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Chapter 16 -Chemical Kinetics Additional Problems
Instructor: Sherman Henzel
16.1 The conversion of CO to CO 2 in a catalytic converter is shown to be zero order with respect
to CO. If the initial concentration of CO is 25.0 g/L, what will be its final concentration after
30.0 minutes? k for this reaction is 0.725 g/L-min.
16.2. How many minutes will it take to reduce the concentration of CO from 15.00 g/L to 0.
g/L in a catalytic converter?
16.3 Given the data below determine the rate for the process A ) products.
time, s [A], M
0 3.
100 3.
200 3.
300 2.
400 2.
500 2.
600 1.
700 1.
(a) How long will it take the concentration of A to decrease from 0.9950 M to 0.0350 M?
(b) If the concentration of A is 2.280 M after 900 s, what was the initial concentration A?
(c) Starting with [A] = 5.25 M, what is [A] after 1200.0 s?
16.4 B ) products is a first order process. If after 80.0 s the concentration of B is 0.7500 M,
what was the initial concentration of B? k = 6.85 x 10
16.5 Given the data below determine the rate for the process B ) products.
time, min [B], M
0 2.
10.0 2.
30.0 2.
90.0 1.
270.0 1.
810.0 0.
2430.0 0.
3000.0 0.
(a) How long will it take the concentration of B to decrease from 0.9950 M to 0.0350 M?
(b) If the concentration of B is 2.280 M after 900.0 min, what was the initial
concentration B?
(c) Starting with [B] = 5.25 M, what is [B] after 1200.0 min?
16.6 C ) products is a second order process. If the initial concentration of C is 0.3750 M, what
will be the concentration of C after 1500.0 seconds? k = 0.00125 L mol
16.7 Given the data below determine the rate for the process C ) products.
time, h [C], M
0 0.
0.50 0.
1.50 0.
5.50 0.
8.50 0.
12.50 0.
15.50 0.
25.50 0.
(a) How long will it take the concentration of A to decrease from 0.9950 M to 0.0350 M?
(b) If the concentration of A is 0.002280 M after 900.0 h, what was the initial
concentration A?
(c) Starting with [A] = 5.25 M, what is [A] after 1200.0 h?
Problem 13.3 (^) y = -0.0026x + 3.
R
2 = 0.
0 100 200 300 400 500 600 700
Time, s
[A], M
1 1 CO
0 725 g
L min
x 30 0 min
25 0 g
L
3 25 g
L
- [ ] (^) 30 0
. .
.. . =
โ
1 2
0 010 g
L
0 725 g
L min
x t
15 00 g
L
14 99 g
L
0 725 g
L min
x t t 20 7 min
... .. ;.
=
โ
โ
โ
16.3 Since a plot of [A] v. time yields a straight line, the reaction is zero order.
[A] (^) t
= -
0.0026 mol L
- s - x t + 3.515 mol L -
( ).
.
. ;.
( ).
. [ ];[ ].
( ) [ ]
.
...
a 0 0350 M
0 0026 mol
L s
x t 0 9950 M t 3 7x 10 s
b 2 280 M
0 0026 mol
L s
x 900 s A A 4 6 M
c A
0 0026 mol
L s
x 1200 0 s 5 25 M 2 1 M
2
2
1200 3
= โ + =
= โ + =
= โ + =
time,min [B], M ln[B] 1/[B]
0 2.500 0.916 0.
10 2.437 0.891 0. 30 2.315 0.839 0. 90 1.984 0.685 0. 270 1.245 0.219 0.
810 0.312 -1.164 3. 2430 4.873E-03 -5.324 205. 3000 1.128E-03 -6.787 886.
Problem 13.
0 500 1000 1500 2000 2500 3000 3500
Time, min
[B], min
Problem 13.5 y = -0.002568x + 0.
R
2 = 1.
-8.
-6.
-4.
-2.
0 500 1000 1500 2000 2500 3000
Time, min
ln[B]
1 4 ln 0 7500
6 85 x 10
s
x 80 0 s ln B
ln B 0 360 B 1 3 M
3
0
0 0
6
4
. (. )
.
. [ ]
[ ]. ; [ ].
=
โ
= =
โ
16.5 Since the plot of ln[B] vs. time yielded a straight line, the reaction is first order.
ln[B] (^) t = -0.00257 min
- x t + 0.
Problem 13.
0
1
0 5 10 15 20 25 30
Time, h
[C], M
Problem 13.
-1.
-1.
-0.
-0.
-0.
-0.
0
0 5 10 15 20 25 30
Time, h
ln[C]
Problem 13.7 (^) y = 0.0895x + 1.
R^2 = 1
0
1
2
3
4
0 5 10 15 20 25 30
time, h
1/[C], M
-
1/[C] (^) t = 0.0895 L mol
- h - + 1.053 M -
1 0 0 0
1
1200 1200
1200
1 0.0895 L 1
(a) x t ; t 308 h [0.0350 M] mol h [0.9950 M]
1 0.0895 L 1 1
(b) x 900.0 h ; 276 M ; [C] 0.00362 M [0.00280 M] mol h [C] [C]
1 0.0895 L 1 1
(c) x 1200.0 h ; 108 M [C] mol h 5.25 M [C]
[C] 0.00930 M
โ
โ
16.8 The slow step determines the rate law:
rate = k 3 [NO
4 ][NO - 2 ]
Problem: NO
4 is not in our original (overall) reaction.
To get rid of NO
4 we use the fast step and solve for NO - 4 :
1 2 2 2 4
1 4 2 2 4 2
1 2 3 2 2 2 2 2 2 2 2 2
k [NO ][O ] k [NO ]
k [NO ] [NO ][O ]; substitute this for [NO ] in the rate law : k
k rate k [NO ][O ][NO ] k[NO ] [O ] k
rate k[NO ] [O ]; if this agees with exp erimental rate, then the mechanism
is plausible.
โ โ
โ โ โ
โ โ โ
โ
16.9 The slow step determines the rate law:
rate = k 5 [NO 2 ][C]
Problem: C is not in our original (overall) reaction.
To get rid of C we use the fast step containing C and solve for [C]:
16.11 There are three ways to solve this type of problem.
a
5
5
E / RT
3.251 x 10 /8.314472 x 325.62 120.1 53) 077
49 49 3.251 x 10 / 8.314472 x 568.91 68.73 49 65 2 65 65
49 30 19 2 65 2
Equation #1: k Ae
0.00258 Ae ; 0.00258 Ae ; 0.00258 A(7. x 10 )
A 3. x 10 ; k 3. x 10 e e 3. x 10
k 3. x 10 (1.4 x 10 ) 5. x 10 s
โ
โ โ โ
โ โ
โ
โ 1
Equation #2:
E (^) a 1 ln(k) x ln(A) R T
1 4533621
5 49 2 45
2 1 2 879 14065 4533621
- x 10 19 1 2 9
3.251 x 10 J / mol 1 ln(k ) x ln(3. x 10 ) 8.314472 J / mol K 568.91 K
ln(k ) 68.72 1. x 10 4. x 10
k e 4. x 10 s
โ
Equation #3:
2 a
1 1 2
k E 1 1 ln k R T T
โ โ =^ โ โ โ
5 2
2 4 1 1
2 4 1
2 2 82444
k 3.251 x 10 J / mol 1 1 ln 0.00258 8.314472 J / mol K 325.52 K 568.91 K
k ln 3.910 x 10 K (0.0030720 K 0.0017577 K )
k ln 3.910 x 10 K (0.0013143 K )
k k ln 51.38 ; 0.00258 0.
โ โ
โ
โ โ โ^ โ
โ โ=^ โ โ โ
โ โ=^ โ
51.38 82444 2 22 7793
19 1 2
k e ; 2.0 x 10 0258 0.
k 5.4 x 10 s
โ
โ โ =^ โ โ=
16.12 There are three ways to solve this type of problem.
a
4
5
E / RT
8.654 x 10 / 8.314472 x 278.96 37.31 17) 5046
13 13 3.251 x 10 / 8.314472 x 568.91 68.73 49 7578 2 7578 65
13 8. 7578
Equation #1: k Ae
0.00186 Ae ; 0.00186 Ae ; 0.00186 A(62. x 10 )
A 2.9 x 10 ; k 2.9 x 10 e e 3. x 10
0.00325 2.9 x 10 (e
โ
โ โ โ
โ โ
โ
4
4
654 x 10 / 8.314472 x T
13 8.654 x 10 / 8.314472 x T 7578 4 4
0241 7531
2 311154 320
ln(0.00325) ln(2.9 x 10 ) ln(e )
8.654 x 10 8.654 x 10 5.729 31. ; 36. 8.314472 x T 8.314472 x T
0.0035 ; T 2.8 x 10 K T
โ = +
Equation #2:
E (^) a 1 ln(k) x ln(A) R T
4 13 7578
4 91 8358 4112
4 1 8358 31125
2 319
8.654 x 10 J / mol 1 ln(0.00325) x ln(2.9 X 10 ) 8.314472 J / mol K T
5.72 1.040 x 10 K x 31. T
1 1 36.75 1.040 x 10 K x ; 0.0035 K T T
T 2.8 x 10 K
โ
Equation #3:
2 a
1 1 2
k E 1 1 ln k R T T
โ โ =^ โ โ โ
4
2
4 1
2
1 1 1 1
2 2
2
0.00325 8.654 x 10 J / mol 1 1 ln 0.00186 8.314472 J / mol K 278.96 K T
0.558 1.040 x 10 K 0.0035847 K T
0.0000536 K 0.0035847 K ; 0.00353112 K 0.0035847 K
T T
T 283 K
โ
โ โ โ โ
โ โ โ^ โ
โ โ=^ โ^ โ โ
