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instructions are in the pdf. The work is to complete the "labreportform" using the image attached. the completed work should NOT be posted online or be searchable.

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Qualitative examination of primary and probe reactions. The number of drops of each reagent is in
ackets. The group should a
ive at a consensus quickly and record your observations.
Well # | Procedure Observation
[i Pp —— Selb hanged om desc/ealorless Fo
1 | Addsior Ry a after adding dees
3 Tenge. Py n
2 | waters:00Kustarch [5127211] Soh “ Er REL
2 | adasior solnkon ged Olorless alter adding
3 | sosmarts.oeki nan) | Yell Silks wh hele prt fore = clare,
TCH on werk from colar liss To Ge Deck
4 |sorsacs ot KIRBY | “oii fos we
5 | sioisarcs,007KI 1735] Siu A om lla dy ble Je
Small-scale kinetic runs:
Reagent ‘Well #1 ‘Well #2 Well #3 ‘Well #4.
5:05 1 drop 1 drop 1 drop 1 drop
Starch 1 drop 1 drop 1 drop 1 drop
KI 2 drops 4 drops 2 drops 2 drops
5:0 2 drops 2 drops 4 drops 3 drops
=. Tero
me IM dd [24490 [22-9056 [33H wes
More on the reverse side.
Room Temperature [2.1]: _5-50°C

‘Time Data for Kinetic Reactions
Experiment Temperature Trial #1 time Trial #2 time
350: (x gees XXXXXXXXXX
" 25.5 °C 05: XXXXXXXXXX: 59-09
min Sec ih Sec
XXXXXXXXXX25s _
# 25:5 ©3:)7. 9,4 0:56: 25
no wo nin see
183.0% \U-73s
a] 25-5 ¢ 093%03:09 83:1: 33
Ns Sect min Sec
YA395-3F XXXXXXXXXX
“ XXXXXXXXXX:35:37 | 95:19: Uo
wins sec min sec
Jorn Way.eq
Low Temp. [I C Qb:43 (a 07:58: 57
XXXXXXXXXX
weep | 35.7 ¢ | ou:lb:86 | 59 55:08
Catalyst 25.5 °c 00: al: 14
sec



Name                                    
Group             
PLEASE ENABLE EDITING BEFORE FILLING OUT THIS REPORT!
Results
Qualitative examination of primary and probe reactions. The number of drops of each reagent is in
ackets. (10 points)
Procedure
Observation
wate
S2O82-/KI [5/2/2]
Add S2O32-
wate
S2O82-/KI/starch [5/2/2/1]
Add S2O32-
S2O32-/starch/S2O82-/KI [3/1/3/3]
S2O32-/starch/S2O82-/KI [2/1/3/3]
S2O32-/starch/S2O82-/KI [1/1/3/3]
Small-scale kinetic runs: (10 points)
Reagent
Well #1
Well #2
Well #3
Well #4
S2O32-
1 drop
1 drop
1 drop
1 drop
Starch
1 drop
1 drop
1 drop
1 drop
KI
2 drops
4 drops
2 drops
2 drops
S2O82-
2 drops
2 drops
4 drops
3 drops
Time
Room Temperature [2.1]:
Δ[I2] = 0.5 [Na2S2O3]initial =
        
Time Data for Kinetic Reactions (10 points)
Experiment
Temperature
Trial #1 time
Trial #2 time
Average time
#1
#2
#3
#4
Low Temp.
High Temp.
Catalyst
Experimental Rates and Determination of the Rate Law (20 points)
Experiment
[I-]initial
[S2O82-]initial
Avg. Δt
Rate (M/s)
#1
#2
#3
#4
Low Temp.
High Temp.
Catalyst
Determine the specific form of the rate equation. (10 points)
Rate = k [KI]x [(NH4)2S2O8]y. You must solve for x and y.
Value for x =
Value for y =
Specific form of the rate law:
Insert your data sheet HERE. (10 points)
Calculation of the rate constants (k) (10 points)
Experiment
Value for k (use co
ect units!)
#1 (room temp)
#2 (room temp)
#3 (room temp)
Average (1-3)
#1 (low temp)
#1 (high temp)
#1 (catalyst)
Calculation of the predicted rate for experiment #4. Show or explain your work! (10 points)
(You may scan and attach a handwritten version.)
Predicted value for the experiment #4 rate:
Actual value for the experiment #4 rate:
Questions (10 points)
1.    How does the rate differ in the presence of a catalyst? Do you think that the catalyst would be more effective at higher concentrations? Do you have any suggestions as to how this catalyst might work (hint: what are the charges on the reactants of interest and the catalyst)? Briefly explain your answer.
1.    Determine the percentage difference between the experimental and the predicted value for the rate in experiment 4.
    
1.    How well do you think you could predict rates at other concentrations? If you have what you consider a large % difference how would you run the experiment to make it better?
1.    
Use the A
henius Equation to calculate the activation energy Ea for this reaction using the experiment #1 data at room temperature and the higher temperature of ~35ºC (remember to convert to Kelvins).
1.    Using this activation energy (Ea) calculate the rate constant at the lower temperature using the room temperature data for experiment #1. How well do the two rate constants compare?

1¾ « @U û 8 8 9 : : NORMAL
General Chemistry II - CHM202
Laboratory 4 – Chemical Kinetics

Purpose
In this laboratory, you will examine the rate of chemical reactions in solution. For a series of reactions run at
oom temperature, you will vary the concentration of reactants to determine the order of the reaction for two
eactants. You will also examine the effect of temperature and a catalyst on the reaction rate. This series of
experiments shows how you determine kinetic parameters using the initial rate method. You will use your
data to calculate the experimental rate law, the rate constants, the order of reaction (overall and reactant
specific) and activation energy, Ea.
General
You have learned previously how to balance chemical equations and determine the amounts of products and
eactants involved in those reactions. You have also measured the heat generated or abso
ed and used these
data along with the stoichiometry data to determine enthalpies of reactions.
In this experiment, you will measure how quickly products are formed from reactants, a
anch of chemistry
known as chemical kinetics. In order to study the rate at which a product is produced from one or more
eactants you need to understand and control a number of experimental variables. Factors that can affect the
ate at which a product can be formed in a chemical reaction are:
1. Solution concentration – more concentrated reactants may form products faster than more dilute
eactants.
2. Temperature – Reactions are almost always faster at higher temperatures, and in order to understand
eaction rates you must carefully control the temperature of our experiments.
3. Catalysis – Some materials can act to increase the reaction rate without being part of the net
observed equation. These materials are called catalysts and may have a large effect on the reaction
ate even when present in small (sometimes undetectable) quantities.
4. Surface area of a solid reactant or catalyst – The more surface area (more finely divided solid) the
faster the reaction if there is a solid-liquid or solid-gas phase interface as a component of the
chemical reaction.
The factors that you will evaluate are concentration, temperature, and catalysts. Since the reaction you will
study involves the reaction of aqueous solutions with no solids present as reactants, you will not be
concerned with surface area (#4 above).
In order to understand the rate at which reactants combine to form products you must have a clear
understanding of the reaction you are studying, and how you can measure the reaction rate. Sometimes this
is straightforward, as in the decomposition of hydrogen peroxide given below:
H2O2(aq) → H2O(l) + O2(g)
In this reaction, there is one gaseous product (oxygen). The other product, water, and the reactant, hydrogen
peroxide, will remain in solution. If you measure the volume of gaseous oxygen produced as a function of
time, you can determine the number of moles of oxygen produced by this reaction as a function of time.
From these data, you can determine the rate law and the rate constant of this reaction.
The reaction that you will study in this laboratory is called an iodine clock reaction. It is called a clock
eaction because the completion of the reaction (similar to a titration endpoint) is signaled by the sudden
appearance of the dark blue starch-iodine complex at a fixed time for the same concentration of reactants. In
order to obtain this dramatic color change and relate that to the chemical reaction rate, you will have to
perform two reactions at the same time. The first reaction (primary reaction) is the reaction of kinetic
interest. This is the reaction of iodide ion with persulfate ion to form iodine and sulfate ion. Is this reaction a
precipitation, acid-base, or oxidation-reduction reaction? How can you tell? If it is an acid-base reaction,
identify the acid and the base, or if it is an oxidation-reduction reaction, identify what is being oxidized and
what is being reduced. The molecular and net ionic equations are as follows:
Molecular: (NH4)2S2O8 + 2KI → I2 + 2(NH4)2SO4
Net Ionic: S2O82- + 2I- → I2 + 2SO42-
(persulfate ion) (iodide ion) (iodine) (sulfate ion)
In order to measure the kinetics you must experimentally determine the rate at which persulfate ion or iodide
ion is consumed or the rate at which iodine or sulfate ion is produced. If you run this reaction in the
laboratory, you will observe that the solution turns yellow-orange which is an indication that iodine is
formed. One experimental possibility to determine the reaction rate would be to determine how dark the
iodine color is as a function of time with an instrument (spectrophotometer). Since starch will form a blue
color with iodine, you may perform the same reaction in the presence of starch. If you do this, you will
observe that a dark blue color forms almost instantaneously. To determine the rate this way would be
difficult because it is hard to tell the difference between dark blue, darker blue and even darker blue! Thus,
the direct method that could be employed in the oxygen production reaction of hydrogen peroxide cannot
eadily be used in this case.
In order to provide a reaction that can be more easily observed you can run a second reaction (probe
eaction) at the same time the first reaction is proceeding. This second reaction will quickly consume all of
the iodine produced in the primary reaction until the probe reaction uses up all of its reagents. When the
probe reaction stops it will be due to the fact that all of the reactants have been used up. At this point, iodine
continues to form from the primary reaction. If you add starch into the system the end effect will be a fast,
dramatic color change from colorless to dark blue (iodine starch complex). This rapid color change is the
eason it is called an iodine clock reaction. The probe reaction that you will use is (precipitation, acid-base
or oxidation-reduction reaction?):
Molecular: 2Na2S2O3 + I2 ( excess from → 2NaI + 2Na2S4O6
primary reaction)
XXXXXXXXXXNet Ionic: 2S2O32- + I2 → 2I- + 2S4O62-
(thiosulfate
ion)
(iodine) (iodide
ion)
(tetrathionate ion)
Because the probe reaction is fast (much faster than the primary reaction) and goes to completion, it gives us
an important piece of information. You can determine the number of moles of iodine (using volume and
molarity) produced by the primary reaction in a measured amount of time. You can measure the time from
the start of the reaction (when you mixed the reactants) to the completion (when the blue color appears).
You may then use the stoichiometry of the reaction to determine the amount of iodine that was produced.
The appearance of the dark blue starch-iodine complex occurs only after all of the thiosulfate has been
consumed and the starch-iodine complex is formed from additional iodine being produced in the primary
eaction. You now have enough information to relate the molar concentration change of a product as a
function of time. Using this, you can now determine other kinetic parameters.
Let's look at some conditions that you could recreate in the lab and see what you would expect to observe for
each case. You are concerned only with three ions present in the solution: thiosulfate ion (S2O32-), persulfate
ion (S2O82-) iodide ion (I-). From the concentrations of these three ions, you can predict the amount of iodine
consumed:
If you always use the same amount of thiosulfate, that amount is less than the amount of iodine that can be
produced, then you will always observe blue color when a known quantity of iodine has been consumed by
all of the available thiosulfate. This amount depends on the amount of thiosulfate initially present in the
solution. The concentration of iodine produced is always one-half of the initial concentration of thiosulfate
(from the equation).
???? =
∆[?2]
∆?
=
0.5[??2?2?3]
???? ?? ???? ?????

The rate of a chemical reaction is the increase in the concentration of product (or the decrease in the
concentration of reactant) over a given time. As discussed above, you now know the concentration of iodine
produced at the appearance of the blue color from the known concentration of thiosulfate used. Note that
you must calculate the initial concentration of all reactants. Using these data, you can determine the
elationship between the concentration of iodine and the time that the blue color appears. This rate in M/sec
(mol/L-sec) is the experimentally measurable chemical rate for the primary reaction.
Now that you know how to measure the rate of a reaction, what can you do to understand the kinetics of this
process better? There are three major experimental goals to be achieved from the experimental
determination of the reaction rate:
1. Determination of the general and specific (experimental) rate laws. Specifically:
a
Answered 2 days After Apr 01, 2024

Solution

Baljit answered on Apr 04 2024
7 Votes
Name                                    
Group             
PLEASE ENABLE EDITING BEFORE FILLING OUT THIS REPORT!
Results
Qualitative examination of primary and probe reactions. The number of drops of each reagent is in
ackets. (10 points)
    Procedure
    Observation
    wate
S2O82-/KI [5/2/2]
    Solution changed from clear colorless to yellow
    Add S2O32-
    Solution turned colorless after adding 5 drops
    wate
S2O82-/KI/starch [5/2/2/1]
    Solution changed from clear colorless to yellow and to a darker shade with black ppt
    Add S2O32-
    Solution turned colorless after adding 8 drops
    S2O32-/starch/S2O82-/KI [3/1/3/3]
    Solution changed from colorless to yellow with black ppt after few second
    S2O32-/starch/S2O82-/KI [2/1/3/3]
    Solution turned colorless to blue black after few second
    S2O32-/starch/S2O82-/KI [1/1/3/3]
    Solution turned colorless to black after few second
Small-scale kinetic runs: (10 points)
    Reagent
    Well #1
    Well #2
    Well #3
    Well #4
    S2O32-
    1 drop
    1 drop
    1 drop
    1 drop
    Starch
    1 drop
    1 drop
    1 drop
    1 drop
    KI
    2 drops
    4 drops
    2 drops
    2 drops
    S2O82-
    2 drops
    2 drops
    4 drops
    3 drops
    Time
    35.15 secs
    24.49 secs
    22.90 secs
    33.71 secs
    Room Temperature [2.1]:
    25.5oC
    Δ[I2] = 0.5 [Na2S2O3]initial =
    
        
Time Data for Kinetic Reactions (10 points)
    Experiment
    Temperature
    Trial #1 time
    Trial #2 time
    Average time
    #1
    25.5oC
    350.68 secs
    359.09 secs
    354.89 secs
    #2
    25.5oC
    197.90 secs
    176.25 secs
    187.07 secs
    #3
    25.5oC
    183.09 secs
    191.33 secs
    187.21 secs
    #4
    25.5oC
    325.37 secs
    314.40 secs
    319.89 secs
    Low Temp.
    14oC
    402.12 secs
    478.57 secs
    440.345 secs
    High Temp.
    35.7oC
    256.86 secs
    295.08 secs
    275.97 secs
    Catalyst
    25.5oC
    21.19 secs
    
    
Experimental Rates and Determination of the Rate Law (20 points)
Now the initial concentration will be the concentration of the reagent used times the volume used divided by the total volume
    Experiment
    [I-]initial
    [S2O82-]initial
    Avg. Δt
    Rate...
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