CHAPTER 14
CHEMICAL KINETICS
PRACTICE EXAMPLES
1A
(E) The rate of consumption for a reactant is expressed as the negative of the change in molarity divided by the time interval. The rate of reaction is expressed as the rate of consumption of a reactant or production of a product divided by its stoichiometric coefficient.
A 0.3187 M 0.3629 M 1min rate of consumption of A =
=
= 8.93 105 M s 1
t
8.25 min
60 s rate of reaction = rate of consumption of A2 =
8.93 105 M s 1
4.46 105 M s 1
2
1B
(E) We use the rate of reaction of A to determine the rate of formation of B, noting from the balanced equation that 3 moles of B form (+3 moles B) when 2 moles of A react (–2 moles A).
(Recall that “M” means “moles per liter.”)
0.5522 M A 0.5684 M A 3moles B rate of B formation=
1.62 104 M s 1
60s
2 moles A
2.50 min
1min
2A
(M)
(a) The 2400-s tangent line intersects the 1200-s vertical line at 0.75 M and reaches 0 M at
3500 s. The slope of that tangent line is thus
0 M 0.75 M slope =
= 3.3 104 M s 1 = instantaneous rate of reaction
3500 s 1200 s
The instantaneous rate of reaction = 3.3 104 M s 1 .
(b)
At 2400 s, H 2 O 2 = 0.39 M. At 2450 s, H 2 O 2 = 0.39 M + rate t
At 2450 s, H 2 O 2
= 0.39 M + 3.3 10 4 mol H 2 O 2 L1s 1 50s
= 0.39 M 0.017 M = 0.37 M
2B
(M) With only the data of Table 14.2 we can use only the reaction rate during the first 400 s,
H 2 O 2 /t = 15.0 104 M s 1 , and the initial concentration, H 2 O 2 0 = 2.32 M.
We calculate the change in H 2 O 2 and add it to H 2 O 2 0 to determine H 2 O 2 100 .
H 2 O 2 = rate of reaction of H 2 O 2 t = 15.0 104 M s 1 100 s = 0.15 M
H 2O2 100 = H 2O 2 0 + H 2O2 = 2.32 M + 0.15 M = 2.17 M
This value differs from the value of 2.15 M determined in text Example 14-2b because the text used the initial rate of reaction 17.1104 M s 1 , which is a bit faster than the average rate over the first 400 seconds.
611
Chapter 14: Chemical Kinetics
3A
(M) We write the equation for each rate, divide them into each other, and solve for n.
R1 = k N 2 O5 1 = 5.45 105 M s 1 = k 3.15 M
n n
R2 = k N 2 O5 2 = 1.35 105 M s 1 = k 0.78 M
n n
n k N 2 O 5 1 k 3.15 M
R1 5.45 105 M s 1 n
3.15
=
= 4.04 =
=
=
= 4.04 n n
5
1
R2 1.35 10 M s
0.78 k N 2 O5 2 k 0.78 M
n n
We kept an extra significant figure (4) to emphasize that the value of n = 1. Thus, the reaction is first-order in N 2 O5 .
3B
2
1
2
(E) For the reaction, we know that rate = k HgCl2 C2 O 4 . Here we will compare Expt. 4 to
Expt. 1 to find the rate.
2
1
2
2
rate 4 k HgCl2 C 2 O 4 rate 4
= 0.025 M 0.045 M = 0.0214 =
=
2
1
2 2 rate1 k HgCl C O
1.8 105 M min 1
0.105 M 0.150 M
2 2 4
The desired rate is rate4 = 0.0214 1.8 105 M min 1 = 3.9 107 M min 1 .
4A
(E) We place the initial concentrations and the initial rates into the rate law and solve for k .
2
2 rate = k A B = 4.78 102 M s 1 = k 1.12 M 0.87 M
4.78 102 M s 1
k=
1.12 M
2
0.87 M
= 4.4 102 M 2 s 1
2
4B
1
2
(E) We know that rate = k HgCl2 C2 O 4 and k = 7.6 103 M 2 min 1 .
Thus, insertion of the starting concentrations and the k value into the rate law yields:
1
2
Rate = 7.6 103 M 2 min 1 0.050 M 0.025 M = 2.4 107 M min 1
5A
(E) Here we substitute directly into the integrated rate law equation. ln A t = kt + ln A 0 = 3.02 103 s1 325 s + ln 2.80 = 0.982 +1.030 = 0.048
b g
At= e
5B
0.048
= 1.0 M
(M) This time we substitute the provided values into text Equation 14.13.
1.49 M
0.443
H 2O2 t
ln
= kt = k 600 s = ln
= 0.443
= 7.38 104 s 1 k=
2.32 M
600 s
H 2O2 0
Now we choose H 2 O2 0 = 1.49 M, H 2 O 2 t = 0.62,
ln
H 2O2 t
2 0
H 2O
= kt = k 1200 s = ln
0.62 M
= 0.88
1.49 M
t = 1800 s 600 s = 1200 s
k=
0.88
= 7.3 104 s 1
1200 s
These two values agree within the limits of the experimental error and thus, the reaction is firstorder in [H2O2].
612
Chapter 14: Chemical Kinetics
6A
(M) We can use the integrated rate equation to find the ratio of the final and initial concentrations. This ratio equals the fraction of the initial concentration that remains at time t.
A t
= kt = 2.95 103 s 1 150 s = 0.443
A 0
A t
= e 0.443 = 0.642; 64.2% of A 0 remains.
A 0
ln
6B
(M) After two-thirds of the sample has decomposed, one-third of the sample remains.
Thus H 2 O 2 t = H 2 O 2 0 3 , and we have
H 2O2 t
2 0
ln
H2O
t=
7A
= kt = ln
H 2O2 0 3
H 2O
2 0
= ln 1/ 3 = 1.099 = 7.30 104 s 1t
1.099
1 min
= 1.51103 s
= 25.1 min
4 1
7.30 10 s
60 s
(M) At the end of one half-life the pressure of DTBP will have been halved, to 400 mmHg. At the end of another half-life, at 160 min, the pressure of DTBP will have halved again, to 200 mmHg.
Thus, the pressure of DTBP at 125 min will be intermediate between the pressure at
80.0 min (400 mmHg) and that at 160 min (200 mmHg). To obtain an exact answer, first we determine the value of the rate constant from the half-life. k= ln
0.693
0.693
=
= 0.00866 min 1 t1/2 80.0 min
PDTBP t
PDTBP 0
= kt = 0.00866 min 1 125 min = 1.08
PDTBP t
= e 1.08 = 0.340
PDTBP 0
PDTBP t = 0.340 PDTBP 0 = 0.340 800 mmHg = 272 mmHg
7B
(M)
(a) We use partial pressures in place of concentrations in the integrated first-order rate equation. Notice first that more than 30 half-lives have elapsed, and thus the ethylene oxide
30
pressure has declined to at most 0.5 = 9 1010 of its initial value.
b g
ln
P30
P
3600 s
= kt = 2.05 104 s 1 30.0 h
= 22.1 30 = e22.1 = 2.4 1010
P0
1h
P0
P30 = 2.4 1010 P0 = 2.4 1010 782 mmHg = 1.9 107 mmHg
613
Chapter 14: Chemical Kinetics
Pethylene oxide initially 782 mmHg 1.9 107 mmHg (~ 0). Essentially all of the ethylene oxide is converted to CH4 and CO. Since pressure is proportional to moles, the final pressure will be twice the initial pressure (1 mole gas 2 moles gas;
782 mmHg 1564 mmHg). The final pressure will be 1.56 103 mmHg.
(b)
(D) We first begin by looking for a constant rate, indicative of a zero-order reaction. If the rate is constant, the concentration will decrease by the same quantity during the same time period. If we choose a 25-s time period, we note that the concentration decreases 0.88 M 0.74 M = 0.14 M
during the first 25 s, 0.74 M 0.62 M = 0.12 M during the second 25 s,
0.62 M 0.52 M =
0.10 M during the third 25 s, and 0.52 M 0.44 M = 0.08 M during the
fourth 25-s period. This is hardly a constant rate and we thus conclude that the reaction is not zero-order. We next look for a constant half-life, indicative of a first-order reaction. The initial concentration of 0.88 M decreases to one half of that value, 0.44 M, during the first 100 s, indicating a 100-s half-life. The concentration halves again to 0.22 M in the second 100 s, another 100-s half-life.
Finally, we note that the concentration halves also from 0.62 M at 50 s to 0.31 M at 150 s, yet another 100-s half-life. The rate is established as first-order. The rate constant is
0.693 0.693 k= =
= 6.93 103 s1 . t1/2 100 s
That the reaction is first-order is made apparent by the fact that the ln[B] vs time plot is a straight line with slope = -k (k = 6.85 103 s1).
Plot of 1/[B] versus
Time
Plot of ln([B]) versus
Time
Plot of [B] versus Time
7
0
0.85
-0.2
0.75
0
200
400
6
-0.4
0.65
-0.6
5
0.45
1/[B] (M -1)
-0.8
0.55
ln([B])
[B] (M)
8A
-1
-1.2
0.35
3
-1.4
0.25
4
-1.6
2
-1.8
0.15
0
100
200
Time(s)
300
1
-2
Time(s)
614
0
100
200
Tim e (s)
300
Chapter 14: Chemical Kinetics
8B
(D) We plot the data in three ways to determine the order. (1) A plot of [A] vs. time is linear if the reaction is zero-order. (2) A plot of ln [A] vs. time will be linear if the reaction is first-order.
(3) A plot of 1/[A] vs. time will be linear if the reaction is second-order. It is obvious from the plots below that the reaction is zero-order. The negative of the slope of the line equals k = 0.083 M 0.250 M 18.00 min = 9.28 103 M/min (k = 9.30 103 M/min using a
graphical approach).
Plot of [A] versus Time
0.25
0
10
Plot of ln([A]) versus Time
-1.5
0.23
0.21
20
12
Plot of 1/[A] versus Time
11
10
-1.7
ln([A])
[A] (M)
0.17
0.15
1/[A] (M-1)
0.19
-1.9
9
8
7
-2.1
0.13
6
0.11
-2.3
5
0.09 y = -0.00930x + 0.2494
4
0.07
0
9A
10
Time (min)
-2.5
20
Time (min)
0
10
Time (min)
20
(M) First we compute the value of the rate constant at 75.0 C with the Arrhenius equation. We know that the activation energy is Ea = 1.06 105 J/mol, and that k = 3.46 105 s1 at 298 K.
The temperature of 75.0 o C = 348.2 K. ln FG
H
IJ
K
FG
H
IJ
K
k2 k2 E 1 1
1.06 105 J / mol
1
1
= ln
= a
=
= 6.14
5 1
1
1 k1 3.46 10 s
R T1 T2
8.3145 J mol K 298.2 K 348.2 K
k2 = 3.46 105 s 1 e +6.14 = 3.46 105 s 1 4.6 102 = 0.016 s 1
t1/2 =
9B
0.693
0.693
=
= 43 s at 75 C
0.016 s 1 k (M) We use the integrated rate equation to determine the rate constant, realizing that one-third remains when two-thirds have decomposed.
N 2 O5 t
5 0
ln
k=
N 2O
= ln
N 2 O5 0 3
N 2O
5 0
1
= ln = kt = k 1.50 h = 1.099
3
1.099
1h
= 2.03 104 s 1
1.50 h 3600 s
615
Chapter 14: Chemical Kinetics
Now use the Arrhenius equation to determine the temperature at which the rate constant is 2.04
104 s1.
IJ
K
FG
H
FG
H
k2
2.04 104 s1
Ea 1 1
1.06 105 J / mol
1
1
ln = ln
= 1.77 =
=
5 1
1
1 k1 3.46 10 s
R T1 T2
8.3145 J mol K 298 K T2
1
1
1.77 8.3145 K 1
=
= 3.22 103 K 1
5
T2 298 K
1.06 10
IJ
K
T2 = 311 K
10A (M) The two steps of the mechanism must add, in a Hess's law fashion, to produce the overall reaction. Overall reaction: CO + NO2 CO 2 + NO
or
CO + NO 2 CO 2 + NO
Second step: NO3 + CO NO 2 + CO 2 or + NO 2 + CO 2 NO3 + CO
2 NO 2 NO + NO 3
First step:
If the first step is the slow step, then it will be the rate-determining step, and the rate of that step
2
will be the rate of the reaction, namely, rate of reaction = k1 NO2 .
10B (M)
(1) The steps of the mechanism must add, in a Hess's law fashion, to produce the overall reaction.
This is done below. The two intermediates, NO 2 F2 g and F(g), are each produced in one step and consumed in the next one.
bg
Fast:
Slow:
Fast:
Net:
(2)
NO 2 g + F2 g NO2 F2 g
NO 2 F2 g NO 2 F g + F g
F g + NO 2 g NO 2 F g
2 NO 2 g + F2 g 2 NO 2 F g
bg bg bg bg bg bg bg bg bg The proposed mechanism must agree with the rate law. We expect the rate-determining step to determine the reaction rate: Rate = k3 NO 2 F2 . To eliminate NO 2 F2 , we recognize that the first elementary reaction is very fast and will have the same rate forward as reverse: R f = k1 NO 2 F2 = k2 NO 2 F2 = Rr . We solve for the concentration of intermediate: NO 2 F2 = k1 NO 2 F2 /k2 . We now substitute this expression for NO 2 F2 into the rate equation: Rate = k1k3 /k2 NO 2 F2 . Thus the predicted rate law agrees with the experimental rate law.
616
Chapter 14: Chemical Kinetics
INTEGRATIVE EXAMPLE
A.
(M)
(a) The time required for the fixed (c) process of souring is three times as long at 3 °C refrigerator temperature (276 K) as at 20 °C room temperature (293 K).
E 1 1 E 1
64 h
1 Ea c / t2 t 4
ln 1 ln
1.10 a a
ln
( 2.10 10 )
3 64 h c / t1 t2 R T1 T2 R 293 K 276 K R
1.10 R
1.10 8.3145 J mol1 K 1
4.4 104 J/mol 44 kJ/mol
2.10 104 K 1
2.10 104 K 1
(b) Use the Ea determined in part (a) to calculate the souring time at 40 °C = 313 K.
E 1 1 t 4.4 104 J/mol 1
1 t1 ln 1 a
1.15 ln
1
1 t2 R T1 T2 8.3145 J mol K 313 K 293 K
64 h
Ea
t1
e 1.15 0.317 t1 0.317 64 h 20. h
64 h
B.
(M) The species A* is a reactive intermediate. Let’s deal with this species by using a steady state approximation. d[A*]/dt = 0 = k1[A]2 – k-1[A*][A] – k2[A*]. Solve for [A*]. k-1[A*][A] + k2[A*] = k1[A]2 k1 [A]2 k 2 k1 [A]2
[A*] =
The rate of reaction is: Rate = k2[A*] = k -1 [A] k 2 k -1 [A] k 2
At low pressures ([A] ~ 0 and hence k2 >> k-1[A]), the denominator becomes ~ k2 and the rate law is k k [A]2
Rate = 2 1
= k1 [A]2 Second-order with respect to [A] k2 At high pressures ([A] is large and k-1[A] >> K2), the denominator becomes ~ k-1[A] and the rate law is k k [A]2 k k [A]
Rate = 2 1
= 2 1
First-order with respect to [A] k -1 [A] k -1
EXERCISES
Rates of Reactions
1.
(M) 2A + B C + 3D
(a)
(b)
(c)
[A]
= 6.2 104 M s1
t
1 [A]
= 1/2(6.2 104 M s1) = 3.1 104 Ms1
2 t
1 [A]
Rate of disappearance of B =
= 1/2(6.2 104 M s1) = 3.1 104 Ms1
2 t
3 [A]
Rate of appearance of D =
= 3(6.2 104 M s1) = 9.3 104 Ms1
2 t
Rate =
617
Chapter 14: Chemical Kinetics
2.
(M) In each case, we draw the tangent line to the plotted curve.
H 2 O 2 1.7 M 0.6 M
=
= 9.2 104 M s 1
(a) The slope of the line is
t
400 s 1600 s
H 2O2 reaction rate =
= 9.2 104 M s 1
t
(b)
Read the value where the horizontal line [H2O2] = 0.50 M intersects the curve,
2150 s or 36 minutes
3.
(E) Rate =
4.
A
0.474 M 0.485 M = 1.0 103 M s 1
=
t
82.4 s 71.5 s
(M)
(b)
5.
(M)
(a)
(b)
Rate =
A
0.1498 M 0.1565 M
=
= 0.0067 M min 1
t
1.00 min 0.00 min
Rate =
(a)
A
0.1433 M 0.1498 M
=
= 0.0065 M min 1
t
2.00 min 1.00 min
The rates are not equal because, in all except zero-order reactions, the rate depends on the concentration of reactant. And, of course, as the reaction proceeds, reactant is consumed and its concentration decreases, so the rate of the reaction decreases.
A = A i + A = 0.588 M 0.013 M = 0.575 M
A = 0.565 M 0.588 M = 0.023 M
t
0.023 M
t = A
=
= 1.0 min
A 2.2 102 M/min time = t + t = 4.40 +1.0 min = 5.4 min
6.
(M) Initial concentrations are HgCl 2 = 0.105 M and C 2 O 4
2
= 0.300 M.
The initial rate of the reaction is 7.1 105 M min 1 . Recall that the reaction is:
2
2 HgCl2 (aq) + C2 O 4 (aq) 2 Cl (aq) + 2 CO 2 (g) + Hg 2 Cl2 (aq) .
The rate of reaction equals the rate of disappearance of C 2 O 2 . Then, after 1 hour, assuming that
4
the rate is the same as the initial rate,
(a)
(b)
mol C 2 O 4 2 2 mol HgCl2
60 min
1 h
0.096 M
2
L s
1h
1 mol C2 O 4
FG
H
mol
60 min
1 h
= 0.296 M
L min
1h
HgCl2 = 0.105 M 7.1 105
C 2 O 2 = 0.300 M 7.1 105
4
IJ
K
618
Chapter 14: Chemical Kinetics
7.
(M)
(a)
Rate =
C
(b)
(c)
8.
(M)
A
t
=
C
2t
= 1.76 105 M s 1
= 2 1.76 105 M s 1 = 3.52 105 M/s
t
A
C
=
= 1.76 105 M s1 Assume this rate is constant.
t
2 t
60 s
A = 0.3580 M + 1.76 105 M s 1 1.00 min
= 0.357 M
1min
A
= 1.76 105 M s 1
t
A
0.3500 M 0.3580 M
t =
=
= 4.5 102 s
5
5
1.76 10 M/s
1.76 10 M/s
(a)
n O 2
5.7 104 mol H 2 O 2
1 mol O 2
= 1.00 L soln
= 2.9 104 mol O 2 /s
t
1 L soln s
2 mol H 2 O 2
(b)
n O 2 mol O 2 60 s
= 2.9 104
= 1.7 102 mol O 2 / min
t
s
1 min
(c)
9.
V O 2 mol O 2 22,414 mL O 2 at STP 3.8 102 mL O 2 at STP
= 1.7 102
=
t min min
1 mol O 2
(M) Notice that, for every 1000 mmHg drop in the pressure of A(g), there will be a corresponding
2000 mmHg rise in the pressure of B(g) plus a 1000 mmHg rise in the pressure of C(g).
(a)
We set up the calculation with three lines of information below the balanced equation: (1) the initial conditions, (2) the changes that occur, which are related to each other by reaction stoichiometry, and (3) the final conditions, which simply are initial conditions + changes.
A(g)
2B(g)
+
C(g)
Initial
1000. mmHg
0. mmHg
0. mmHg
Changes
–1000. mmHg
+2000. mmHg
+1000. mmHg
Final
0. mmHg
2000. MmHg
1000. mmHg
Total final pressure = 0. mmHg + 2000. mmHg +1000. mmHg = 3000. mmHg
(b)
Initial
Changes
Final
A(g)
1000. mmHg
–200. mmHg
800 mmHg
2B(g)
+
0. mmHg
+400. mmHg
400. mmHg
C(g)
0. mmHg
+200. mmHg
200. mmHg
Total pressure = 800. mmHg + 400. mmHg + 200. mmHg = 1400. mmHg
619
Chapter 14: Chemical Kinetics
10.
(M)
(a) We will use the ideal gas law to determine N 2 O5 pressure
1 mol N 2 O5
1.00 g× 108.0 g nRT
P{N 2 O5 } =
=
V
(b)
(c)
L atm
×0.08206 mol K × 273 + 65 K
760 mmHg
×
= 13 mmHg
15 L
1 atm
After 2.38 min, one half-life passes. The initial pressure of N 2 O5 decreases by half to 6.5 mmHg. From the balanced chemical equation, the reaction of 2 mol N 2 O5 g produces 4 mol
bg
bg
bg
NO 2 g and 1 mol O 2 g . That is, the consumption of 2 mol of reactant gas produces 5 mol of product gas. When measured at the same temperature and confined to the same volume, pressures will behave as amounts: the reaction of 2 mmHg of reactant produces 5 mmHg of product.
5 mmHg(product)
Ptotal = 13 mmHg N 2 O5 (initially) 6.5 mmHg N 2 O5 (reactant) + 6.5 mmHg(reactant)
2 mmHg(reactant)
(13 6.5 16) mmHg 23 mmHg
Method of Initial Rates
11.
(M)
(a) From Expt. 1 to Expt. 3, [A] is doubled, while [B] remains fixed. This causes the rate to
6.75 104 M s1 increases by a factor of
= 2.01 2 .
3.35 104 M s1
Thus, the reaction is first-order with respect to A.
From Expt. 1 to Expt. 2, [B] doubles, while [A] remains fixed. This causes the rate to
1.35 103 M s1
= 4.03 4 . increases by a factor of
3.35 104 M s1
Thus, the reaction is second-order with respect to B.
(b) Overall reaction order order with respect to A order with respect to B = 1 2 = 3. The reaction is third-order overall.
2
(c) Rate = 3.35 104 M s1 = k 0.185 M 0.133 M
b
k=
12.
3.35 10
4
Ms
1
b0.185 Mgb0.133 Mg
2
gb
g
= 0.102 M 2 s1
(M) From Expt. 1 and Expt. 2 we see that [B] remains fixed while [A] triples. As a result, the initial rate increases from 4.2 103 M/min to 1.3 102 M/min, that is, the initial reaction rate triples. Therefore, the reaction is first-order in [A]. Between Expt. 2 and Expt. 3, we see that [A] doubles, which would double the rate, and [B] doubles. As a consequence, the initial rate goes from 1.3 102 M/min to 5.2 102 M/min, that is, the rate quadruples. Since an additional doubling of the rate is due to the change in [B], the reaction is first-order in [B]. Now we determine the value of the rate constant.
620
Chapter 14: Chemical Kinetics
Rate = k A
1
B
1
Rate
5.2 102 M / min k= =
= 5.8 103 L mol 1min 1
A B
3.00 M 3.00 M
c
h
The rate law is Rate = 5.8 103 L mol 1min 1 A
13.
1
B .
(M) From Experiment 1 to 2, [NO] remains constant while [Cl2] is doubled. At the same time the initial rate of reaction is found to double. Thus, the reaction is first-order with respect to [Cl2], since dividing reaction 2 by reaction 1 gives 2 = 2x when x = 1. From Experiment 1 to 3, [Cl2] remains constant, while [NO] is doubled, resulting in a quadrupling of the initial rate of reaction.
Thus, the reaction must be second-order in [NO], since dividing reaction 3 by reaction 1 gives
4 = 2x when x = 2. Overall the reaction is third-order: Rate = k [NO]2[Cl2]. The rate constant may be calculated from any one of the experiments. Using data from Exp. 1,
k=
14.
1
Rate
2.27 105 M s 1
=
= 5.70 M2 s1
[NO]2 [Cl2 ] (0.0125 M) 2 (0.0255 M)
(M)
(a) From Expt. 1 to Expt. 2, [B] remains constant at 1.40 M and [C] remains constant at
1.00 M, but [A] is halved 0.50 . At the same time the rate is halved 0.50 . Thus, the
b
b
g
g
x
reaction is first-order with respect to A, since 0.50 = 0.50 when x = 1.
From Expt. 2 to Expt. 3, [A] remains constant at 0.70 M and [C] remains constant at
1.00 M, but [B] is halved 0.50 , from 1.40 M to 0.70 M. At the same time, the rate is
b
b
g
g
quartered 0.25 . Thus, the reaction is second-order with respect to B, since 0.50 y = 0.25 when y = 2 .
From Expt. 1 to Expt. 4, [A] remains constant at 1.40 M and [B] remains constant at
1.40 M, but [C] is halved 0.50 , from 1.00 M to 0.50 M. At the same time, the rate is increased by a factor of 2.0.
1
1
Rate4 = 16 Rate3 = 16 Rate 2 = 4 Rate 2 = 4 Rate1 = 2 Rate1.
4
2
Thus, the order of the reaction with respect to C is 1 , since 0.5z = 2.0 when z = 1 .
(b)
rate5 = k 0.70 M 0.70 M 0.50 M
1
k
11
2
2
1.40 M
1 12
2
1.40 M
2 1 1
2
1
1
1.00 M
1
621
2
1+2 1
1
= rate1
2
This is based on rate1 = k 1.40 M 1.40 M 1.00 M
1
2
1.40 M 1.40 M 1.00 M
= k
2 2 2
1
2
1
1
= rate1 = 1 rate1
4
2
Chapter 14: Chemical Kinetics
First-Order Reactions
15.
The rate of the reaction does decrease as more and more of B and C are formed, but not because more and more of B and C are formed. Rather, the rate decreases because the concentration of A must decrease to form more and more of B and
C.
FALSE
The time required for one half of substance A to react—the half-life—is independent of the quantity of A present.
(E)
(a)
FALSE
(b)
17.
TRUE
(b)
16.
(E)
(a)
TRUE
For first-order reactions, a plot of ln [A] or log [A] vs. time yields a straight line.
A graph of [A] vs. time yields a curved line.
The rate of formation of C is related to the rate of disappearance of A by the stoichiometry of the reaction.
(M)
(a) Since the half-life is 180 s, after 900 s five half-lives have elapsed, and the original quantity of A has been cut in half five times.
5
final quantity of A = 0.5 initial quantity of A = 0.03125 initial quantity of A
About 3.13% of the original quantity of A remains unreacted after 900 s. or More generally, we would calculate the value of the rate constant, k, using ln 2 0.693
=
= 0.00385 s 1 Now ln(% unreacted) = -kt = -0.00385 s-1×(900s) = k=
180 s t1/2 3.465
(% unreacted) = 0.0313 × 100% = 3.13% of the original quantity.
b g
(b)
18.
Rate = k A = 0.00385 s1 0.50 M = 0.00193 M / s
(M)
(a) The reaction is first-order, thus
A t
0.100 M ln = kt = ln
= 54 min(k )
0.800 M
A 0
k=
2.08
= 0.0385 min 1
54 min
We may now determine the time required to achieve a concentration of 0.025 M
A t
3.47
0.025 M ln = kt = ln
= 0.0385 min 1 (t )
= 90. min t= 0.0385 min 1
0.800 M
A 0
(b)
Since we know the rate constant for this reaction (see above),
1
Rate = k A = 0.0385 min 1 0.025 M = 9.6 104 M/min
622
Chapter 14: Chemical Kinetics
19.
(M)
(a) The mass of A has decreased to one fourth of its original value, from 1.60 g to
1 1 1
0.40 g. Since
= , we see that two half-lives have elapsed.
4 2 2
Thus, 2 t1/2 = 38 min, or t1/2 = 19 min .
A t
0.693
= 0.036 min 1 ln
= kt = 0.036 min 1 60 min = 2.2
(b) k = 0.693/t1/2 =
19 min
A 0
A t
=
A 0
20.
(M)
A t
A 0
e 2.2 = 0.11 or A t = A 0 e kt = 1.60 g A 0.11 = 0.18 g A
= kt = ln
0.632 M
= 0.256
0.816 M
k=
0.256
= 0.0160 min 1
16.0 min
(a)
ln
(b)
t1/2 =
(c)
We need to solve the integrated rate equation to find the elapsed time. ln (d)
ln
0.693
0.693
=
= 43.3 min
0.0160 min 1 k A t
A 0
= kt = ln
t=
1.245
= 77.8 min
0.0160 min 1
A
A = e kt which in turn becomes
= kt becomes
A0
A 0
A = A
0
21.
0.235 M
= 1.245 = 0.0160 min 1 t
0.816 M
e kt = 0.816 M exp 0.0160 min 1 2.5 h
60 min
1h
= 0.816 0.0907 = 0.074 M
(M) We determine the value of the first-order rate constant and from that we can calculate the half-life. If the reactant is 99% decomposed in 137 min, then only 1% (0.010) of the initial concentration remains.
A t
4.61
0.010 ln = kt = ln
= 4.61 = k 137min
= 0.0336 min 1 k= 137 min
1.000
A 0
t1/2 =
0.0693
0.693
=
= 20.6 min k 0.0336 min 1
623
Chapter 14: Chemical Kinetics
(E) If 99% of the radioactivity of
32
22.
P is lost, 1% (0.010) of that radioactivity remains. First we
0.693 0.693 compute the value of the rate constant from the half-life. k =
=
= 0.0485 d 1 t1/2 14.3 d
Then we use the integrated rate equation to determine the elapsed time.
At
At
1
1
0.010
ln
= kt t = ln
=
ln
= 95 days
1
A0 k A0
0.0485 d
1.000
23.
(D)
(a)
(b)
35
100 [A]0
3
1 ln
= ln(0.35) = kt = (4.81 10 min )t t = 218 min.
[A]0
Note: We did not need to know the initial concentration of acetoacetic acid to answer the question. Let’s assume that the reaction takes place in a 1.00L container.
10.0 g acetoacetic acid
1 mol acetoacetic acid
102.090 g acetoacetic acid
= 0.09795 mol acetoacetic acid.
After 575 min. (~ 4 half lives, hence, we expect ~ 6.25% remains as a rough approximation), use integrated form of the rate law to find [A]t = 575 min.
[A]t
3
1 ln
= kt = (4.81 10 min )(575 min) = 2.766
[A]0
[A]t
[A]t
= e2.766 = 0.06293 (~ 6.3% remains)
= 0.063 [A]t = 6.2 103
[A]0
0.09795 moles moles. [A]reacted = [A]o [A]t = (0.098 6.2 103) moles = 0.092 moles acetoacetic acid. The stoichiometry is such that for every mole of acetoacetic acid consumed, one mole of CO2 forms. Hence, we need to determine the volume of 0.0918 moles CO2 at 24.5 C (297.65
K) and 748 torr (0.984 atm) by using the Ideal Gas law.
L atm
0.0918 mol 0.08206
297.65 K
K mol nRT
= 2.3 L CO2
V=
=
P
0.984 atm
24.
(M)
ln
(a)
A t
A 0
= kt = ln
2.5 g
= 3.47 = 6.2 104 s 1t
80.0 g
3.47
= 5.6 103 s 93 min
6.2 104 s 1
We substituted masses for concentrations, because the same substance (with the same molar mass) is present initially at time t , and because it is a closed system. t= 624
Chapter 14: Chemical Kinetics
(b)
25.
(D)
(a)
1 mol N 2 O5
1 mol O 2
= 0.359 mol O 2
108.0 g N 2 O5 2 mol N 2 O5
L atm
0.359 mol O 2 0.08206
(45 273) K nRT mol K
V
9.56 L O 2
1 atm
P
745 mmHg
760 mmHg amount O 2 = 77.5 g N 2 O5
If the reaction is first-order, we will obtain the same value of the rate constant from several sets of data.
A t
0.497 M
0.188
ln
= kt = ln
= k 100 s = 0.188, k =
= 1.88 103 s 1
A 0
0.600 M
100 s
A t
0.344 M
= kt = ln
= k 300 s = 0.556,
0.600 M
A 0
A t
0.285 M ln = kt = ln
= k 400 s = 0.744,
0.600 M
A 0
A t
0.198 M ln = kt = ln
= k 600 s = 1.109,
0.600 M
A 0
A t
0.094 M ln = kt = ln
= k 1000 s = 1.854,
0.600 M
A 0 ln (b)
(c)
26.
(D)
(a)
k=
0.556
= 1.85 103 s 1
300 s
k=
0.744
= 1.86 103 s 1
400 s
k=
1.109
= 1.85 103 s 1
600 s
k=
1.854
= 1.85 103 s 1
1000 s
The virtual constancy of the rate constant throughout the time of the reaction confirms that the reaction is first-order.
For this part, we assume that the rate constant equals the average of the values obtained in part (a).
1.88 +1.85 +1.86 +1.85 k= 103 s1 = 1.86 103 s1
4
We use the integrated first-order rate equation:
A 750 = A 0 exp kt = 0.600 M exp 1.86 103 s 1 750 s
A 750 = 0.600 M e 1.40 = 0.148 M
If the reaction is first-order, we will obtain the same value of the rate constant from several sets of data.
0.167
P
264 mmHg ln t = kt = ln
= k 390 s = 0.167 , k= = 4.28 104 s1
390 s
P0
312 mmHg
0.331
P
224 mmHg ln t = kt = ln
= k 777 s = 0.331 , k= = 4.26 104 s1
777 s
312 mmHg
P0
0.512
P
187 mmHg ln t = kt = ln
= k 1195 s = 0.512 , k= = 4.28 104 s1
1195 s
312 mmHg
P0
625
Chapter 14: Chemical Kinetics
Pt
78.5 mmHg
1.38
= kt = ln
= k 3155 s = 1.38 , k= = 4.37 104 s 1
P0
312 mmHg
3155 s
The virtual constancy of the rate constant confirms that the reaction is first-order. ln (b)
For this part we assume the rate constant is the average of the values in part
(a): 4.3 104 s1 .
(c)
At 390 s, the pressure of dimethyl ether has dropped to 264 mmHg. Thus, an amount of dimethyl ether equivalent to a pressure of 312 mmHg 264 mmHg = 48 mmHg has decomposed. For each 1 mmHg pressure of dimethyl ether that decomposes, 3 mmHg of pressure from the products is produced. Thus, the increase in the pressure of the products is
3 48 = 144 mmHg. The total pressure at this point is 264 mmHg +144 mmHg = 408 mmHg. Below, this calculation is done in a more systematic fashion:
b
bCH g Obgg
3 2
Initial
312 mmHg
Changes – 48 mmHg
Final
264 mmHg
g
bg
CH 4 g
bg
+
0 mmHg
+ 48 mmHg
48 mmHg
H2 g
bg
+
CO g
0 mmHg
+ 48 mmHg
48 mmHg
0 mmHg
+ 48 mmHg
48 mmHg
Ptotal = PDME + Pmethane + Phydrogen + PCO
264 mmHg 48 mmHg 48 mmHg 48 mmHg 408 mmHg
(d) This question is solved in the same manner as part (c). The results are summarized below.
bCH g Obgg
3 2
bg
CH 4 g
Initial
312 mmHg
0 mmHg
Changes 312 mmHg 312 mmHg
Final
0 mmHg
312 mmHg
Ptotal = PDME + Pmethane + Phydrogen + PCO
bg
H2 g
bg
+ CO g
0 mmHg
312 mmHg
312 mmHg
0 mmHg
+ 312 mmHg
312 mmHg
0 mmHg 312 mmHg 312 mmHg 312 mmHg 936 mmHg
P
(e) We first determine PDME at 1000 s. ln 1000 = kt = 4.3 104 s1 1000 s = 0.43
P0
P
1000
= e 0.43 = 0.65
P = 312 mmHg 0.65 = 203 mmHg
1000
P0
Then we use the same approach as was used for parts (c) and (d)
CH3 2 O g
Initial
312 mmHg
Changes 109 mmHg
CH 4 g
+
H2 g
0 mmHg
109 mmHg
0 mmHg
109 mmHg
Final
203 mmHg
109 mmHg
Ptotal = PDME + Pmethane +Phydrogen +PCO
109 mmHg
+
CO g
0 mmHg
109 mmHg
109 mmHg
= 203 mmHg +109 mmHg +109 mmHg +109 mmHg = 530. mmHg
626
Chapter 14: Chemical Kinetics
Reactions of Various Orders
27.
(M)
(a) Set II is data from a zero-order reaction. We know this because the rate of set II is constant.
0.25 M/25 s = 0.010 M s1 . Zero-order reactions have constant rates of reaction.
(b)
A first-order reaction has a constant half-life. In set I, the first half-life is slightly less than
75 sec, since the concentration decreases by slightly more than half (from 1.00 M to 0.47 M) in 75 s. Again, from 75 s to 150 s the concentration decreases from 0.47 M to 0.22 M, again by slightly more than half, in a time of 75 s. Finally, two half-lives should see the concentration decrease to one-fourth of its initial value. This, in fact, is what we see. From
100 s to 250 s, 150 s of elapsed time, the concentration decreases from 0.37 M to 0.08 M,
i.e., to slightly less than one-fourth of its initial value. Notice that we cannot make the same statement of constancy of half-life for set III. The first half-life is 100 s, but it takes more than 150 s (from 100 s to 250 s) for [A] to again decrease by half.
(c)
For a second-order reaction,1/ A t 1/ A 0 = kt . For the initial 100 s in set III, we have
1
1
= 1.0 L mol1 = k 100 s, k = 0.010 L mol1 s 1
0.50 M 1.00 M
For the initial 200 s, we have
1
1
= 2.0 L mol1 = k 200 s, k = 0.010 L mol1 s 1
0.33 M 1.00 M
Since we obtain the same value of the rate constant using the equation for second-order kinetics, set III must be second-order.
28.
(E) For a zero-order reaction (set II), the slope equals the rate constant: k = A /t = 1.00 M/100 s = 0.0100 M/s
29.
(M) Set I is the data for a first-order reaction; we can analyze those items of data to determine the half-life. In the first 75 s, the concentration decreases by a bit more than half. This implies a halflife slightly less than 75 s, perhaps 70 s. This is consistent with the other time periods noted in the answer to Review Question 18 (b) and also to the fact that in the 150-s interval from 50 s to 200 s, the concentration decreases from 0.61 M to 0.14 M, which is a bit more than a factor-of-four decrease. The factor-of-four decrease, to one-fourth of the initial value, is what we would expect for two successive half-lives. We can determine the half-life more accurately, by obtaining a value of k from the relation ln A t / A 0 = kt followed by t1/2 = 0.693 / k For instance,
ln(0.78/1.00) = -k (25 s); k = 9.94 ×10-3 s-1. Thus, t1/2 = 0.693/9.94 ×10-3 s-1 = 70 s.
30.
(E) We can determine an approximate initial rate by using data from the first 25 s.
A
0.80 M 1.00 M
Rate =
=
= 0.0080 M s 1
t
25 s 0 s
627
Chapter 14: Chemical Kinetics
31.
(M) The approximate rate at 75 s can be taken as the rate over the time period from 50 s to 100 s.
A
0.00 M 0.50 M
(a) Rate II =
=
= 0.010 M s1
t
100 s 50 s
(b)
Rate I =
A
0.37 M 0.61 M
=
= 0.0048 M s1
t
100 s 50 s
A
0.50 M 0.67 M
=
= 0.0034 M s1
t
100 s 50 s
Alternatively we can use [A] at 75 s (the values given in the table) in the m relationship Rate = k A , where m = 0 , 1, or 2.
(c)
Rate III =
(a)
Rate II = 0.010 M s 1 0.25 mol/L = 0.010 M s 1
(b)
Since t1/2 70s, k 0.693 / 70s 0.0099s 1
0
Rate I 0.0099 s 1 (0.47 mol/L)1 0.0047 M s 1
(c)
32.
Rate III = 0.010 L mol1 s 1 0.57 mol/L = 0.0032 M s 1
2
(M) We can combine the approximate rates from Exercise 31, with the fact that 10 s have elapsed, and the concentration at 100 s.
(a) A II = 0.00 M There is no reactant left after 100 s.
(b)
AI= A
(c)
A
III
100
= A
b
g
c
h
10 s rate = 0.37 M 10 s 0.0047 M s1 = 0.32 M
100
b
g
c
h
10 s rate = 0.50 M 10 s 0.0032 M s1 = 0.47 M
33.
(E) Substitute the given values into the rate equation to obtain the rate of reaction.
2
0
2
0
Rate = k A B = 0.0103 M 1min 1 0.116 M 3.83 M = 1.39 10 4 M / min
34.
(M)
(a) A first-order reaction has a constant half-life. Thus, half of the initial concentration remains after 30.0 minutes, and at the end of another half-life—60.0 minutes total—half of the concentration present at 30.0 minutes will have reacted: the concentration has decreased to one-quarter of its initial value. Or, we could say that the reaction is 75% complete after two half-lives—60.0 minutes.
c
(b)
gb
hb
g
A zero-order reaction proceeds at a constant rate. Thus, if the reaction is 50% complete in
30.0 minutes, in twice the time—60.0 minutes—the reaction will be 100% complete. (And in one-fifth the time—6.0 minutes—the reaction will be 10% complete. Alternatively, we can say that the rate of reaction is 10%/6.0 min.) Therefore, the time required for the
60.0 min reaction to be 75% complete = 75%
= 45 min.
100%
628
Chapter 14: Chemical Kinetics
(M) For reaction: HI(g) 1/2 H2(g) 1/2 I2(g) (700 K)
Time
(s)
0
100
200
300
400
[HI] (M)
1/[HI](M1)
ln[HI]
y = 0.00118x + 0.997
1.00
0.90
0.81
0.74
0.68
0
0.105
0.211
0.301
0.386
1.00
1.11
1.235
1.35
1.47
1.40
1.20
From data above, a plot of 1/[HI] vs. t yields a straight line. The reaction is second-order in HI at 700 K. Rate = k[HI]2. The slope of the line = k = 0.00118 M1s1
(D)
(a)
1.00
0
250
Time(s)
500
We can graph 1/[ArSO2H] vs. time and obtain a straight line. We can also graph [ArSO2H] vs. time and ln([ArSO2H]) vs. time to demonstrate that they do not yield a straight line.
Only the plot of 1/[ArSO2H] versus time is shown.
Plot of 1/[ArSO H] versus Time
Plot of 1/[ArSO2H]2versus Time
50
-1
1/[ArSO2H] (M )
1/[ArSO2H] (M-1)
36.
Plot of 1/[HI] vs time
1.60
1/[HI] (M-1)
35.
30 y = 0.137x + 9.464
10
0
50
100
150
Time (min)
200
250
300
The linearity of the line indicates that the reaction is second-order.
(b)
We solve the rearranged integrated second-order rate law for the rate constant, using the longest time interval. k= (c)
1
1
= kt
At A0
F
GH
I
JK
1 1
1
=k
t At A0
1
1
1
1
1
= 0.137 L mol min
300 min 0.0196 M 0.100 M
We use the same equation as in part (b), but solved for t , rather than k .
1 1
1
1
1
1
t=
=
= 73.0 min
1
1 k A t A 0 0.137 L mol min 0.0500 M 0.100 M
629
Chapter 14: Chemical Kinetics
(d)
(e)
We use the same equation as in part (b), but solve for t , rather than k .
1 1
1
1
1
1
t=
=
= 136 min
A A 0.137 L mol1min 1 0.0350 M 0.100 M
k
t 0
(M)
(a) Plot [A] vs t, ln[A] vs t, and 1/[A] vs t and see which yields a straight line.
0.8
0
y = -0.0050x + 0.7150
0.7
-0.5 0
R 2 = 1.0000
0.6
150
-1
0.5
ln[A]
[A] (mol/L)
0.4
0.3
-1.5
-2
-2.5
0.2
-3
0.1
y = -0.0191x
- 0.1092
R2 = 0.9192
-3.5
0
Time(s)
0
Time(s)
150
20 y = 0.1228x
- 0.9650
R2 = 0.7905
15
1/[A]
37.
We use the same equation as in part (b), but solve for t , rather than k .
1 1
1
1
1
1
t=
=
= 219 min
1
1 k A t A 0 0.137 L mol min 0.0250 M 0.100 M
10
5
0
0
Time(s)
150
Clearly we can see that the reaction is zero-order in reactant A with a rate constant of 5.0 ×10-3.
(b)
The half-life of this reaction is the time needed for one half of the initial [A] to react.
0.358 M
Thus, A = 0.715 M 2 = 0.358 M and t1/2 =
= 72 s.
5.0 103 M / s
630
Chapter 14: Chemical Kinetics
38.
(D)
(a)
We can either graph 1/ C 4 H 6 vs. time and obtain a straight line, or we can determine the second-order rate constant from several data points. Then, if k indeed is a constant, the reaction is demonstrated to be second-order. We shall use the second technique in this case.
First we do a bit of algebra.
1
1
= kt
At A0
F
GH
I
JK
1 1
1
=k
t At A0
FG
IJ
H
K
1
FG 1 1 IJ = 0.875 L mol min k= 24.55 min H 0.0124 M 0.0169 M K
1
FG 1 1 IJ = 0.892 L mol min k= 42.50 min H 0.0103 M 0.0169 M K
1
FG 1 1 IJ = 0.870 L mol min k= 68.05 min H 0.00845 M 0.0169 M K k= 1
1
1
= 0.843 L mol 1min 1
12.18 min 0.0144 M 0.0169 M
1
1
1
1
1
(b)
1
The fact that each calculation generates similar values for the rate constant indicates that the reaction is second-order.
The rate constant is the average of the values obtained in part (a).
0.843 + 0.875 + 0.892 + 0.870 k= L mol1min 1 = 0.87 L mol1min 1
4
(c)
(d)
39.
We use the same equation as in part (a), but solve for t , rather than k .
1 1
1
1
1
1
= 2.0 102 min
t=
=
1
1
A A 0.870 L mol min 0.00423 M 0.0169 M k
0 t We use the same equation as in part (a), but solve for t , rather than k .
1 1
1
1
1
1
2
t=
=
= 1.6 10 min
1
1 k A t A 0 0.870 L mol min 0.0050 M 0.0169 M
(E)
(a)
(b)
A
1.490 M 1.512 M
=
= +0.022 M / min
t
1.0 min 0.0 min
A
2.935 M 3.024 M initial rate =
=
= +0.089 M / min
t
1.0 min 0.0 min initial rate =
b g
When the initial concentration is doubled 2.0 , from 1.512 M to 3.024 M, the initial rate
b g
quadruples 4.0 . Thus, the reaction is second-order in A (since 2.0 x = 4.0 when x = 2 ).
631
Chapter 14: Chemical Kinetics
40.
(M)
(a) Let us assess the possibilities. If the reaction is zero-order, its rate will be constant. During the first 8 min, the rate is 0.60 M 0.80 M /8 min = 0.03 M/min . Then, during the first
24 min, the rate is 0.35 M 0.80 M /24 min = 0.019 M/min . Thus, the reaction is not zero-order. If the reaction is first-order, it will have a constant half-life that is consistent with its rate constant. The half-life can be assessed from the fact that 40 min elapse while the concentration drops from 0.80 M to 0.20 M, that is, to one-fourth of its initial value.
Thus, 40 min equals two half-lives and t1/2 = 20 min .
This gives k = 0.693 / t1/2 = 0.693 / 20 min = 0.035 min 1 . Also
At
0.827
0.35 M kt = ln
= ln
= 0.827 = k 24 min k= = 0.034 min 1
A0
0.80 M
24 min
The constancy of the value of k indicates that the reaction is first-order.
(b)
The value of the rate constant is k = 0.034 min 1 .
(c)
Reaction rate = ln A
A 0
1
2
1
(rate of formation of B) = k A First we need [A] at t = 30 . min
= kt = 0.034 min 1 30. min = 1.02
A
A 0
= e 1.02 = 0.36
A = 0.36 0.80 M = 0.29 M rate of formation of B = 2 0.034 min 1 0.29 M = 2.0 102 M min 1
41.
(M) The half-life of the reaction depends on the concentration of A and, thus, this reaction cannot be first-order. For a second-order reaction, the half-life varies inversely with the reaction rate: t1/2 = 1/ k A 0 or k = 1/ t1/2 A 0 . Let us attempt to verify the second-order nature of this
c
h
c
h
reaction by seeing if the rate constant is fixed.
1
k=
= 0.020 L mol1min 1
1.00 M 50 min
1
k=
= 0.020 L mol1min 1
2.00 M 25 min
1
k=
= 0.020 L mol 1 min 1
0.50 M 100 min
The constancy of the rate constant demonstrates that this reaction indeed is second-order. The rate
2
equation is Rate = k A and k = 0.020 L mol 1min 1 .
632
Chapter 14: Chemical Kinetics
42.
(M)
(a) The half-life depends on the initial NH 3 and, thus, the reaction cannot be first-order. Let us attempt to verify second-order kinetics.
1
1 k= for a second-order reaction k =
= 42 M 1min 1
NH3 0 t1/2
0.0031 M 7.6 min
1
1
= 180 M 1min 1 k= = 865 M 1min 1
0.0015 M 3.7 min
0.00068 M 1.7min
The reaction is not second-order. But, if the reaction is zero-order, its rate will be constant.
A 0 / 2 0.0031 M 2
Rate =
=
= 2.0 104 M/min t1/2 7.6 min
0.0015 M 2
Rate =
= 2.0 104 M/min
3.7 min
0.00068 M 2
Rate =
2.0 104 M / min
Zero-order reaction
1.7 min k= (b)
43.
44.
The constancy of the rate indicates that the decomposition of ammonia under these conditions is zero-order, and the rate constant is k = 2.0 104 M/min.
[A]0
1
Second-order: t1/2 = k[A]0 2k
A zero-order reaction has a half life that varies proportionally to [A]0, therefore, increasing [A]0 increases the half-life for the reaction. A second-order reaction's half-life varies inversely proportional to [A]0, that is, as [A]0 increases, the half-life decreases. The reason for the difference is that a zero-order reaction has a constant rate of reaction (independent of [A]0). The larger the value of [A]0, the longer it will take to react. In a second-order reaction, the rate of reaction increases as the square of the [A]0, hence, for high [A]0, the rate of reaction is large and for very low [A]0, the rate of reaction is very slow. If we consider a bimolecular elementary reaction, we can easily see that a reaction will not take place unless two molecules of reactants collide. This is more likely when the [A]0 is large than when it is small.
(M) Zero-order: t1/2 =
(M)
(a)
[A]0 0.693
=
k
2k
Hence,
[A]0
= 0.693 or [A]0 = 1.39 M
2
(b)
[A]0
1
,
2k k[A]0 Hence,
[A]0 2
= 1 or [A]02 = 2.00 M
2
(c)
0.693
1
1
, Hence, 0.693 = or [A]0 = 1.44 M k k[A]o
[A]0
633
[A]0 = 1.414 M
Chapter 14: Chemical Kinetics
Collision Theory; Activation Energy
45.
(M)
(a) The rate of a reaction depends on at least two factors other than the frequency of collisions.
The first of these is whether each collision possesses sufficient energy to get over the energy barrier to products. This depends on the activation energy of the reaction; the higher it is, the smaller will be the fraction of successful collisions. The second factor is whether the molecules in a given collision are properly oriented for a successful reaction. The more complex the molecules are, or the more freedom of motion the molecules have, the smaller will be the fraction of collisions that are correctly oriented.
(b)
(c)
46.
Although the collision frequency increases relatively slowly with temperature, the fraction of those collisions that have sufficient energy to overcome the activation energy increases much more rapidly. Therefore, the rate of reaction will increase dramatically with temperature. The addition of a catalyst has the net effect of decreasing the activation energy of the overall reaction, by enabling an alternative mechanism. The lower activation energy of the alternative mechanism, (compared to the uncatalyzed mechanism), means that a larger fraction of molecules have sufficient energy to react. Thus the rate increases, even though the temperature does not.
(M)
(a) The activation energy for the reaction of hydrogen with oxygen is quite high, too high, in fact, to be supplied by the energy ordinarily available in a mixture of the two gases at ambient temperatures. However, the spark supplies a suitably concentrated form of energy to initiate the reaction of at least a few molecules. Since the reaction is highly exothermic, the reaction of these first few molecules supplies sufficient energy for yet other molecules to react and the reaction proceeds to completion or to the elimination of the limiting reactant. (b)
47.
A larger spark simply means that a larger number of molecules react initially. But the eventual course of the reaction remains the same, with the initial reaction producing enough energy to initiate still more molecules, and so on.
(M)
(a) The products are 21 kJ/mol closer in energy to the energy activated complex than are the reactants. Thus, the activation energy for the reverse reaction is
84 kJ / mol 21 kJ / mol = 63 kJ / mol.
634
Chapter 14: Chemical Kinetics
Potential Energy (kJ)
(b) The reaction profile for the reaction in Figure 14-10 is sketched below.
A ....B
Transiton State
Ea(reverse) = 63 kJ
Ea(forward) = 84 kJ
Products
C+D
H = +21 kJ
Reactants
A+B
Progress of Reaction
48.
(M) In an endothermic reaction (right), Ea must be larger than the H for the reaction. For an exothermic reaction (left), the magnitude of Ea may be either larger or smaller than that of H .
In other words, a small activation energy can be associated with a large decrease in the enthalpy, or a large Ea can be connected to a small decrease in enthalpy.
products
reactants
ΔH0
(E)
(a)
There are two intermediates (B and C).
(b)
There are three transition states (peaks/maxima) in the energy diagram.
(c)
The fastest step has the smallest Ea, hence, step 3 is the fastest step in the reaction with step 2 a close second.
(d)
Reactant A (step 1) has the highest Ea, and therefore the slowest smallest constant
(e)
Endothermic; energy is needed to go from A to B.
(f)
Exothermic; energy is released moving from A to D.
635
Chapter 14: Chemical Kinetics
50.
(E)
(a)
There are two intermediates (B and C).
(b)
There are three transition states (peaks/maxima) in the energy diagram.
(c)
The fastest step has the smallest Ea, hence, step 2 is the fastest step in the reaction.
(d)
Reactant A (step 1) has the highest Ea, and therefore the slowest smallest constant
(e)
Endothermic; energy is needed to go from A to B.
(f)
Endothermic, energy is needed to go from A to D.
Effect of Temperature on Rates of Reaction
51.
(M)
1
1 k E 1 1
5.4 104 L mol 1 s1 E a
=
ln 1 = a
= ln
2
1 1 k2 R T2 T1
2.8 10 L mol s
R 683 K 599 K
FG
H
IJ
K
FG
H
IJ
K
3.95R = Ea 2.05 104
3.95 R
Ea =
= 1.93 104 K 1 8.3145 J mol 1 K 1 = 1.60 105 J / mol = 160 kJ / mol
4
2.05 10
52.
(M)
E 1 1 k 5.0 103 L mol1 s 1
1.60 105 J/mol 1
1
ln 1 = a = ln
=
2
1 1
1
1 k2 R T2 T1
2.8 10 L mol s
8.3145 J mol K 683 K T
1
1
1.72
1
1
1.72 = 1.92 104
=
= 8.96 105
683 K T
683 K T 1.92 104
1
= 8.96 105 +1.46 103 = 1.55 103
T = 645 K
T
53.
(D)
(a)
First we need to compute values of ln k and 1/ T . Then we plot the graph of ln k versus
1/T.
T , C
0 C
10 C
20 C
30 C
T,K
273 K
283 K
293 K
303 K
1
1/ T , K
0.00366
0.00353
0.00341
0.00330
1
6
5
4 k, s
5.6 10
3.2 10
1.6 10
7.6 104 ln k
12.09
10.35
8.74
7.18
636
Chapter 14: Chemical Kinetics
0.00325
1/T (K-1)
Plot of ln k versus 1/T
0.00345
0.00365
-7.1
-8.1
ln k
-9.1
-10.1
-11.1
y = -13520x + 37.4
-12.1
(b)
The slope = Ea / R .
J
1 kJ kJ 1.352 104 K
= 112 mol K
1000 J mol 6 1
We apply the Arrhenius equation, with k = 5.6 10 s at 0 C (273 K), k = ? at 40 C
(313 K), and Ea = 113 103 J/mol.
Ea = R slope = 8.3145
(c)
E 1 1 k = a
6 1
R T1 T2
5.6 10 s k e6.306 = 548 =
5.6 106 s 1
ln
t1/2 =
(D)
(a)
k = 548 5.6 106 s 1 = 3.07 103 s 1
0.693
0.693
=
= 2.3 10 2 s
3
1 k 3.07 10 s
Here we plot ln k vs. 1/ T . The slope of the straight line equals Ea / R . First we tabulate the data to plot. (the plot is shown below).
T , C
T ,K
1/ T , K1 k , M1s1 ln k
15.83
288.98
0.0034604
5.03 105
9.898
32.02
305.17
0.0032769
3.68 104
7.907
59.75
332.90
0.0030039
6.71 103
5.004
90.61
363.76
0.0027491
0.119
2.129
Plot of ln k versus 1/T
0.0027
0.0029
-1
1/T (K )
0
-2
-4
ln k
54.
112 103 J/mol 1
1
=
= 6.306
1
1
8.3145 J mol K 273 K 313 K
-6
-8
-10
y = -10890x + 27.77
-12
637
0.0031
0.0033
0.0035
Chapter 14: Chemical Kinetics
The slope of this graph = 1.09 104 K = Ea / R
J
J
1 kJ kJ Ea = 1.089 104 K 8.3145
= 9.054 104
= 90.5 mol K mol 1000 J mol (b)
We calculate the activation energy with the Arrhenius equation using the two extreme data points. E 1 1 E k 0.119
1
1
ln 2 = ln
= +7.77 = a = a
5
k1
R T1 T2 R 288.98 K 363.76 K
5.03 10
Ea
7.769 8.3145 J mol1 K 1
Ea =
= 9.08 104 J/mol
4
1
R
7.1138 10 K
Ea = 91 kJ/mol. The two Ea values are in quite good agreement, within experimental limits.
= 7.1138 104 K 1
(c)
We apply the Arrhenius equation, with Ea = 9.080 104 J/mol, k = 5.03 105 M 1 s1 at 15.83 C (288.98 K), and k = ? at 100.0 C (373.2 K).
Ea 1 1 90.80 103 J/mol
1
1 k ln =
=
5
1 1
1
1
5.03 10 M s
R T1 T 2 8.3145 J mol K 288.98 K 373.2 K ln k
5
1
1
e8.528 = 5.05 103 =
= 8.528
5.03 10 M s k = 5.05 103 5.03 105 M 1 s 1 = 0.254 M 1 s 1
55.
5.03 10
M 1 s 1
(M) The half-life of a first-order reaction is inversely proportional to its rate constant: k = 0.693 / t1/2 . Thus we can apply a modified version of the Arrhenius equation to find Ea.
(a)
(b)
t1/2 1 Ea 1 1 k2 46.2 min Ea 1
1
= ln
=
=
= ln k1 2.6 min
R 298 K 102 + 273 K
t1/2 2 R T1 T2
1
1
E
2.88 8.3145 J mol K
1 kJ
2.88 = a 6.89 104 Ea =
= 34.8 kJ/mol
4
1
R
6.89 10 K
1000 J
10.0 min
34.8 103 J / mol 1
1
1
1
ln
=
= 1.53 = 4.19 103
1
1
T 298
46.2 min 8.3145 J mol K T 298
ln
FG
H
1
1.53
1
= 3.65 104
=
T 298 4.19 103
56.
k
5
IJ
K
1
= 2.99 103
T
FG
H
IJ
K
T = 334 K = 61 o C
(M) The half-life of a first-order reaction is inversely proportional to its rate constant: k = 0.693 / t1/2 . Thus, we can apply a modified version of the Arrhenius equation.
(a)
t1/2 1 Ea 1 1 k2 22.5 h Ea 1
1
= ln
=
=
= ln k1 1.5 h
R 293 K 40 + 273 K
t1/2 2 R T1 T2
1
1
E
2.71 8.3145 J mol K
1 kJ
2.71 = a 2.18 104 , Ea =
= 103 kJ/mol
4
1
R
2.18 10 K
1000 J
ln
638
Chapter 14: Chemical Kinetics
(b)
b
The relationship is k = A exp Ea / RT
g
103 10 J mol1
13 1
40.9
3.5 105 s 1 k = 2.05 1013 s 1exp
= 2.05 10 s e
1
1
8.3145 J mol K 273 + 30 K
3
57.
(M)
(a) It is the change in the value of the rate constant that causes the reaction to go faster. Let k1 be the rate constant at room temperature, 20 C (293 K). Then, ten degrees higher (30° C or 303 K), the rate constant k2 = 2 k1 .
E 1 1 E 1 k2 2 k1
1
4
1 Ea
= ln
= 0.693 = a = a
= 1.13 10 K k1 k1
R T1 T2 R 293 303 K
R
0.693 8.3145 J mol 1 K 1
= 5.1 104 J / mol = 51 kJ / mol
Ea =
4
1
1.13 10 K
Since the activation energy for the depicted reaction (i.e., N2O + NO N2 + NO2) is 209 kJ/mol, we would not expect this reaction to follow the rule of thumb. ln (b)
58.
(M) Under a pressure of 2.00 atm, the boiling point of water is approximately 121 C or 394 K.
Under a pressure of 1 atm, the boiling point of water is 100 C or 373 K. We assume an activation energy of 5.1 104 J / mol and compute the ratio of the two rates.
Rate 2 E a 1 1
5.1 104 J / mol
1
1
ln
=
=
= 0.88
1
1
Rate1
R T1 T2
8.3145 J mol K 373 394 K
FG
H
IJ
K
FG
H
IJ
K
Rate 2 = e0.88 Rate1 = 2.4 Rate1 . Cooking will occur 2.4 times faster in the pressure cooker.
Catalysis
59.
(E)
(a)
(b)
60.
Although a catalyst is recovered unchanged from the reaction mixture, it does “take part in the reaction.” Some catalysts actually slow down the rate of a reaction. Usually, however, these negative catalysts are called inhibitors.
The function of a catalyst is to change the mechanism of a reaction. The new mechanism is one that has a different (lower) activation energy (and frequently a different A value), than the original reaction.
(M) If the reaction is first-order, its half-life is 100 min, for in this time period [S] decreases from
1.00 M to 0.50 M, that is, by one half. This gives a rate constant of k = 0.693 / t1/2 = 0.693 / 100 min = 0.00693 min 1 .
The rate constant also can be determined from any two of the other sets of data.
A 0
1.00 M
0.357
= ln
= 0.357 = k 60 min
= 0.00595 min 1 kt = ln k= 0.70 M
60 min
A t
This is not a very good agreement between the two k values, so the reaction is probably not firstorder in [A]. Let's try zero-order, where the rate should be constant.
639
Chapter 14: Chemical Kinetics
0.90 M 1.00 M
0.50 M 1.00 M
= 0.0050 M/min
Rate =
= 0.0050 M/min
20 min
100 min
0.20 M 0.90 M
0.50 M 0.90 M
Rate =
= 0.0050 M/min
Rate =
= 0.0050 M/min
160 min 20 min
100 min 20 min
Thus, this reaction is zero-order with respect to [S].
Rate =
61.
(E) Both platinum and an enzyme have a metal center that acts as the active site. Generally speaking, platinum is not dissolved in the reaction solution (heterogeneous), whereas enzymes are generally soluble in the reaction media (homogeneous). The most important difference, however, is one of specificity. Platinum is rather nonspecific, catalyzing many different reactions. An enzyme, however, is quite specific, usually catalyzing only one reaction rather than all reactions of a given class.
62.
(E) In both the enzyme and the metal surface cases, the reaction occurs in a specialized location: either within the enzyme pocket or on the surface of the catalyst. At high concentrations of reactant, the limiting factor in determining the rate is not the concentration of reactant present but how rapidly active sites become available for reaction to occur. Thus, the rate of the reaction depends on either the quantity of enzyme present or the surface area of the catalyst, rather than on how much reactant is present (i.e., the reaction is zero-order). At low concentrations or gas pressures the reaction rate depends on how rapidly molecules can reach the available active sites.
Thus, the rate depends on concentration or pressure of reactant and is first-order.
63.
(E) For the straight-line graph of Rate versus [Enzyme], an excess of substrate must be present.
64.
(E) For human enzymes, we would expect the maximum in the curve to appear around 37C, i.e., normal body temperature (or possibly at slightly elevated temperatures to aid in the control of diseases (37 - 41 C). At lower temperatures, the reaction rate of enzyme-activated reactions decreases with decreasing temperature, following the Arrhenius equation. However, at higher temperatures, these temperature sensitive biochemical processes become inhibited, probably by temperature-induced structural modifications to the enzyme or the substrate, which prevent formation of the enzyme-substrate complex.
Reaction Mechanisms
65.
(E) The molecularity of an elementary process is the number of reactant molecules in the process.
This molecularity is equal to the order of the overall reaction only if the elementary process in question is the slowest and, thus, the rate-determining step of the overall reaction. In addition, the elementary process in question should be the only elementary step that influences the rate of the reaction. 66.
(E) If the type of molecule that is expressed in the rate law as being first-order collides with other molecules that are present in much larger concentrations, the reaction will seem to depend only on the amount of those types of molecules present in smaller concentration, since the larger concentration will be essentially unchanged during the course of the reaction. Such a situation is quite common, and has been given the name pseudo first-order. It is also possible to have molecules which, do not participate directly in the reaction— including product molecules—
640
Chapter 14: Chemical Kinetics
strike the reactant molecules and impart to them sufficient energy to react. Finally, collisions of the reactant molecules with the container walls may also impart adequate energy for reaction to occur. 67.
(M) The three elementary steps must sum to give the overall reaction. That is, the overall reaction is the sum of step 1 step 2 step 3. Hence, step 2 = overall reaction step 1 step 3. Note that all species in the equations below are gases.
2 NO + 2 H 2 N 2 + 2 H 2 O
Overall: 2 NO + 2H 2 N 2 + 2 H 2 O
First:
2 NO N 2 O2
N 2 O2 2NO
Third
N 2 O + H 2 N 2 + H 2 O or N 2 + H 2 O N 2 O + H 2
H 2 + N 2O2 H 2O + N 2O
The result is the second step, which is slow:
The rate of this rate-determining step is: Rate = k2 H2 N2O2
Since N 2 O 2 does not appear in the overall reaction, we need to replace its concentration with the concentrations of species that do appear in the overall reaction. To do this, recall that the first step is rapid, with the forward reaction occurring at the same rate as the reverse reaction.
2
k1 NO = forward rate = reverse rate = k 1 N 2 O 2 . This expression is solved for N 2 O 2 , which then is substituted into the rate equation for the overall reaction.
2
k1 NO k k
2
N 2O2 =
Rate = 2 1 H 2 NO k 1 k 1
The reaction is first-order in H 2 and second-order in [NO]. This result conforms to the experimentally determined reaction order.
68.
k
1
I 2 (g) 2 I(g)
k
(M) Proposed mechanism:
-1
k2
2 I(g) + H 2 (g) 2 HI(g)
Observed rate law:
Rate = k[I2][H2]
I2(g) + H2(g) 2 HI(g)
The first step is a fast equilibrium reaction and step 2 is slow. Thus, the predicted rate law is
Rate = k2[I]2[H2]. In the first step, set the rate in the forward direction for the equilibrium equal to the rate in the reverse direction. Then solve for [I]2.
Rateforward = Ratereverse
Use: Rateforward = k1[I2] and Ratereverse = k-1[I]2
From this we see: k1[I2] = k-1[I]2. Rearranging (solving for [I]2) k [I ] k [I ]
Substitute into Rate = k2[I]2[H2] = k2 1 2 [H2] = kobs[I2][H2]
[I]2 = 1 2 k-1 k-1
Since the predicted rate law is the same as the experimental rate law, this mechanism is plausible. 641
Chapter 14: Chemical Kinetics
69.
k
1
Cl2 (g) 2 Cl(g)
k
(M) Proposed mechanism:
Observed rate law:
-1
k2
2 Cl(g) + 2 NO(g) 2 NOCl(g)
Rate = k[Cl2][NO]2
Cl2(g) + 2NO(g) 2 NOCl(g)
The first step is a fast equilibrium reaction and step 2 is slow. Thus, the predicted rate law is
Rate = k2[Cl]2[NO]2 In the first step, set the rate in the forward direction for the equilibrium equal to the rate in the reverse direction. Then express [Cl]2 in terms of k1, k-1 and [Cl2]. This mechanism is almost certainly not correct because it involves a tetra molecular second step.
Rateforward = Ratereverse
Use: Rateforward = k1[Cl2] and Ratereverse = k-1[Cl]2
From this we see: k1[Cl2] = k-1[Cl]2. Rearranging (solving for [Cl]2) k [Cl ] k [Cl ]
Substitute into Rate = k2[Cl]2[NO]2 = k2 1 2 [NO]2 = kobs[Cl2][NO2]2
[Cl]2 = 1 2 k-1 k-1 k1
There is another plausible mechanism. Cl2 (g) + NO(g) NOCl(g) + Cl(g)
k-1 k1
Cl(g) + NO(g) NOCl(g)
k-1
Cl2(g) + 2NO(g) 2 NOCl(g)
Rateforward = Ratereverse
Use: Rateforward = k1[Cl2][NO] and Ratereverse = k-1[Cl][NOCl]
From this we see: k1[Cl2][NO] = k-1[Cl][NOCl]. Rearranging (solving for [Cl]) k [Cl ][NO] k k [Cl ][NO]2
Substitute into Rate = k2[Cl][NO] = 2 1 2
[Cl] = 1 2 k -1 [NOCl] k-1 [NOCl]
If [NOCl], the product is assumed to be constant (~ 0 M using method of initial rates), then k2 k1
constant = k obs Hence, the predicted rate law is kobs [Cl2 ][NO]2 which agrees with k-1 [NOCl] the experimental rate law. Since the predicted rate law agrees with the experimental rate law, both this and the previous mechanism are plausible, however, the first is dismissed as it has a tetramolecular elementary reaction (extremely unlikely to have four molecules simultaneously collide). 70.
(M) A possible mechanism is:
Step 1:
Step 2:
O3 O 2 + O fast k 3
O + O3 2 O 2 slow
The overall rate is that of the slow step: Rate = k 3 O O 3 . But O is a reaction intermediate, whose concentration is difficult to determine. An expression for [O] can be found by assuming that the forward and reverse “fast” steps proceed with equal speed.
2
k1 O3 k1 O3 k3 k1 O3
Rate = k3 O3 =
Rate1 = Rate 2 k1 O3 = k2 O 2 O O = k2 O 2 k2 O 2 k2 O 2
Then substitute this expression into the rate law for the reaction. This rate equation has the same form as the experimentally determined rate law and thus the proposed mechanism is plausible.
642
Chapter 14: Chemical Kinetics
71.
(M)
k1
S1 S2 S1 : S2
k
1
S1 : S2
k2
S1 : S2
d S1 : S2
k1 S1 S2 k 1 S1 : S2 k 2 S1 : S2
dt
k1 S1 S2 k 1 k 2 S1 : S2
S1 : S2
d S1 : S2 dt 72.
k1
S1 S2 k 1 k 2
k 2 S1 : S2
k 2 k1 S1 S2 k 1 k 2
(M) k1
CH3 2 CO (aq) + OH CH3C O CH 2 (aq) + H 2O (l)
k
1
k2
CH 3C O CH (aq) + CH 3 2 CO (aq) Pr od
2
We note that CH3C(O)CH2– is an intermediate species. Using the steady state approximation, while its concentration is not known during the reaction, the rate of change of its concentration is zero, except for the very beginning and towards the end of the reaction. Therefore,
d CH 3C O CH 2
k CH CO OH k CH C O CH H O
1
3 2
2 2
1 3 dt
k 2 CH 3C O CH 2 CH 3 2 CO 0
Rearranging the above expression to solve for CH3C(O)CH2– gives the following expression k1 CH 3 2 CO OH
CH3C O CH 2
k H O k CH CO
2
3 2
1 2
The rate of formation of product, therefore, is: d Pr od
k 2 CH 3C O CH 2 CH 3 2 CO
dt
k 2 CH 3 2 CO
k1 CH 3 2 CO OH
k 1 H 2 O k 2 CH 3 2 CO
k 2 k1 CH 3 2 CO OH
CH 3 2 CO k 1 H 2 O k 2
2
643
Chapter 14: Chemical Kinetics
INTEGRATIVE AND ADVANCED EXERCISES
73.
(M) The data for the reaction starting with 1.00 M being first-order or second-order as well as that for the first-order reaction using 2.00 M is shown below
Time
(min)
0
5
10
15
25
[A]o = 1.00 M
(second order)
1.00
0.63
0.46
[A]o = 1.00 M
(first order)
1.00
0.55
0.30
[A]o = 2.00 M
(second order)
2.00
0.91
0.59
[A]o = 2.00 M
(first order)
2.00
1.10
0.60
0.36
0.25
0.165
0.05
0.435
0.286
0.33
0.10
Clearly we can see that when [A]o = 1.00 M, the first-order reaction concentrations will always be lower than that for the second-order case (assumes magnitude of the rate constant is the same). If, on the other hand, the concentration is above 1.00 M, the second-order reaction decreases faster than the first-order reaction (remember that the half-life shortens for a second-order reaction as the concentration increases, whereas for a first-order reaction, the half-life is constant).
From the data, it appears that the crossover occurs in the case where [A]0 = 2.00 M at just over 10 minutes. Second-order at 11 minutes:
1
1 0.12
(11 min)
[A] 2 M min
[A] = 0.549 M
0.12 ln[A] ln(2)
[A] = 0.534 M
(11 min)
min
A quick check at 10.5 minutes reveals,
1
1 0.12(10.5 min)
Second-order at 10.5 minutes:
[A] = 0.568 M
[A] 2
M min
0.12(10.5 min)
First-order at 10.5 minutes: ln[A] ln(2)
[A] = 0.567 M
M min
Hence, at approximately 10.5 minutes, these two plots will share a common point (point at which the concentration versus time curves overlap).
First-order at 11 minutes:
74. (M)
(a) The concentration vs. time graph is not linear. Thus, the reaction is obviously not zero-order
(the rate is not constant with time). A quick look at various half lives for this reaction shows the ~2.37 min (1.000 M to 0.5 M), ~2.32 min (0.800 M to 0.400 M), and ~2.38 min(0.400 M to 0.200 M). Since the half-life is constant, the reaction is probably first-order.
(2.37 2.32 2.38)
0.693
0.693
2.36 min k
0.294 min 1
3
t1/2
2.36 min
–1
or perhaps better expressed as k = 0.29 min due to imprecision.
(b) average t1/ 2
644
Chapter 14: Chemical Kinetics
(c) When t 3.5 min, [A] 0.352 M. Then, rate = k[A] = 0.294 min –1 0.352 M 0.103 M/min.
(d) Slope
[A]
t
Rate
0.1480 M 0.339 M
6.00 min 3.00 min
0.0637 M / min
Rate 0.064 M/min.
(e) Rate = k[A] = 0.294 min–1 × 1.000 M = 0.294 M/min.
75. (M) The reaction being investigated is:
2 MnO 4- (aq) + 5 H 2O 2 (aq) + 6 H + (aq) 2 Mn 2+ (aq) +8 H 2O(l) + 5O 2 (g)
We use the stoichiometric coefficients in this balanced reaction to determine [H2O2].
0.1000 mmol MnO-4 5 mmol H 2 O 2
37.1 mL titrant
1 mL titrant
2 mmol MnO-4
[H 2 O 2 ]
1.86 M
5.00 mL
76.
(D) We assume in each case that 5.00 mL of reacting solution is titrated.
2.32 mmol H 2 O2 2 mmol MnO-4
1 mL titrant volume MnO-4 5.00 mL
1 mL
5 mmol H 2 O 2 0.1000 mmol MnO-4
At 200 s
At 400 s
At 600 s
At 1200 s
At 1800 s
At 3000 s
20.0 2.32 M H 2 O 2 46.4 mL 0.1000 M MnO-4 titrant at 0 s
Vtitrant = 20.0 × 2.01 M H2O2 = 40.2 mL
38.5
33
Vtitrant = 20.0 × 1.72 M H2O2 = 34.4 mL
27.5
Vtitrant = 20.0 × 1.49 M H2O2 = 29.8 mL
22
16.5
Vtitrant = 20.0 × 0.98 M H2O2 = 19.6 mL
11
Tangent line
Vtitrant = 20.0 × 0.62 M H2O2 = 12.5.5mL
4
Vtitrant = 20.0 × 0.25 M H2O2 = 5.000mL
0
700
1400
2100
2800
The graph of volume of titrant vs. elapsed time is given above. This graph is of approximately the same shape as Figure 14-2, in which [H2O2] is plotted against time. In order to determine the rate, the tangent line at 1400 s has been drawn on the graph. The intercepts of the tangent line are at 34 mL of titrant and 2800 s. From this information we determine the rate of the reaction.
34 mL
1L
0.1000 mol MnO-4 5 mol H 2 O 2
2800 s 1000 mL
1 L titrant
2 mol MnO-4
Rate
6.1 104 M/s
0.00500 L sample
This is the same as the value of 6.1 × 10–4 obtained in Figure 14-2 for 1400 s. The discrepancy is due, no doubt, to the coarse nature of our plot.
645
Chapter 14: Chemical Kinetics
77.
(M) First we compute the change in [H2O2]. This is then used to determine the amount, and ultimately the volume, of oxygen evolved from the given quantity of solution. Assume the O2(g) is dry.
[H 2 O 2 ]
60 s
[H 2 O 2 ]
t 1.7 103 M/s 1.00 min
0.102 M
t
1 min
0.102 mol H 2 O 2
1 mol O 2 amount O 2 0.175 L soln
0.00892 mol O 2
1L
2 mol H 2 O 2
L atm
0.00892 mol O 2 0.08206
(273 24) K nRT mol K
Volume O 2
0.22 L O 2
1 atm
P
757 mmHg
760 mmHg
78. (M) We know that rate has the units of M/s, and also that concentration has the units of M. The generalized rate equation is Rate k A 0 . In terms of units, this becomes
M/s = {units of k} M0. Therefore {Units of k}
M/s
M10 s 1
M0
79. (M)
(a) Comparing the third and the first lines of data, [I-] and [OH-]stay fixed, while [OCl–] doubles. Also the rate for the third kinetics run is one half of the rate found for the first run.
Thus, the reaction is first-order in [OCl-]. Comparing the fourth and fifth lines, [OCl–] and
[I-] stay fixed, while [OH-] is halved. Also, the fifth run has a reaction rate that is twice that of the fourth run. Thus, the reaction is minus first-order in [OH-]. Comparing the third and second lines of data, [OCl-] and [OH-] stay fixed, while the [I-] doubles. Also, the second run has a reaction rate that is double that found for the third run. Thus, the reaction is firstorder in [I-].
(b) The reaction is first-order in [OCl-] and [I-] and minus first-order in [OH-]. Thus, the overall order = 1 + 1 – 1 = 1. The reaction is first-order overall.
Rate k
(c)
[OCl ][I ]
[OH ]
using data from first run:
Rate [OH ] 4.8 104 M/s 1.00 M
60. s 1
[OCl ][I ]
0.0040 M 0.0020 M
80. (M) We first determine the number of moles of N2O produced. The partial pressure of N2O(g) in the “wet” N2O is 756 mmHg – 12.8 mmHg = 743 mmHg.
1 atm
743 mmHg
0.0500 L
760 mmHg
PV
amount N 2 O
0.00207 mol N 2 O
RT 0.08206 L atm mol 1 K 1 (273 15) K
Now we determine the change in [NH2NO2].
646
Chapter 14: Chemical Kinetics
1 mol NH 2 NO 2
1 mol N 2 O
[NH 2 NO 2 ]
0.0125 M
0.165 L soln
0.693
[NH 2 NO 2 ]final 0.105 M–0.0125 M 0.093 M
0.00563 min 1 k 123 min
[A]t
1
1
0.093 M ln
22 min elapsed time t ln
1
0.00563 min
0.105 M k [A]0
0.00207 mol N 2 O
81. (D) We need to determine the partial pressure of ethylene oxide at each time in order to determine the order of the reaction. First, we need the initial pressure of ethylene oxide. The pressure at infinite time is the pressure that results when all of the ethylene oxide has decomposed. Because two moles of product gas are produced for every mole of reactant gas, this infinite pressure is twice the initial pressure of ethylene oxide. Pinitial = 249.88 mmHg 2 = 124.94 mmHg. Now, at each time we have the following.
(CH 2 ) 2 O(g) CH 4 (g) CO(g)
Initial: 124.94 mmHg Changes: –x mmHg +x mmHg +x mmHg Final: 124.94 + x mmHg
Thus, x = Ptot – 124.94 and PEtO = 124.94 – x = 124.94 – (Ptot – 124.94) = 249.88 – Ptot
Hence, we have, the following values for the partial pressure of ethylene oxide.
t, min
0
10
20
40
60
100
200
PEtO, mmHg 124.94
110.74
98.21 77.23 60.73 37.54 11.22
For the reaction to be zero-order, its rate will be constant.
(110.74 124.94) mmHg
The rate in the first 10 min is: Rate
1.42 mmHg/min
10 min
(77.23 124.94) mmHg
The rate in the first 40 min is: Rate
1.19 mmHg/min
40 min
We conclude from the non-constant rate that the reaction is not zero-order. For the reaction to be first-order, its half-life must be constant. From 40 min to 100 min—a period of 60 min—the partial pressure of ethylene oxide is approximately halved, giving an approximate half-life of 60 min.
And, in the first 60 min, the partial pressure of ethylene oxide is approximately halved. Thus, the reaction appears to be first-order. To verify this tentative conclusion, we use the integrated firstorder rate equation to calculate some values of the rate constant.
1 P
1
110.74 mmHg
0.0121 min 1 ln k ln
10 min 124.94 mmHg t P0 k 1
100 min
ln
37.54 mmHg
124.94 mmHg
0.0120 min 1
k
1
60 min
ln
60.73 mmHg
124.94 mmHg
0.0120 min 1
The constancy of the first-order rate constant suggests that the reaction is first-order.
Pt
1 P
kt Elapsed time is computed as: t ln t
P0
k P0
We first determine the pressure of DTBP when the total pressure equals 2100 mmHg.
82. (M) For this first-order reaction
ln
647
Chapter 14: Chemical Kinetics
Reaction:
C8 H18O 2 (g) 2C3 H 6 O(g) C2 H 6 (g) [Equation 15.16]
Initial:
Changes:
800.0 mmHg
–x mmHg
2x mmHg
x mmHg
x mmHg
Final:
(800.0 x) mmHg
2x mmHg
Total pressure (800.0–x) 2x x 800.0 2 x 2100. x 650. mmHg P{C8 H18O 2 (g)} 800. mmHg 650. mmHg 150. mmHg
1 P
1
150. mmHg t ln t ln 192 min 1.9 102 min
3
1 k P0
8.7 10 min
800. mmHg
83. (D) If we compare Experiment 1 with Experiment 2, we notice that [B] has been halved, and also that the rate, expressed as [A]/ t , has been halved. This is most evident for the times 5 min,
10 min, and 20 min. In Experiment 1, [A] decreases from 1.000 × 10–3 M to 0.779 × 10–3 M in 5 min, while in Experiment 2 this same decrease in [A] requires 10 min. Likewise in Experiment 1,
[A] decreases from 1.000 × 10–3 M to 0.607 M× 10–3 in 10 min, while in Experiment 2 the same decrease in [A] requires 20 min. This dependence of rate on the first power of concentration is characteristic of a first-order reaction. This reaction is first-order in [B]. We now turn to the order of the reaction with respect to [A]. A zero-order reaction will have a constant rate. Determine the rate (0.951 1.000) 10 3 M
After over the first minute:
Rate
4.9 10 5 M/min
1 min
(0.779 1.000) 10 3 M
After over the first five minutes:
Rate
4.4 10 5 M/min
5 min
(0.368 1.000) 10 3 M
After over the first twenty minutes: Rate
3.2 10 5 M/min
20 min
This is not a very constant rate; we conclude that the reaction is not zero-order. There are no clear half-lives in the data with which we could judge the reaction to be first-order. But we can determine the value of the first-order rate constant for a few data.
1
[A]
1
0.951 mM k ln ln
0.0502 min 1 t [A]0
1 min 1.000 mM
1
0.607 mM
1
0.368 mM ln k ln 0.0499 min 1
0.0500 min 1
10 min 1.000 mM
20 min 1.000 mM
The constancy of the first-order rate constant indicates that the reaction indeed is first-order in [A].
(As a point of interest, notice that the concentrations chosen in this experiment are such that the reaction is pseudo-zero-order in [B]. Here, then it is not necessary to consider the variation of [B] with time as the reaction proceeds when determining the kinetic dependence on [A].) k 648
Chapter 14: Chemical Kinetics
84. (M) In Exercise 79 we established the rate law for the iodine-hypochlorite ion reaction:
Rate k[OCl ][I ][OH ]1 . In the mechanism, the slow step gives the rate law;
Rate k3 [I ][HOCl] . We use the initial fast equilibrium step to substitute for [HOCl] in this rate equation. We assume in this fast step that the forward rate equals the reverse rate.
k1[OCl ][H 2 O] k1[HOCl][OH ]
Rate k2 [I ]
k1[OCl ][H 2 O]
[HOCl] k1[OH ]
k1[OCl ][H 2 O] k2 k1[H 2 O] [OCl ][I ]
[OCl ][I ]
k k1[OH ] k1 [OH ]
[OH ]
This is the same rate law that we established in Exercise 79. We have incorporated [H2O] in the rate constant for the reaction because, in an aqueous solution, [H2O] remains effectively constant during the course of the reaction. (The final fast step simply involves the neutralization of the acid
HOI by the base hydroxide ion, OH–.)
85. (M) It is more likely that the cis-isomer, compound (I), would be formed than the trans-isomer, compound (II). The reason for this is that the reaction will involve the adsorption of both
CH3CCCH3 and H2 onto the surface of the catalyst. These two molecules will eventually be adjacent to each other. At some point, one of the bonds in the CC bond will break, the H–H bond will break, and two C–H bonds will form. Since these two C–H bonds form on the same side of the carbon chain, compound (I) will be produced. In the sketches below, dotted lines (…) indicate bonds forming or breaking.
86. (D) Hg2
2+
C C CH3
CH3
···
CH3
H ··· H
C ··· C CH3
···
H H
+ Tl 2 Hg + Tl Experimental rate law = k
3+
Possible mechanism:
2+
+
CH3
C C CH3
[Hg 2 2+ ][Tl3+ ]
[Hg 2+ ]
k
1
Hg22+ + Tl3+ Hg2+ + HgTl3+ (fast)
k1 k2
(slow)
HgTl3+ Hg2+ + Tl3+
2+
3+
2+
+
Hg2 + Tl 2 Hg + Tl
Rate = k2[HgTl3+]
k1[Hg22+][Tl3+] = k-1 [Hg2+][HgTl3+]
Rate = k2[HgTl3+] =
H H
rearrange [HgTl3+ ] =
k2 k1 [Hg 2 2+ ][Tl3+ ]
[Hg 2 2+ ][Tl3+ ]
= kobs k-1 [Hg 2+ ]
[Hg 2+ ]
649
k1 [Hg 2 2+ ][Tl3+ ] k-1 [Hg 2+ ]
Chapter 14: Chemical Kinetics
87.
(M)
ΔCCl 3
rateformation ratedisappearance 0
Δt
so rateformation rate decomposition
k 2 [Cl(g)][CHCl 3 ] k 3[CCl 3 ][[Cl(g)] and, simplifying, [CCl 3 ]
k2
[CHCl 3 ] k3 k
since rate k 3[CCl 3 ][Cl(g)] k3 2 [CHCl 3 ] [Cl(g)] k 2 [CHCl 3 ][Cl(g)]
k3
1/ 2
k
We know: [Cl(g)] 1 [Cl 2 (g)]
k -1
1/2
k
then rateoverall = k 2 [CHCl 3 ] 1 [Cl 2 (g)]
k -1
1/2
k and the rate constant k will be: k k2 1
k -1
88.
1/2
4.8 103
(1.3 10 )
3
3.6 10
2
0.015
(D) d[A] d[A]
Rearrange:
-kt = dt [A]3
Integrate using the limits time (0 to t) and concentration ([A]0 to [A]t )
Rate = k[A]3 -
[A]t
1 1
1 1
2
2 [A]t 2 [A]o 2
0
1
1
1
1
Rearrange :
kt
Simplify : kt
2
2
2
2[A]t
2[A]0 2
2[A]t 2[A]o
1
1
Multiply through by 2 to give the integrated rate law:
2kt
2
[A]t
[A]o 2 t -k t =
d[A]
[A]3
[A]o
kt (k (0))
To derive the half life(t1 2 ) substitute t =t1 2 and [A]t =
1
[A]0
2
2kt1 2
2
2kt1 2
1
1
4
2
2
[A]o
[A]o 2
[A]o
4
1
3
4
2
2
[A]o [A]o
[A]o 2
[A]o
2
Collect terms
Solve for t1 2
t1 2
3
2k[A]o 2
89. (D) Consider the reaction: A + B products (first-order in A, first-order in B). The initial concentration of each reactant can be defined as [A]0 and [B]0 Since the stoichiometry is 1:1, we can define x as the concentration of reactant A and reactant B that is removed (a variable that changes with time).
The [A]t = ([A]o – x) and [B]t = ([B]o – x).
Algebra note: [A]o – x = – (x – [A]o) and [B]o – x = – (x – [B]o).
650
Chapter 14: Chemical Kinetics
As well, the calculus requires that we use the absolute value of |x – [A]o| and |x – [B]o| when taking the integral of the reciprocal of |x – [A]o| and |x – [B]o|
Rate = -
d[A]
dt
=-
d[B] dx
=
k [A]t [B]t =k ([A]o -x)([B]o -x) dt dt
or
dx
([A]o -x)([B]o -x)
= kdt
In order to solve this, partial fraction decomposition is required to further ease integration:
1
1
kdt
([B]o -[A]o ) ([A]o -x) ([B]o -x)
Note: this is a constant
dx
1
From the point of view of integration, a further rearrangement is desirable (See algebra note above).
-1
-1
kdt
([B]o -[A]o ) x-[A]o x-[B]o
t dx
1
-ln x-[A]o - (-ln x-[B]o ) = kt + C
([B]o -[A]o ) x-[A]o = [A]o -x and x-[B]o = [B]o -x
Substitute and simplify
Integrate both sides
1
([B]o -x)
1
ln
= kt + C Determine C by setting x = 0 at t = 0
([B]o -[A]o ) [A]o -x
[B]o
([B]o -x)
[B]o
1
1
Hence:
ln
ln
= kt +
ln
([B]o -[A]o ) [A]o
([B]o -[A]o ) ([A]o -x)
([B]o -[A]o ) [A]o
([B]o -x)
[B]o
Multiply both sides by ([B]o -[A]o ), hence, ln
= ([B]o -[A]o )×kt + ln
[A]o
([A]o -x)
([B]o -x)
([A] -x)
([B]o -x) [B]o
= ln [A]o ([B]o -x) o ln
-ln
=([B]o -[A]o )×kt = ln
[B]o
([A]o -x) [A]o
[B]o ([A]o -x)
[A]o
C=
1
[A]o ×[B]t
= ([B]o -[A]o )×kt
[B]o ×[A]t
Set ([A]o -x) =[A]t and ([B]o -x) =[B]t to give ln
651
Chapter 14: Chemical Kinetics
(D) Let 250-2x equal the partial pressure of CO(g) and x be the partial pressure of CO2(g).
2CO CO 2 C(s)
Ptot =PCO + PCO2 = 250 – 2x +x = 250 – x
250-2x
x
Ptot
[torr]
250
238
224
210
Time
[sec]
0
398
1002
1801
PCO2
PCO [torr]
0
12
26
40
250
226
198
170
The plots that follow show that the reaction appears to obey a second-order rate law.
Rate = k[CO]2
250
0.006
5.52
240
230
220
5.32
210
0.005
-1
200
1/PCO (mmHg )
ln PCO
PCO (mmHg)
90.
190
180
R = 0.9867
170
R2 = 0.9965
5.12
2
0
0
1000 time (S)
PCO
250
226
198
170
1000
1500
2000
ZERO ORDER PLOT
T
0
398
1002
1801
500
time (S)
1st ORDER PLOT
T
0
398
1002
1801
lnPCO
5.521461
5.420535
5.288267
5.135798
652
2000
R2 = 1
0.004
0
1000 time (S)
2nd ORDER PLOT
T
0
398
1002
1801
1/CO
0.004
0.004425
0.005051
0.005882
2000
(Best correlation coefficient)
Chapter 14: Chemical Kinetics
91. (D) Let 100-4x equal the partial pressure of PH3(g), x be the partial pressure of P4(g)and 6x be the partial pressure of H2 (g)
4 PH 3 (g) P 4 ( g )
6 H 2 (g)
100 4x
x
6x
Ptot =PPH3 + PP4 + PH2 = 100 4x + x + 6x = 100+ 3x
Ptot
[torr]
100
150
167
172
Time [sec]
PP4 [torr]
PPH3 [torr]
0
40
80
120
0
50/3
67/3
72/3
100
100-(4)(50/3)
100-(4)(67/3)
100-(4)(72/3)
The plots to follow show that the reaction appears to obey a first-order rate law.
Rate = k[PH3]
100
R2 = 0.8652
4.35
90
2
R = 0.8369
80
0.210
2
R = 0.9991
3.85
0.160
50
1/PPH3 (mmHg -1 )
3.35
60
2.85
ln PPH3
PPH3 (mmHg)
70
40
0.110
2.35
30
0.060
20
1.85
10
-50
0.010
1.35
0
50
time (S)
150
ZERO ORDER PLOT
T
0
40
80
120
PPH3
100
33.3
10.7
4
0
0
100 time (S)
1st ORDER PLOT
T
0
40
80
120
lnPPH3
4.61
3.51
2.37
1.39
653
50
100
200 time (S)
2nd ORDER PLOT
T
0
40
80
120
1/PPH3
0.010
0.030
0.093
0.250
150
Chapter 14: Chemical Kinetics
92.
(D) Consider the following equilibria. k1 KI k2
E + S ES E + P
E + I EI
k -1
d[P]
= k 2 [ES] dt Product production
Use the steady state approximation for [ES]
d[ES]
= k1 [E][S] k -1 [ES] k 2 [ES] = k1 [E][S] [ES] k -1 k 2 = 0 dt k k2 k [E][S] [E][S] solve for [ES] Keep in mind K M = -1
[ES] = 1
=
k1 k -1 k 2
KM
[E][I]
[E][I]
Formation of EI: K I =
[EI] =
[EI]
KI
[E o ] = [E] + [ES] + [EI] = [E] +
[S]
[I]
[E o ] = [E] 1 +
+
KM
KI
From above: [ES] =
[ES] =
[E][S] [E][I]
+
KM
KI
Solve for [E]
[I]K M
K M + [S] +
KI
d[P]
= k 2 [ES] =
Remember
dt
[E o ]
[S]
[I]
+
1 +
KM
KI
[S]
[E o ]
KM
[ES] =
[S]
[I]
+
1 +
KM
KI
[E][S]
KM
[E o ][S]
[E] =
=
multiplication by
[E o ][S]
[I]
K M 1 +
+ [S]
KI
k 2 [E o ][S]
[I]
K M 1 +
+ [S]
KI
If we substitute k 2 [E o ] = Vmax then
d[P]
=
dt
Vmax [S]
[I]
K M 1 +
+ [S]
KI
Vmax [S] decreases;
[I]
K M 1 +
+ [S]
KI
i.e., the rate of product formation decreases as [I] increases.
Thus, as [I] increases, the ratio
654
KM affords KM
Chapter 14: Chemical Kinetics
93.
(D) In order to determine a value for KM, we need to rearrange the equation so that we may obtain a linear plot and extract parameters from the slope and intercepts. k [E ][S]
K M +[S]
KM
1
[S]
V= 2 o
=
=
+
K M +[S]
V
k 2 [E o ][S] k 2 [E o ][S] k 2 [E o ][S]
KM
1
1
=
+
V
k 2 [E o ][S] k 2 [E o ]
Plot
We need to have this in the form of y = mx+b
1
1
on the y-axis and on the x-axis. See result below:
V
[S]
1
[V]
x-intercept
= -1
KM
pe lo s
K M o]
[E
= k2
y-intercept
= 1 k2[Eo] 1
[S]
The plot of 1/V vs 1/[S] should yield a slope of y-intercept of
0=
KM and a k 2 [E o ]
1
. The x-intercept = -1/KM k 2 [E o ]
KM 1
1
[S] k [E ] k 2 [E o ]
2 o 1
[S]
=
-k 2 [E o ] k 2 [E o ] K M
-1
KM
To find the value of KM take the negative inverse of x-intercept.
To find k2, invert the y-intercept and divide by the [Eo].
94. (M)
a) The first elementary step HBr O 2 k1 HOOBr is rate-determining if the reaction obeys reaction rate = k [HBr][O2] since the rate of this step is identical to that of the experimental rate law. b) No, mechanisms cannot be shown to be absolutely correct, only consistent with experimental observations. c) Yes; the sum of the elementary steps (3 HBr + O2 HOBr + Br2 + H2O) is not consistent with the overall stoichiometry (since HOBr is not detected as a product) of the reaction and therefore cannot be considered a valid mechanism.
95.
(M)
(a) Both reactions are first-order, because they involve the decomposition of one molecule.
(b) k2 is the slow reaction.
(c) To determine the concentration of the product, N2, we must first determine how much reactant remains at the end of the given time period, from which we can calculate the amount of reactant consumed and therefore the amount of product produced. Since this is a first-order reaction, the concentration of the reactant, N2O after time t is determined as follows:
A t A 0 e kt
N 2O0.1 2.0 M exp 25.7 s1 0.1 s 0.153 M N 2O remaining
655
Chapter 14: Chemical Kinetics
The amount of N2O consumed = 2.0 M – 0.153 M =1.847 M
1 M N2
0.9235 M N 2
[N 2 ] 1.847 M NO
2 M NO
(d) The process is identical to step (c).
N 2O0.1 4.0 M exp 18.2 s1 0.025 s 2.538 M N 2O remaining
The amount of N2O consumed = 4.0 M – 2.538 M =1.462 M
1 M N 2O
0.731 M N 2O
[N 2 O] 1.462 M NO
2 M NO
FEATURE PROBLEMS
96.
(D)
(a)
To determine the order of the reaction, we need C 6 H 5 N 2 Cl at each time. To determine
this value, note that 58.3 mL N 2 g evolved corresponds to total depletion of C6 H 5 N 2 Cl , to C 6 H 5 N 2 Cl = 0.000 M .
F
GH
bg
Thus, at any point in time, C 6 H 5 N 2 Cl = 0.071 M volume N 2 g
I b g JK
0.071 M C 6 H 5 N 2 Cl
58.3 mL N 2 g
0.071 M C6 H 5 N 2 Cl
Consider 21 min: C6 H 5 N 2 Cl = 0.071 M 44.3 mL N 2
= 0.017 M
58.3 mL N 2 g
The numbers in the following table are determined with this method.
Time, min
VN2 , mL
0
0
C H
71
6
(b)
5
N 2 Cl , mM
3
6
9
12
15
18
21
24
27
30
10.8 19.3 26.3 32.4 37.3 41.3 44.3 46.5 48.4 50.4 58.3
58
47
39
32
26
21
17
14
12
10
[The concentration is given in thousandths of a mole per liter (mM).] time(min) 0 3 6 9 12 15 18 21 24
27
0
30
T(min)
3 3 3
3
3
3
3
3
3
3
[C6H5N2Cl](mM) 71 58 47 39 32 26
21 17
14
12 10
[C6H5N2Cl](mM)
-13 -11 -8
-1
Reaction Rate (mM min )
-7
4.3 3.7 2.7 2.3
656
-6
-5
2.0
1.7
-4
1.3
-3
-2
1.0
0.7
-2
0.7
Chapter 14: Chemical Kinetics
The two graphs are drawn on the same axes.
Plot of [C 6H 5N 2Cl] and VN versus time
[C6H5N2Cl (mM) or VN2 (mL)
(c)
(m
L)
2
60
mL of N2
N
2
Cl]
(m
M)
or
V
2
N5
H6
[C
40
20
[C6H5N2Cl]
0
0
(d)
5
10
15
time (min)
20
25
30
The rate of the reaction at t = 21 min is the slope of the tangent line to the
C 6 H 5 N 2 Cl curve. The tangent line intercepts the vertical axis at about
C 6 H 5 N 2 Cl = 39 mM and the horizontal axis at about 37 min
39 103 M
= 1.05 103 M min 1 = 1.1 103 M min 1
37 min
The agreement with the reported value is very good.
Reaction rate =
(e)
The initial rate is the slope of the tangent line to the C 6 H 5 N 2 Cl curve at t = 0 . The intercept with the vertical axis is 71 mM, of course. That with the horizontal axis is about
13 min.
71103 M
Rate =
= 5.5 103 M min 1
13 min
(f)
The first-order rate law is Rate = k C6 H 5 N 2 Cl , which we solve for k: k= Rate
C6 H5 N 2Cl
k0 =
5.5 103 M min 1
= 0.077 min 1
3
71 10 M
1.1103 M min 1 k21 =
= 0.065 min 1
3
17 10 M
An average value would be a reasonable estimate: k avg = 0.071 min 1
(g)
The estimated rate constant gives one value of the half-life:
0.693
0.693 t1/2 =
=
= 9.8 min
0.071 min 1 k The first half-life occurs when C6 H 5 N 2 Cl drops from 0.071 M to 0.0355 M. This occurs at about 10.5 min.
(h)
The reaction should be three-fourths complete in two half-lives, or about 20 minutes.
657
Chapter 14: Chemical Kinetics
(i)
The graph plots ln C6 H 5 N 2 Cl (in millimoles/L) vs. time in minutes.
0
Time (min)
Plot of ln[C6H5N2Cl] versus time
10
20
30
ln[C6H5N2Cl]
-2.75
-3.25
-3.75
y = -0.0661x - 2.66
-4.25
-4.75
The linearity of the graph demonstrates that the reaction is first-order.
(j)
k = slope = 6.61102 min 1 = 0.0661 min 1 t1/2 =
97.
(D)
(a)
0.693
= 10.5 min , in good agreement with our previously determined values.
0.0661 min 1
In Experiments 1 & 2, [KI] is the same (0.20 M), while
bNH g S O
4 2
2
8
is halved, from 0.20
M to 0.10 M. As a consequence, the time to produce a color change doubles (i.e., the rate is
2
halved). This indicates that reaction (a) is first-order in S2 O8 . Experiments 2 and 3 produce a similar conclusion. In Experiments 4 and 5,
bNH g S O
4 2
2
8
is the same (0.20 M)
while [KI] is halved, from 0.10 to 0.050 M. As a consequence, the time to produce a color change nearly doubles, that is, the rate is halved. This indicates that reaction (a) is also firstorder in I . Reaction (a) is (1 + 1) second-order overall.
(b)
The blue color appears when all the S2 O 2 has been consumed, for only then does reaction
3
2
(b) cease. The same amount of S2 O 3 is placed in each reaction mixture.
0.010 mol Na 2 S2 O3 1 mol S2 O32
1L
amount S2 O32 = 10.0 mL
= 1.0 104 mol
1000 mL
1L
1 mol Na 2 S2 O3
Through stoichiometry, we determine the amount of each reactant that reacts before this
2
amount of S2 O3 will be consumed.
658
Chapter 14: Chemical Kinetics
amount S2 O8
2
1.0 10
4
mol S2 O3
2
= 5.0 105 mol S2 O8
2
amount I = 5.0 105 mol S2 O8
1 mol I3
1 mol S2 O8
2
2 mol S2 O3
1 mol I3
2
2
2 mol I
= 1.0 104 mol I
2
1 mol S2 O8
Note that we do not use “3 mol I ” from equation (a) since one mole has not been oxidized; it simply complexes with the product I 2 . The total volume of each solution is 25.0 mL + 25.0 mL +10.0 mL + 5.0 mL = 65.0 mL , or 0.0650 L.
The amount of S2 O8
2
that reacts in each case is 5.0 105 mol and thus
5.0 105 mol
= 7.7 104 M
0.0650 L
2
S2 O8
+7.7 104 M
Thus, Rate1 =
=
= 3.7 105 M s1
21 s
t
S2 O8 2 =
(c)
S2 O8
2
+7.7 104 M
= 1.8 105 M s1
42 s
t
To determine the value of k , we need initial concentrations, as altered by dilution.
25.0 mL
25.0 mL
2
S2 O8 = 0.20 M
I = 0.20 M
= 0.077 M
= 0.077 M
1
1
65.0 mL total
65.0 mL
For Experiment 2, Rate 2 =
1
2
Rate1 = 3.7 105 M s1 = k S2 O8
=
1
b
I = k 0.077 M
3.7 105 M s 1 k= = 6.2 103 M 1 s 1
0.077 M 0.077 M
25.0 mL
2
S2 O8 = 0.10 M
= 0.038 M
2
65.0 mL total
g b0.077 Mg
1
1
25.0 mL
I = 0.20 M
= 0.077 M
2
65.0 mL
2
Rate 2 = 1.8 105 M s 1 = k S2 O8 I = k 0.038 M 0.077 M
5
1
1.8 10 M s k= = 6.2 103 M 1 s1
0.038 M 0.077 M
1
(d)
1
1
1
First we determine concentrations for Experiment 4.
25.0 mL
25.0 mL
S2 O8 2 = 0.20 M
I = 0.10
= 0.077 M
= 0.038 M
4
4
65.0 mL total
65.0 mL
We have two expressions for Rate; let us equate them and solve for the rate constant.
2
S2 O8 +7.7 104 M
1
2 1
=
Rate 4 =
= k S2O8 I = k 0.077 M 0.038 M
4
4
t
t
k=
7.7 104 M
0.26 M 1
=
t 0.077 M 0.038 M
t
659
k3 =
0.26 M 1
= 0.0014 M 1 s 1
189 s
Chapter 14: Chemical Kinetics
k13 =
k33 =
(e)
0.26 M 1
= 0.0030 M 1 s 1
88 s
0.26 M 1
= 0.012 M 1 s 1
21 s
k24 =
0.26 M 1
= 0.0062 M 1 s 1
42 s
We plot ln k vs. 1/ T The slope of the line = Ea / R .
Plot of ln k versus 1/T
0.00325
-1
1/T (K )
0.0034
0.00355
-3.5
-4
ln k
-4.5
-5
y = -6135x + 16
-5.5
-6
Ea = +6135 K 8.3145 J mol1 K 1 = 51.0 103 J/mol = 51.0 kJ/mol
The scatter of the data permits only a two significant figure result: 51 kJ/mol
(f)
For the mechanism to agree with the reaction stoichiometry, the steps of the mechanism must sum to the overall reaction, in the manner of Hess's law.
2
3
(slow) I + S2 O8 IS2 O8
3
(fast) IS2 O8 2 SO 2 + I +
4
(fast) I + + I I 2
(fast) I 2 + I I 3
2
(net) 3 I + S2 O8 2 SO 2 + I 3
4
3
Each of the intermediates cancels: IS2 O8 is produced in the first step and consumed in the second, I + is produced in the second step and consumed in the third, I 2 is produced in the third step and consumed in the fourth. The mechanism is consistent with the stoichiometry.
The rate of the slow step of the mechanism is
2
Rate1 = k1 S2 O8
1
I
1
This is exactly the same as the experimental rate law. It is reasonable that the first step be slow since it involves two negatively charged species coming together. We know that like charges repel, and thus this should not be an easy or rapid process.
660
Chapter 14: Chemical Kinetics
SELF-ASSESSMENT EXERCISES
98.
(E)
(a) [A]0: Initial concentration of reactant A
(b) k: Reaction rate constant, which is the proportionality constant between reaction rate and reactant concentration
(c) t1/2: Half-life of the reaction, the amount of time that the concentration of a certain reactant is reduced by half
(d) Zero-order reaction: A reaction in which the rate is not dependent on the concentration of the reactant (e) Catalyst: A substance which speeds up the reaction by lowering the activation energy, but it does not itself get consumed
99.
(E)
(a) Method of initial rates: A study of the kinetics of the reaction by measuring the initial reaction rates, used to determine the reaction order
(b) Activated complex: Species that exist in a transitory state between the reactants and the products (c) Reaction mechanism: Sequential elementary steps that show the conversion of reactant(s) to final product(s)
(d) Heterogeneous Catalyst: A catalyst which is in a different physical phase than the reaction medium (e) Rate-determining step: A reaction which occurs more slowly than other reactions in a mechanism and therefore usually controls the overall rate of the reaction
100. (E)
(a) First-order and second-order reactions: In a first-order reaction, the rate of the reaction depends on the concentration of only one substrate and in a 1-to-1 manner (doubling the concentration of the reactant doubles the rate of the reaction). In a second-order reaction, the rate depends on two molecules reacting with each other at the elementary level.
(b) Rate law and integrated rate law: Rate law describes how the rate relates to the concentration of the reactants and the overall rate of a reaction, whereas the integrated rate law expresses the concentration of a reactant as a function of time
(c) Activation energy and enthalpy of reaction: Activation energy is the minimum energy required for a particular reaction to take place, whereas enthalpy of reaction is the amount of heat generated (or consumed) by a reaction when it happens
(d) Elementary process and overall reaction: Individual steps of a reaction mechanism, which describes any molecular event that significantly alters a molecule’s energy or geometry or produces a new molecule
(e) Enzyme and substrate: An enzyme is a protein that acts as a catalyst for a biological reaction.
A substrate is the reactant that is transformed in the reaction (in this context, by the enzyme).
101. (E) The answer is (c). The rate constant k is only dependent on temperature, not on the concentration of the reactants
661
Chapter 14: Chemical Kinetics
102. (E) The answers are (b) and (e). Because half-life is 75 seconds, the quantity of reactant left at two half-lives (75 + 75 = 150) equals one-half of the level at 75 seconds. Also, if the initial concentration is doubled, after one half-life the remaining concentration would have to be twice as much as the original concentration.
103. (E) The answer is (a). Half-life t½ = 13.9 min, k = ln 2/t½ = 0.050 min-1. Rate of a first-order reaction is as follows: d A
k A 0.050 min 1 0.40 M 0.020 M min 1 dt 104. (E) The answer is (d). A second-order reaction is expressed as follows: d A
2
k A dt If the rate of the reaction when [A] =0.50 is k(0.50)2 = k(0.25). If [A] = 0.25 M, then the rate is k(0.0625), which is ¼ of the rate at [A] =0.50.
105. (M) The answer is (b). Going to slightly higher temperatures broadens the molecular speed distribution, which in turn increases the fraction of molecules at the high kinetic energy range
(which are those sufficiently energetic to make a reaction happen).
106. (E) The answer is (c). Since the reaction at hand is described as an elementary one, the rate of the reaction is k[A][B].
107. (E) We note that from the given data, the half-life of the reaction is 100 seconds (at t = 0, [A] =
0.88 M/s, whereas at t = 100, [A] = 0.44 M/s). Therefore, the rate constant k is: k = ln 2/100 s = 0.00693 s-1. We can now calculate instantaneous rate of the reaction: d[A]/dt = (0.00693 s-1)(0.44 M) = 3.0×10-3 M·s-1
108. (M)
(a) For a first-order reaction, ln 2 0.693
0.0231 min 1 k t1/2
30
ln[A]t ln[A]0 kt
ln 0.25 0 ln[A]t ln[A]0
60.0 min
k
0.0231 min 1
(b) For a zero-order reaction,
0.5 0.5
0.0167 min 1 k t1/2 30 t [A]t [A]0 kt t [A]t [A]0
0.25 1.00
45.0 min
k
0.0167 min 1
662
Chapter 14: Chemical Kinetics
109. (M) The reaction is second-order, because the half-life doubles with each successive half-life period. 110. (M)
(a) The initial rate = ΔM/Δt = (1.204 M – 1.180 M)/(1.0 min) = 0.024 M/min
(b) In experiment 2, the initial concentration is twice that of experiment A. For a second-order reaction: Rate = k [Aexp 2]2 = k [2×Aexp 1]2 = 4 k [Aexp 1]2
This means that if the reaction is second order, its initial rate of experiment 2 will be 4 times that of experiment 1 (that is, 4 times as many moles of A will be consumed in a given amount of time). The initial rate is 4×0.024 M/min = 0.096 M/s. Therefore, at 1 minute, [A] = 2.408 –
0.0960 = 2.312 M.
(c) The half-life of the reaction, obtained from experiment 1, is 35 minutes. If the reaction is first-order, then k = ln 2/35 min = 0.0198 min-1.
For a first-order reaction,
A t A 0 e kt
A 35min 2.408 exp 0.0198 min -1 30 min 1.33 M
111. (D) The overall stoichiometry of the reaction is determined by adding the two reactions with each other: A + 2B C + D
(a) Since I is made slowly but is used very quickly, its rate of formation is essentially zero. The amount of I at any given time during the reaction can be expressed as follows: d I dt I
0 k1 A B k 2 B I k1 A k2 Using the above expression for [I], we can now determine the overall reaction rate law: d C k k 2 I B k 2 1 A B k1 A B dt k2
(b) Adding the two reactions given, we still get the same overall stoichiometry as part (a).
However, with the given proposed reaction mechanisms, the rate law for the product(s) is given as follows:
663
Chapter 14: Chemical Kinetics
d B2 dt B2
k1 B k 1 B2 k 2 A B 0
2
k1 B
2
k 1 k 2 A
Therefore, d C dt k 2 A B2
k 2 k1 A B
2
k 1 k 2 A
which does not agree with the observed reaction rate law.
112. (M) The answer is (b), first-order, because only in a first-order reaction is the half-life independent of the concentration of the reacting species.
113. (E) The answer is (a), zero-order, because in a zero-order reaction the relationship between concentration and time is: [A]t = kt + [A]0
114. (M) The answer is (d). The relationship between rate constant (and thus rate) between two reactions can be expressed as follows:
Ea 1 1 k2 exp
R T T
k1
2
1
If T2 is twice T1, the above expression gets modified as follows:
k Ea 1
1 ln 2
R T1 2T1
k1
R k 2 T1 1
,
ln
E a k1 2T1
R
k Ea T 1 ln 2 1
2T1
k1
For reasonably high temperatures,
R
k Ea
T
1 ln 2 1
2T1 2
k1
Therefore,
R
k 2 Ea
1/2
e 1.64 k1
664
Chapter 14: Chemical Kinetics
115. (E) The answer is (c), remain the same. This is because for a zero-order reaction,
d[A]/dt = k[A]0 = k. Therefore, the reaction rate is independent of the concentration of the reactant. 116. (M) The overarching concept for this concept map is kinetics as a result of successful collision.
The subtopics are collision theory, molecular transition theory. Molecular speed and orientation derive from collision theory. Transition complexes and partial bonds fall under the molecular transition theory heading. Deriving from the collision theory is another major topic, the
Arrhenius relationship. The Arrhenius relationship encompasses the ideas of activation energy,
Arrhenius collision factor, and exponential relationship between temperature and rate constant.
665 