FLASK B [1] (M₂) 0.100MM (NH4)2S₂O (buret) mL mL mL M (Ahw Exp 18: Kinetics (lodine Clock) To find the rate of the reaction and the average rate constant using Runs 1-5 FLASK A RUN 0.200 M 0.200M TotalVol KI KCI (buret) gr.cylinder) (V2) [0]- = 0.00500M 250 mL BEAKER+FlaskA+FlaskB TotalVol. [S,0,0.00500M Starch TotalVol. [S,0,1 (M.) NOV 0.100M (NH4)2SO4 (gr. cylinder) mL (K) mL M (pipet). mdrops mL 1 20.00 о 20.00 0.2000 20.00 0 20.00 0.1000 10.00 3 M 50.00 0.001000 2 20.00 0 20.00 0.2000 10.00 10.00 20.00 0.0500 10.00 3 50.00 0.001000 3 20.00 0 20.00 0.2000 6.670 13.30 19.97 0.03340 10.00 3 49.97 0.001001 4 10.00 10.00 20.00 0.1000 20.00 ° 20.00 0.1000 10.00 3 50.00 0.001000 5 6.670 13.30 19.97 0.06680 20.00 0 20.00 0.1000 10.00 3 49.97 0.001001 RUN Time T T rate = 0.000545 "k In (k) °C K mol/Ls t T(K) 1 33.54 25.0 298.2 1.491E-05 0.0007454 0.003354 -7.202 2 143.52 25.0 298.2 3.484E-06 3 298.79 25.0 298.2 1.673E-06 4 67.00 25.0 298.2 7.463E-06 5 99.38 25.0 298.2 5.031E-06 average k 0.003354 0.0003484 0.0002505 0.003354 -8.292 0.0007463 0.003354 -7.200 0.0007532 0.003354 0.0005687 -7.962 -7.191 rate-k [1] [S,O,1," "k-rate [1].[S₂O,³], rate = k[1] [4] M-KMM Ist Onder rate hud T- To find energy of activation (uncatalyzed) using Runs 1, 7, 9 (Plot 1/T vs Ink) K = ' RUN Time T T rate-d[S,O, 1/2dt 1 In (k) F MA = S °C K mol/Ls T(K) Runs 1-5 25.0 298.2 average k= 0.0005687 0.003354 -7.472 7 40.28 45.0 318.2 1.241E-05 0.0006207 0.003143 -7.385 9 63.62 5.0 278.2 7.859E-06 0.0003930 0.003595 -7.842 rate= Note: [1] and [S,O for Runs 6 through 14 are same as those for Run 1 Arrhenius Equation In(k) (-E,/R)*(1/T) + In A From Graph From Graph y=-1023.6x-4.1226 E/R slope 1023.6, R = 8.3145 J/Kmol M 2nd Order k ([1]' [5.0] rate law M = k MM² k = = Ms 1 1023.6 R=1023.6 K 8.3145 J/Kmol (1 kJ / 1000 J) = In A intercept -4.1226 4.1226) 8.5107 kJ/mol E, From Graph A = Run 1 [1]。 = 0.016202 Sample Calculation Run 1 [S,O,] Sample Calculation M1V1 = M2V2 M1V1 = M2V2 0.200M 20.00mL = M2 20.00mL M2=> 0.2000 M 0.100M 20.00mL M2 20.00mL M2=> 0.100 M Run 1 [S,O,] Sample Calculation M1V1 = M2V2 Run 1 rate AIS,0,1/2st [S₂O,21= 0.001000 M 0.00500M 10.00mL M2 50.00mL rate [0.001000 M]/2t M2= 0.001000 M rate 0.0005000 M/t To Find the order of reactants and to find the overall order of reacation: rate = k [1]. [S₂O,1, rate = k [1]. [S,O,1 Run 1 1.49076E-05k (0.2000m (0.1000)" Run 5 5.0312E-06k (0.06680m (0.1000) Run 2 3.48384E-06k (0.2000) (0.0500)" Run 4 7.4627E-06 = k (0.1000) (0.1000)" Run 1 = 1.49076E-05 k (0.2000m (0.1000)" Run 5 5.0312E-06 k (0.06680 (0.1000) Run 2 3.48384E-06 k (0.2000) (0.0500)" Run 4 7.4627E-06 k (0.1000) (0.1000)" Run 1 Run 2 4.28 2 Run 5 0.67 0.668 Run 4 √n= m= 0.97718 = 1 2.0972972 = 2 (experimental error because n should be = 1) OVERALL Order Of Rx = 3 (should be 2) rate = k [I], [S₂O,1,2 Data: Determination of Rate Law Run Tempature Time 20.0°C 02:20 2 20.0°C 03:21 3 20.0°C 05:10 ५ 20.0°C 01:49 5 20.0°C 02:57 Determination of Activation Energy (uncatalyzed) Temperature Time Run 6 30.0°C 01:18 7 40.0°C 01:09 8 11.9°C 03:15 9 4.5°C 04:00 Run Determination of Activation Energy I Catalyzed) Temperature Time 10 20.0% 01:24 11 30.0°C 00:46 12 که یه یه 31.0°0 00:31 13 17.0% 00:47 14 6.5°℃ 02:33
FLASK B [1] (M₂) 0.100MM (NH4)2S₂O (buret) mL mL mL M (Ahw Exp 18: Kinetics (lodine Clock) To find the rate of the reaction and the average rate constant using Runs 1-5 FLASK A RUN 0.200 M 0.200M TotalVol KI KCI (buret) gr.cylinder) (V2) [0]- = 0.00500M 250 mL BEAKER+FlaskA+FlaskB TotalVol. [S,0,0.00500M Starch TotalVol. [S,0,1 (M.) NOV 0.100M (NH4)2SO4 (gr. cylinder) mL (K) mL M (pipet). mdrops mL 1 20.00 о 20.00 0.2000 20.00 0 20.00 0.1000 10.00 3 M 50.00 0.001000 2 20.00 0 20.00 0.2000 10.00 10.00 20.00 0.0500 10.00 3 50.00 0.001000 3 20.00 0 20.00 0.2000 6.670 13.30 19.97 0.03340 10.00 3 49.97 0.001001 4 10.00 10.00 20.00 0.1000 20.00 ° 20.00 0.1000 10.00 3 50.00 0.001000 5 6.670 13.30 19.97 0.06680 20.00 0 20.00 0.1000 10.00 3 49.97 0.001001 RUN Time T T rate = 0.000545 "k In (k) °C K mol/Ls t T(K) 1 33.54 25.0 298.2 1.491E-05 0.0007454 0.003354 -7.202 2 143.52 25.0 298.2 3.484E-06 3 298.79 25.0 298.2 1.673E-06 4 67.00 25.0 298.2 7.463E-06 5 99.38 25.0 298.2 5.031E-06 average k 0.003354 0.0003484 0.0002505 0.003354 -8.292 0.0007463 0.003354 -7.200 0.0007532 0.003354 0.0005687 -7.962 -7.191 rate-k [1] [S,O,1," "k-rate [1].[S₂O,³], rate = k[1] [4] M-KMM Ist Onder rate hud T- To find energy of activation (uncatalyzed) using Runs 1, 7, 9 (Plot 1/T vs Ink) K = ' RUN Time T T rate-d[S,O, 1/2dt 1 In (k) F MA = S °C K mol/Ls T(K) Runs 1-5 25.0 298.2 average k= 0.0005687 0.003354 -7.472 7 40.28 45.0 318.2 1.241E-05 0.0006207 0.003143 -7.385 9 63.62 5.0 278.2 7.859E-06 0.0003930 0.003595 -7.842 rate= Note: [1] and [S,O for Runs 6 through 14 are same as those for Run 1 Arrhenius Equation In(k) (-E,/R)*(1/T) + In A From Graph From Graph y=-1023.6x-4.1226 E/R slope 1023.6, R = 8.3145 J/Kmol M 2nd Order k ([1]' [5.0] rate law M = k MM² k = = Ms 1 1023.6 R=1023.6 K 8.3145 J/Kmol (1 kJ / 1000 J) = In A intercept -4.1226 4.1226) 8.5107 kJ/mol E, From Graph A = Run 1 [1]。 = 0.016202 Sample Calculation Run 1 [S,O,] Sample Calculation M1V1 = M2V2 M1V1 = M2V2 0.200M 20.00mL = M2 20.00mL M2=> 0.2000 M 0.100M 20.00mL M2 20.00mL M2=> 0.100 M Run 1 [S,O,] Sample Calculation M1V1 = M2V2 Run 1 rate AIS,0,1/2st [S₂O,21= 0.001000 M 0.00500M 10.00mL M2 50.00mL rate [0.001000 M]/2t M2= 0.001000 M rate 0.0005000 M/t To Find the order of reactants and to find the overall order of reacation: rate = k [1]. [S₂O,1, rate = k [1]. [S,O,1 Run 1 1.49076E-05k (0.2000m (0.1000)" Run 5 5.0312E-06k (0.06680m (0.1000) Run 2 3.48384E-06k (0.2000) (0.0500)" Run 4 7.4627E-06 = k (0.1000) (0.1000)" Run 1 = 1.49076E-05 k (0.2000m (0.1000)" Run 5 5.0312E-06 k (0.06680 (0.1000) Run 2 3.48384E-06 k (0.2000) (0.0500)" Run 4 7.4627E-06 k (0.1000) (0.1000)" Run 1 Run 2 4.28 2 Run 5 0.67 0.668 Run 4 √n= m= 0.97718 = 1 2.0972972 = 2 (experimental error because n should be = 1) OVERALL Order Of Rx = 3 (should be 2) rate = k [I], [S₂O,1,2 Data: Determination of Rate Law Run Tempature Time 20.0°C 02:20 2 20.0°C 03:21 3 20.0°C 05:10 ५ 20.0°C 01:49 5 20.0°C 02:57 Determination of Activation Energy (uncatalyzed) Temperature Time Run 6 30.0°C 01:18 7 40.0°C 01:09 8 11.9°C 03:15 9 4.5°C 04:00 Run Determination of Activation Energy I Catalyzed) Temperature Time 10 20.0% 01:24 11 30.0°C 00:46 12 که یه یه 31.0°0 00:31 13 17.0% 00:47 14 6.5°℃ 02:33
Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter7: Reaction Rates And Chemical Equilibrium
Section: Chapter Questions
Problem 7.54P
Related questions
Question
Completeness and correctness (calculations)
• Printout of excel spreadsheet included, with calculations (formulas) indicated
• Activation energies
Using the data please change the sample data/calculation to match the data I provided
Please please answer everything super super fast it's important and urgent
![FLASK B
[1]
(M₂)
0.100MM
(NH4)2S₂O
(buret)
mL
mL
mL
M
(Ahw
Exp 18: Kinetics (lodine Clock)
To find the rate of the reaction and the average rate constant using Runs 1-5
FLASK A
RUN 0.200 M 0.200M TotalVol
KI
KCI
(buret) gr.cylinder) (V2)
[0]-
= 0.00500M
250 mL BEAKER+FlaskA+FlaskB
TotalVol. [S,0,0.00500M Starch TotalVol. [S,0,1
(M.) NOV
0.100M
(NH4)2SO4
(gr. cylinder)
mL
(K)
mL
M
(pipet).
mdrops
mL
1
20.00
о
20.00
0.2000
20.00
0
20.00 0.1000
10.00
3
M
50.00 0.001000
2
20.00
0
20.00
0.2000
10.00
10.00
20.00
0.0500
10.00
3
50.00
0.001000
3
20.00
0
20.00
0.2000
6.670
13.30
19.97
0.03340 10.00
3
49.97 0.001001
4
10.00
10.00
20.00
0.1000
20.00
°
20.00
0.1000
10.00
3
50.00 0.001000
5
6.670
13.30
19.97
0.06680
20.00
0
20.00
0.1000
10.00
3
49.97 0.001001
RUN Time
T
T
rate = 0.000545 "k
In (k)
°C
K
mol/Ls t
T(K)
1
33.54
25.0
298.2
1.491E-05
0.0007454
0.003354
-7.202
2
143.52
25.0
298.2
3.484E-06
3
298.79
25.0
298.2
1.673E-06
4
67.00
25.0
298.2
7.463E-06
5
99.38
25.0
298.2
5.031E-06
average k
0.003354
0.0003484
0.0002505 0.003354 -8.292
0.0007463 0.003354 -7.200
0.0007532 0.003354
0.0005687
-7.962
-7.191
rate-k [1] [S,O,1,"
"k-rate
[1].[S₂O,³],
rate = k[1] [4]
M-KMM
Ist Onder
rate hud
T-
To find energy of activation (uncatalyzed) using Runs 1, 7, 9 (Plot 1/T vs Ink)
K =
'
RUN Time
T
T
rate-d[S,O, 1/2dt
1
In (k)
F
MA
=
S
°C
K
mol/Ls
T(K)
Runs 1-5
25.0
298.2
average k=
0.0005687
0.003354
-7.472
7
40.28
45.0
318.2
1.241E-05
0.0006207
0.003143
-7.385
9
63.62
5.0
278.2
7.859E-06
0.0003930 0.003595
-7.842
rate=
Note: [1] and [S,O
for Runs 6 through 14 are same as those for Run 1
Arrhenius Equation
In(k) (-E,/R)*(1/T) + In A
From Graph
From Graph
y=-1023.6x-4.1226
E/R slope 1023.6, R = 8.3145 J/Kmol
M
2nd Order
k ([1]' [5.0] rate law
M = k MM²
k =
=
Ms
1
1023.6 R=1023.6 K 8.3145 J/Kmol (1 kJ / 1000 J) =
In A intercept -4.1226
4.1226)
8.5107 kJ/mol
E,
From Graph
A =
Run 1
[1]。
= 0.016202
Sample Calculation
Run 1
[S,O,] Sample Calculation
M1V1 = M2V2
M1V1 = M2V2
0.200M 20.00mL = M2 20.00mL
M2=>
0.2000 M
0.100M 20.00mL M2 20.00mL
M2=>
0.100 M
Run 1
[S,O,] Sample Calculation
M1V1 = M2V2
Run 1
rate AIS,0,1/2st
[S₂O,21= 0.001000 M
0.00500M 10.00mL M2 50.00mL
rate
[0.001000 M]/2t
M2=
0.001000 M
rate
0.0005000 M/t
To Find the order of reactants and to find the overall order of reacation:
rate = k [1]. [S₂O,1,
rate = k [1]. [S,O,1
Run 1
1.49076E-05k (0.2000m (0.1000)"
Run 5
5.0312E-06k (0.06680m (0.1000)
Run 2
3.48384E-06k (0.2000) (0.0500)"
Run 4
7.4627E-06 = k (0.1000)
(0.1000)"
Run 1
=
1.49076E-05 k (0.2000m (0.1000)"
Run 5
5.0312E-06 k (0.06680
(0.1000)
Run 2
3.48384E-06
k (0.2000)
(0.0500)"
Run 4
7.4627E-06 k (0.1000)
(0.1000)"
Run 1
Run 2
4.28
2
Run 5
0.67
0.668
Run 4
√n=
m=
0.97718 = 1
2.0972972 = 2
(experimental error because n should be = 1)
OVERALL Order Of Rx = 3 (should be 2)
rate = k [I], [S₂O,1,2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdab67aeb-c4d9-43f7-88d0-dfd6e7af2d5d%2Fbb721f1b-c3ee-4269-b07b-9496294e4cb3%2Ftou4x2nu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:FLASK B
[1]
(M₂)
0.100MM
(NH4)2S₂O
(buret)
mL
mL
mL
M
(Ahw
Exp 18: Kinetics (lodine Clock)
To find the rate of the reaction and the average rate constant using Runs 1-5
FLASK A
RUN 0.200 M 0.200M TotalVol
KI
KCI
(buret) gr.cylinder) (V2)
[0]-
= 0.00500M
250 mL BEAKER+FlaskA+FlaskB
TotalVol. [S,0,0.00500M Starch TotalVol. [S,0,1
(M.) NOV
0.100M
(NH4)2SO4
(gr. cylinder)
mL
(K)
mL
M
(pipet).
mdrops
mL
1
20.00
о
20.00
0.2000
20.00
0
20.00 0.1000
10.00
3
M
50.00 0.001000
2
20.00
0
20.00
0.2000
10.00
10.00
20.00
0.0500
10.00
3
50.00
0.001000
3
20.00
0
20.00
0.2000
6.670
13.30
19.97
0.03340 10.00
3
49.97 0.001001
4
10.00
10.00
20.00
0.1000
20.00
°
20.00
0.1000
10.00
3
50.00 0.001000
5
6.670
13.30
19.97
0.06680
20.00
0
20.00
0.1000
10.00
3
49.97 0.001001
RUN Time
T
T
rate = 0.000545 "k
In (k)
°C
K
mol/Ls t
T(K)
1
33.54
25.0
298.2
1.491E-05
0.0007454
0.003354
-7.202
2
143.52
25.0
298.2
3.484E-06
3
298.79
25.0
298.2
1.673E-06
4
67.00
25.0
298.2
7.463E-06
5
99.38
25.0
298.2
5.031E-06
average k
0.003354
0.0003484
0.0002505 0.003354 -8.292
0.0007463 0.003354 -7.200
0.0007532 0.003354
0.0005687
-7.962
-7.191
rate-k [1] [S,O,1,"
"k-rate
[1].[S₂O,³],
rate = k[1] [4]
M-KMM
Ist Onder
rate hud
T-
To find energy of activation (uncatalyzed) using Runs 1, 7, 9 (Plot 1/T vs Ink)
K =
'
RUN Time
T
T
rate-d[S,O, 1/2dt
1
In (k)
F
MA
=
S
°C
K
mol/Ls
T(K)
Runs 1-5
25.0
298.2
average k=
0.0005687
0.003354
-7.472
7
40.28
45.0
318.2
1.241E-05
0.0006207
0.003143
-7.385
9
63.62
5.0
278.2
7.859E-06
0.0003930 0.003595
-7.842
rate=
Note: [1] and [S,O
for Runs 6 through 14 are same as those for Run 1
Arrhenius Equation
In(k) (-E,/R)*(1/T) + In A
From Graph
From Graph
y=-1023.6x-4.1226
E/R slope 1023.6, R = 8.3145 J/Kmol
M
2nd Order
k ([1]' [5.0] rate law
M = k MM²
k =
=
Ms
1
1023.6 R=1023.6 K 8.3145 J/Kmol (1 kJ / 1000 J) =
In A intercept -4.1226
4.1226)
8.5107 kJ/mol
E,
From Graph
A =
Run 1
[1]。
= 0.016202
Sample Calculation
Run 1
[S,O,] Sample Calculation
M1V1 = M2V2
M1V1 = M2V2
0.200M 20.00mL = M2 20.00mL
M2=>
0.2000 M
0.100M 20.00mL M2 20.00mL
M2=>
0.100 M
Run 1
[S,O,] Sample Calculation
M1V1 = M2V2
Run 1
rate AIS,0,1/2st
[S₂O,21= 0.001000 M
0.00500M 10.00mL M2 50.00mL
rate
[0.001000 M]/2t
M2=
0.001000 M
rate
0.0005000 M/t
To Find the order of reactants and to find the overall order of reacation:
rate = k [1]. [S₂O,1,
rate = k [1]. [S,O,1
Run 1
1.49076E-05k (0.2000m (0.1000)"
Run 5
5.0312E-06k (0.06680m (0.1000)
Run 2
3.48384E-06k (0.2000) (0.0500)"
Run 4
7.4627E-06 = k (0.1000)
(0.1000)"
Run 1
=
1.49076E-05 k (0.2000m (0.1000)"
Run 5
5.0312E-06 k (0.06680
(0.1000)
Run 2
3.48384E-06
k (0.2000)
(0.0500)"
Run 4
7.4627E-06 k (0.1000)
(0.1000)"
Run 1
Run 2
4.28
2
Run 5
0.67
0.668
Run 4
√n=
m=
0.97718 = 1
2.0972972 = 2
(experimental error because n should be = 1)
OVERALL Order Of Rx = 3 (should be 2)
rate = k [I], [S₂O,1,2
Transcribed Image Text:Data:
Determination of Rate Law
Run Tempature
Time
20.0°C
02:20
2
20.0°C
03:21
3
20.0°C
05:10
५
20.0°C
01:49
5
20.0°C
02:57
Determination of Activation Energy (uncatalyzed)
Temperature Time
Run
6
30.0°C
01:18
7
40.0°C
01:09
8
11.9°C
03:15
9
4.5°C
04:00
Run
Determination of Activation Energy I Catalyzed)
Temperature Time
10
20.0% 01:24
11
30.0°C
00:46
12
که
یه
یه
31.0°0
00:31
13
17.0%
00:47
14
6.5°℃
02:33
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