For this experiment, the reaction is a pseudo-first-order reaction in A.! 2

a. ln = -k t, ln = -k × 8.0 s, k = 0.12 s! ! ! -1

For the reaction: k = k[B] , k = 0.12 s /(3.0 mol/L) = 1.3 × 10 L /mol Cs! 2 -1 2 -2 2 2

1/2b. t = = 5.8 s

c. ln = -0.12 s × 13.0 s, = e = 0.21, [A] = 2.1 × 10 M-1 -0.12(13.0) -3

reacted reactedd. [A] = 0.010 M - 0.0021 M = 0.008 M; [C] = 0.008 M × = 0.016 M . 0.02 M

remaining[C] = 2.0 M - 0.02 M = 2.0 M; As expected, the concentration of C basically remains

o o constant during this experiment since [C] >> [A] .

Reaction Mechanisms


45. For elementary reactions, the rate law can be written using the coefficients in the balanced equation todetermine orders.

3 3a. Rate = k[CH NC] b. Rate = k[O ][NO]

3 3c. Rate = k[O ] d. Rate = k[O ][O]

246. The observed rate law for this reaction is: Rate = k[NO] [H ]. For a mechanism to be plausible, the sum2

of all the steps must give the overall balanced equation (true for all the proposed mechanisms in thisproblem), and the rate law derived from the mechanism must agree with the observed mechanism. Ineach mechanism (I - III), the first elementary step is the rate-determining step (the slow step), so thederived rate law for each mechanism will be the rate of the first step. The derived rate laws follow:

2Mechanism I: Rate = k[H ] [NO]2 2

2Mechanism II: Rate = k[H ][NO]

2Mechanism III: Rate = k[H ][NO]2

Only in Mechanism III does the derived rate law agree with the observed rate law. Thus, onlyMechanism III is a plausible mechanism for this reaction.

47. A mechanism consists of a series of elementary reactions where the rate law for each step can bedetermined using the coefficients in the balanced equations. For a plausible mechanism, the rate lawderived from a mechanism must agree with the rate law determined from experiment. To derive therate law from the mechanism, the rate of the reaction is assumed to equal the rate of the slowest stepin the mechanism.

4 9Since step 1 is the rate-determining step, the rate law for this mechanism is: Rate = k[C H Br]. Toget the overall reaction, we sum all the individual steps of the mechanism.

Summing all steps gives:

4 9 4 9 C H Br ÷ C H + Br+ -

4 9 2 4 9 2 C H + H O ÷ C H OH+ +

4 9 2 2 4 9 3 C H OH + H O ÷ C H OH + H O+ +

4 9 2 4 9 3 C H Br + 2 H O ÷ C H OH + Br + H O- +

Intermediates in a mechanism are species that are neither reactants nor products, but that areformed and consumed during the reaction sequence. The intermediates for this mechanism are

4 9 4 9 2C H and C H OH .+ +

48. Since the rate of the slowest elementary step equals the rate of a reaction, then:

2Rate = rate of step 1 = k[NO ]2

The sum of all steps in a plausible mechanism must give the overall balanced reaction. Summing allsteps gives:


2 2 3NO + NO ÷ NO + NO

3 2 2 NO + CO ÷ NO + CO

2 2 NO + CO ÷ NO + CO

Temperature Dependence of Rate Constants and the Collision Model

a49. In the following plot, R = reactants, P = products, E = activation energy and RC = reaction coordinatewhich is the same as reaction progress. Note for this reaction that )E is positive since the productsare at a higher energy than the reactants.

50. When )E is positive, the products are at a higher energy relative to reactants and, when )E isnegative, the products are at a lower energy relative to reactants.


51.

The activation energy for the reverse reaction is:

a, reverseE = 216 kJ/mol + 125 kJ/mol = 341 kJ/mol

a, reverse a, forward52. When )E is negative, then E > E (see energy profile in Exercise 12.51). When )E ispositive (the products have higher energy than the reactants as represented in the energy profile for

a, forward a, reverseExercise 12.49), then E > E . Therefore, this reaction has a positive )E value.

a a53. The Arrhenius equation is: k = A exp (-E /RT) or in logarithmic form, ln k = -E /RT + ln A.

aHence, a graph of ln k vs. 1/T should yield a straight line with a slope equal to -E /R since thelogarithmic form of the Arrhenius equation is in the form of a straight line equation, y = mx + b. Note:We carried extra significant figures in the following ln k values in order to reduce round off error.

T (K) 1/T (K ) k (s ) ln k-1 -1

4.9 × 10338 2.96 × 10 -3 -5.32-3

318 3.14 × 10 5.0 × 10 -7.60-3 -4

298 3.36 × 10 3.5 × 10 -10.26-3 -5

aSlope = = -1.2 × 10 K = -E /R4

a aE = -slope × R = 1.2 × 10 K × , E = 1.0 × 10 J/mol = 1.0 × 10 kJ/mol4 5 2

a54. From the Arrhenius equation in logarithmic form (ln k = -E /RT + ln A), a graph of ln k vs. 1/T

ashould yield a straight line with a slope equal to -E /R and a y-intercept equal to ln A.


a aa. slope = -E /R, E = 1.10 × 10 K × = 9.15 × 10 J/mol = 91.5 kJ/mol4 4

b. The units for A are the same as the units for k(s ).-1

y-intercept = ln A, A = e = 3.54 × 10 s33.5 14 -1

c. a aln k = -E /RT + ln A or k = A exp(-E /RT)

k = 3.54 × 10 s × exp = 3.24 × 10 s14 -1 -2 -1

a55. k = A exp(-E /RT) or ln k = + ln A (the Arrhenius equation)

For two conditions: (Assuming A is temperature independent.)

1 1 2 2 aLet k = 2.0 × 10 s , T = 298 K; k = ?, T = 348 K; E = 15.0 × 10 J/mol3 -1 3

ln = 0.87

2 ln = 0.87, = e = 2.4, k = 2.4(2.0 × 10 ) = 4.8 × 10 s0.87 3 3 -1

56. For two conditions: ln (Assuming A factor is T independent.)

a0.57 = (2.5 × 10 ), E = 1.9 × 10 J/mol = 19 kJ/mol-4 4

1 a57. ln ; = 7.00, T = 295 K, E = 54.0 × 10 J/mol3

ln (7.00) = = 3.00 × 10-4

2= 3.09 × 10 , T = 324 K = 51°C -3


2 158. ; Since the rate doubles, then k = 2 k .

aln (2.00) = , E = 5.3 × 10 J/mol = 53 kJ/mol4

3 2 359. H O (aq) + OH (aq) ÷ 2 H O(l) should have the faster rate. H O and OH will be electrostatically+ - + -

2 aattracted to each other; Ce and Hg will repel each other (so E is much larger).4+ 2+

60. Carbon cannot form the fifth bond necessary for the transition state because of the small atomic sizeof carbon and because carbon doesn’t have low energy d orbitals available to expand the octet.

Catalysts

61. a. NO is the catalyst. NO is present in the first step of the mechanism on the reactant side, but it isnot a reactant since it is regenerated in the second step.

2b. NO is an intermediate. Intermediates also never appear in the overall balanced equation. In amechanism, intermediates always appear first on the product side while catalysts always appearfirst on the reactant side.

ac. k = A exp(-E /RT);

= e = 2.30.85

The catalyzed reaction is 2.3 times faster than the uncatalyzed reaction at 25°C.

62. The mechanism for the chlorine catalyzed destruction of ozone is:

3 2 O + Cl ÷ O + ClO (slow)

2ClO + O ÷ O + Cl (fast)

3 2 O + O ÷ 2 O

Since the chlorine atom-catalyzed reaction has a lower activation energy, then the Cl-catalyzed rateis faster. Hence, Cl is a more effective catalyst. Using the activation energy, we can estimate theefficiency with which Cl atoms destroy ozone as compared to NO molecules (see Exercise 12.61c).

At 25°C: = e = 523.96

At 25°C, the Cl-catalyzed reaction is roughly 52 times faster than the NO-catalyzed reaction, assuming the frequency factor A is the same for each reaction.63. The reaction at the surface of the catalyst is assumed to follow the steps:


2 2Thus, CH DSCH D should be the product. If the mechanism is possible, then the reaction mustbe:

2 4 2 2 2C H + D ÷ CH DCH D

If we got this product, then we could conclude that this is a possible mechanism. If we got some other

3 2product, e.g., CH CHD , then we would conclude that the mechanism is wrong. Even though thismechanism correctly predicts the products of the reaction, we cannot say conclusively that this is thecorrect mechanism; we might be able to conceive of other mechanisms that would give the sameproducts as our proposed one.

64. a. W since it has a lower activation energy than the Os catalyst.

w w a uncat uncat a w uncatb. k = A exp[-E (W)/RT]; k = A exp[-E (uncat)/RT]; Assume A = A

= 1.41 × 1030

The W-catalyzed reaction is approximately 10 times faster than the uncatalyzed reaction.30

2 2c. Since [H ] is in the denominator of the rate law, then H decreases the rate of the reaction. For

3 2the decomposition to occur, NH molecules must be adsorbed on the surface of the catalyst. If H

3is also adsorbed on the catalyst surface, then there are fewer sites for NH molecules to beadsorbed and the rate decreases.

Additional Exercises

2 265. Rate = k[NO] [O ] ; comparing the first two experiments, [O ] is unchanged, [NO] is tripled, andx y

the rate increases by a factor of nine. Therefore, the reaction is second order in NO (3 = 9). 2

2The order of O is more difficult to determine. Comparing the second and third experiments;

, 1.74 = 0.694 (2.50) , 2.51 = 2.50 , y = 1y y

2Rate = k[NO] [O ]; From experiment 1: 2


2.00 × 10 molecules/cm Cs = k (1.00 × 10 molecules/cm ) (1.00 × 10 molecules/cm )16 3 18 3 2 18 3

meank = 2.00 × 10 cm /molecules Cs = k-38 6 2

Rate =

= 5.68 × 10 molecules/cm Cs18 3

66. The pressure of a gas is directly proportional to concentration. Therefore, we can use the pressure

2 2data to solve the problem since Rate = -)[SO Cl ]/)t % - /)t.

Assuming a first order equation, the data and plot follow.

Time (hour) 0.00 1.00 2.00 4.00 8.00 16.00

(atm) 4.93 4.26 3.52 2.53 1.30 0.34

ln 1.595 1.449 1.258 0.928 0.262 -1.08

2 2Since the ln vs. time plot is linear, the reaction is first order in SO Cl .

a. Slope of ln(P) vs. t plot is -0.168 hour = -k, k = 0.168 hour = 4.67 × 10 s-1 -1 -5 -1

Since concentration units don’t appear in first-order rate constants, this value of k determined fromthe pressure data will be the same as if concentration data in molarity units were used.

1/2b. t = = 4.13 hour


c. = -kt = -0.168 h (20.0 hr) = -3.36, = e = 3.47 × 10-1 -3.36 -2

Fraction left = 0.0347 = 3.47%

2 567. From 338 K data, a plot of ln[N O ] vs. t is linear and the slope = -4.86 × 10 (plot not included).-3

2 5This tells us the reaction is first order in N O with k = 4.86 × 10 at 338 K.-3

2 5From 318 K data, the slope of ln[N O ] vs t plot is equal to -4.98 × 10 , so k = 4.98 × 10 at 318 K.-4 -4

aWe now have two values of k at two temperatures, so we can solve for E .

aE = 1.0 × 10 J/mol = 1.0 × 10 kJ/mol5 2

a a68. The Arrhenius equation is: k = A exp (-E /RT) or in logarithmic form, ln k = -E /RT + ln A. Hence,

aa graph of ln k vs. 1/T should yield a straight line with a slope equal to -E /R since the logarithmicform of the Arrhenius equation is in the form of a straight line equation, y = mx + b. Note: Wecarried one extra significant figure in the following ln k values in order to reduce round off error.

T (K) 1/T (K ) k (L/molCs) ln k-1

195 5.13 × 10 1.08 × 10 20.80-3 9

230. 4.35 × 10 2.95 × 10 21.81-3 9

260. 3.85 × 10 5.42 × 10 22.41-3 9

298 3.36 × 10 12.0 × 10 23.21-3 9

369 2.71 × 10 35.5 × 10 24.29-3 9

From "eyeballing" the line on the graph:

slope = = -1.35 × 10 K = 3


aE = 1.35 × 10 K × = 1.12 × 10 J/mol = 11.2 kJ/mol3 4

aFrom a graphing calculator: slope = -1.43 × 10 K and E = 11.9 kJ/mol3

69. At high [S], the enzyme is completely saturated with substrate. Once the enzyme is completelysaturated, the rate of decomposition of ES can no longer increase, and the overall rate remainsconstant.

a70. k = A exp (-E /RT); = = exp

2.50 × 10 = 3

a,catln (2.50 × 10 ) × 2.58 × 10 J/mol = -E + 5.00 × 10 J/mol3 3 4

= 5.00 × 10 J/mol - 2.02 × 10 J/mol = 2.98 × 10 J/mol = 29.8 kJ/mol4 4 4

Challenge Problems

71. Rate = k[I ] [OCl ] [OH ] ; Comparing the first and second experiments:- x - y - z

, 2.0 = 2.0 , x = 1x

Comparing the first and third experiments:

, 2.0 = 2.0 , y = 1y

Comparing the first and sixth experiments:

, 1/2 = 2.0 , z = -1z

Rate = ; The presence of OH decreases the rate of the reaction.-

For the first experiment:

, k = 60.3 s = 60. s-1 -1

meanFor all experiments, k = 60. s .-1


72. For second order kinetics:

a. = (0.250 L/molCs)t + , = 0.250 × 180. s +

= 145 M , [A] = 6.90 × 10 M-1 -3

Amount of A that reacted = 0.0100 - 0.00690 = 0.0031 M

2[A ] = (3.1 × 10 M) = 1.6 × 10 M -3 -3

b. After 3.00 minutes (180. s): [A] = 3.00 [B], 6.90 × 10 M = 3.00 [B], [B] = 2.30 × 10 M-3 -3

2, k = 2.19 L/molCs

1/2c. t = = 4.00 × 10 s2

73. a. We check for first-order dependence by graphing ln [concentration] vs. time for each set of data.The rate dependence on NO is determined from the first set of data since the ozone concentration

3is relatively large compared to the NO concentration, so [O ] is effectively constant.

Time (ms) [NO] (molecules/cm ) ln [NO]3

0 6.0 × 10 20.218

100. 5.0 × 10 20.038

500. 2.4 × 10 19.308

700. 1.7 × 10 18.958

1000. 9.9 × 10 18.417


Since ln [NO] vs. t is linear, the reaction is first order with respect to NO.

We follow the same procedure for ozone using the second set of data. The data and plot are:

3 3Time (ms) [O ] (molecules/cm ) ln [O ]3

0 1.0 × 10 23.0310

50. 8.4 × 10 22.859

1 00. 7.0 × 10 22.679

200. 4.9 × 10 22.319

300. 3.4 × 10 21.959

3The plot of ln [O ] vs. t is linear. Hence, the reaction is first order with respect to ozone.

3b. Rate = k[NO][O ] is the overall rate law.

c. For NO experiment, Rate = kN[NO] and kN = -(slope from graph of ln [NO] vs. t).

kN = -slope = - = 1.8 s-1

3 3For ozone experiment, Rate = kO[O ] and kO = -(slope from ln [O ] vs. t).

kO = -slope = - = 3.6 s-1

3 3d. From NO experiment, Rate = k[NO][O ] = kN[NO] where kN = k[O ].

kN = 1.8 s = k(1.0 × 10 molecules/cm ), k = 1.8 × 10 cm /moleculesCs-1 14 3 -14 3

3 3We can check this from the ozone data. Rate = kO[O ] = k[NO][O ] where kO = k[NO].


kO = 3.6 s = k(2.0 × 10 molecules/cm ), k = 1.8 × 10 cm /moleculesCs-1 14 3 -14 3

Both values of k agree.

74. On the energy profile to the right, R =

areactants, P = products, E = activation energy, )E = overall energy change for the reaction and I = intermediate.

a - d. See plot to the right.

e. This is a two-step reaction since an intermediate plateau appears between the reactant and theproducts. This plateau represents the energy of the intermediate. The general reaction mechanismfor this reaction is:

R ÷ I I ÷ P R ÷ P

In a mechanism, the rate of the slowest step determines the rate of the reaction. The activationenergy for the slowest step will be the largest energy barrier that the reaction must overcome.Since the second hump in the diagram is at the highest energy, the second step has the largestactivation energy and will be the rate-determining step (the slow step).

75. a. If the interval between flashes is 16.3 sec, then the rate is:

1 flash/16.3 s = 6.13 × 10 s = k-2 -1

Interval k T

16.3 s 6.13 × 10 s 21.0°C (294.2 K)-2 -1

13.0 s 7.69 × 10 s 27.8°C (301.0 K)-2 -1

a; Solving: E = 2.5 × 10 J/mol = 25 kJ/mol4

b. = 0.30

k = e × 6.13 × 10 = 8.3 × 10 s ; Interval = 1/k = 12 seconds0.30 -2 -2 -1

c. T Interval 54-2(Intervals)

21.0 °C 16.3 s 21 °C27.8 °C 13.0 s 28 °C30.0 °C 12 s 30. °C


This rule of thumb gives excellent agreement to two significant figures.

76. We need the value of k at 500. K.

2 = e , k = 1.0 × 10 L/molCs22.2 -2

Since the decomposition reaction is an elementary reaction, then the rate law can be written using the

2coefficients in the balanced equation. For this reaction: Rate = k[NO ] . To solve for the time, we2

must use the integrated rate law for second-order kinetics. The major problem now is converting unitsso they match. Rearranging the ideal gas law gives n/V = P/RT. Substituting P/RT for concentrationunits in the second-order integrated rate law equation:

t = = 1.1 × 10 s3

chapter twelve chemical kinetics - … · chapter twelve chemical kinetics questions 9. ... 13. a...

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