1. home and garden

Chemistry 102
______________________________________________________________________________
EXPERIMENT 1
REACTION RATE, RATE LAW, AND ACTIVATION ENERGY
THE IODINE “CLOCK” REACTION
NAME: ______________________ Date: _____________ Partner: ______________________
67/67
PART I: DETERMINATION OF THE RATE LAW
1.
Initial Concentrations of Reactants
For each kinetic run, calculate the initial concentration of the reactants:
[I-] and [S2O82-].
These calculations should be done prior to performing the
experiment!
Since the reaction takes place in a total volume of 50.00 mL, this volume must be
taken into account in calculating the initial concentration of the two reactants.
For example, in Run 1, since the 20.00 mL of 0.200 M KI added reacts in a total
volume of 50.00 mL, the initial concentration of [I-]0 can be calculated as follows:
20.00 mL
[I-]0 = 0.200 M KI
= 0.0800 M KI
50.00 mL
Similarly, in Run 1, the initial concentration of [S2O82-]0 is calculated as follows:
20.00 mL
[S2O82-]0 = 0.100 M (NH4)2S2O8
= 0.0400 M (NH4)2S2O8
50.00 mL
On the next page, carry out similar calculations for all other initial values of the
two reactants and complete the appropriate columns in DATA TABLE III.
You are required to:
 Show all your calculations
 Include units in your calculations, and
 Express all measured quantities (including your answer) in the appropriate
number of significant figures.
2015 www./proffenyes.com
13
Chemistry 102
______________________________________________________________________________
EXPERIMENT 1
REACTION RATE, RATE LAW, AND ACTIVATION ENERGY
THE IODINE “CLOCK” REACTION
DATA TABLE I
[I-]0
[S2O82-]0
20.00 mL
Run 1
0.200 M
20.00 mL
= 0.0800 M
0.100 M
50.00 mL
= 0.0400 M
50.00 mL
10.00 mL
Run 2
0.200 M
Same as in Run 1
= 0.0400 M
50.00 mL
10.00 mL
Run 3
Same as in Run 1
0.100 M
= 0.0200 M
50.00 mL
5.00 mL
Run 4
Same as in Run 1
0.100 M
= 0.0100 M
50.00 mL
(10)
2 . Reaction Rates
In order to determine the Reaction Rates, the following quantities must be known:
 Initial concentration of [S2O32-] added and completely used up (see below)
 Concentration of [I2] produced, at the time the deep blue color (tcolor) appears.
(see below); Enter this value in the Data Table I on the next page.
 The time in seconds when the deep blue color (tcolor) appears for each Reaction
Run (determined experimentally). Enter these data in DATA TABLE III.
DATA TABLE II
[S2O32-]
2-
[S2O3 ]added
[I2]produced
=
2
All Runs
10.00 mL
0.00500 M
50.00 mL
(2)
12
2015 www./proffenyes.com
0.00100 M
= 0.00100 M
14
= 0.000500 M
2
Chemistry 102
______________________________________________________________________________
EXPERIMENT 1
REACTION RATE, RATE LAW, AND ACTIVATION ENERGY
THE IODINE “CLOCK” REACTION
DATA TABLE III
Initial Concentrations and corresponding Reaction Rates
RATE*: Express Rate as (A x 10-6) throughout the entire experiment
Run
Nr.
Temp.
(0C)
(Room
Temp)
[I-]0
[S2O82-]0
TIME
(tcolor)
RATE
Expressed as A x 10-6
[I2]produced
tcolor
(M)
(M)
(s)
1
23
0.0800 M
0.0400 M
41.0
(M . s-1)
12.2 x 10 - 6
2
23
0.0400 M
0.0400 M
79.0
6.33 x 10 - 6
3
23
0.0800 M
0.0200 M
84.0
5.95 x 10 - 6
4
23
0.0800 M
0.0100 M
168.0
2.98 x 10 - 6
(8)
8
2015 www./proffenyes.com
15
Chemistry 102
______________________________________________________________________________
EXPERIMENT 1
REACTION RATE, RATE LAW, AND ACTIVATION ENERGY
THE IODINE “CLOCK” REACTION
3. Reaction Orders
The general formula for the Rate Law for the reaction studied is:
Rate1 = k [I-]m [S2O82-]n
Rate2 = k [I-]m [S2O82-]n
When performing your calculations you are required to:
 Show all calculations neatly, in a well-organized manner and in detail
(follow the format of the example shown on page 4 or the sample calculation
presented in your textbook (page 605)
 Round off your answer to an integer.
a. Reaction Order with respect to [I-]
From Run 1 and Run 2
Rate 1
12.2 x 10 – 6 M . s-1
k [0.0800 M]m [0.0800]n
=
=
Rate 2
6.33 x 10 - 6 M . s-1
k [0.0400 M]m [0.0800]n
1.93 = 2.00m
log 1.93
log 1.93 = log (2.00m)
log 1.93 = m log 2.00
m=
= 0.950
log 2.00
(4)
Value of “m” rounded off to an integer =
1
b. Reaction Order with respect to [S2O82-]
From Run 3 and Run 4
Rate 3
5.95 x 10 – 6 M . s-1
k [0.0800 M]m [0.0200]n
=
=
-6
-1
Rate 4
2.98 x 10 M . s
k [0.0800 M]m [0.0100]n
2.00 = 2.00n
log 2.00
log 2.00 = log (2.00m)
log 2.00 = m log 2.00
n=
= 1.000
log 2.00
(4)
Value of “n” rounded off to an integer =
8
2015 www./proffenyes.com
16
1
Chemistry 102
______________________________________________________________________________
EXPERIMENT 1
REACTION RATE, RATE LAW, AND ACTIVATION ENERGY
THE IODINE “CLOCK” REACTION
4. Rate Constant “k” at Room Temperature
The Rate Constant, “k” can be calculated by substituting the known values of reactant
concentrations, the reaction orders and the corresponding reaction rates in the formula
of the Rate Law. Refer to the example on page 5.
(1)
Rate = k [I- ] [S2 O2- ]
Write the equation for the Rate Law:
DATA TABLE IV
Calculate “k” for Reactions 1 through 6. Show your calculations and include units!
Run 1
Run 2
[ I- ] = 0.0400 M [S2 O2- ] = 0.0400 M
[ I- ] = 0.0800 M [S2 O2- ] = 0.0400 M
Rate1 = 12.2 x 10 – 6 M . s-1
Rate1 = 6.33 x 10 – 6 M . s-1
12.2 x 10 – 6 M . s-1
6.33 x 10 – 6 M . s-1
k=
k=
[0.0800 M] [0.0400 M]
[0.0400 M] [0.0400 M]
k1 = 3.81 x 10-3 M-1 . s-1
Run 3
[ I- ] = 0.0800 M [S2 O2- ] = 0.0200 M
Rate1 = 5.95 x 10 – 6 M . s-1
kk21 == 3.96 x 10-3 M-1 . s-1
Run 4
[ I- ] = 0.0800 M [S2 O2- ] = 0.0100 M
Rate1 = 2.98 x 10 – 6 M . s-1
5.95 x 10 – 6 M . s-1
2.98 x 10 – 6 M . s-1
k=
k=
[0.0800 M] [0.0200 M]
[0.0800 M] [0.0100 M]
k3 = 3.72 x 10-3 M-1 . s-1
(8)
(2)
k4 =
3.72 x 10-3 M-1 . s-1
DATA TABLE V
Summary of Rate Constants “k” for Reaction Runs at Room Temperature
1
2
3
4
-3
-3
-3
3.81 x 10
3.96 x 10
3.72 x 10
3.72 x 10-3
Units
Run Nr.
k
(M-1 . s-1)
k
3.80 x 10-3
(Average)
(M-1 . s-1 )
Write the complete form of the Rate Law:
 Include the formulas of both Reactants. Do not include numerical values
 Include the respective Reaction Orders for both Reactants
 Include the experimentally determined numerical value of “k”, expressed in the correct units.
(2)
13
RATE = 3.80 x 10-3 M-1 . s-1 [I- ] [S2 O2-]
2015 www./proffenyes.com
17
Chemistry 102
______________________________________________________________________________
EXPERIMENT 1
REACTION RATE, RATE LAW, AND ACTIVATION ENERGY
THE IODINE “CLOCK” REACTION
PART II: THE EFFECT OF TEMPERATURE ON REACTION RATE
To determine the quantitative relationship between Reaction Rate and Temperature you will
be plotting an Arrhenius Plot (ln k versus 1/T) to determine the following Kinetic Parameters:
1. The Activation Energy (Ea)
2. The ratio by which the reaction rate increases when the temperature is increased by 10 0C
(arbitrarily chosen from 200C to 300C)
1. Determination of the Activation Energy, Ea
 Summarize your experimental data for the Reactions Runs with the same concentration of
all reactants (as used in Reaction Run # 1), but run at different temperatures. In this
manner, the corresponding reaction rates for these runs will be affected by temperature
only.
DATA TABLE VI
Run Temp.
Initial
Final
Average
Time
RATE
Nr.
Range Temperature Temperature
Temperature
(tcolor)
[I2]produced
of
Of
of
tcolor
Reactants
Reaction
Reaction
Mixture
Mixture
0
0
( C)
( C)
(0C)
(s)
(M . s-1)
5
About
4
249
2.01 x 10-6
00C
6
About
11
112
4.46 x 10-6
100C
1
Room
23
41.0
12.2 x 10-6
Temp.
4.2
7
About
41
16.0
x 10-6
400C
(4)
Calculate the Rate Constants for Runs 5, 6 & 7 from the respective initial concentrations
of reactants and the corresponding reaction rates.
(8)
(Calculation do not need to be shown)
Run
Recorded
[I-]0
[S2O82-]0
RATE
Rate Constant
Nr.
Average
k
Temp.
(M-1 . s-1)
(M)
(M)
(K)
(M . s-1)
(Expressed as A x 10-3)
-6
5
About
277
0.0800
0.0400 2.01 x 10
0.628 x 10-3
00C
6
About
284
0.0800
0.0400 4.46 x 10-6
1.39 x 10-3
100C
1
Room
296
0.0800
0.0400 12.2 x 10-6
Calculated Average
Temp
3.80 x 10-3
7
About
314
0.0800
0.0400 31.2 x 10-6
9.75 x 10-3
0
40 C
12
2015 www./proffenyes.com
18
Chemistry 102
______________________________________________________________________________
EXPERIMENT 1
REACTION RATE, RATE LAW, AND ACTIVATION ENERGY
THE IODINE “CLOCK” REACTION
 Calculate and collect the data that will be used for plotting ln k versus 1/T (The
Arrhenius Plot)
DATA TABLE VIII
(Calculations do not need to be shown)
Run
1
ln k
Nr.
T
(K-1)
5
About 3.61 x 10-3
-7.373
00C
6
About 3.52 x 10-3
- 6.578
100C
1
Room 3.38 x 10-3
- 5.573
Temp
7
About 3.18 x 10-3
- 4.630
400C
(4)
 Plot a graph of ln k vs. 1/T
A sample graph is attached to your Report Form. Please follow the format, scale, data
reporting style and all the other details included in the sample graph.
Calculate the slope of the graph and show all the calculations on the graph.
 Calculate the Activation Energy (Ea) from the slope of the graph
 Please show calculations and include units.
J
3
Slope: - 6.5 x 10 K
Recall: R = 8.314
mol . K
Ea
J
Slope = - Ea = (Slope) x (R)
- Ea = (- 6.5 x 103 K ) (8.314
)
R
mol . K
1 kJ
Ea = 5.4 x 104 J/mol x
= 54 kJ/mol
103 J
Ea = 5.4 x 104 J/mol
Ea = 54 kJ/mol
(4)
8
2015 www./proffenyes.com
19
Chemistry 102
______________________________________________________________________________
EXPERIMENT 1
REACTION RATE, RATE LAW, AND ACTIVATION ENERGY
THE IODINE “CLOCK” REACTION
2. Determination of the effect of 100C temperature increase on Reaction Rate
 Collect your data:
t1 = 200C
T1 = 293 K 1/T1 = 3.41 x 10-3 K-1
t2 = 300C
T2 = 303 K
1/T2 = 3.30 x 10-3 K-1
 Read the corresponding Values for ln k1 and ln k2 from your Arrhenius plot
Indicate these values and the source of these readings on your graph
(as shown on Figure 2, page 8)
ln k1 = - 5.95
ln k2 = - 5.25
 Follow the guidelines given below Figure 2 (page 8) to calculates the ratio between
the Reaction Rate at 300C and the Reaction Rate at 200C
 Show all your calculations neatly, in a well-organized manner and in detail.
 Include units in your calculations
 Round off your answer to the nearest integer.
k1 = anti ln (- 5.95) = 2.6 x 10-3 M-1 . s-1
k2 = anti ln (- 5.25) = 5.2 x 10-3 M-1 . s-1
Rate2 (300C)
k2 [[I-] [S2O82-]
=
0
Rate1 (20 C)
5.2 x 10-3 M-1.s-1 [0.0800 M][0.0400 M]
=
-
2-
k1 [[I ] [S2O8 ]
= 2.0 = 2
-3
-1 -1
2.6 x 10 M .s [0.0800 M][0.0400 M]
(4)
State your conclusion:
(2)
A temperature increase of 100C approximately doubled the reaction rate.
6
Bibliography:
1. Nivaldo J. Tro, “Chemistry: A Molecular Approach”, Third Edition
2. R.A.D. Wentworth “Experiments in General Chemistry”, Sixth Edition
3. James M. Postma & all, “Chemistry in the Laboratory”, Seventh Edition
2015 www./proffenyes.com
20
Spring 2015
Experiment # 1 – The Iodine “Clock” Reaction
Post-Lab Quiz
21/21
Name: ________________________________
Instructions:
1. You may consult your Laboratory Notebook.
2. You may NOT consult any other materials (report forms, syllabus, notes, etc.)
3. Please show all your calculations clearly and neatly in the spaces provided.
No credit is earned by unsupported answers.
4. All numerical answers should be given in in the appropriate units and to the proper number of
significant figures.
1. Write a balanced Net Ionic Equation representing the reaction whose rate is being analyzed. Include
state designations.
Underneath each species involved in the reaction, indicate its name:
(2)
(1)
NET IONIC EQUATION: 2 I- (aq)
NAMES:
+
iodide ion
S2O82-(aq)
persulfate ion
I2(aq)
2SO42-(aq)
+
iodine
sulfate ion
2. What is the purpose of adding Na2S2O3 (S2O32-) to the reaction mixture?
(2)
In order to measure the rate of formation of Iodine (I2), the reaction in which the I2 is formed
(see above) is coupled with a much faster reaction that consumes most of the I 2
I2
+
2S2O322I+
S4O62- (Reaction 2)
Reaction 2 immediately consumes the I2 generated in the first reaction until all of the S2O32(thiosulfate) is used up. When the S2O32- (thiosulfate) is consumed, I2 builds up and reacts with
starch to form the the deep blue Starch-Iodine complex.
The appearance of the deep-blue complex tells us that at this point in time (tcolor)
sufficient I2 has been produced to use up all of the S2O32- (thiosulfate) ion
originally added.
3. Write a net ionic equation that illustrates the chemical reaction in which the
S2O32- (thiosulfate) ion is involved. Include State designations.
(2)
I2(aq) +
2S2O32-(aq)
2I-(aq) +
S4O62-(aq) (Reaction 2)
(1) 4. Is the S2O32- ion a reactant in the reaction whose rate is being analyzed? NO
5. What is the numerical relationship between:
 The concentration of the [I2) produced
and
 The concentration of the [S2O32-] originally added and used up
at the time the blue color appears (tcolor)
Hint: Consider the stoichiometry (mole ratio) of the equation in 2. (b) above.
[S2O32-] originally added and used up
(2)
[I2] produced =
10
2
1
Spring 2015
6. The following experimental data are available:
20.00 mL of 0.200 M KI are mixed with 10.00 mL of 0.00500 M Na2S2O3 in a 250 mL reaction flask. A
few drops of soluble starch are added to this mixture.
5.00 mL of 0.100 M (NH4)2S2O8 are mixed with 15.0 mL of 0.100 (NH4)2SO4 in a
50 mL reaction flask.
When the contents of the two flasks are mixed a deep blue color forms after 78.2 seconds.
(a) Write the chemical formula of Reactant 1 in Ionic Form: I Calculate the molarity of Reactant 1 after mixing the contents of the two flasks. Show
calculations and include units.
(2)
20.00 mL
[I ] = 0.200 M x
= 0.0800 M
50.00 mL
(b) Write the chemical formula of Reactant 2 in Ionic Form: S2O82 Calculate the molarity of Reactant 2 after mixing the contents of the two flasks. Show
calculations and include units.
(2)
5.00 mL
[ S2O82-] = 0.100 M x
= 0.0100 M
50.00 mL
(c) Calculate the molarity of Na2S2O3 added after mixing the contents of the two flasks.
Show calculations and include units.
(2)
10.00 mL
2[ S2O3 ] = 0.00500 M x
= 0.00100 M
50.00 mL
(d) Calculate the concentration of the [I2] produced at the time the blue color
appears
0.00100 M
(1)
[I2] produced =
= 0.000500 M
2
(e) Calculate the Rate of Reaction based on your data above, and express it in the proper units.
0.000500 M
(1)
= 6.39 x10 -6 M/s
Rate =
78.2 s
(f) Calculate the Rate Constant, k, based on your data above and express it in the proper units.
Assume that the reaction is of the first order with respect to both reactants.
(3)
6.39 x10 -6 M/s
Rate
k =
= 7.99 x 10-3 M-1 s-1
=
[I-] [S2O82-]
(0.0800 M) (0.0100 M
11
2