# SM Chapter 15

July 17, 2017 | Author: 李承家 | Category: Catalysis, Activation Energy, Chemical Kinetics, Unit Processes, Chemical Reactions

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CHAPTER 15 CHEMICAL KINETICS

Reaction Rates 10.

The reaction rate is defined as the change in concentration of a reactant or product per unit time. Consider the general reaction: aA  products where rate =

 d[A] dt

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

a

[A]

b c 0

t1

t2

time

An instantaneous rate is the slope of a tangent line to the graph of [A] vs. t. We can determine the instantaneous rate at any time during the reaction. On the plot, tangent lines at t  0 and t = t1 are drawn. The slope of these tangent lines would be the instantaneous rates at t  0 and t = t1. We call the instantaneous rate at t  0 the initial rate. The average rate is measured over a period of time. For example, the slope of the dashed line connecting points a and c is the average rate of the reaction over the entire length of time 0 to t2 (average rate = Δ[A]/Δt). An average rate is determined over some time period, whereas an instantaneous rate is determined at one specific time. The rate that is largest is generally the initial rate. At t  0, the slope of the tangent line is greatest, which means the rate is largest at t  0. The initial rate is used by convention so that the rate of reaction only depends on the forward reaction; at t  0, the reverse reaction is insignificant because no products are present yet.

613

614 11.

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CHEMICAL KINETICS

Using the coefficients in the balanced equation to relate the rates:

d[ NH3 ] d[H 2 ] d[ N 2 ] d[ N 2 ] 3 and  2 dt dt dt dt So :

d[ NH3 ] 1 d[H 2 ] 1 d[ NH3 ] 2 d[H 2 ]  or  3 dt 2 dt dt 3 dt

Ammonia is produced at a rate equal to 2/3 of the rate of consumption of hydrogen. 12.

The coefficients in the balanced reaction relate the rate of disappearance of reactants to the rate of production of products. From the balanced reaction, the rate of production of P 4 will be 1/4 the rate of disappearance of PH3, and the rate of production of H2 will be 6/4 the rate of disappearance of PH3. By convention, all rates are given as positive values. Rate  

Δ[PH3 ]  (0.0048 mol/ 2.0 L)  = 2.4 × 103 mol L1 s1 Δt s

Δ[P4 ] 1 Δ[PH3 ]  = 2.4 × 103/4 = 6.0 × 104 mol L1 s1 Δt 4 Δt Δ[H 2 ] 6 Δ[PH3 ]  = 6(2.4 × 103)/4 = 3.6 × 103 mol L1 s1 Δt 4 Δt 13.

0.0120/0.0080 = 1.5; reactant B is used up 1.5 times faster than reactant A. This corresponds to a 3 to 2 mole ratio between B and A in the balanced equation. 0.0160/0.0080 = 2; product C is produced twice as fast as reactant A is used up. So the coefficient for C is twice the coefficient for A. A possible balanced equation is 2A + 3B  4C.

14.

1.24  1012 cm3 1L 6.022  1023 molecules = 7.47 × 108 L mol1 s1   3 molecules s mol 1000 cm

15.

mol  mol   k Rate = k[Cl] [CHCl3],  Ls  L 

16.

a. The units for rate are always mol L1 s1.

1/ 2

1/2

c. Rate = k[A],

mol  mol   k  Ls  L 

k must have units of s-1. e. L2 mol2 s1

 mol  1/2 -1/2 1   ; k must have units of L mol s .  L  b. Rate = k; k has units of mol L1 s1. 2

d. Rate = k[A] ,

mol  mol   k  Ls  L 

k must have units L mol1 s1.

2

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615

Rate Laws from Experimental Data: Initial Rates Method 17.

a. In the first two experiments, [NO] is held constant and [Cl2] is doubled. The rate also doubled. Thus the reaction is first order with respect to Cl2. Or mathematically: Rate = k[NO]x[Cl2]y 0.36 k (0.10) x (0.20) y (0.20) y , 2.0 = 2.0y, y = 1   x y y 0.18 k (0.10) (0.10) (0.10)

We can get the dependence on NO from the second and third experiments. Here, as the NO concentration doubles (Cl2 concentration is constant), the rate increases by a factor of four. Thus, the reaction is second order with respect to NO. Or mathematically: 1.45 k (0.20) x (0.20) (0.20) x , 4.0 = 2.0x, x = 2; so, Rate = k[NO]2[Cl2].   0.36 k (0.10) x (0.20) (0.10) x

Try to examine experiments where only one concentration changes at a time. The more variables that change, the harder it is to determine the orders. Also, these types of problems can usually be solved by inspection. In general, we will solve using a mathematical approach, but keep in mind that you 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: 2

0.18 mol  0.10 mol   0.10 mol  2 2 1  k    , k = 180 L mol min L min L L     From the other experiments: k = 180 L2 mol2 min1 (second exp.); k = 180 L2 mol2 min1 (third exp.) The average rate constant is kmean = 1.8 × 102 L2 mol2 min1. 18.

Rate = k[NO]x[O2]y; Comparing the first two experiments, [O2] is unchanged, [NO] is tripled, and the rate increases by a factor of nine. Therefore, the reaction is second order in NO (3 2 = 9). The order of O2 is more difficult to determine. Comparing the second and third experiments: 3.13  1017 k (2.50  1018 ) 2 (2.50  1018 ) y  1.80  1017 k (3.00  1018 ) 2 (1.00  1018 ) y 1.74 = 0.694(2.50)y, 2.51 = 2.50y, y = 1 Rate = k[NO]2[O2]; from experiment 1: 2.00 × 1016 molecules cm3 s1 = k(1.00 × 1018 molecules/cm3)2  (1.00 × 1018 molecules/cm3)

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k = 2.00 × 1038 cm6 molecules2 s1 = kmean 2

Rate =

 6.21  1018 molecules  7.36  1018 molecules 2.00  1038 cm6       cm3 cm3 molecules2 s  

Rate = 5.68 × 1018 molecules cm3 s1 19.

a. Rate = k[ClO2]x[OH]y; From the first two experiments: 2.30 × 101 = k(0.100)x(0.100)y and 5.75 × 102 = k(0.0500)x(0.100)y (0.100) x = 2.00x, x = 2 (0.0500) x

Dividing the two rate laws: 4.00 =

Comparing the second and third experiments: 2.30 × 101 = k(0.100)(0.100)y and 1.15 × 101 = k(0.100)(0.0500)y Dividing: 2.00 =

(0.100) y = 2.0y, y = 1 y (0.050)

The rate law is: Rate = k[ClO2]2[OH] 2.30 × 101 mol L1 s1 = k(0.100 mol/L)2(0.100 mol/L), k = 2.30 × 102 L2 mol2 s1 = kmean 2

b. Rate = 20.

2.30  102 L2  0.175 mol  0.0844 mol  = 0.594 mol L1 s1   2 L L mol s  

a. Rate = k[Hb]x[CO]y Comparing the first two experiments, [CO] is unchanged, [Hb] doubles, and the rate doubles. Therefore, x = 1, and the reaction is first order in Hb. Comparing the second and third experiments, [Hb] is unchanged, [CO] triples, and the rate triples. Therefore, y = 1 and the reaction is first order in CO. b. Rate = k[Hb][CO] c. From the first experiment: 0.619 µmol L1 s1 = k(2.21 µmol/L)(1.00 µmol/L), k = 0.280 L µmol1 s1 The second and third experiments give similar k values, so kmean = 0.280 L µmol1 s1. d. Rate = k[Hb][CO] =

0.280 L 3.36 μmol 2.40 μmol   = 2.26 µmol L1 s1 μmol s L L

CHAPTER 15 21.

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617

a. Rate = k[NOCl]n; using experiments two and three: 2.66  104 k (2.0  1016 ) n  , 4.01 = 2.0n, n = 2; Rate = k[NOCl]2 3 16 n 6.64  10 k (1.0  10 ) 2

 3.0  1016 molecules  5.98  104 molecules   , k = 6.6 × 1029 cm3 molecules1 s1  k 3   cm3 s cm  

b.

The other three experiments give (6.7, 6.6, and 6.6) × 1029 cm3 molecules1 s1, respectively. The mean value for k is 6.6 × 1029 cm3 molecules1 s1. 6.6  1029 cm3 1L 6.022  1023 molecules 4.0  108 L =   molecules s mol mol s 1000 cm3

c. 22.

Rate = k[N2O5]x; the rate laws for the first two experiments are: 2.26 × 103 = k(0.190)x and 8.90 × 104 = k(0.0750)x Dividing the two rate laws: 2.54 = k 

(0.190) x = (2.53)x, x = 1; Rate = k[N2O5] (0.0750) x

Rate 8.90  104 mol L1 s 1 = 1.19 × 102 s1  [ N 2 O5 ] 0.0750 mol / L

The other experiments give similar values for k. kmean = 1.19 × 102 s1 23.

Rate = k[I]x[OCl]y[OH]z; Comparing the first and second experiments: 18.7  103 k (0.0026) x (0.012) y (0.10) z , 2.0 = 2.0x, x = 1  9.4  103 k (0.0013) x (0.012) y (0.10) z

Comparing the first and third experiments: 9.4  103 k (0.0013)(0.012) y (0.10) z , 2.0 = 2.0y, y = 1  4.7  103 k (0.0013)(0.0060) y (0.10) z

Comparing the first and sixth experiments: 4.8  103 k (0.0013)(0.012)(0.20) z , 1/2 = 2.0z, z = 1  9.4  103 k (0.0013)(0.012)(0.10) z

Rate =

k[I  ][OCl  ] ; the presence of OH decreases the rate of the reaction.  [OH ]

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For the first experiment: 9.4  103 mol (0.0013 mol/ L) (0.012 mol/ L) , k = 60.3 s 1 = 60. s 1 k Ls (0.10 mol/ L)

For all experiments, kmean = 60. s 1 .

24.

 [A]  Rate2 k[A]2x    2  x Rate1 k[A]1  [A]1 

x

The rate doubles as the concentration quadruples: 2 = (4)x, x = 1/2 The order is 1/2 (the square root of the concentration of reactant). For a reactant that has an order of 1 and the reactant concentration is doubled:

Rate2 1  (2) 1  Rate1 2 The rate will decrease by a factor of 1/2 when the reactant concentration is doubled for a 1 order reaction. Negative orders are seen for substances that hinder or slow down a reaction. 25.

Rate = k[H2SeO 3 ]x [H ] y [I ]z ; comparing the first and second experiments:

3.33  107 7

1.66  10

k (2.0  104 ) x (2.0  102 ) y (2.0  102 ) z 4 x

2 y

2 z

k (1.0  10 ) (2.0  10 ) (2.0  10 )

, 2.01 = 2.0x, x = 1

Comparing the first and fourth experiments:

6.66  107 1.66  107

k (1.0  104 )(4.0  102 ) y (2.0  102 ) z k (1.0  104 )(2.0  102 ) y (2.0  102 ) z

, 4.01 = 2.0y, y = 2

Comparing the first and sixth experiments:

13.2  107 1.66  107

k (1.0  104 )(2.0  102 ) 2 (4.0  102 ) z k (1.0  104 )(2.0  102 ) 2 (2.0  102 ) z

7.95 = 2.0 z , log(7.95) = z log(2.0), z = Rate = k[H2SeO3][H+]2[I−]3

log (7.95) = 2.99 ≈ 3 log(2.0)

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619

Experiment 1: 2

 1.0  104 mol   2.0  102 mol   2.0  102 mol  1.66  107 mol      k     Ls L L L     

3

k = 5.19 × 105 L5 mol-5 s-1 = 5.2 × 105 L5 mol-5 s-1 = kmean

Integrated Rate Laws 26.

Zero order:

 d[A]  k, dt [ A ]t

[ A]

[ A ]t

t

[ A ]0

0

 d[A]    k dt t

  kt , [A]t  [A]0   kt, [A]t = kt + [A]0

[ A ]0

0

First order:

 d[A]  k[A], dt [ A ]t

ln[A]

[ A ]t

t

d[A]  [A]    k dt [ A ]0 0

  kt, ln[A]t  ln[A]0   kt, ln[A]t = kt + ln[A]0

[ A ]0

Second order:

 d[A]  k[A]2 , dt

27.

1 [ A]

[ A ]t

[ A ]t

  kt, 

[ A ]0

Zero order: t1/2 =

t

d[A]  [A]2    k dt [ A ]0 0

[ A ]0 ; 2k

1 1 1 1    kt,  kt  [A]t [A]0 [ A]t [A]0 first order: t1/2 =

ln 2 1 ; second order: t1/2 = k[A]0 k

For a first-order reaction, if the first half-life equals 20. s, the second half-life will also be 20. s because the half-life for a first-order reaction is concentration-independent. The second half-life for a zero-order reaction will be 1/2(20.) = 10. s. This is because the half-life for a zero-order reaction has a direct relationship with concentration (as the concentration decreases by a factor of 2, the half-life decreases by a factor of 2). Because a second-order reaction has an inverse relationship between t1/2 and [A]0, the second half-life will be 40. s (twice the first half-life value).

620 28.

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a. Because the 1/[A] versus time plot was linear, the reaction is second order in A. The slope of the 1/[A] versus time plot equals the rate constant k. Therefore, the rate law, the integrated rate law, and the rate constant value are: Rate = k[A]2;

1 1 = kt + ; k = 3.60 × 10-2 L mol-1 s-1 [ A ]0 [A]

b. The half-life expression for a second-order reaction is: t1/2 = For this reaction: t1/2 =

1 k[A]0

1 2

3.60  10

1

L mol s

1

3

 2.80  10 mol/ L

= 9.92 × 103 s

Note: We could have used the integrated rate law to solve for t1/2, where [A] = (2.80 × 103 /2) mol/L. c. Because the half-life for a second-order reaction depends on concentration, we must use the integrated rate law to solve.

1 1 1 3.60  102 L 1 = kt + ,   t 4 [A] [ A ]0 7.00  10 M mol s 2.80  103 M 1.43 × 103  357 = (3.60 × 102)t, t = 2.98 × 104 s 29.

a. Because the ln[A] versus time plot was linear, the reaction is first order in A. The slope of the ln[A] versus time plot equals k. Therefore, the rate law, the integrated rate law, and the rate constant value are: Rate = k[A]; ln[A] = kt + ln[A]0; k = 2.97 × 102 min1 b. The half-life expression for a first order rate law is: t1/2 =

0.6931 ln 2 0.6931  , t1/2 = = 23.3 min k k 2.97  10 2 min 1

c. 2.50 × 103 M is 1/8 of the original amount of A present initially, so the reaction is 87.5% complete. When a first-order reaction is 87.5% complete (or 12.5% remains), then the reaction has gone through 3 half-lives: 100%  50.0%  25.0%  12.5%; t = 3 × t1/2 = 3 × 23.3 min = 69.9 min t1/2 t1/2 t1/2 Or we can use the integrated rate law:  2.50  103 M   [ A]  2 1     kt, ln ln  2.00  102 M  = (2.97 × 10 min )t [ A ] 0    

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CHEMICAL KINETICS

t= 30.

621

ln(0.125) = 70.0 min  2.97  10 2 min 1

a. Because the [C2H5OH] versus time plot was linear, the reaction is zero order in C2H5OH. The slope of the [C2H5OH] versus time plot equals -k. Therefore, the rate law, the integrated rate law, and the rate constant value are: Rate = k[C2H5OH]0 = k; [C2H5OH] = kt + [C2H5OH]0; k = 4.00 × 105 mol L1 s1 b. The half-life expression for a zero-order reaction is t1/2 = [A]0/2k. t1/2 =

[C 2 H 5OH]0 1.25  102 mol/ L = 156 s  2k 2  4.00  105 mol L1 s 1

Note: We could have used the integrated rate law to solve for t1/2, where [C2H5OH] = (1.25 × 102/2) mol/L. c. [C2H5OH] = kt + [C2H5OH]0 , 0 mol/L = (4.00 × 105 mol L1 s1)t + 1.25 × 102 mol/L t= 31.

1.25  102 mol/ L = 313 s 4.00  105 mol L1 s 1

The first assumption to make is that the reaction is first order. For a first order reaction, a graph of ln[H2O2] versus time will yield a straight line. If this plot is not linear, then the reaction is not first order, and we make another assumption. Time (s)

[H2O2] (mol/L)

0 120. 300. 600. 1200. 1800. 2400. 3000. 3600.

1.00 0.91 0.78 0.59 0.37 0.22 0.13 0.082 0.050

ln[H2O2] 0.000 0.094 0.25 0.53 0.99 1.51 2.04 2.50 3.00

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

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The plot of ln[H2O2] versus time is linear. Thus the reaction is first order. The differential  d[H 2 O 2 ] rate law and integrated rate law are Rate = = k[H2O2] and ln[H2O2] = kt + dt ln[H2O2]0. We determine the rate constant k by determining the slope of the ln[H2O2] versus time plot (slope = k). Using two points on the curve gives: slope = k =

Δy 0  (3.00)  = 8.3 × 104 s1, k = 8.3 × 104 s1 Δx 0  3600.

To determine [H2O2] at 4000. s, use the integrated rate law, where [H2O2]0 = 1.00 M.

 [H 2 O 2 ]   = kt ln[H2O2] = kt + ln[H2O2]0 or ln   [H 2 O 2 ]0 

 [H O ]  ln  2 2  = 8.3 × 104 s1 × 4000. s, ln[H2O2] = 3.3, [H2O2] = e3.3 = 0.037 M  1.00  32.

The first assumption to make is that the reaction is first order. For a first-order reaction, a graph of ln[C4H6] versus t should yield a straight line. If this isn't linear, then try the second order plot of 1/[C4H6] versus t. The data and the plots follow: Time

195

604

1246

2180

[C4H6]

1.6 × 102

1.5 × 102

1.3 × 102

1.1 × 102

ln[C4H6] 1/[C4H6]

4.14 62.5

4.20 66.7

4.34 76.9

4.51 90.9

6210 s 0.68 × 102 M 4.99 147 M 1

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

The natural log plot is not linear, so the reaction is not first order. Because the second order plot of 1/[C4H6] versus t is linear, we can conclude that the reaction is second order in butadiene. The differential rate law is:

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CHEMICAL KINETICS

623

Rate = k[C4H6]2 For a second-order reaction, the integrated rate law is

1 1 = kt + . [C 4 H 6 ]0 [C 4 H 6 ]

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 = 33.

144 L / mol  73 L / mol = 1.4 × 102 L mol1 s1 6000. s  1000. s

From the data, the pressure of C2H5OH decreases at a constant rate of 13 torr for every 100. s. Since the rate of disappearance of C2H5OH is not dependent on concentration, the reaction is zero order in C2H5OH. k=

13 torr 1 atm  = 1.7 × 104 atm/s 100. s 760 torr

The rate law and integrated rate law are:

 1 atm   = kt + 0.329 atm Rate = k = 1.7 × 104 atm/s; PC2 H5OH = kt + 250. torr   760 torr  At 900. s: PC2 H5OH = 1.7 × 104 atm/s × 900. s + 0.329 atm = 0.176 atm = 0.18 atm = 130 torr

34.

a. Because the 1/[A] versus time plot is linear with a positive slope, the reaction is second order with respect to A. The y intercept in the plot will equal 1/[A]0. Extending the plot, the y intercept will be about 10, so 1/10 = 0.1 M = [A]0. b. The slope of the 1/[A] versus time plot will equal k. Slope = k =

(60  20) L / mol = 10 L mol1 s1 (5  1) s

1 10 L 1 1   9s = kt + = 100, [A] = 0.01 M [ A ]0 mol s 0.1 M [A] c. For a second-order reaction, the half-life does depend on concentration: t1/2 = First half-life: t1/2 =

1 10 L 0.1 mol  mol s L

=1s

Second half-life ([A]0 is now 0.05 M): t1/2 = 1/(10 × 0.05) = 2 s Third half-life ([A]0 is now 0.025 M): t1/2 = 1/(10 × 0.025) = 4 s

1 k[A]0

624 35.

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Assume the reaction is first order and see if the plot of ln[NO2] versus time is linear. If this isn’t linear, try the second order plot of 1/[NO2] versus time. The data and plots follow. Time (s)

[NO2] (M)

ln[NO2]

0 1.20 × 103 3.00 × 103 4.50 × 103 9.00 × 103 1.80 × 104

0.500 0.444 0.381 0.340 0.250 0.174

0.693 0.812 0.965 1.079 1.386 1.749

1/[NO2] (M -1) 2.00 2.25 2.62 2.94 4.00 5.75

The plot of 1/[NO2] versus time is linear. The reaction is second order in NO2. The differential rate law and integrated rate law are Rate = k[NO2]2 and

1 1 = kt + . [ NO 2 ] [ NO 2 ]0

The slope of the plot 1/[NO2] versus time gives the value of k. Using a couple of points on the plot: slope = k =

Δy (5.75  2.00) M 1 = 2.08 × 104 L mol1 s1  Δx (1.80  104  0) s

To determine [NO2] at 2.70 × 104 s, use the integrated rate law, where 1/[NO2]0= 1/0.500 M = 2.00 M 1. 1 1 1 2.08  104 L = kt + , = × 2.70 × 104 s + 2.00 M 1 [ NO 2 ] [ NO 2 ]0 [ NO 2 ] mol s

1 = 7.62, [NO2] = 0.131 M [ NO 2 ]

CHAPTER 15 36.

CHEMICAL KINETICS

625

a. First, assume the reaction to be first order with respect to O. Hence a graph of ln[O] versus t would be linear if the reaction is first order. t (s) 0 10. × 103 20. × 103 30. × 103

[O] (atoms/cm3)

ln[O]

5.0 × 109 1.9 × 109 6.8 × 108 2.5 × 108

22.33 21.37 20.34 19.34

Because the graph is linear, we can conclude the reaction is first order with respect to O. b. The overall rate law is: Rate = k[NO2][O] Because NO2 was in excess, its concentration is constant. Thus for this experiment, the rate law is Rate: k[O], where k = k[NO2]. In a typical first-order plot, the slope equals k. For this experiment, the slope equals k = k[NO2]. From the graph: slope =

19.34  22.23 = 1.0 × 102 s1, k = slope = 1.0 × 102 s1 3 (30.  10  0) s

To determine k, the actual rate constant: k = k[NO2], 1.0 × 102 s1 = k(1.0 × 1013 molecules/cm3) k = 1.0 × 1011 cm3 molecules1 s1 37.

a. We check for first-order dependence by graphing ln[concentration] versus time for each set of data. The rate dependence on NO is determined from the first set of data since the ozone concentration is relatively large compared to the NO concentration, so it is effectively constant.

626

CHAPTER 15 Time (ms) 0 100. 500. 700. 1000.

[NO] (molecules/cm3) 8

6.0 × 10 5.0 × 108 2.4 × 108 1.7 × 108 9.9 × 107

CHEMICAL KINETICS

ln[NO] 20.21 20.03 19.30 18.95 18.41

Because ln[NO] versus 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: Time (ms)

[O3] (molecules/cm3)

ln[O3]

0 50. 100. 200. 300.

1.0 × 1010 8.4 × 109 7.0 × 109 4.9 × 109 3.4 × 109

23.03 22.85 22.67 22.31 21.95

The plot of ln[O3] versus t is linear. Hence the reaction is first order with respect to ozone. b. Rate = k[NO][O3] is the overall rate law.

CHAPTER 15

CHEMICAL KINETICS

627

c. For NO experiment, Rate = k[NO] and k = (slope from graph of ln[NO] versus t). k = slope = 

18.41  20.21 = 1.8 s1 (1000.  0)  103 s

For ozone experiment, Rate = k[O3] and k = (slope from ln[O3] versus t). k = slope = 

(21.95  23.03) = 3.6 s1 (300.  0)  103 s

d. From NO experiment, Rate = k[NO][O3] = k[NO] where k = k[O3]. k = 1.8 s1 = k(1.0 × 1014 molecules/cm3), k = 1.8 × 1014 cm3 molecules1 s1 We can check this from the ozone data. Rate = k[O3] = k[NO][O3], where k = k[NO]. k = 3.6 s1 = k(2.0 × 1014 molecules/cm3), k = 1.8 × 1014 cm3 molecules1 s1 Both values of k agree. 38.

This problem differs in two ways from previous problems: 1. A product is measured instead of a reactant. 2. Only the volume of a gas is given and not the concentration. We can find the initial concentration of C6H5N2Cl from the amount of N2 evolved after infinite time when all the C6H5N2Cl has decomposed (assuming the reaction goes to completion). n=

PV 1.00 atm  (58.3  103 L)  = 2.20 × 103 mol N2 0.08206L atm RT  323 K K mol

Because each mole of C6H5N2Cl that decomposes produces one mole of N2, the initial concentration (t = 0) of C6H5N2Cl was: 2.20  103 mol = 0.0550 M 40.0  103 L

We can similarly calculate the moles of N2 evolved at each point of the experiment, subtract that from 2.20 × 103 mol to get the moles of C6H5N2Cl remaining, and then calculate [C6H5N2Cl] at each time. We would then use these results to make the appropriate graph to determine the order of the reaction. Because the rate constant is related to the slope of the straight line, we would favor this approach to get a value for the rate constant. There is a simpler way to check for the order of the reaction that saves doing a lot of math. The quantity (V  Vt ), where V = 58.3 mL N2 evolved and Vt = mL of N2 evolved at time t, will be proportional to the moles of C6H5N2Cl remaining; (V  Vt ) will also be

628

CHAPTER 15

CHEMICAL KINETICS

proportional to the concentration of C6H5N2Cl. Thus we can get the same information by using (V  Vt ) as our measure of [C6H5N2Cl]. If the reaction is first order, a graph of ln(V  Vt ) versus t would be linear. The data for such a graph are: t (s)

Vt (mL)

0 6 9 14 22 30.

0 19.3 26.0 36.0 45.0 50.4

(V  Vt )

ln(V  Vt )

58.3 39.0 32.3 22.3 13.3 7.9

4.066 3.664 3.475 3.105 2.588 2.07

We can see from the graph that this plot is linear, so the reaction is first order. The differential rate law is d[C6H5N2Cl]/dt = Rate = k[C6H5N2Cl], and the integrated rate law is ln[C6H5N2Cl] = kt + ln[C6H5N2Cl]0. From separate data, k was determined to be 6.9 × 102 s1. 39.

Because [V]0 > > [AV]0, the concentration of V is essentially constant in this experiment. We have a pseudo-first-order reaction in AV: Rate = k[AV][V] = k[AV], where k = k[V]0 The slope of the ln[AV] versus time plot is equal to k. k = slope = 0.32 s1; k 

40.

k 0.32 s 1   1.6 L mol-1 s-1 [V]0 0.20 mol/L

 [A]  ln 2 0.6931  = kt; k = ln  = 4.85 × 102 d1 t 1/ 2 14.3 d  [A]0  If [A]0 = 100.0, then after 95.0% completion, [A] = 5.0.

CHAPTER 15

CHEMICAL KINETICS

629

 5.0  2 1 ln  = -4.85 × 10 d × t, t = 62 days 100 . 0   41.

Comparing experiments 1 and 2, as the concentration of AB is doubled, the initial rate increases by a factor of 4. The reaction is second order in AB. Rate = k[AB]2, 3.20 × 103 mol L1 s 1 = k1(0.200 M)2 k = 8.00 × 102 L mol1 s 1 = kmean For a second-order reaction: 1 1  t1/2 = = 12.5 s 2 1 1 k[AB]0 8.00  10 L mol s  1.00 mol/ L

42.

Box a has 8 NO2 molecules. Box b has 4 NO2 molecules, and box c has 2 NO2 molecules. Box b represents what is present after the first half-life of the reaction, and box c represents what is present after the second half-life. a. For first order kinetics, t1/2 = 0.693/k; the half-life for a first order reaction is concentration independent. Therefore, the time for box c, the time it takes to go through two half-lives, will be 10 + 10 = 20 minutes. b. For second order kinetics, t1/2 = 1/k[A]0; the half-life for a second order reaction is inversely proportional to the initial concentration. So if the first half-life is 10 minutes, the second half-life will be 20 minutes. For a second order reaction, the time for box c will be 10 + 20 = 30 minutes. c. For zero order kinetics, t1/2 = [A]0/2k; the half-life for a zero order reaction is directly related to the initial concentration. So if this reaction was zero order, then the second half-life would decrease from 10 min to 5 min. The time for box c will be 10 + 5 = 15 minutes if the reaction is zero order.

43.

For a first-order reaction, the integrated rate law is ln([A]/[A]0) = kt. Solving for k:

 0.250 mol/ L   = k × 120. s, k = 0.0116 s1 ln  1.00 mol/ L 

 0.350 mol/ L   = 0.0116 s1 × t, t = 150. s ln  2.00 mol/ L  44.

Successive half-lives increase in time for a second-order reaction. Therefore, assume reaction is second order in A. t1/2 =

1 1 1 , k= = = 1.0 L mol1 min1 t1/ 2 [A]0 10.0 min(0.10 M ) k[A]0

630

CHAPTER 15

a.

CHEMICAL KINETICS

1 1 1.0 L 1  = kt +  80.0 min + = 90. M 1, [A] = 1.1 × 102 M [A] [A]0 0.10 M mol min

b. 30.0 min = 2 half-lives, so 25% of original A is remaining. [A] = 0.25(0.10 M) = 0.025 M 45.

a. The integrated rate law for this zero-order reaction is [HI] = kt + [HI]0.  1.20  104 mol      25 min  60 s   0.250 mol [HI] = kt + [HI]0, [HI] =      Ls min  L  

[HI] = 0.18 mol/L + 0.250 mol/L = 0.07 M b. [HI] = 0 = kt + [HI]0, kt = [HI]0, t = t=

46.

0.250 mol/L 1.20  104 mol L-1 s -1

[HI]0 k

= 2080 s = 34.7 min

a. The integrated rate law for a second order reaction is 1/[A] = kt + 1/[A]0, and the halflife expression is t1/2 = 1/k[A]0. We could use either to solve for t1/2. Using the integrated rate law:

1 1 1.11 L / mol 0.555 L = k × 2.00 s + , k= = (0.900/2)mol/L 0.900 mol/ L 2.00 s mol s b. 47.

8.9 L / mol 1 1 = 0.555 L mol1 s1 × t + , t= = 16 s 0.100 mol/ L 0.900 mol/ L 0.555 L mol1 s 1

a. When a reaction is 75.0% complete (25.0% of reactant remains), this represents two halflives (100% → 50%→ 25%). The first-order half-life expression is t1/2 = (ln 2)/k. Because there is no concentration dependence for a first-order half-life, 320. s = two halflives, t1/2 = 320./2 = 160. s. This is both the first half-life, the second half-life, etc. b. t1/2 =

ln 2 ln 2 ln 2 , k   = 4.33 × 103 s 1 k t 1/ 2 160. s

At 90.0% complete, 10.0% of the original amount of the reactant remains, so [A] = 0.100[A]0.

 [ A]  0.100[A]0 ln(0.100)    kt, ln   (4.33  103 s 1 )t, t   532 s ln  [A]0  4.33  103 s 1  [A]0  48.

Because [B]0 >> [A]0, the B concentration is essentially constant during this experiment, so rate = k[A] where k = k[B]2. For this experiment, the reaction is a pseudo-first-order reaction in A.

CHAPTER 15

a.

CHEMICAL KINETICS

631

 3.8  103 M   [A]   = k × 8.0 s, k = 0.12 s 1  = kt, ln  ln  2    [A]0   1.0  10 M 

For the reaction: k = k[B]2, k = 0.12 s1/(3.0 mol/L)2 = 1.3 × 102 L2 mol2 s1 b. t1/2 =

c.

ln 2 0.693  = 5.8 s k' 0.12 s 1

  [A] [ A]  = 0.12 s 1 × 13.0 s, = e0.12(13.0) = 0.21 ln  2 2  1 . 0  10 M 1 . 0  10  

[A] = 2.1 × 103 M d. [A] reacted = 0.010 M  0.0021 M = 0.008 M [C] reacted = 0.008 M ×

2 mol C = 0.016 M ≈ 0.02 M 1 mol A

[C]remaining = 2.0 M  0.02 M = 2.0 M; as expected, the concentration of C basically remains constant during this experiment since [C]0 >> [A]0. 49.

a. Because [A]0