613
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 = dt ] A [ d
If we graph [A] vs. t, it would usually look like the solid 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 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.
[A] 0 a c b t1 t2 time
11. Using the coefficients in the balanced equation to relate the rates: dt ] N [ d 2 dt ] NH [ d and dt ] N [ d 3 dt ] H [ d 2
2 3
2 dt ] H [ d 3 2 dt ] NH [ d or dt ] NH [ d 2 1 dt ] H [ d 3 1 : So 2
3 3
2Ammonia 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 P4 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 s ) L 0 . 2 / mol 0048 . 0 ( t Δ ] PH [ Δ 3
= 2.4 × 103 mol L1 s1 t Δ ] PH [ Δ 4 1 t Δ ] P [ Δ 4 3
= 2.4 × 103/4 = 6.0 × 104 mol L1 s1 t Δ ] PH [ Δ 4 6 t Δ ] H [ Δ 2 3
= 6(2.4 × 103)/4 = 3.6 × 103 mol L1 s113. 0.0120/0.0080 = 1.5; reactant B is used up 1.5 times faster than reactant A. This corres-ponds 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. mol molecules 10 022 . 6 cm 1000 L 1 s molecules cm 10 24 . 1 23 3 3 12 = 7.47 × 108 L mol1 s1 15. Rate = k[Cl]1/2[CHCl3],
L mol L mol k s L mol 1/2; k must have units of L1/2 mol-1/2 s1.
16. a. The units for rate are always mol L1 s1. b. Rate = k; k has units of mol L1 s1.
c. Rate = k[A],
L mol k s L mol d. Rate = k[A]2, 2 L mol k s L mol
k must have units of s-1. k must have units L mol1 s1. e. L2 mol2 s1
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 y y x x ) 10 . 0 ( ) 20 . 0 ( ) 10 . 0 ( ) 10 . 0 ( k ) 20 . 0 ( ) 10 . 0 ( k 18 . 0 36 . 0 y y , 2.0 = 2.0y, y = 1
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:
x x x x ) 10 . 0 ( ) 20 . 0 ( ) 20 . 0 ( ) 10 . 0 ( k ) 20 . 0 ( ) 20 . 0 ( k 36 . 0 45 . 1
, 4.0 = 2.0x, x = 2; so, Rate = k[NO]2[Cl2].
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:
L mol 10 . 0 L mol 10 . 0 k min L mol 18 . 0 2 , k = 180 L2 mol2 min1From the other experiments:
k = 180 L2 mol2 min1 (second exp.); k = 180 L2 mol2 min1 (third exp.) The average rate constant is kmean = 1.8 × 10
2
L2 mol2 min1. 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 (32 = 9). The order of O2 is more difficult to determine. Comparing the second and third
experiments: y y ) 10 00 . 1 ( ) 10 00 . 3 ( k ) 10 50 . 2 ( ) 10 50 . 2 ( k 10 80 . 1 10 13 . 3 18 2 18 18 2 18 17 17 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 cm3 s1 = k(1.00 × 1018 molecules/cm3)2
k = 2.00 × 1038 cm6 molecules2 s1 = kmean Rate = s molecules cm 10 00 . 2 2 6 38 3 18 2 3 18 cm molecules 10 36 . 7 cm molecules 10 21 . 6 Rate = 5.68 × 1018 molecules cm3 s1 19. a. Rate = k[ClO2] x
[OH]y; From the first two experiments:
2.30 × 101 = k(0.100)x(0.100)y and 5.75 × 102 = k(0.0500)x(0.100)y
Dividing the two rate laws: 4.00 = x
x ) 0500 . 0 ( ) 100 . 0 ( = 2.00x, x = 2
Comparing the second and third experiments:
2.30 × 101 = k(0.100)(0.100)y and 1.15 × 101 = k(0.100)(0.0500)y Dividing: 2.00 = y y ) 050 . 0 ( ) 100 . 0 ( = 2.0y, y = 1
The rate law is: Rate = k[ClO2] 2
[OH]
2.30 × 101 mol L1 s1 = k(0.100 mol/L)2(0.100 mol/L), k = 2.30 × 102 L2 mol2 s1 = kmean b. Rate = L mol 0844 . 0 L mol 175 . 0 s mol L 10 30 . 2 2 2 2 2 = 0.594 mol L1 s1 20. 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 L1 s1 = k(2.21 µmol/L)(1.00 µmol/L), k = 0.280 L µmol1 s1 The second and third experiments give similar k values, so kmean = 0.280 L µmol
1 s1. d. Rate = k[Hb][CO] = L mol μ 40 . 2 L mol μ 36 . 3 s mol μ L 280 . 0 = 2.26 µmol L1 s1
21. a. Rate = k[NOCl]n; using experiments two and three: , ) 10 0 . 1 ( k ) 10 0 . 2 ( k 10 64 . 6 10 66 . 2 16 16 3 4 n n 4.01 = 2.0n, n = 2; Rate = k[NOCl]2 b. 2 3 16 3 4 cm molecules 10 0 . 3 k s cm molecules 10 98 . 5
, k = 6.6 × 1029 cm3 molecules1 s1The other three experiments give (6.7, 6.6, and 6.6) × 1029 cm3 molecules1 s1
, respectively. The mean value for k is 6.6 × 1029 cm3 molecules1 s1.
c. mol molecules 10 022 . 6 cm 1000 L 1 s molecules cm 10 6 . 6 23 3 3 29 = s mol L 10 0 . 4 8
22. Rate = k[N2O5]x; the rate laws for the first two experiments are:
2.26 × 103 = k(0.190)x and 8.90 × 104 = k(0.0750)x
Dividing the two rate laws: 2.54 = x
x ) 0750 . 0 ( ) 190 . 0 ( = (2.53)x, x = 1; Rate = k[N2O5] k L / mol 0750 . 0 s L mol 10 90 . 8 ] O N [ Rate 4 1 1 5 2 = 1.19 × 102 s1
The other experiments give similar values for k. kmean = 1.19 × 10 2
s1 23. Rate = k[I]x[OCl]y[OH]z; Comparing the first and second experiments:
z y x z y x ) 10 . 0 ( ) 012 . 0 ( ) 0013 . 0 ( k ) 10 . 0 ( ) 012 . 0 ( ) 0026 . 0 ( k 10 4 . 9 10 7 . 18 3 3 , 2.0 = 2.0x, x = 1
Comparing the first and third experiments:
z y z y ) 10 . 0 ( ) 0060 . 0 )( 0013 . 0 ( k ) 10 . 0 ( ) 012 . 0 )( 0013 . 0 ( k 10 7 . 4 10 4 . 9 3 3 , 2.0 = 2.0y, y = 1
Comparing the first and sixth experiments:
z z ) 10 . 0 )( 012 . 0 )( 0013 . 0 ( k ) 20 . 0 )( 012 . 0 )( 0013 . 0 ( k 10 4 . 9 10 8 . 4 3 3 , 1/2 = 2.0z, z = 1 Rate = ] OH [ ] OCl ][ I [ k
For the first experiment: ) L / mol 10 . 0 ( ) L / mol 012 . 0 ( ) L / mol 0013 . 0 ( k s L mol 10 4 . 9 3 , k = 60.3 s1= 60. s1
For all experiments, kmean = 60. s1.
24. x x x
1 2 1 2 1 2 ] A [ ] A [ ] A [ k ] A [ k Rate RateThe 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:
2 1 ) 2 ( Rate Rate 1 1 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[H2SeO3]x[H]y[I]z; comparing the first and second experiments:
z y x z y x ) 10 0 . 2 ( ) 10 0 . 2 ( ) 10 0 . 1 ( k ) 10 0 . 2 ( ) 10 0 . 2 ( ) 10 0 . 2 ( k 10 66 . 1 10 33 . 3 2 2 4 2 2 4 7 7
, 2.01 = 2.0x, x = 1Comparing the first and fourth experiments:
z y z y ) 10 0 . 2 ( ) 10 0 . 2 )( 10 0 . 1 ( k ) 10 0 . 2 ( ) 10 0 . 4 )( 10 0 . 1 ( k 10 66 . 1 10 66 . 6 2 2 4 2 2 4 7 7
, 4.01 = 2.0y, y = 2Comparing the first and sixth experiments:
z z ) 10 0 . 2 ( ) 10 0 . 2 )( 10 0 . 1 ( k ) 10 0 . 4 ( ) 10 0 . 2 )( 10 0 . 1 ( k 10 66 . 1 10 2 . 13 2 2 2 4 2 2 2 4 7 7
7.95 = 2.0z, log(7.95) = z log(2.0), z = ) 0 . 2 log( ) 95 . 7 ( log = 2.99 ≈ 3 Rate = k[H2SeO3][H + ]2[I−]3Experiment 1: 3 2 2 2 4 7 L mol 10 0 . 2 L mol 10 0 . 2 L mol 10 0 . 1 k s L mol 10 66 . 1
k = 5.19 × 105 L5 mol-5 s-1 = 5.2 × 105 L5 mol-5 s-1 = kmean
Integrated Rate Laws
26. Zero order: t d k ] A [ d , k dt ] A [ d t 0 ] A [ ] A [ t 0
, kt ] A [ ] A [ , kt ] A [ t 0 t 0 ] A [ ] A [ t 0
[A]t = kt + [A]0 First order: t d k ] A [ ] A [ d ], A [ k dt ] A [ d t 0 ] A [ ] A [ t 0
, kt ] A ln[ ] A ln[ , kt ] A ln[ t 0 ] A [ ] A [ t 0
ln[A]t = kt + ln[A]0 Second order: t d k ] A [ ] A [ d , ] A [ k dt ] A [ d t 0 ] A [ ] A [ 2 2 t 0
0 t 0 t ] A [ ] A [ [A] 1 kt ] A [ 1 , kt ] A [ 1 ] A [ 1 ] A [ 1,
kt
t 0
27. Zero order: t1/2 = ; k 2 ] A [ 0 first order: t1/2 = ; k 2 ln second order: t1/2 = 0 ] A [ k 1For 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
28. 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; ] A [ 1 = kt + 0 ] A [ 1 ; k = 3.60 × 10-2 L mol-1 s-1 b. The half-life expression for a second-order reaction is: t1/2 =
0 ] A [ k 1
For this reaction: t1/2 =
L / mol 10 80 . 2 s mol L 10 60 . 3 1 3 1 1 2 = 9.92 × 10 3 s
Note: We could have used the integrated rate law to solve for t1/2, where
[A] = (2.80 × 103 /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.
] A [ 1 = kt + 0 ] A [ 1 , M M 3 2 4 10 80 . 2 1 t s mol L 10 60 . 3 10 00 . 7 1 1.43 × 103 357 = (3.60 × 102)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
min1 b. The half-life expression for a first order rate law is:
t1/2= , k 6931 . 0 k 2 ln
t1/2= 2 1 min 10 97 . 2 6931 . 0 = 23.3 minc. 2.50 × 103 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:
M M 2 3 0 2.00 10 10 50 . 2 ln , kt ] A [ ] A [ ln = (2.97 × 102 min1)tt = 2 1 min 10 97 . 2 ) 125 . 0 ln( = 70.0 min
30. 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 L1 s1
b. The half-life expression for a zero-order reaction is t1/2 = [A]0/2k.
t1/2 = 5 1 1 2 0 5 2 s L mol 10 00 . 4 2 L / mol 10 25 . 1 k 2 ] OH H C [ = 156 s
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 L1 s1)t + 1.25 × 102 mol/L t = 5 1 1 2 s L mol 10 00 . 4 L / mol 10 25 . 1 = 313 s
31. 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 [H2O2] ln[H2O2] (s) (mol/L) 0 1.00 0.000 120. 0.91 0.094 300. 0.78 0.25 600. 0.59 0.53 1200. 0.37 0.99 1800. 0.22 1.51 2400. 0.13 2.04 3000. 0.082 2.50 3600. 0.050 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.
The plot of ln[H2O2] versus time is linear. Thus the reaction is first order. The differential
rate law and integrated rate law are Rate = dt ] O H [ d 2 2 = k[H2O2] and ln[H2O2] = kt + 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 = . 3600 0 ) 00 . 3 ( 0 x Δ y Δ
= 8.3 × 104 s1, k = 8.3 × 104 s1To determine [H2O2] at 4000. s, use the integrated rate law, where [H2O2]0 = 1.00 M.
ln[H2O2] = kt + ln[H2O2]0 or ln 0 2 2 2 2 ] O H [ ] O H [ = kt ln 00 . 1 ] O H [ 2 2 = 8.3 × 104 s1 × 4000. s, ln[H2O2] = 3.3, [H2O2] = e 3.3 = 0.037 M
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 6210 s [C4H6] 1.6 × 10 2 1.5 × 102 1.3 × 102 1.1 × 102 0.68 × 102 M ln[C4H6] 4.14 4.20 4.34 4.51 4.99 1/[C4H6] 62.5 66.7 76.9 90.9 147 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
Rate = k[C4H6] 2
For a second-order reaction, the integrated rate law is
] H C [ 1 6 4 = kt + 0 6 4H ] C [ 1 .
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 = s . 1000 s . 6000 mol / L 73 mol / L 144 = 1.4 × 102 L mol1 s1
33. 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 = torr 760 atm 1 s . 100 torr 13 = 1.7 × 104 atm/s The rate law and integrated rate law are:
Rate = k = 1.7 × 104 atm/s; PC2H5OH= kt + 250. torr torr 760 atm 1 = kt + 0.329 atm At 900. s: OH H C2 5
P = 1.7× 104 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 = s ) 1 5 ( mol / L ) 20 60 ( = 10 L mol1 s1 ] A [ 1 = kt + 0 ] A [ 1 M 1 . 0 1 s 9 s mol L 10
= 100, [A] = 0.01 Mc. For a second-order reaction, the half-life does depend on concentration: t1/2 = 0 ] A [ k 1 First half-life: t1/2 = L mol 1 . 0 s mol L 10 1 = 1 s
Second half-life ([A]0 is now 0.05 M): t1/2 = 1/(10 × 0.05) = 2 s
35. 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] 1/[NO2] (M -1 ) 0 0.500 0.693 2.00 1.20 × 103 0.444 0.812 2.25 3.00 × 103 0.381 0.965 2.62 4.50 × 103 0.340 1.079 2.94 9.00 × 103 0.250 1.386 4.00 1.80 × 104 0.174 1.749 5.75
The plot of 1/[NO2] versus time is linear. The reaction is second order in NO2. The
differ-ential rate law and integrated rate law are Rate = k[NO2]2 and
] NO [ 1 2 = kt + 0 2] NO [ 1 .
The slope of the plot 1/[NO2] versus time gives the value of k. Using a couple of points on
the plot: slope = k = s ) 0 10 80 . 1 ( ) 00 . 2 75 . 5 ( x Δ y Δ 4 1 M = 2.08 × 104 L mol1 s1 To determine [NO2] at 2.70 × 10 4
s, use the integrated rate law, where 1/[NO2]0= 1/0.500 M =
2.00 M 1. ] NO [ 1 2 = kt + 0 2] NO [ 1 , ] NO [ 1 2 = s mol L 10 08 . 2 4 × 2.70 × 104 s + 2.00 M 1 ] NO [ 1 2 = 7.62, [NO2] = 0.131 M
36. 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) [O] (atoms/cm3) ln[O]
0 5.0 × 109 22.33
10. × 103 1.9 × 109 21.37
20. × 103 6.8 × 108 20.34
30. × 103 2.5 × 108 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 = s ) 0 10 . 30 ( 23 . 22 34 . 19 3 = 1.0 × 10 2 s1, k = slope = 1.0 × 102 s1
To determine k, the actual rate constant: k = k[NO2], 1.0 × 10
2
s1 = k(1.0 × 1013 molecules/cm3) k = 1.0 × 1011 cm3 molecules1 s1
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.
Time (ms) [NO] (molecules/cm3) ln[NO] 0 6.0 × 108 20.21 100. 5.0 × 108 20.03 500. 2.4 × 108 19.30 700. 1.7 × 108 18.95 1000. 9.9 × 107 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/cm 3 ) ln[O3] 0 1.0 × 1010 23.03 50. 8.4 × 109 22.85 100. 7.0 × 109 22.67 200. 4.9 × 109 22.31 300. 3.4 × 109 21.95
The plot of ln[O3] versus t is linear. Hence the reaction is first order with respect to
ozone.
c. For NO experiment, Rate = k[NO] and k = (slope from graph of ln[NO] versus t). k = slope = s 10 ) 0 . 1000 ( 21 . 20 41 . 18 3 = 1.8 s1
For ozone experiment, Rate = k[O3] and k = (slope from ln[O3] versus t).
k = slope = s 10 ) 0 . 300 ( ) 03 . 23 95 . 21 ( 3 = 3.6 s1
d. From NO experiment, Rate = k[NO][O3] = k[NO] where k = k[O3].
k = 1.8 s1 = k(1.0 × 1014 molecules/cm3), k = 1.8 × 1014 cm3 molecules1 s1
We can check this from the ozone data. Rate = k[O3] = k[NO][O3], where k = k[NO].
k = 3.6 s1 = k(2.0 × 1014 molecules/cm3), k = 1.8 × 1014 cm3 molecules1 s1 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 = K 323 mol K atm L 08206 . 0 ) L 10 3 . 58 ( atm 00 . 1 RT PV 3 = 2.20 × 103 mol N2
Because each mole of C6H5N2Cl that decomposes produces one mole of N2, the initial
concentration (t = 0) of C6H5N2Cl was: L 10 0 . 40 mol 10 20 . 2 3 3 = 0.0550 M
We can similarly calculate the moles of N2 evolved at each point of the experiment, subtract
that from 2.20 × 103 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
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) (V Vt ) ln(V Vt ) 0 0 58.3 4.066 6 19.3 39.0 3.664 9 26.0 32.3 3.475 14 36.0 22.3 3.105 22 45.0 13.3 2.588 30. 50.4 7.9 2.07
We can see from the graph that this plot is linear, so the reaction is first order. The differ-ential 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
s1.
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 s1; mol/L 20 . 0 s 32 . 0 ] V [ k k 1 0 1.6 L mol-1 s-1 40. 0 ] A [ ] A [ ln = kt; k = d 3 . 14 6931 . 0 t 2 ln 2 / 1
= 4.85 × 102 d1 If [A]0 = 100.0, then after 95.0% completion, [A] = 5.0. 0 . 100 0 . 5 ln = -4.85 × 102 d1 × t, t = 62 days
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 × 103 molL1s1 = k1(0.200 M) 2
k = 8.00 × 102 Lmol1s1 = kmean
For a second-order reaction: t1/2 = L / mol 00 . 1 s mol L 10 00 . 8 1 ] AB [ k 1 1 1 2 0
= 12.5 s42. 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
concen-tration 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
in-versely 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:
L / mol 00 . 1 L / mol 250 . 0 ln = k × 120. s, k = 0.0116 s1 L / mol 00 . 2 L / mol 350 . 0 ln = 0.0116 s1 × t, t = 150. s
44. Successive half-lives increase in time for a second-order reaction. Therefore, assume reaction is second order in A.
t1/2 = 0 ] A [ k 1 , k = 0 2 / 1 [A] t 1 = ) 10 . 0 min( 0 . 10 1 M = 1.0 L mol 1 min1
a. ] A [ 1 = kt + min mol L 0 . 1 ] A [ 1 0
80.0 min + M 10 . 0 1 = 90. M 1, [A] = 1.1 × 102 M 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.
[HI] = kt + [HI]0, [HI] =
L mol 250 . 0 min s 60 min 25 s L mol 10 20 . 1 4
[HI] = 0.18 mol/L + 0.250 mol/L = 0.07 M
b. [HI] = 0 = kt + [HI]0, kt = [HI]0, t =
k ] HI [ 0 t = 1 -1 -4 s L mol 10 1.20 mol/L 0.250 = 2080 s = 34.7 min
46. a. The integrated rate law for a second order reaction is 1/[A] = kt + 1/[A]0, and the half-
life expression is t1/2 = 1/k[A]0. We could use either to solve for t1/2. Using the
inte-grated rate law:
mol/L (0.900/2) 1 = k × 2.00 s + L / mol 900 . 0 1 , k = s 00 . 2 mol / L 11 . 1 = s mol L 555 . 0 b. L / mol 100 . 0 1 = 0.555 L mol1 s1 × t + L / mol 900 . 0 1 , t = 1 1 s mol L 555 . 0 mol / L 9 . 8 = 16 s
47. a. When a reaction is 75.0% complete (25.0% of reactant remains), this represents two half- lives (100% → 50%→ 25%). The first-order half-life expression is t1/2 = (ln 2)/k.
Be-cause there is no concentration dependence for a first-order life, 320. s = two half-lives, t1/2 = 320./2 = 160. s. This is both the first half-life, the second half-life, etc.
b. t1/2 = s . 160 2 ln t 2 ln k , k 2 ln 2 / 1
= 4.33 × 103s1At 90.0% complete, 10.0% of the original amount of the reactant remains, so [A] = 0.100[A]0. ln
1 3 1 3 0 0 0 4.33 10 s ) 100 . 0 ln( t , t ) s 10 33 . 4 ( ] A [ ] A [ 100 . 0 ln , kt ] A [ ] A [ 532 s48. 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.
a. 0 ] A [ ] A [ ln = kt, M M 2 3 10 0 . 1 10 8 . 3 ln = k× 8.0 s, k = 0.12 s1
For the reaction: k = k[B]2, k = 0.12 s1/(3.0 mol/L)2 = 1.3 × 102L2 mol2 s1
b. t1/2 = 1 s 12 . 0 693 . 0 ' k 2 ln
= 5.8 s c. M 2 10 0 . 1 ] A [ ln = 0.12 s1 × 13.0 s, 2 10 0 . 1 ] A [ = e 0.12(13.0) = 0.21 [A] = 2.1 × 103M d. [A] reacted = 0.010 M 0.0021 M = 0.008 M [C] reacted = 0.008 M × A mol 1 C mol 2 = 0.016 M ≈ 0.02 M[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 << [B]0 or [C]0, the B and C concentrations remain constant at 1.00 M for
this experiment. Thus, rate = k[A]2[B][C] = k[A]2 where k = k[B][C]. For this pseudo-second-order reaction:
] A [ 1 = kt + M 5 0 3.26 10 1 , ] A [ 1 = k(3.00 min) + 1.00 10 4 M 1
k = 6890 L mol–1 min–1 = 115 L mol–1 s–1
k = k[B][C], k = ) 00 . 1 )( 00 . 1 ( s mol L 115 k , ] C [ ] B [ k 1 1 M M = 115 L3 mol–3 s–1
b. For this pseudo-second-order reaction:
Rate = k[A]2, t1/2 = ) L / mol 10 00 . 1 ( s mol L 115 1 ] A [ k 1 4 1 1 0
' = 87.0 s c. ] A [ 1 = kt + 0 ] A [ 1 = 115 L mol–1 s–1 × 600. s + L / mol 10 00 . 1 1 4 = 7.90 × 10 4 L/molFrom the stoichiometry in the balanced reaction, 1 mol of B reacts with every 3 mol of A. Amount A reacted = 1.00 × 10–4 M – 1.27 × 10–5 M = 8.7 × 10–5 M
Amount B reacted = 8.7 × 10–5 mol/L ×
A mol 3 B mol 1 = 2.9 × 10–5 M [B] = 1.00 M 2.9 × 10–5 M = 1.00 M
As we mentioned in part a, the concentration of B (and C) remain constant because the A concentration is so small compared to the B (or C) concentration.
50. The integrated rate law for each reaction is:
ln[A] = -4.50 × 104 s1(t) + ln[A]0 and ln[B] = 3.70 × 10 3
s1(t) + ln[B]0
Subtracting the second equation from the first equation (ln[A]0 = ln[B]0):
ln[A] ln[B] = 4.50 × 104(t) + 3.70 × 103(t), ] B [ ] A [ ln = 3.25 × 103(t) When [A] = 4.00 [B], ln(4.00) = 3.25 × 103(t), t = 427 s.
51. The consecutive half-life values of 24 hours, then12 hours, show a direct relationship with concentration; as the concentration decreases, the half-life decreases. Assuming the drug reaction is either zero, first, or second order, only a zero order reaction shows this direct relationship between half-life and concentration. Therefore, assume the reaction is zero order in the drug. 5 3 1/2 0 0 1/2 4.2 10 h) 2(24 mol/L 10 2.0 2t [A] k , 2k [A] t mol L-1 h-1
Reaction Mechanisms
52. For a mechanism to be acceptable, the sum of the elementary steps must give the overall balanced equation for the reaction, and the mechanism must give a rate law that agrees with the experimentally determined rate law. A mechanism can never be proven absolutely. We can only say it is possibly correct if it follows the two requirements just described.
Most reactions occur by a series of steps. If most reactions were one step, then all reactants would appear in the overall rate law, and these orders would be the coefficients in the balanced equation. This is not the case.
53. In a unimolecular reaction, a single reactant molecule decomposes to products. In a bimolecular reaction, two molecules collide to give products. The probability of the simultaneous collision of three molecules with enough energy and the proper orientation is very small, making termolecular steps very unlikely.
54. a. An elementary step (reaction) is one for which the rate law can be written from the molecularity, i.e., from coefficients in the balanced equation.
b. The molecularity is the number of species that must collide to produce the reaction represented by an elementary step in a reaction mechanism.
c. The mechanism of a reaction is the series of proposed elementary reactions that may occur to give the overall reaction. The sum of all the steps in the mechanism gives the balanced chemical reaction.
d. An intermediate is a species that is neither a reactant nor a product but that is formed and consumed in the reaction sequence.
e. The rate-determining step is the slowest elementary reaction in any given mechanism. 55. For elementary reactions, the rate law can be written using the coefficients in the balanced
equation to determine the orders.
a. Rate = k[CH3NC] b. Rate = k[O3][NO]
c. Rate = k[O3] d. Rate = k[O3][O]
e. Rate = k
146C or Rate = kN, where N = the number of C146 atoms (convention)56. The rate law is Rate = k[NO]2[Cl2]. If we assume the first step is rate-determining, we would
expect the rate law to be Rate = k1[NO][Cl2]. This isn't correct. However, if we assume the
second step to be rate-determining, Rate = k2[NOCl2][NO]. To see if this agrees with
experiment, we must substitute for the intermediate NOCl2 concentration. Assuming a fast-
equilibrium first step (rate reverse = rate forward): k-1[NOCl2] = k1[NO][Cl2], [NOCl2] =
1 1
k k
[NO][Cl2]; substituting into the rate equation:
Rate = 1 1 2 k k k [NO]2[Cl2] = k[NO] 2 [Cl2] where k = 1 1 2 k k k
This is a possible mechanism with the second step the rate-determining step because the derived rate law agrees with the experimentally determined rate law.
57. Let's determine the rate law for each mechanism. If the rate law derived from the mechanism is the same as the experimental rate law, then the mechanism is possible (assuming the sum of all the steps in the mechanism gives the overall balanced equation). When deriving rate laws from a mechanism, we must substitute for all intermediate concentrations.
a. Rate = k1[NO][O2]; not possible
b. Rate = k2[NO3][NO] and k1[NO][O2] = k1[NO3] or [NO3] = 1 1 k k [NO][O2] Rate = 1 1 2 k k k [NO]2[O2]; possible
c. Rate = k1[NO] 2
; not possible d. Rate = k2[N2O2] and [N2O2] = 1 1 k k [NO]2 Rate = 1 1 2 k k k
[NO]2 ; not possible
Only the mechanism in b is consistent, so only mechanism b is a possible mechanism for this reaction.
58. From experiment (Exercise 31), we know the rate law is Rate = k[H2O2]. A mechanism
consists of a series of elementary reactions where the rate law for each step can be determined using the coefficients in the balanced equation for each respective step. For a plausible mechanism, the rate law derived from a mechanism must agree with the rate law determined from experiment. To derive the rate law from the mechanism, the rate of the reaction is assumed to equal the rate of the slowest step in the mechanism.
This mechanism will agree with the experimentally determined rate law only if step 1 is the slow step (called the rate-determining step). If step 1 is slow, then Rate = k[H2O]2 which
agrees with experiment.
Another important property of a mechanism is that the sum of all steps must give the overall balanced equation. Summing all steps gives:
H2O2 2 OH
H2O2 + OH H2O + HO2
HO2 + OH H2O+ O2 ____________________________________
2 H2O2 2 H2O+ O2
59. A mechanism consists of a series of elementary reactions in which the rate law for each step can be determined using the coefficients in the balanced equations. For a plausible mecha-nism, the rate law derived from a mechanism must agree with the rate law determined from experiment. To derive the rate law from the mechanism, the rate of the reaction is assumed to equal the rate of the slowest step in the mechanism.
Because step 1 is the rate determining step, the rate law for this mechanism is Rate =
k[C4H9Br]. To get the overall reaction, we sum all the individual steps of the mechanism.
Summing all steps gives: C4H9Br C4H9 + + Br C4H9 + + H2O C4H9OH2 + C4H9OH2 + + H2O → C4H9OH + H3O + ___________________________________________________________ C4H9Br + 2 H2O C4H9OH + Br + H3O +
Intermediates in a mechanism are species that are neither reactants nor products but that are formed and consumed during the reaction sequence in the mechanism. The intermediates for this mechanism are C4H9
+
and C4H9OH2 +
60. a. This rate law occurs when the first step is assumed rate determining. Rate = k1[Br2] = k[Br2]
b. This rate law occurs when the second step is assumed rate-determining and the first step is a fast- equilibrium step.
Rate = k2[Br][H2]; from the fast-equilibrium first step (rate forward = rate reverse):
k1[Br2] = k-1[Br] 2 , [Br] = (k1/k-1) 1/2 [Br2] 1/2
Substituting into the rate equation: Rate = k2(k1/k-1) 1/2 [Br2] 1/2 [H2] = k[H2][Br2] 1/2 c. From a, k = k1; from b, k = k2(k1/k-1) 1/2 .
61. Rate = k2[I][HOCl]; from the fast-equilibrium first step:
k1[OCl] = k-1[HOCl][OH], [HOCl] =
] OH [ k ] OCl [ k 1 1
; substituting into the rate equation:
Rate = ] OH [ ] OCl ][ I [ k ] OH [ k ] OCl ][ I [ k k 1 1 2
62. a. Rate = k3[COCl][Cl2]; from the fast-equilibrium reactions 1 and 2:
2 2 k k ] CO ][ Cl [ ] COCl [
, [COCl] = 2 2 k k [CO][Cl] 2 / 1 2 1 1 1 1 2 2 ] Cl [ k k ] Cl [ , k k ] Cl [ ] Cl [
Thus [COCl] = 2 / 1 1 1 2 2 k k k k [CO][Cl2] 1/2; Substituting into rate law:
Rate = 2 / 1 1 1 2 2 3 k k k k k [CO][Cl2] 3/2 = k[CO][Cl2] 3/2
b. Cl and COCl are intermediates. 63. Rate = k3[Br][H2BrO3
+
]; we must substitute for the intermediate concentration. Because steps 1 and 2 are fast-equilibrium steps, rate forward reaction = rate reverse reaction. k2[HBrO3][H + ] = k-2[H2BrO3 + ]; k1[BrO3][H + ] = k-1[HBrO3]
[HBrO3] = 1 1 k k [BrO3 ][H+]; [H2BrO3 + ] = 2 2 k k [HBrO3][H + ] = 1 2 1 2 k k k k [BrO3 ][H+]2 Rate = 1 2 1 2 3 k k k k k [Br][BrO3 -][H+]2 = k[Br][BrO3 ][H+]2
64. Rate = k[BrO3][SO3 2
][H+]; first step: SO3 2
+ H+ HSO3 (fast)
The rate law contains BrO3, SO3
2 and H+
. In order to incorporate all these ions into the rate law, the rate-determining step could contain BrO3
and HSO3
(the intermediate produced in the first step). A possible second step is:
BrO3
+ HSO3
products (slow) A likely choice is: BrO3 + HSO3 SO4
2 + HBrO
2 (slow)
Followed by: HBrO2 + SO3 2
HBrO + SO4 2
(fast); HBrO + SO32 SO42 + H+ + Br (fast)
All species in this mechanism (HSO3, HBrO2, and HBrO) are known substances. This
mechanism also gives the correct experimentally determined rate law.
Rate = k2[BrO3
] [HSO3
]; assuming reaction one is a fast equilibrium step:
k1[SO3 2][H+ ] = k-1[HSO3 ], [HSO3 ] = 1 1 k k [SO3 2][H+ ] Rate = 1 1 2 k k k [BrO3][SO3 2][H+ ] = k[BrO3][SO3 2][H+ ] 65. a. Rate = dt ] E [ d = k2[B*]; assume dt *] B [ d = 0, then k1[B] 2 = k-1[B][B*] + k2[B*]. [B*] = 2 1 2 1 k ] B [ k ] B [ k
; the rate law is: Rate = dt ] E [ d = 2 1 2 2 1 k ] B [ k ] B [ k k
b. When k2 << k-1[B], then Rate =
dt ] E [ d = ] B [ k ] B [ k k 1 2 2 1 = [B]. k k k 1 2 1
The reaction is first order when the rate of the second step is very slow (when k2 is very
small).
c. Collisions between B molecules only transfer energy from one B to another. This occurs at a much faster rate than the decomposition of an energetic B molecule (B*).
66. Rate = dt ] O [ d 3
dt ] O [ d
= 0, so k1[M][O3] = k-1[M][O2][O] + k2[O][O3]
k1[M][O3] k-1[M][O2][O] = k2[O][O3]
Substitute this expression into the rate law: Rate =
dt ] O [ d 3 = 2k2[O][O3]
Rearranging the steady-state approximation for [O]: [O] =
] O [ k ] O ][ M [ k ] M ][ O [ k 3 2 2 1 3 1
Substituting into the rate law: Rate = dt ] O [ d 3 = ] O [ k ] O ][ M [ k ] M [ ] O [ k k 2 3 2 2 1 2 3 1 2 67. a. MoCl5 b. Rate = dt ] NO [ d 2 = k2[NO3 ][MoCl5
] (Only the last step contains NO2
.)
We use the steady-state assumption to substitute for the intermediate concentration in the rate law. The steady-state approximation assumes that the concentration of an intermediate remains constant; i.e., d[intermediate]/dt = 0. To apply the steady-state assumption, we write rate laws for all steps where the intermediate is produced and equate the sum of these rate laws to the sum of the rate laws where the intermediate is consumed. Applying the steady-state approximation to MoCl5:
dt ] MoCl [ d 5 = 0, so k1[MoCl6 2] = k
-1[MoCl5][Cl] + k2[NO3][MoCl5]
[MoCl5 ] = ] NO [ k ] Cl [ k ] MoCl [ k 3 2 1 2 6 1 ; Rate = dt ] NO [ d 2 = ] NO [ k ] Cl [ k ] MoCl ][ NO [ k k 3 2 1 2 6 3 2 1
Temperature Dependence of Rate Constants and the Collision Model
68. a. Thermodynamics is concerned with only the initial and final states of a reaction, and not the pathway required to get from reactants to products. It is the kinetics of a reaction that concentrates on the pathway and speed of a reaction. For these plots, the reaction with the smallest activation energy will be fastest. The activation energy is the energy reactants must have in order to overcome the energy barrier to convert to products. So the fastest reaction is reaction 5 with the smallest activation. Reactions 1, 2, and 4 all should have the same speed because they have the same activation energy. (This assumes all other kinetic factors are the same.) Reaction 3 will be the slowest since it has the largest activation energy.
b. If the products have lower energy than the reactants, then the reaction is exothermic. Reaction 1, 2, 3, and 5 are exothermic. If the products have higher energy than the
reactants, then the reaction is endothermic. Only reaction 4 is endothermic. Note that it is the thermodynamics of a reaction that dictates whether a reaction is exothermic or endothermic. The kinetics plays no factor in these designations.
c. The thermodynamics of reaction determine potential energy change. In an exothermic reaction, some of the potential energy stored in chemical bonds is converted to thermal energy (heat is released as the potential energy decreases). The exothermic reaction with the greatest loss of potential energy is the reaction with the largest energy difference (E) between reactants and products. Reactions 2 and 5 both have the same largest E values, so reactions 2 and 5 have the greatest change in potential energy. Reactions 1 and 3 both have the same smallest E values, so reactions 1 and 3 have the smallest change in potential energy.
69. Two reasons are:
1) The collision must involve enough energy to produce the reaction; i.e., the collision energy must be equal to or exceed the activation energy.
2) The relative orientation of the reactants when they collide must allow formation of any new bonds necessary to produce products.
70. a. The larger the activation energy, the slower is the rate.
b. The higher the temperature, the more molecular collisions there are with sufficient energy to convert to products, and the faster is the rate.
c. The greater the frequency of collisions, the greater are the opportunities for molecules to react, and hence the greater is the rate.
b. For a reaction to occur, it is the reactive portion of each molecule that must be involved in a collision. Only some of all the possible collisions have the correct orientation to convert reactants to products.
71. a. T2 > T1; as temperature increases, the distribution of collision energies shifts to the right.
That is, as temperature increases, there are fewer collision energies with small energies and more collisions with large energies.
b. As temperature increases, more of the collisions have the required activation energy necessary to convert reactants into products. Hence, the rate of the reaction increases with increasing temperature.
72. H3O +
(aq) + OH(aq) 2 H2O(l) should have the faster rate. H3O +
and OH will be electro-statically attracted to each other; Ce4+ and Hg2
2+
will repel each other (so Ea is much larger).
73. From the Arrhenius equation in logarithmic form (ln k = Ea/RT + ln A), a graph of ln k
versus. 1/T should yield a straight line with a slope equal to Ea/R and a y intercept equal to
ln A. a. Slope = Ea/R, Ea = 1.10 × 10 4 K × mol K J 3145 . 8 = 9.15 × 104 J/mol = 91.5 kJ/mol
b. The units for A are the same as the units for k (s1). y intercept = ln A, A = e33.5 = 3.54 × 1014 s1 c. ln k = Ea/RT + ln A or k = A exp(Ea/RT) k = 3.54 × 1014 s1 × K 298 mol K J 3145 . 8 mol / J 10 15 . 9 exp 1 1 4 = 3.24 × 102 s1 74. ; T 1 T 1 R E k k ln 2 1 a 1 2
because the rate doubles, assume k2 = 2k1.ln(2.00) = , K 308 1 K 298 1 mol K J 3145 . 8 E 1 1 a Ea = 5.3 × 10 4 J/mol = 53 kJ/mol 75. k = A exp(Ea/RT) or ln k = RT Ea
+ ln A (the Arrhenius equation)
For two conditions:
2 1 a 1 2 T 1 T 1 R E k kln (Assuming A is temperature independent.)
Let k1 = 3.52 × 7 10 L mol1s1, T1 = 555 K; k2 = ?, T2 = 645 K; Ea = 186 × 10 3 J/mol
645K 1 K 555 1 mol K J 3145 . 8 mol / J 10 86 . 1 10 52 . 3 k ln 1 1 5 7 2 = 5.6 7 2 10 52 . 3 k = e 5.6 = 270, k2 = 270(3.52 × 107) = 9.5 × 105 L mol1 s 1 76.
2 1 a 1 2 T 1 T 1 R E k k ln ;
K . 720 1 K . 660 1 mol K J 3145 . 8 E s 10 2 . 7 s 10 7 . 1 ln a 1 1 1 4 1 2 Ea = 2.1 × 10 5 J/mol For k at 325oC (598 K):
K . 720 1 K 598 1 mol K J 3145 . 8 mol / J 10 1 . 2 k s 10 7 . 1 ln 1 1 5 1 2 , k = 1.3 × 105s1For three half-lives, we go from 100% 50% 25% 12.5%. After three half-lives, 12.5% of the original amount of C2H5I remains. Partial pressures are directly related to gas
concentrations in mol/L:
I H C2 5
77. 1 2 2 1 a 1 2 k k ; T 1 T 1 R E k k ln
= 7.00, T1 = 295 K, Ea = 54.0 × 10 3 J/mol ln(7.00) = 2 2 1 1 3 T 1 K 295 1 , T 1 K 295 1 mol K J 3145 . 8 mol / J 10 0 . 54 = 3.00 × 10 4 2 T 1 = 3.09 × 103, T2 = 324 K = 51°C78. A graph of ln k versus 1/T should be linear with slope = Ea/R.
T (K) 1/T (K1) k (s1) ln k 338 2.96 × 103 4.9 × 103 5.32 318 3.14 × 103 5.0 × 104 7.60 298 3.36 × 103 3.5 × 105 10.26 Slope = 3 3 10 00 . 3 10 40 . 3 ) 85 . 5 ( 76 . 10 = 1.2 × 104 K = Ea/R Ea = slope × R = 1.2 × 10 4 K × 8.3145 J K1 mol1, Ea = 1.0 × 10 5 J/mol = 1.0 × 102 kJ/mol 79. The Arrhenius equation is k = Aexp(Ea/RT) or, in logarithmic form, ln k = Ea/RT + ln A.
Hence a graph of ln k versus 1/T should yield a straight line with a slope equal to Ea/R since
the logarithmic form of the Arrhenius equation is in the form of a straight-line equation, y = mx + b. Note: We carried one extra significant figure in the following ln k values in order to reduce round-off error.
T (K) 1/T (K1) k (L mol1 s1) ln k 195 5.13 × 103 1.08 × 109 20.80 230. 4.35 × 103 2.95 × 109 21.81 260. 3.85 × 103 5.42 × 109 22.41 298 3.36 × 103 12.0 × 109 23.21 369 2.71 × 103 35.5 × 109 24.29
Using a couple of points on the plot:
slope = 3 3 3 10 00 . 2 70 . 2 10 00 . 3 10 00 . 5 65 . 23 95 . 20
= 1.35 × 103 K = R Ea Ea = 1.35 × 10 3K × 8.3145 J K1 mol1 = 1.12 × 104 J/mol = 11.2 kJ/mol
From the best straight line (by calculator): slope = 1.43 × 103 K and Ea = 11.9 kJ/mol
80. When ΔE is positive, the products are at a higher energy relative to reactants, and when ΔE is negative, the products are at a lower energy relative to reactants.
81. In the following reaction profiles R = reactants, P = products, Ea = activation energy, ΔE =
overall energy change for the reaction, and RC = reaction coordinate, which is the same as reaction progress.
The second reaction profile represents a two-step reaction since an intermediate plateau appears between the reactants and the products. This plateau (see I in plot) represents the energy of the intermediate. The general reaction mechanism for 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 activation energy 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 largest activation energy and will be the rate determining step (the slow step).
82.
The activation energy for the reverse reaction is:
Ea, reverse = 216 kJ/mol + 125 kJ/mol = 341 kJ/mol
Ea, reverse 125 kJ/mol 216 kJ/mol P RC E R
83.
The activation energy for the reverse reaction is ER in the diagram.
ER = 167 28 = 139 kJ/mol
Catalysis
84. The slope of the ln k versus 1/T plot is equal to Ea/R. Because Ea for the catalyzed reaction
will be smaller than Ea for the uncatalyzed reaction, the slope of the catalyzed plot should be
less negative.
85. a. W because it has a lower activation energy than the Os catalyst.
b. kw = Aw exp[Ea(W)/RT]; kuncat = Auncat exp[Ea(uncat)/RT]; assume Aw = Auncat. RT ) uncat ( E RT ) W ( E exp k k a a uncat w K 298 mol K J 3145 . 8 mol / J 000 , 335 mol / J 000 , 163 exp k k 1 1 uncat w = 1.41 × 1030
The W-catalyzed reaction is approximately 1030 times faster than the uncatalyzed reaction.
c. Because [H2] is in the denominator of the rate law, the presence of H2 decreases the rate
of the reaction. For the decomposition to occur, NH3 molecules must be adsorbed on the
surface of the catalyst. If H2 is also adsorbed on the catalyst surface, then there are fewer
sites for NH3 molecules to be adsorbed, and the rate decreases.
86. 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,
thus increasing the rate of the reaction.
A heterogeneous catalyst is in a different phase from 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 catalyzed reaction most likely will have a different rate law.
87. a. The blue plot is the catalyzed pathway. The catalyzed pathway has the lower activation. This is why the catalyzed pathway is faster.
b. E1 represents the activation energy for the uncatalyzed pathway.
c. E2 represents the energy difference between the reactants and products. Note that E2 is
the same for both the catalyzed and the uncatalyzed pathways. It is the activation energy that is different for a catalyzed pathway versus an uncatalyzed pathway.
d. Because the products have a higher total energy as compared to reactants, this is an endothermic reaction.
88. a. NO is the catalyst. NO is present in the first step of the mechanism on the reactant side, but it is not a reactant since it is regenerated in the second step and does not appear in the overall balanced equation.
b. NO2 is an intermediate. Intermediates also never appear in the overall balanced equation.
In a mechanism, intermediates always appear first on the product side, whereas catalysts always appear first on the reactant side.
c. k = A exp(Ea/RT);
RT (cat) E (un) E exp (un)/RT ] E [ exp A (cat)/RT ] E [ exp A k k a a a a un cat
K 298 mol K J 3145 . 8 mol / J 2100 exp k k 1 1 un cat = e0.85 = 2.3The catalyzed reaction is approximately 2.3 times faster than the uncatalyzed reaction at 25°C.
89. The mechanism for the chlorine catalyzed destruction of ozone is: O3 + Cl O2 + ClO (slow)
ClO + O O2 + Cl (fast)
_________________________ O3 + O 2 O2
Because the chlorine atom-catalyzed reaction has a lower activation energy, the Cl-catalyzed rate is faster. Hence Cl is a more effective catalyst. Using the activation energy, we can estimate the efficiency that Cl atoms destroy ozone compared to NO molecules (see Exercise 15.88c). At 25°C:
mol / J ) 298 3145 . 8 ( mol / J ) 900 , 11 2100 ( exp RT ) NO ( E RT ) Cl ( E exp k k a a NO Cl = e3.96 = 52At 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.
90. Assuming the catalyzed and uncatalyzed reactions have the same form and orders, and be- cause concentrations are assumed equal, the rates will be equal when the k values are equal. k = A exp(Ea/RT), kcat = kun when Ea,cat/RTcat = Ea,un/RTun
, T mol K J 3145 . 8 mol / J 10 00 . 7 K 293 mol K J 3145 . 8 mol / J 10 20 . 4 un 1 1 4 1 1 4 Tun = 488 K = 215 o C 91. Rate = k[A]x dt ] A [ d
Assuming the catalyzed and uncatalyzed reaction have the same form and orders and because concentrations are assumed equal, rate 1/t, where t = time.
cat cat un un cat t Δ yr 2400 t Δ t Δ Rate Rate
and
un cat rate rate un cat k k
un cat Rate Rate
un cat k k (un)/RT ] a E [ exp A (cat)/RT ] a E [ exp A = RT (un) E (cat) E exp a a
un cat k k K . 600 mol K J 3145 . 8 mol / J 10 84 . 1 mol / J 10 90 . 5 exp 1 1 5 4 = 7.62 × 1010 cat un cat un cat cat un t Δ yr 2400 , k k rate rate t Δ t Δ
= 7.62 × 1010, tcat = 3.15 × 8 10 yr 1 s92. The reaction at the surface of the catalyst follows the steps:
Thus CH2D‒CH2D should be the product.
93. At high [S], the enzyme is completely saturated with substrate. Once the enzyme is com-pletely saturated, the rate of decomposition of ES can no longer increase, and the overall rate remains constant. metal surface CH2DCH2D(g) DCH2 CH2 D D D H H H C C H D D D D
Additional Exercises
94. One experimental method to determine rate laws is the method of initial rates. Several experiments are carried out using different initial concentrations of reactants, and the initial rate is determined for each experiment. The results are then compared to see how the initial rate depends on the initial concentrations. This allows the orders in the rate law to be determined. The value of the rate constant is determined from the experiments once the orders are known.
The second experimental method utilizes the fact that the integrated rate laws can be put in the form of a straight-line equation. Concentration versus time data are collected for a reactant as a reaction is run. This data are then manipulated and plotted to see which manipulation gives a straight line. From the straight-line plot we get the order of the reactant, and the slope of the line is mathematically related to k, the rate constant.
95. The most common method to experimentally determine the differential rate law is the method of initial rates. Once the differential rate law is determined experimentally, the integrated rate law can be derived. However, sometimes it is more convenient and more accurate to collect concentration versus time data for a reactant. When this is the case, then we do “proof” plots to determine the integrated rate law. Once the integrated rate law is determined, the differential rate law can be determined. Either experimental procedure allows determination of both the integrated and the differential rate law; and which rate law is determined by experiment and which is derived is usually decided by which data are easiest and most accurately collected. 96. Rate = dt ] P [ d
= k2[ES]; apply the steady-state approximation to ES,
dt ] ES [ d = 0.
k1[E][S] = k-1[ES] + k2[ES]; [E]T = [E] + [ES], so [E] = [E]T [ES]
Substituting: k1[S]([E]T [ES]) = (k-1 + k2)[ES], k1[E]T[S] = (k-1 + k2 + k1[S]) [ES]
[ES] = ] S [ k k k ] S [ ] E [ k 1 2 1 T 1
; substituting into the rate equation, Rate = k2[ES]:
Rate = dt ] P [ d = ] S [ k k k ] S [ ] E [ k k 1 2 1 T 2 1 97. Rate = dt ] Cl [ d 2 = k2[NO2Cl][Cl]; Assume dt ] Cl [ d = 0, then: k1[NO2Cl] = k-1[NO2][Cl] + k2[NO2Cl][Cl], [Cl] =
] Cl NO [ k ] NO [ k ] Cl NO [ k 2 2 2 1 2 1 Rate = dt ] Cl [ d 2 = ] Cl NO [ k ] NO [ k ] Cl NO [ k k 2 2 2 1 2 2 2 1
98. We need the value of k at 500. K;
2 1 a 1 2 T 1 T 1 R E k k ln
500K 1 K 273 1 mol K J 3145 . 8 mol / J 10 11 . 1 s mol L 10 3 . 2 k ln 1 1 5 1 1 12 2 = 22.2 12 2 10 3 . 2 k = e 22.2 , k2 = 1.0 × 10 2 L mol1 s1Because the decomposition reaction is an elementary reaction, the rate law can be written using the coefficients in the balanced equation. For this reaction, Rate = k[NO2]
2
. To solve for the time, we must use the integrated rate law for second-order kinetics. The major problem now is converting units so they match. Rearranging the ideal gas law gives n/V = P/RT. Substituting P/RT for concentration units in the second-order integrated rate equation:
0 0 0 0 0 2 2 P P P P k RT t , kt P RT P RT , RT / P 1 kt RT / P 1 , ] NO [ 1 kt ] NO [ 1 t = atm 5 . 2 atm 5 . 1 atm 5 . 1 atm 5 . 2 s mol L 10 0 . 1 ) K . 500 )( mol K atm L 08206 . 0 ( 1 1 2 1 1 = 1.1 × 103 s 99. a. T (K) 1/T (K1) k (min1) ln k 298.2 3.353 × 103 178 5.182 293.5 3.407 × 103 126 4.836 290.5 3.442 × 103 100. 4.605A plot of ln k versus 1/T gives a straight line (plot not included). The equation for the straight line is:
ln k = -6.48 × 103(1/T) + 26.9
For the ln k versus 1/T plot, slope = Ea/R = 6.48 × 10 3 K. 6.48 × 103 K = Ea/8.3145 J K1 mol 1 , Ea = 5.39 × 10 4 J/mol = 53.9 kJ/mol b. ln k = 6.48 × 103(1/288.2) + 26.9 = 4.42, k = e4.42 = 83 min1
About 83 chirps per minute per insect. Note: We carried extra significant figures. c. k gives the number of chirps per minute. The number or chirps in 15 s is k/4.
T (°C) T (°F) k (min1) 42 + 0.80(k/4)
25.0 77.0 178 78° F
20.3 68.5 126 67°F
17.3 63.1 100. 62°F
15.0 59.0 83 59°F
100. a. If the interval between flashes is 16.3 s, then the rate is: 1 flash/16.3 s = 6.13 × 102 s1 = k Interval k T 16.3 s 6.13 × 102 s1 21.0°C (294.2 K) 13.0 s 7.69 × 102 s1 27.8°C (301.0 K)
2 1 a 1 2 T 1 T 1 R E k k ln ; solving: Ea = 2.5 × 10 4 J/mol = 25 kJ/mol b.
303.2K 1 K 2 . 294 1 mol K J 3145 . 8 mol / J 10 5 . 2 10 13 . 6 k ln 1 1 4 2 = 0.30 k = e0.30 × 6.13 × 102 = 8.3 × 102 s1; interval = 1/k = 12 seconds. c. T Interval 54-2(Intervals) 21.0 °C 16.3 s 21 °C 27.8 °C 13.0 s 28 °C 30.0 °C 12 s 30. °CThis rule of thumb gives excellent agreement to two significant figures.
101. k = A exp(Ea/RT);
RT E E exp ) RT / E ( exp A ) RT / E ( exp A k k a,cat a,uncat uncat , a uncat cat , a cat uncat cat 2.50 × 103 =
K . 310 mol K J 3145 . 8 mol / J 10 00 . 5 E exp k k 1 1 4 cat , a uncat cat ln(2.50 × 103) × 2.58 × 103 J/mol = Ea, cat + 5.00 × 10 4 J/mol cat , aE = 5.00 × 104 J/mol 2.02 × 104 J/mol = 2.98 × 104 J/mol = 29.8 kJ/mol
102. From 338 K data, a plot of ln[N2O5] versus t is linear (plot not included). The integrated rate
law is: ln[N2O5] = (4.86 × 10 3 )t 2.30; k = 4.86 × 103 s1 at 338 K From 318 K data: ln[N2O5] = (4.98 × 104)t 2.30; k = 4.98 × 10 4 s1 at 318 K
K 338 1 K 318 1 mol K J 3145 . 8 E 10 98 . 4 10 86 . 4 ln , T 1 T 1 R E k k ln 4 a 1 1 3 2 1 a 1 2 Ea = 1.0 × 105 J/mol = 1.0 × 102 kJ/mol103. Rate = k[DNA]x[CH3I] y
; comparing the second and third experiments:
1 , 00 . 2 00 . 2 , ) 200 . 0 ( ) 100 . 0 ( k ) 200 . 0 ( ) 200 . 0 ( k 10 40 . 6 10 28 . 1 4 3
x x y x y xComparing the first and second experiments:
1 , 00 . 2 00 . 2 , ) 100 . 0 )( 100 . 0 ( k ) 200 . 0 )( 100 . 0 ( k 10 20 . 3 10 40 . 6 4 4
y y y yThe rate law is Rate = k[DNA][CH3I].
Mechanism I is possible because the derived rate law from the mechanism (Rate = k[DNA][CH3I]) agrees with the experimentally determined rate law. The derived rate law for
Mechanism II will equal the rate of the slowest step. This is step 1 in the mechanism giving a derived rate law that is Rate = k[CH3I]. Because this rate law does not agree with
experi-ment, Mechanism II would not be a possible mechanism for the reaction.
104. a. From the plot, the activation energy for the reverse reaction = Ea, forward + (ΔG°) =
Ea, forward ΔG° (ΔG° is a negative number as shown in the diagram).
kf = A exp RT Ea and kr = A exp
RT ) G E ( exp A RT E exp A k k , RT ) G E ( o a a r f o aIf the A factors are equal:
RT ) G E ( RT E exp k k a a o r f = exp RT G Δ o From ΔG° = RT ln K, K = exp RT G Δ o ; because K and r f k kare both equal to the same expression, K = kf/kr.
b. A catalyst will lower the activation energy for both the forward and reverse reactions (but not change ΔG°). Therefore, a catalyst must increase the rate of both the forward and reverse reactions.
105. Carbon cannot form the fifth bond necessary for the transition state because of the small atomic size of carbon and because carbon doesn’t have low-energy d orbitals available to expand the octet.
106. The pressure of a gas is directly proportional to concentration. Therefore, we can use the pressure data to solve the problem because d[SO2Cl2]/dt is proportional to dPSO2Cl2/dt.
