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![Page 1: Homework #1 Chapter 15 - people.chem.ucsb.edu · Chapter 15 Chemical Kinetics 8. Arrhenius Equation = đžđ âđ
đ Therefore, k depends only on temperature. The rate of the](https://reader030.vdocuments.mx/reader030/viewer/2022040108/5e25997d78143f59f304bf06/html5/thumbnails/1.jpg)
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Homework #1
Chapter 15 Chemical Kinetics
8. Arrhenius Equation
đ = đŽđđžđ
đ đâ Therefore, k depends only on temperature. The rate of the reaction depends on all of these items (a-d).
14. a) đđđ
đżâđ b) đđđ
đżâđ c) 1
đ
d) đż
đđđâđ e) đż2
đđđ2âđ
15. Rate has units of đđđ
đżâđ therefore, the units of the rate constant must compensate for any
missing/additional units.
đż1
2â
đđđ1
2â âđ
18. a) General Rate Law đ đđĄđ = đ[đđ]đ„[đ¶đ2]đŠ
Experiment [NO]o (M) [Cl2]o (M) Initial Rate ( đđđ
đżâđđđ)
1 0.10 0.10 0.18
2 0.10 0.20 0.36
3 0.20 0.20 1.45
From experiments 1 and 2 ([NO]o is kept constant) as the initial concentration of Cl2 is
doubled, the initial rate of the reaction doubles (0.36
0.18=2.0); therefore, the reaction is first
order with respect to Cl2. From experiments 2 and 3 ([Cl2]o is kept constant) as the initial concentration of NO is
doubled, the initial rate of the reaction quadruples(1.45
0.36=4.0); therefore, the reaction is
second order with respect to NO. đ đđĄđ = đ[đđ]2[đ¶đ2]
b)
đ đđĄđ = đ[đđ]2[đ¶đ2]
0.18 đđđ
đżâđđđ= đ(0.10 đ)2(0.10 đ)
đ = 180 đż2
đđđ2âđđđ
19. a) General Rate Law
đ đđĄđ = đ[đŒâ]đ„[đđ¶đâ]đŠ Experiment [I-]o (M) [OCl-]o (M) Initial Rate (đđđ
đżâđ )
1 0.12 0.18 7.91Ă10-2
2 0.060 0.18 3.95Ă10-2
3 0.030 0.090 9.88Ă10-2
4 0.24 0.090 7.91Ă10-2
From experiments 1 and 2 ([OCl-]o is kept constant) as the initial concentration of I- is
doubled, the initial rate of the reaction doubles (7.91Ă10â2
3.95Ă10â2=2.0); therefore, the reaction is
first order with respect to I-. đ đđĄđ = đ[đŒâ][đđ¶đâ]đŠ
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đ Therefore, k depends only on temperature. The rate of the](https://reader030.vdocuments.mx/reader030/viewer/2022040108/5e25997d78143f59f304bf06/html5/thumbnails/2.jpg)
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You must find the order of OCl- the long way. Divide the 2 equations (in which the [OCl-] changes) by each other.
7.91 Ă 10â2 đđđđżâđ
9.88 Ă 10â3 đđđđżâđ
=đ(0.12)(0.18 đ)đŠ
đ(0.030)(0.090 đ)đŠ
2.0 = 2.0đŠ đŠ = 1
đ đđĄđ = đ[đŒâ][đđ¶đâ] b)
đ đđĄđ = đ[đŒâ][đđ¶đâ]
7.91 Ă 10â2 đđđ
đżâđ = đ(0.12 đ)(0.18 đ)
đ = 3.7đż
đđđâđ
c)
đ đđĄđ = đ[đŒâ][đđ¶đâ] = (3.7 đż
đđđâđ )(0.15 đ)(0.15 đ) = 0.082 đđđ
đżâđ
20. General Rate Law
đ đđĄđ = đ[đ2đ5]đ„ Experiment [N2O5]o (M) Initial Rate (đđđ
đżâđ )
1 0.0750 8.90Ă10-4
2 0.190 2.26Ă10-3
3 0.275 3.26Ă10-3
4 0.410 4.85Ă10-3
You must solve this problem the long way.
8.90 Ă 10â4 đđđ
đżâđ = đ(0.0750 đ)đ„
2.26 Ă 10â3 đđđ
đżâđ = đ(0.190 đ)đ„
Divide the 2 equations by each other.
8.90 Ă 10â4 đđđđżâđ
2.26 Ă 10â3 đđđđżâđ
=đ(0.0750 đ)đ„
đ(0.190 đ)đ„
0.394 = 0.395đ„ đ„ = 1
đ đđĄđ = đ[đ2đ4]
8.90 Ă 10â4 đđđ
đżâđ = đ(0.0750 đ)
đ = 0.0119 1
đ
21. a) General Rate Law
đ đđĄđ = đ[đđđ¶đ]đ„
Experiment [NOCl]o (đđđđđđąđđđ
đđ3 ) Initial Rate (đđđđđđąđđđ
đđ3âđ )
1 3.0Ă1016 5.98Ă104
2 2.0Ă1016 2.66Ă104
3 1.0Ă1016 6.64Ă103
4 4.0Ă1016 1.06Ă105
From experiments 2 and 3, as the initial concentration of NOCl is doubled, the initial rate
of the reaction quadruples (2.66Ă104
6.64Ă103=4.0); therefore, the reaction is second order with
respect to NOCl.
đ đđĄđ = đ[đđđ¶đ]2
b) 5.98 Ă 104 đđđđđđąđđđ
đđ3âđ = đ(3.0 Ă 1016 đđđđđđąđđđ
đđ3 )2
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đ = 6.6 Ă 10â29 đđ3
đđđđđđąđđđ âđ
c) đ = 6.6 Ă 10â29 đđ3
đđđđđđąđđđ âđ (6.022Ă1023 đđđđđđąđđđ
1 đđđ)( 1 đż
103 đđ3) = 4.0 Ă 10â8 đż
đđđâđ
22. a) General Rate Law
đ đđĄđ = đ[đ»đ]đ„[đ¶đ]đŠ
Experiment [Hb]o (đđđđ
đż) [CO]o (đđđđ
đż) Initial Rate (đđđđ
đżâđ )
1 2.21 1.00 0.619
2 4.42 1.00 1.24
3 4.42 3.00 3.71
From experiments 1 and 2 ([CO]o is kept constant) as the initial concentration of Hb is
doubled, the initial rate of the reaction doubles( 1.24
0.619=2.00); therefore, the reaction is
first order with respect to Hb. You cannot use the shortcut to calculate the order with respect to [CO]
Find 2 equations where the concentration of Hb is constant
1.24 đđđđ
đżâđ = đ (4.42
đđđđ
đż) (1.00
đđđđ
đż)
đŠ
3.71 đđđđ
đżâđ = đ (4.42
đđđđ
đż) (3.00
đđđđ
đż)
đŠ
Divide the 2 equations by each other.
1.24 đđđđđżâđ
= đ(4.42 đđđđđż
)(1.00 đđđđđż
)đŠ
3.71 đđđđđżâđ = đ(4.42 đđđđ
đż )(3.00 đđđđđż )
đŠ
0.33 =(1.00 đđđđ
đż)
đŠ
(3.00 đđđđđż )
đŠ = (0.33)đŠ
đŠ = 1
The reaction is 1st order with respect to CO b) đ đđĄđ = đ[đ»đ][đ¶đ]
c) đ đđĄđ = đ[đ»đ][đ¶đ]
0.619 đđđđ
đżâđ = đ (2.21
đđđđ
đż) (1.00
đđđđ
đż)
đ = 0.280 đż
đđđđâđ
d) đ đđĄđ = 0.280 đż
đđđđâđ [đ»đ][đ¶đ] = (0.280 đż
đđđđâđ ) (3.36 đđđđ
đż)(2.40 đđđđ
đż) = 2.26 đđđđ
đżâđ
23. a) General
đ đđĄđ = đ[đ¶đđ2]đ„[đđ»â]đŠ
Experiment [ClO2]o (M) [OH-]o (M) Initial Rate (đđđ
đżâđ )
1 0.0500 0.100 5.75Ă10-2
2 0.100 0.100 2.30Ă10-1
3 0.100 0.0500 1.15Ă10-1
From experiments 1 and 2 ([OH-]o is kept constant) as the initial concentration of ClO2 is
doubled, the initial rate of the reaction quadruples(2.30Ă10â1
5.75Ă10â2=4.00); therefore, the
reaction is second order with respect to ClO2. From experiments 2 and 3 ([ClO2]o is kept constant) as the initial concentration of OH- is
doubled, the initial rate of the reaction doubles(2.30Ă10â1
1.15Ă10â1=2.00); therefore, the reaction is
first order with respect to OH-. đ đđĄđ = đ[đ¶đđ2]2[đđ»â]
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Find k
5.75 Ă 10â2 đđđ
đżâđ = đ(0.0500 đ)2(0.100 đ)
đ = 230.đż2
đđđ2âđ
đ đđĄđ = 230.đż2
đđđ2âđ [đ¶đđ2]2[đđ»â]
b) đ đđĄđ = 230. đż2
đđđ2âđ (0.175 đ)2(0.0844 đ) = 0.594 đđđ
đżâđ
30. a) Because the plot of 1
[đŽ] verse time was linear the rate law is a second order rate law.
đ đđĄđ = đ[đŽ]2
âđ[đŽ]
đđĄ= đ[đŽ]2
1
[đŽ]2đ[đŽ] = âđđđĄ
â«1
[đŽ]2đ[đŽ]
[đŽ]
[đŽ]°
= â« âđđđĄđĄ
0
â1
[đŽ]+
1
[đŽ]°= âđđĄ + 0
1
[đŽ]= đđĄ +
1
[đŽ]° (integrated rate law)
The rate constant is the slope of the line đ = 3.60 Ă 10â2 đż
đđđâđ
đ đđĄđ = 3.60 Ă 10â2 đż
đđđâđ [đŽ]2 (rate law)
b) The half-life is the amount of time it takes for œ of the original concentration to react.
1
[đŽ]= đđĄ +
1
[đŽ]°
1
0.5(2.80 Ă 10â3 đ)= (3.60 Ă 10â2
đż
đđđâđ ) đĄ +
1
2.80 Ă 10â3 đ
đĄ = 9920 đ
c) 1
[đŽ]= đđĄ +
1
[đŽ]°
1
7.00 Ă 10â4 đ= (3.60 Ă 10â2
đż
đđđâđ ) đĄ +
1
2.80 Ă 10â3 đ
đĄ = 29,800 đ
31. a) Because the plot of ln[A] verses t is linear, this is a 1st order rate law. đ đđĄđ = đ[đŽ]
âđ[đŽ]
đđĄ= đ[đŽ]
1
[đŽ]đ[đŽ] = âđđđĄ
â«1
[đŽ]đ[đŽ]
[đŽ]
[đŽ]°
= â â« đđđĄđĄ
0
đđ[đŽ] â đđ[đŽ]° = âđđĄ + 0 đđ[đŽ] = âđđĄ + đđ[đŽ]° (integrated rate law)
The rate constant is the negative slope of the line đ = 2.97 Ă 10â2 1
đđđ
đ đđĄđ = 2.97 Ă 10â2 1
đđđ[đŽ] (rate law)
b) đđ[đŽ] = âđđĄ + đđ[đŽ]°
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đđ (1
2(2.00 Ă 10â2 đ)) = â (2.97 Ă 10â2
1
đđđ) đĄ + đđ(2.00 Ă 10â2 đ)
đĄ = 23.3 đđđ c)
đđ[đŽ] = âđđĄ + đđ[đŽ]°
đđ(2.50 Ă 10â3 đ) = â (2.97 Ă 10â2 1
đđđ) đĄ + đđ(2.00 Ă 10â2 đ)
đĄ = 70.0 đđđ
32. a) Because the plot of [C2H5OH] verses t is linear, this is a 0th order rate law
đ đđĄđ = đ
âđ[đ¶2đ»5đđ»]
đđĄ= đ
đ[đ¶2đ»5đđ»] = âđđđĄ
â« đ[đ¶2đ»5đđ»]đ¶2đ»5đđ»
[đ¶2đ»5đđ»]°
= â â« đđđĄđĄ
0
[đ¶2đ»5đđ»] â [đ¶2đ»5đđ»]° = âđđĄ + 0 [đ¶2đ»5đđ»] = âđđĄ + [đ¶2đ»5đđ»]° (integrated rate law)
The rate constant equals the negative slope of the line; đ = 4.00 Ă 10â5 đđđ
đżâđ
đ đđĄđ = 4.00 Ă 10â5 đđđ
đżâđ (rate law)
b) [đ¶2đ»5đđ»] = âđđĄ + [đ¶2đ»5đđ»]°
1
2(1.25 Ă 10â2 đ) = â(4.00 Ă 10â5 đđđ
đżâđ )đĄ + (1.25 Ă 10â2 đ)
đĄ = 156 đ
c) Since this is a 0th order rate law (which means that the rate is independent of concentration), the amount of time that it takes to decompose the total amount of C2H5OH is just double the half-life.
đĄ = 2(156 đ ) = 312 đ Or
0 = â(4.00 Ă 10â5 đđđ
đżâđ )đĄ + (1.25 Ă 10â2 đ)
đĄ = 312 đ 33. Plot [H2O2] vs. t, ln[H2O2] vs. t, and [H2O2]-1 vs. t
Since the plot of ln[H2O2] vs. t results in a linear relationship, the reaction is 1st order.
0
0.5
1
1.5
0 1000 2000 3000 4000
[H2
O2
]
time (s)
-3.5
-2.5
-1.5
-0.5
0.5
0 2000 4000
ln[H
2O
2]
time (s)
0
5
10
15
20
25
0 2000 4000
[H2
O2
]^-1
time (s)
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If you do not have a graphing calculator look at the spacing between the evenly spaces times to determine the order.
Time (s) [H2O2] Difference
between points ln[H2O2]
Difference between points
[H2O2]-1 Difference
between points
0 1.00 0.59-1.00=-0.41 0.00 -0.53-0.00=-0.53 1.00 1.7-1.00=0.7
600 0.59 -0.22 -0.53 -0.46 1.7 1.0
1200 0.37 -0.15 -0.99 -0.5 2.7 1.8
1800 0.22 -0.09 -1.5 -0.5 4.5 3.3
2400 0.13 -0.05 -2.0 -0.5 7.7 4.3
3000 0.082 -2.5 12
Since the all the differences between ln[H2O2] are all approximately -0.5 this will be a straight line and the reaction is 1st order. Differential Rate Law
đ đđĄđ = đ[đ»2đ2] Integrated Rate Law
đđ[đ»2đ2] = âđđĄ + đđ[đ»2đ2]° The slope of the line will equal âk.
đ đđđđ =đđđ đ
đđąđ=
đđ(0.050 đ)âđđ(1.00 đ)
3,600 đ â0 đ = â8.3 Ă 10â4 1
đ = âđ
đ = 8.3 Ă 10â4 1
đ
Calculate the concentration of H2O2 after 4000. s.
đđ[đ»2đ2] = â (8.3 Ă 10â4 1
đ ) (4,000 đ ) + đđ(1.00 đ)
[đ»2đ2] = 0.036 đ 35. Plot
[A] vs. t, ln[A] vs. t, and [A]-1 vs. t
Since the plot of [A]-1 vs. t is linear, this is a 2nd order rate law
If you do not have a graphing calculator look at the spacing between the evenly spaces times to determine the order. You must have at least 3 points.
Time (s) [NO2] Difference
between points ln[NO2]
Difference between points
[NO2]-1 Difference
between points
0 0.500 0.25-0.50=-0.25 -0.693 -1.39-0.693=-0.69 2.00 4.00-2.00=2.00
9.00Ă103 0.250 -0.076 -1.39 -0.36 4.00 1.75
1.80Ă104 0.174 -1.75 5.75
This problem it is a little hard to determine. (I will not give you something this hard on an exam.) The difference between the difference between the [NO2] points is 0.17 over a concentration
range of 0.326 making the difference between differences 52% (0.17
0.326100% = 52%) of the
0
0.1
0.2
0.3
0.4
0.5
0.6
0 5000 10000 15000
[NO
2]
time (s)
0
2
4
6
8
0 5000 10000 15000
[NO
2]^
-1
time (s)
-2
-1.5
-1
-0.5
0 5000 10000 15000
ln[N
O2
]
time (s)
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range. The difference between the difference between the ln[NO2] is 0.33 over a ln[NO2] range
of -1.06 making the difference between differences 31% (0.33
1.06100% = 33%) of the range. The
difference between the difference between [NO2]-1 is 0.25 over a range of 3.75 making the
difference between differences 6% (0.25
3.75100% = 6%)of the range. The graph [NO2]-1 has the
smallest difference between differences, when you take the size of the range into account, the order is 2nd order. Differential Rate Law
đ đđĄđ = đ[đđ2]2
Integrated Rate Law 1
[đđ2]= đđĄ +
1
[đđ2]°
The slope of the line is the rate constant
đ đđđđ =đđđ đ
đđąđ=
1
0.174 đâ
1
0.500 đ
1.80Ă104 đ â0 đ = 2.08 Ă 10â4 đż
đđđâđ = đ
1
[đđ2]= (2.08 Ă 10â4
đż
đđđâđ ) đĄ +
1
[đđ2]°
1
[đđ2]= (2.08 Ă 10â4
đż
đđđâđ ) (2.70 Ă 104 đ ) +
1
0.500 đ= 7.62
đż
đđđ
[đđ2] = 0.131 đ 36. a) Plot [O] vs. t, ln[O] vs. t, and [O]-1 vs. t
Since ln[O] vs. t is linear, the rate expression is 1st order with respect to O.
If you do not have a graphing calculator look at the spacing between the evenly spaces times to determine the order.
Time (s) [O] Difference
between points ln[O]
Difference between points
[O]-1 Difference
between points
0 5.0Ă109 1.9Ă109-5.0Ă109
=-3.1Ă109 22 21-22=-1.0 2.0Ă10-10 5.2Ă10-10-2.0Ă10-10
= 3.2Ă10-10
0.010 1.9Ă109 -1.2Ă109 21 -1.0 5.2Ă10-10 9.8Ă10-10
0.020 6.8Ă108 -4.3Ă108 20. -1.0 1.5Ă10-9 2.5Ă10-9
0.030 2.5Ă108 19 4.0Ă10-9
Since the all the differences between ln[H2O2] are all approximately -1.0 this will be a straight line and the reaction is 1st order. Differential Rate Law
b) Overall Rate Law đ đđĄđ = đ[đ][đđ2]
Because the concentration of [NO2] >>[O] this can be turned into a pseudo rate law đ đđĄđ = đâČ[đ]
0.E+00
2.E+09
4.E+09
6.E+09
0 0.01 0.02 0.03 0.04
[O]
time
19
20
21
22
23
0 0.02 0.04
ln[O
]
time
0.E+00
1.E-09
2.E-09
3.E-09
4.E-09
5.E-09
0 0.02 0.04
[O]^
-1
time
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đâČ = đ[đđ2] Determine k' using the 1st order integrated rate law
đđ[đ] = âđâČđĄ + đđ[đ]° Therefore, the slope of the line is âk'
đ đđđđ =đđđ đ
đđąđ=
đđ(2.5Ă108 đđĄđđđ
đđ3 )âđđ(5.0Ă109 đđĄđđđ
đđ3 )
3.0Ă10â2 đ â0 đ = â1.0 Ă 102 1
đ = âđâČ
đâČ = 1.0 Ă 102 1
đ
Find k đâČ = đ[đđ2]
(1.0 Ă 1021
đ ) = đ[đđ2]
đ = 1.0 Ă 10â11 đđ3
đđđđđđąđđđ âđ
38. a) The order of the reaction with respect to A is 2nd order because the plot of [A]-1 vs. t is
linear. The integrated rate law is.
1
[đŽ]= đđĄ +
1
[đŽ]°
Therefore, the [A]0-1 is the y intercept. [A]0
-1 = 10 đż
đđđ and [A]0 = 0.1 M
b) 1
[đŽ]= đđĄ +
1
[đŽ]°
đ = đ đđđđ =đđđ đ
đđąđ=
70 đżđđđ
â20 đżđđđ
6 đ â0 đ = 10 đż
đđđâđ
1
[đŽ]= (10
đż
đđđâđ ) (9 đ ) + 10
đż
đđđ= 100
đż
đđđ
[đŽ] = 0.01 đ c) Calculate the time for the 1st half-life ([A] 0.1 0.05)
1
0.05 đ= (10
đż
đđđâđ ) đĄ +
1
0.1 đ
đĄ = 1 đ Calculate the time for the 2nd half-life
1
0.025 đ= (10
đż
đđđâđ ) đĄ +
1
0.05 đ
đĄ = 2 đ Calculate the time for the 3rd half-life
1
0.0125 đ= (10
đż
đđđâđ ) đĄ +
1
0.025 đ
đĄ = 4 đ 39. a) Plot [NO] vs. t, ln[NO] vs. t, and, [NO]-1 vs. t
Since the plot of ln[NO] vs. t is linear, the rate law is 1st order with respect to NO.
0.E+00
2.E+08
4.E+08
6.E+08
8.E+08
0 500 1000 1500
[NO
]
time
18
18.5
19
19.5
20
20.5
0 500 1000 1500
ln[N
O]
time
0.E+00
5.E-09
1.E-08
2.E-08
0 500 1000 1500
[NO
]^-1
time
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If you do not have a graphing calculator look at the spacing between the evenly spaces times to determine the order.
Time (s) [NO] Difference
between points ln[NO]
Difference between points
[NO]-1 Difference
between points
0 6.0Ă108 2.4Ă108-6.0Ă108
=-3.6Ă108 20 19-20=-1.0 1.7Ă10-9 4.2Ă10-9-1.7Ă10-9
= 2.5Ă10-9
500 2.4Ă108 -1.4Ă108 19 -1.0 4.2Ă10-9 5.8Ă10-9
1000 9.9Ă107 18 1.0Ă10-8
Since the all the differences between ln[NO] are all approximately -1.0 this will be a straight line and the reaction is 1st order.
Plot [O3] vs. t, ln[O3] vs. t, and [O3]-1 vs. t
Since plot of ln[O3] vs. t is linear, the rate law is 1st order with respect to O3
If you do not have a graphing calculator look at the spacing between the evenly spaces times to determine the order.
Time (s) [O3] Difference
between points ln[O3]
Difference between points
[O3]-1 Difference
between points
0 1.0Ă1010 7.0Ă109-1.0Ă1010
=-3.0Ă109 23.0 22.7-23.0=-0.3 1.0Ă10-10 1.4Ă10-10-1.0Ă10-10
= 4.0Ă10-11
100 7.0Ă109 -2.1Ă109 22.7 -0.4 1.4Ă10-10 6.0Ă10-11
200 4.9Ă109 -1.5Ă109 22.3 -0.4 2.0Ă10-10 9.0Ă10-11
300 3.4Ă109 21.9 2.9Ă10-10
Since the all the differences between ln[O3] are all approximately -0.4 this will be a straight line and the reaction is 1st order. b) Rate Law
đ đđĄđ = đ[đđ][đ3] c) When the [NO] data was taken the concentration of [O3]>>[NO]. Therefore, a pseudo
rate law can be written. đ đđĄđ = đâČ[đđ] đâČ = đ[đ3]
Since the reaction was first order with respect to [NO] the slope of the line equals -k'
đâČ = âđ đđđđ =đđđ đ
đđąđ=
đđ(9.9Ă107 đđđđđđąđđđ
đđ3 )âđđ(6.0Ă108 đđđđđđąđđđ
đđ3 )
1000. đđ â0 đđ = 0.0018 1
đđ = 1.8 1
đ
When the [O3] data was taken the concentration of [NO]>>[O3]. Therefore, a pseudo rate law can be written.
đ đđĄđ = đâČâČ[đ3] đâČâČ = đ[đđ]
Since the reaction was first order with respect to [O3] the slope of the line equals -k''
0.E+00
5.E+09
1.E+10
2.E+10
0 200 400
[O3
]
Time
21.5
22
22.5
23
23.5
0 200 400
ln[O
3]
Time
0.E+00
1.E-10
2.E-10
3.E-10
4.E-10
0 100 200 300 400
[O3
]^-1
Time
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10
đâČâČ = âđ đđđđ =đđđ đ
đđąđ=
đđ(3.4 Ă 109 đđđđđđąđđđ đđ3 ) â đđ(1.0 Ă 1010 đđđđđđąđđđ
đđ3 )
300. đđ â 0 đđ = 0.0036
1
đđ = 3.6
1
đ d) You can solve this using either of the two data sets. In both data sets the
reactions are pseudo 1st order system. Overall rate = pseudo 1st order rate Using data set 1 [O3] held constant
đ[đđ][đ3] = đâČ[đđ]
đ =đâČ
[đ3]=
1.8 1đ
1.0 Ă 1014 đđđđđđąđđđ đđ3
= 1.8 Ă 10â14 đđ3
đđđđđđąđđđ âđ
= 1.8 Ă 10â17 đđ3
đđđđđđąđđđ âđđ
Using data set 2 [NO] held constant đ[đđ][đ3] = đâČâČ[đ3]
đ =đâČâČ
[đđ]=
3.6 1đ
2.0 Ă 1014 đđđđđđąđđđ đđ3
= 1.8 Ă 10â14 đđ3
đđđđđđąđđđ âđ
= 1.8 Ă 10â17 đđ3
đđđđđđąđđđ âđđ
43. đđ[đđ»3] = âđđĄ + đđ[đđ»3]đ đđ(0.250 đ) = âđ(120. đ ) + đđ(1.00 đ)
đ = 0.0116 1
đ
đđ[đđ»3] = â (0.0116 1
đ ) đĄ + đđ[đđ»3]đ
đđ(0.350 đ) = â (0.0116 1
đ ) đĄ + đđ(2.00 đ)
đĄ = 151 đ 47. a) Calculate k
đđ[đŽ] = âđđĄ + đđ[đŽ]đ đđ(0.250[đŽ]đ) = âđ(320. đ ) + đđ[đŽ]đ
đđ (0.250[đŽ]đ
[đŽ]đ) = âđ(320. đ )
đ = 0.00433 1
đ
Calculate time for 1st half life
đđ ( 1
2[đŽ]đ) = â (0.00433
1
đ ) đĄ + đđ[đŽ]đ
đđ ( 12[đŽ]đ
[đŽ]đ) = â (0.00433
1
đ ) đĄ
đĄ = 159 đ Calculate time for the 2nd half life
đđ ( 1
4[đŽ]đ) = â (0.00433
1
đ ) đĄ + đđ (
1
2[đŽ]đ)
đđ ( 14[đŽ]đ
12
[đŽ]đ
) = â (0.00433 1
đ ) đĄ
đĄ = 159 đ b)
đđ( 0.100[đŽ]đ) = â(0.00433 1
đ )đĄ + đđ[đŽ]đ
đđ ( 0.100[đŽ]đ
[đŽ]đ) = â (0.00433
1
đ ) đĄ
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đĄ = 532 đ 48. Since the time of the half-life doubles every successive half-life, the reaction is a 2nd order
reaction
a) 1
[đŽ]= đđĄ +
1
[đŽ]đ
1
0.05 đ= đ(10.0 đđđ) +
1
0.10 đ
đ = 1.0 đż
đđđâđđđ
1
[đŽ]= (1.0
đż
đđđâđđđ) (80.0 đđđ) +
1
0.10 đ= 90.
đż
đđđ
[đŽ] = 0.011 đ
b)
1
[đŽ]= (1.0 đż
đđđâđđđ)(30.0 đđđ) +
1
0.10 đ= 40. đż
đđđ
[đŽ] = 0.025 đ 52. The problem wants you to find the time when [đŽ] = 4.00[đ”]. Known
đđŽ = 4.50 Ă 10â4 1
đ
đđ” = 3.70 Ă 10â3 1
đ
Both 1st order reactions Initially
[đŽ]đ = [đ”]đ đđ[đŽ] = âđđĄ + đđ[đŽ]đ đđ[đ”] = âđđĄ + đđ[đ”]đ
Plug in[đ”]đ for[đŽ]đ and[đŽ] = 4.00[đ”] for[đŽ]and solve for [đ”] đđ(4.00[đ”]) = âđđŽđĄ + đđ[đ”]đ
4.00[đ”] = đâđđŽđĄ+đđ[đ”]đ
[đ”] =đâđđŽđĄ+đđ[đ”]đ
4.00
Plug into equation for [đ”] and solve for t đđ[đ”] = âđđ”đĄ + đđ[đ”]đ
đđ (đâđđŽđĄ+đđ[đ”]đ
4.00) = âđđ”đĄ + đđ[đ”]đ
đđ(đâđđŽđĄ+đđ[đ”]đ) â đđ(4.00) = âđđ”đĄ + đđ[đ”]đ
âđđŽđĄ + đđ[đ”]đ â đđ(4.00) = âđđ”đĄ + đđ[đ”]đ âđđŽđĄ â đđ(4.00) = âđđ”đĄ âđđ(4.00) = (âđđ” + đđŽ)đĄ
đĄ =âđđ(4.00)
âđđ” + đđŽ=
âđđ(4.00)
â3.70 Ă 10â3 1đ
+ 4.50 Ă 10â4 1đ
= 427 đ
54. The concentrations of B and C are so much larger than A, therefore, during the course of the
reaction they do not effectively change making this a pseudo 1st order reaction. The rate of the pseudo 1st order reaction is equal to the rate of the overall reaction times [đ”]2. a) đđđĄđ = đ[đŽ][đ”]2 = đâČ[đŽ] đâČ = đ[đ”]2
đđ[đŽ] = âđâČđĄ + đđ[đŽ]đ đđ(3.8 Ă 10â3 đ) = âđâČ(8.0 đ ) + đđ(1.0 Ă 10â2 đ)
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12
đâČ = 0.12 1
đ
đâČ = đ[đ”]2
đ =đâČ
[đ”]2=
0.12 1đ
(3.0 đ)2= 0.013
đż2
đđđ2âđ
b)
đđ[đŽ] = â(0.12 1
đ )đĄ + đđ[đŽ]đ
đđ(0.0050 đ) = â (0.12 1
đ ) đĄ + đđ(1.0 Ă 10â2 đ)
đĄ = 5.8 đ c)
đđ[đŽ] = â(0.12 1
đ )đĄ + đđ[đŽ]đ
đđ[đŽ] = â (0.12 1
đ ) (13.0 đ ) + đđ(1.0 Ă 10â2 đ)
[đŽ] = 0.0021 đ d) The concentration of C is the [đ¶]đ â 2â[đŽ]đ â[đŽ] = 1.0 Ă 10â2 đ â 0.0021 đ = 0.008 đ
[đ¶] = [đ¶]đ â 2â[đŽ] = 2.0 đ â 2(0.008 đ) = 2.0 đ This should not be a surprising answer because in order to have a pseudo 1st order reaction with respect to A, the concentration of C must be essentially constant.
56. a) Elementary Step: An individual reaction in a proposed reaction mechanism. The rate law
can be written from the coefficients in the balanced equation b) Molecularity: The number of reactant molecules (or free atoms) taking part in an
elementary reaction. This is also the number of species that must collide in order to produce the reaction represented by an elementary reaction.
c) Reaction Mechanism: The pathway that is proposed for an overall reaction and accounts for the experimental rate law.
d) Intermediate: A species that is produced and consumed during a reaction but does not appear in the overall chemical equation.
e) Rate Determining Step: The elementary reaction that governs the rate of the overall reaction. This is the slowest elementary reaction that occurs in a mechanism.
59. For elementary reactions the rate is just equal to the rate constant times the concentration of
the reactants. a) đ đđĄđ = đ[đ¶đ»3đđ¶] b) đ đđĄđ = đ[đ3][đđ] c) đ đđĄđ = đ[đ3] d) đ đđĄđ = đ[đ3][đ] e) đ đđĄđ = đ[ đ¶6
14 ] 60. The rate law found in problem 33 was
đ đđĄđ = đ[đ»2đ2] In order for this to be the rate, the first step would have to be the rate determining step. Overall Reaction 2H2O2 2H2O + O2
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61. đ đđĄđ = đ[đ¶4đ»9đ”đ] Overall Reaction C4H9Br + 2H2O Br- + C4H9OH + H3O+ Intermediates C4H9
+ and C4H9OH2+
63. The rate of a reaction is dependent on the slowest step. a) đ đđĄđ = đ[đđ][đ2] The mechanism is not consistent with rate law.
b) đ đđĄđ = đ2[đđ3][đđ] NO3 is an intermediate, therefore, it needs to be eliminated from the rate equation. Use equilibrium to eliminate
đ1[đđ][đ2] = đâ1[đđ3]
[đđ3] =đ1[đđ][đ2]
đâ1
đ đđĄđ = đ2[đđ3][đđ] = đ2[đđ]đ1[đđ][đ2]
đâ1=
đ2đ1[đđ]2[đ2]
đâ1= đ[đđ]2[đ2]
The mechanism is consistent with rate law. c) đ đđĄđ = đ[đđ]2 The mechanism is not consistent with rate law.
d) In order for a mechanism to be plausible the elementary reactions must add up to the overall reaction. These elementary reactions do not add up to the overall reaction. In fact the middle reaction is not even balanced.
65. đ đđĄđ = đ3[đ”đâ][đ»2đ”đđ3
+] H2BrO3
+ is an intermediate therefore, needs to be eliminated from the rate law. Use equilibrium to eliminate the intermediate.
đ2[đ»đ”đđ3][đ»+] = đâ2[đ»2đ”đđ3+]
[đ»2đ”đđ3+] =
đ2[đ»đ”đđ3][đ»+]
đâ2
đ đđĄđ = đ3[đ”đâ][đ»2đ”đđ3+] =
đ2đ3[đ”đâ][đ»đ”đđ3][đ»+]
đâ2
HBrO3 is an intermediate therefore, needs to be eliminated from the rate law. Use equilibrium to eliminate the intermediate.
đ1[đ”đđ3â][đ»+] = đâ1[đ»đ”đđ3]
[đ»đ”đđ3] =đ1[đ”đđ3
â][đ»+]
đâ1
đ đđĄđ =đ2đ3[đ”đâ][đ»đ”đđ3][đ»+]
đâ2=
đ1đ2đ3[đ”đâ][đ”đđ3â][đ»+]2
đâ1đâ2= đ[đ”đâ][đ”đđ3
â][đ»+]2
68. a) đ đđĄđ = đ3[đ¶đđ¶đ][đ¶đ2]
COCl is an intermediate, therefore, it needs to be eliminated from the rate law. Use equilibrium to eliminate the intermediate.
đ2[đ¶đ][đ¶đ] = đâ2[đ¶đđ¶đ]
[đ¶đđ¶đ] =đ2[đ¶đ][đ¶đ]
đâ2
đ đđĄđ = đ3[đ¶đđ¶đ][đ¶đ2] =đ3đ2[đ¶đ][đ¶đ][đ¶đ2]
đâ2
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14
Cl is an intermediate, therefore, it needs to be eliminated from the rate law. Use equilibrium to eliminate the intermediate.
đ1[đ¶đ2] = đâ1[đ¶đ]2
[đ¶đ] =đ1
12â
[đ¶đ2]1
2â
đâ1
12â
đ đđĄđ =đ3đ2[đ¶đ][đ¶đ][đ¶đ2]
đâ2=
đ3đ2[đ¶đ][đ¶đ2]đ1
12â
[đ¶đ2]1
2â
đâ2đâ1
12â
=đ3đ2đ1
12â
[đ¶đ][đ¶đ2]3
2â
đâ2đâ1
12â
đ đđĄđ = đ[đ¶đ][đ¶đ2]3
2â b) COCl and Cl are intermediates
69. a) MoCl5- is an intermediate
b) đ[đđ2
â]
đđĄ= đ2[đđ3
â][đđđ¶đ5â]
Use the steady state approximation to eliminate the intermediate, rate of formation = rate of consumption
đ1[đđđ¶đ62â] = đâ1[đđđ¶đ5
â][đ¶đâ] + đ2[đđ3â][đđđ¶đ5
â]
[đđđ¶đ5â] =
đ1[đđđ¶đ62â]
đâ1[đ¶đâ] + đ2[đđ3â]
đ[đđ2â]
đđĄ= đ2[đđ3
â][đđđ¶đ5â] =
đ1đ2[đđ3â][đđđ¶đ6
2â]
đâ1[đ¶đâ] + đ2[đđ3â]
70. Need to calculate the rate of decomposition of O3
âđ[đ3]
đđĄ= đ1[đ][đ3] + đ2[đ][đ3] â đâ1[đ2][đ][đ]
Note: Rate of decomposition=ways O3 used up â ways O3 forms O is an intermediate therefore need to remove it from the rate law.
Use the same steady state approximation to solve for [O] đ1[đ][đ3] = đâ1[đ2][đ][đ] + đ2[đ][đ3]
[đ] =đ1[đ][đ3]
đâ1[đ2][đ] + đ2[đ3]
âđ[đ3]
đđĄ= đ1[đ][đ3] + đ2[đ][đ3] â đâ1[đ2][đ][đ]
âđ[đ3]
đđĄ= đ1[đ][đ3] +
đ1đ2[đ][đ3]2
đâ1[đ2][đ] + đ2[đ3]â
đ1đâ1[đ2][đ]2[đ3]
đâ1[đ2][đ] + đ2[đ3]
â
đ[đ3]
đđĄ=
đ1[đ][đ3](đâ1[đ2][đ] + đ2[đ3]) + đ1đ2[đ][đ3]2 â đ1đâ1[đ2][đ]2[đ3]
đâ1[đ2][đ] + đ2[đ3]
âđ[đ3]
đđĄ=
đ1đâ1[đ2][đ]2[đ3] + đ1đ2[đ][đ3]2 + đ1đ2[đ][đ3]2 â đ1đâ1[đ2][đ]2[đ3]
đâ1[đ2][đ] + đ2[đ3]
â
đ[đ3]
đđĄ=
2đ1đ2[đ][đ3]2
đâ1[đ2][đ] + đ2[đ3]
72. a) As the activation energy decreases the rate of the reaction goes up.
b) As the temperature increases the molecules/atoms move faster and more collisions occur, therefore, the faster the rate of the reaction.
c) As the frequency of collisions increases the rate of the reaction goes up.
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d) As the orientation of collisions is more specific the rate of the reaction goes down.
79.
đđ (đ1
đ2) =
đžđ
đ (
1
đ2â
1
đ1)
đđ (đ1
3.52 Ă 10â7 đżđđđâđ
) =186 đđœ
đđđ
0.0083145 đđœđđđâđŸ
(1
555 đŸâ
1
645 đŸ)
đ1 = 9.75 Ă 10â5 đż
đđđâđ
80.
đđ (đ1
đ2) =
đžđ
đ (
1
đ2â
1
đ1)
đđ (7.2 Ă 10â4 1
đ
1.7 Ă 10â2 1đ
) =đžđ
8.3145 đœđđđâđŸ
(1
720. đŸâ
1
660. đŸ)
đžđ = 210,000 đœ
đđđ
đđ (đ1
đ2) =
đžđ
đ (
1
đ2â
1
đ1)
đđ (đ1
1.7 Ă 10â2 đżđđđâđ
) =210,000 đœ
đđđ
8.3145 đœđđđâđŸ
(1
720. đŸâ
1
598 đŸ)
đ1 = 1.3 Ă 10â5 1
đ
After each half-life the number of moles of iodoethane is halved. Therefore after 3 half-lives
there is 81 the number of moles of iodoethane that there was initially. For a gas, PV=nRT,
therefore, since the volume, gas constant, and temperature are constant đ1
đ1=
đ2
đ2
894 đĄđđđ
đ„=
đ218
đ„
đ2 = 112 đĄđđđ
81.
đđ (đ1
đ2) =
đžđ
đ (
1
đ2â
1
đ1)
đđ (7đ„
đ„) =
5.40 Ă 104 đœđđđ
8.3145 đœđđđâđŸ
(1
295 đŸâ
1
đ1)
đ1 = 324 đŸ = 51â
83. a)
đđ(đ) = âđžđ
đ (
1
đ) + đđ(đŽ)
Therefore, when the ln(k) is plotted vs. 1
đthe slope of the line equalsâ
đžđ
đ
âđžđ
đ = â
đžđ
8.3145 đœđđđâđŸ
= â1.10 Ă 104 đŸ
đžđ = 9.15 Ă 104 đœ
đđđ= 91.5
đđœ
đđđ
b) When the ln(k) is plotted vs. T1 the y intercept of the line equals Aln
đđ(đŽ) = 33.5
đŽ = 3.54 Ă 1014 1
đ
c)
đđ(đ) = âđžđ
đ (
1
đ) + đđ(đŽ)
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đđ(đ) = â91,50 đœ
đđđ
8.3145 đœđđđâđŸ
(1
298 đŸ) + đđ (3.54 Ă 1014
1
đ ) = â3.43
đ = 0.0324 1
đ
84. đđ(đ) = âđžđ
đ (
1
đ) + đđ(đŽ)
Therefore if you plot the ln(k) vs. 1
đ the slope of the line will be â
đžđ
đ
đ đđđđ =đđđ đ
đđąđ=
đđ(3.5 Ă 10â5 1đ ) â đđ(4.9 Ă 10â3 1
đ )
1298 đŸ
â1
338 đŸ
= â12,400 đŸ
â12,400 đŸ =đžđ
đ =
đžđ
0.0083145 đđœđđđâđŸ
đžđ = 103 đđœ
đđđ
86. a) b)
c)
-12
-10
-8
-6
-4
-2
0
0.0028 0.003 0.0032 0.0034
ln(k
)
T^-1
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87.
The second graph is a two-step problem. 1st: reactants intermediates 2nd: intermediates products. Since the second reaction has the greater activation energy it will be the slower reaction and will determine the rate of the reaction. Therefore, we label the activation energy of the 2nd step.
88.
The activation energy for the reverse reaction is the -ÎE + Ea.
â(â216 đđœ
đđđ) + 125
đđœ
đđđ= 341
đđœ
đđđ
90. A catalyst increases the reaction rate because it provides another pathway with a lower
activation energy for the reaction to proceed by. Homogeneous catalysts are present in the same phase as the reactants and heterogeneous catalyst are present in a different phase from the reactants. The uncatalyzed and catalyzed reactions most likely will have different rate laws because they have different pathways that they occur by.
94. a) The catalyst is the species that is initially added to the reaction but not used in the
overall reaction. Therefore NO is the catalyst. b) An intermediate is a species that is formed in one of the steps of the reaction but not
one of the reactants or products. Therefore NO2 is an intermediate.
c)
đđ(đ) = âđžđ
đ (
1
đ) + đđ(đŽ)
đđ(đđđđĄ) â đđ(đđąđđđđĄ) = âđžđ(đđđĄ)
đ (
1
đ) + đđ(đŽ) + â
đžđ(đąđđđđĄ)
đ (
1
đ) â đđ(đŽ)
đđ (đđđđĄ
đđąđđđđĄ) =
1
đ đ(âđžđ(đđđĄ) + âđžđ(đąđđđđĄ)) = 0.848
đđ (đđđđĄ
đđąđđđđĄ) =
1
(8.3145 đœđđđâđŸ)(298 đŸ)
(â11,900 đœ + 14,000 đœ) = 0.848
đđđđĄ
đđąđđđđĄ= 2.3