To determine the rate law for a chemical reaction among hydrogen peroxide, iodide and acid, specifically by observing how changing each of the concentrations of H2O2, I- and H+ affects the rate of reaction
Name: Tonny, Chan Kar Yu
Date of Experiment: 3rd December 2004, Group: B1
Title: Experiment 3 - Chemical Kinetics
Objective:
. To determine the rate law for a chemical reaction among hydrogen peroxide, iodide and acid, specifically by observing how changing each of the concentrations of H2O2, I- and H+ affects the rate of reaction.
2. To examine the effects of temperature and a catalyst on the rate of reaction.
Chemicals and Apparatus:
0.8M hydrogen peroxide containing 0.001M sulphuric acid, 0.05M sodium thiosulphate(Na2S2O3), 0.05M potassium iodide, solid potassium iodide, 2.0M sulphuric acid, 0.05 acetic acid & sodium acetate buffer, 0.30M acetic acid, 0.02 Mo(VI) catalyst (containing 1.76g/L of ammonium hyptamolybdate ([NH4]6Mo7O24•4H2O)), 0.1% of freshly prepared starch solution, water bath at 40, electronic balance, 5.00cm3 pipette, 25cm3 pipette, 50.0cm3 burette, white tile, 125cm3 conical flask, 250cm3 conical flask, 250cm3 beaker, 10cm3 measuring cylinder, glass rod, 50cm3 measuring cylinder, thermometer and stop watch.
Background:
A description of the sequence of steps by which a chemical reaction takes place is called a reaction mechanism. Many studies go into trying to determine possible mechanisms. The rates at which reactions take place provide important insights.
The rate law for a chemical reaction is a quantitative expression involving constants related to the nature of the chemical reaction and the concentrations of reactants. In order for reactants to react, they must come in contact with one another (or at least come nearby). The probability of a collision is related to a function of the concentration. So, for the general rate law:
Rate=k [A]a[B]b[C]c
[A], [B] and [C] represent concentrations of reactants and catalysts, and a, b, and c represent exponents that may or may not be related to the coefficients of the corresponding balanced chemical equation.
The quantities in brackets are read as concentration and are raised to an appropriate power. Multiplied together with the constant (k), they give the rate of the reaction.
The numerical values of a, b, and c must be determined by experimentation. These numbers determine the order of the reaction. Added together they give the over-all order of the reaction. It was the purpose of this experiment to determine the order of H2O2, a reactant in the iodine clock reaction.
The reaction to be studied in this experiment was the acid buffered oxidation of iodide to triiodide by hydrogen peroxide
H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) › I3- (aq)+ 2 H2O (aq) --- (I)
I3- (aq) + 2 S2O32- (aq) › 3 I- (aq) + S4O62- (aq) --- (II)
2 I3- (aq) + starch › starch-I5-complex + I- (aq) --- (III)
The first equation indicates that, in an acidic solution, iodide ions were oxidized by hydrogen peroxide to triiodide ions. These triiodide ions were reduced back to iodide ions by thiosulfate ions, as indicated in equation (II). This reaction was much faster than the reaction of equation (I); it consumed triiodide ions as fast as they were formed. This prevented any readily apparent reaction of equation (III). However, after all the thiosulfate ions had been consumed by the reaction of equation (II), triiodide ions reacted with starch to form the blue starch-pentaiodide complex.
By varying the concentration of each of the three reactants (H2O2, I- and H+), we will be able to determine the order of the reaction with respect to each reactant and the rate law of the reaction, which of the form:
Rate = k [H2O2]x[I-]y[H+]z
By knowing the reaction times (?t) and the concentrations of H2O2 of two sepatate reaction mixtures (mixtures A & B), the reaction order of H2O2, x, can be calculated.
x = log(?t2/?t1) / log ([H2O2]1/[H2O2]2)
The same method is used to obtain the reaction order with respect to I- (mixture A & C), and H+ (mixture of A & D).
Procedures:
Part I - Standardization of H2O2 Solution
. A burette was rinsed with deionized water and 0.05M Na2S2O3 solution.
2. The stopcock of the burette was closed and the sodium thiosulphate solution was poured into it until the liquid level was near to the zero mark. The stopcock of the burette was opened to allow the titrant to fill up the tip and then adjust the liquid level near zero.
3. The initial burette reading was recorded in Table 1. The reading should be accurate to 2 decimal places.
4. 1.00cm3 of the ~0.8M H2O2 solution was pipetted into a clean 125 cm3 conical flask.
5. 25cm3 of the deionized water was measured with a 50cm3 measuring cylinder. It was poured into the conical flask.
6. 10cm3 of 2.0M sulphuric acid was measured with a 10cm3 clean measuring cylinder. It was poured into the conical flask.
7. 1.000g of solid KI and 3 drops of ammonium molybdate catalyst was added into the conical flask.
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3. The initial burette reading was recorded in Table 1. The reading should be accurate to 2 decimal places.
4. 1.00cm3 of the ~0.8M H2O2 solution was pipetted into a clean 125 cm3 conical flask.
5. 25cm3 of the deionized water was measured with a 50cm3 measuring cylinder. It was poured into the conical flask.
6. 10cm3 of 2.0M sulphuric acid was measured with a 10cm3 clean measuring cylinder. It was poured into the conical flask.
7. 1.000g of solid KI and 3 drops of ammonium molybdate catalyst was added into the conical flask.
8. The solution mixture was swirled until the KI dissolved.
9. The reaction mixture was titrated in the conical flask with the sodium thiosulphate solution until pale yellow was just turned.
0. 3 drops of freshly prepared starch solution was added to the conical flask.
1. The reaction mixture was continued to titrate until it just changed from dark blue to colourless.
2. The given sodium thiosulphate solution was added to the burette through a filter funnel if the volume remained was not enough to carry out another titration.
3. Step 6-13 were repeated to obtain 2 sets of consistent results. However, sodium thiosulphate solution was stopped draining at about 3cm3 less than the estimated value. Then the sodium thiosulphate solution was added drop by drop until the reaction mixture in conical flask just changed from dark blue to colourless.
Part II - Reaction Rate Measurements
The information necessary to determined the effects of the concentration of H2O2, I- and H+ on the rate of reaction by six reaction mixtures, as well as the effect of temperature and a catalyst. The temperature and reagent volumes to be used for each reaction mixture were specified by Table 2.
Table2
Reaction Mixture
Temp. /
Water / cm3
0.05M Buffera / cm3
0.3M Acetic acid / cm3
0.05M KI / cm3
0.1% Starch / cm3
0.05M Na2S2O3 / cm3
0.01M Mo(VI) catalyst / cm3
~0.8M H2O2 / cm3
A
Rm. Temp.
75
30
0
25
5
5
0
0
B
Rm. Temp.
80
30
0
25
5
5
0
5
C
Rm. Temp.
50
30
0
50
5
5
0
0
D
Rm. Temp.
30
30
45
25
5
5
0
0
E
Rm. Temp. +10
75
30
0
25
5
5
0
0
F
Rm. Temp.
70
30
0
25
5
5
5
0
. All the reactants were placed in mixture A in the order in which they appear in Table 2 except the H2O2 in a 250 cm3 conical flask.
2. The temperature of the solution mixture was read and recorded.
3. The stop-watch was got ready.
4. Specific amount of standardized H2O2 solution was pipetted to the conical flask and stirred continuously.
5. The stop-watch was started when half of the H2O2 solution had been drained from the pipette.
6. The conical flask was swirled to mix the contents well.
7. The solution mixture was watched carefully for the sudden appearance of the blue colour, and the timer was stopped when blue colour appeared.
8. The time for the reaction was recorded in Table 3. The solution was discarded and the conical flask was rinsed.
9. Step 1-8 were repeated to obtain 2 sets of consistent results.
0. When on set of experiment (2 duplicates) was finished, all the steps were repeated for other mixture (B-F) on Table 2.
During the preparation of Mixture E, all the reactants were placed in a 250cm3 conical flask except the H2O2. Then the conical flask was put in a water bath. Until the temperature of the reaction mixture was held constant, H2O2 solution was pipetted to it.
Results:
Table1: Standardization of H2O2 solution. (Report the reading to 2 decimal places)
Data
Trial 1
Trial 2
Trial 3
Burette final reading / cm3
29.80
47.70
38.50
Burette initial reading / cm3
7.30
8.40
4.80
Volume of 0.05M Na2S2O3(aq) used / cm3
22.50
29.30
33.70
Average titre: 31.5 cm3
Table3
Trial 1
Reaction Mixture
Time / s
Temperature /
A
18 (1:58:53)
22.1
B
241 (4:01:31)
22.4
C
68 (1:08:46)
22.4
D
98 (3:18:60)
22.4
E
74 (1:14:02)
34.1
F
69 (1:09:05)
22.4
Discussion:
(a) the order of reaction with respect to hydrogen peroxide, H2O2
Average titre of sodium thiosulpfate solution was 31.5 cm3.
H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) › I3- (aq)+ 2 H2O (aq) --- (I)
I3- (aq) + 2 S2O32- (aq) › 3 I- (aq) + S4O62- (aq) --- (II)
Number of mole of Na2S2O3 = 0.001575 mol
Number of mole of H2O2 = 0.0007875 mol
Molarity of H2O2 = 0.7875 mol dm-3
Reaction Mixture
Time / s
H2O2 concentration in the reaction mixture / M
A
18
0.0525
B
241
0.02625
Rate = k [H2O2]x [I-]y [H+]z
The order of the reaction with respect to hydrogen peroxide concentration, x, could be determined by measuring the rate of iodine formation when the concentration of the iodide was held constant while varying the initial concentration of the hydrogen peroxide in case A and B.
2.04237 = (2) x
In 2.04237 = x In 2
x = 1.02
The order of reaction with respect to hydrogen peroxide was 1.
(b) the order of the reaction with respect to iodide, I-
Reaction Mixture
Time / s
I- concentration in the reaction mixture / M
A
18
0.008333
C
68
0.01667
Rate = k [H2O2]x [I-]y [H+]z
The order of the reaction with respect to iodide, y, could be determined by measuring the rate of iodine formation when the concentration of the hydrogen peroxide was held constant while varying the initial concentration of the iodide ion in case A and C.
0.5763 = (0.4999) y
In 0.5763 = y In 0.4999
y = 1.15
The order of reaction with respect to iodide was 1.
(c) the order of reaction with respect to H+
Reaction Mixture
Time /s
Acetic acid concentration
Sodium acetate concentration
Calculated H+ concentration / M
A
18
0.01
0.01
.8 x 10-5
D
98
0.1
0.01
.8 x 10-4
Concentration of acetic acid in A = [(0.05)(30/1000)] / (150/1000)
= 0.01 M
Concentration of sodium acetate in A = [0.05 x (30/1000)] / (150/1000)
= 0.01 M
Ka =1.8 x 10-5 = [H+] (0.01) / 0.01
[H+] = 1.8 x 10-5 M
Concentration of acetic acid in D = [0.05 x (30/1000) + 0.3 x (45/1000)] / (150/1000)
= 0.1 M
Concentration of sodium acetate in D = [0.05 x (30/1000)] / (150/1000)
= 0.01 M
.8 x 10-5 = [H+] (0.01) / (0.1)
[H+] = 1.8 x 10-4 M
The reaction order:
x = ln (?t2/?t1) / ln ([H2O2]1/ [H2O2]2)
= ln (118/198) / ln (1.8 x 10-5/1.8 x 10-4)
=0.22
The order of reaction with respect to H+ was 0.
2 (a)(b) From my measurements, the reaction orders with respect to each of the three reactants H2O2, I- and H+ were 1, 1 and 0.
So the rate expression for the uncatalyzed reaction was
Rate = k [H2O2] 1 [I-] 1
And the total reaction order in the reaction was two.
2 (c) The rate law is inconsistent with the idea that the mechanism of the reaction was
H2O (aq) + 2 I-(aq) + 2 H+ (aq) --> I2 (aq) + 2 H2O (l) , the rate law only included the rate determining steps of hydrogen peroxide and iodide.
2 (d) the first term in this rate law results from a rate determining nucleophilic attacked of the iodide anion upon the peroxide:
I- + H-O-O-H --> H-O-I + H-O- [slow; rate determining]
The second step involved the formation of the iodine dimer and is a very fast step:
I-+ H-O-I + --> I2 + OH- [fast]
The HOI (aq) then reacted with I- to form I2 as above. And only hydrogen peroxide and iodide were involved in the rate determining step.
3. Effect on Temperature
Reaction Mixture
Time / s
Temperature /?
A
18
22.1
E
74
34.1
The reaction ratio = (rate E / rate A)
= [(1/74) / (1/118)]
=1.5946
According to Arrhenius Theory, the rate constant of an experiment should equal to
k = A exp [Ea/RT]
where A is a proportionality constant, Ea is the activation energy of the process, R is the ideal gas constant (8.314J/K mole), and T is the temperature in Kelvin.
By measuring the rate constant at different temperatures, it is possible to calculate the activation energy of the reaction. As a general rule of thumb, an increase of temperature by 10? usually results in a doubling of the reaction rate.
4. Effect on Catalyst
Reaction Mixture
Time / s
Mo(VI) concentration in the reaction mixture / M
A
18
0
F
69
3.333 x 10-4
The reaction ratio = (rate F / rate A)
= [(1/69) / (1/118)]
=1.7101
When there was an addition of catalyst, the reaction rate was increase; it was because the catalyst was a substance that entered into the process of a reaction. It might undergo temporary change in the molecular structure, but ultimately the catalyst molecule was restored to its original structure before the product had been formed. The catalyst reduced the energy of activation so that product could be formed at a lower temperature. It was able to do this because was providing a different mechanism for the reaction with an "easier" rate determining step.
5. Standardization of H2O2 solution
H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) › I3- (aq) + 2 H2O (aq)
2 S2O32- (aq) + I3- (aq) --> 3 I- (aq) + S4O62- (aq)
In Part I, the use of Na2S2O3 was the unique reactant for iodine in the second equation; 2 [S2O32-] = [I3-] = 3 [I-], then the amount of iodine formed by H2O2 and KI were reaction with the sodium thiosulfate solution, when colour of the solution turned colourless, all I2 in the solution were reacted, with this, we could indirectly find the moles of hydrogen peroxide.
6. To do the kinetics experiment, we needed to determine the rate at which species were consumed or generated. In the experiment we would use a chemical indicator that provided a visible signal when the reaction had progress a certain amount, when a specified amount of reactants had been consumed. Starch formed a complex with iodine that is a deep blue colour. If starch was present in the reaction system, the colour of the solution would indicate the presence of molecular iodine. Therefore, if starch was present when the above reaction was performed, the solution should immediately turn blue since iodine was generated directly. So, if performed this way, no kinetic information could be obtained since the colour change would do nothing more that show that the reaction had started.
To extract the rate data, Na2S2O3 was added to the reaction mixture, thiosulfate reacted with molecular iodine very rapidly to regenerate I- and the new compound S4O62-.
I3- + 2 S2O32- + 2 H+ › I3- + 2 H2O
By adding a known amount of thiosulphate, the colour change would then signal the point at which the thiosulfate had been consumed, provided that the amount of thiosulfate added was small relative to the amounts of peroxide and iodide initially present as the thiosulfate ion was the limiting reactant in this clock reaction. At that time of the colour change, the I2 was generated by the formation of the blue iodine-starch complex. Since the stoichiometry of both reactions was known, the amount of thiosulfate reacted could be correlated to the amount of peroxide consumed in a given time.
Further Discussion:
The rate of the reaction was influenced greatly by small amounts of contaminants; extreme care must be taken to ensure the cleanliness of all glassware. Rinse several times with small quantities of distilled water and drain thoroughly.
Hydrogen peroxide oxidized iodide ion (I-) according to the equation
H2O2 (aq) + 3 I- + 2 H+ › 2 H2O + I3- ------ (1)
A general expression for the rate of reaction could be written as
Rate = k [H2O2]x [I-]y [H+]z ------ (2)
In the experiment the concentration of the hydronium ion was maintained at a constant value (approximately 1.8 x 10-5 M) by using a buffer solution, acetic acid and sodium acetate (CH3COOH and NaCH3COO). Therefore, the expression for the rate of the reaction simplifies to
Rate = k' [H2O2]x [I-]y ------ (3)
where k' is the rate constant at a given hydronium ion concentration and is equation to k[H+]z. The purpose of the experiment was to establish numerical values for k', x and y, from which the rate law for the reaction was determined.
Because reaction (1) itself had no visible sign of completion, we employed a secondary reaction that involved a colour change. A small known quantity of sodium thiosulfate and starch indicator was added to the reaction mixture. The thiosulfate ion did not react to any appreciable extent with any of the reactants. However, it did react very rapidly with the iodine (I2) produced in reaction (1) according to the equation
I2 + 2 S2O32- › 2 I- + S4O62- ------ (4)
As the iodine produced by reaction (1) forms, it was immediately consumed by the thiosulfate ion. When the small quantity of thiosulfate present was consumed, iodine (I2) would accumulate and formed an intensely blue coloured complex with the starch.
Therefore, by noting the time required for the appearance of the blue complex after mixing the hydrogen peroxide and the iodide ion, the rate of the reaction could be established. In this experiment the rate was expressed as the rate of formation of
iodine (I2) in units of mol L-1 s-1.
The concentration of thiosulfate ion used to measure the reaction rate was very small compared to the concentrations of hydrogen peroxide and the iodide ion. Therefore, it could be assumed that the concentrations of the reactants remain constant at their initial values during the reaction time period.
The initial of thiosulfate ion was the same in every reaction in this experiment since the initial rate of every reaction was approximately [S2O32-] / t and [S2O32-] was the same for every reaction; relative rates would be proportional to 1/t. The time between the start of the reaction and the colour change and it could be determined from the timer readings.
The order of the reaction with respect to hydrogen peroxide concentration, x, can be determined by measuring the rate of iodine formation when the concentration of the iodide ion was held constant while varying the initial concentration of the hydrogen peroxide. In this case, the rate expression becomes
rate = k'' [H2O2]x ------ (5)
where k" = k' [I-]y = k [I-]y [H+]z, with [I-] held constant.
In a similar fashion, the order of the reaction with respect to iodide ion concentration, y, could be determined by measuring the rate of iodine formation when the concentration of hydrogen peroxide was held constant while varying the initial concentration of the iodide ion. The rate expression becomes
rate = k''' [I-]y ------ (6)
where k''' = k' [H2O2]x = k [H2O2]x [H+]x, with [H2O2] held constant.
the first term in this rate law results from a rate determining nucleophilic attacked of the iodide anion upon the peroxide:
I- + H-O-O-H --> H-O-I + H-O- [slow; rate determining]
The second step involved the formation of the iodine dimer and is a very fast step:
I-+ H-O-I + --> I2 + OH- [fast]
The HOI (aq) then reacted with I- to form I2 as above. Iodine was only slightly soluble in water and it was thought that they predominant aqueous species was I3-. Because water made a better leaving group than OH-; therefore, an acid-catalyzed mechanism was favored over the uncatalyzed mechanism, so 0.001 M H2SO4 was contained by hydrogen peroxide.
If it was not for the catalyst, the reaction would occur so slowly that practical product formation would not be feasible. From a molecular point of view, the catalyst provides a surface for the reaction molecules to position themselves with one another so that when they do collide they will do so much more effectively.
Conclusion:
Iodine Clock Reaction was produced by a solution of hydrogen peroxide and sulphuric acid was added to a solution containing iodide ion, thiosulfate ion, and starch. After some time, the dark blue colour of a triiodide-starch complex was observed. The influences of initial reactant concentrations on the reaction were examined, as were the influence of volume and temperature.
In the part I of the experiment, standardization of H2O2 solution, there would be a positive error, which was, the value of trial 2 was larger than that of trial 1, if the glassware were not rinsed clear.
The stirring was required to mix the solution thoroughly, and we should stir the solution in each set of reaction mixtures in the same amplitude to ensure the stirring effect for all reaction mixtures were the same.
Hazards:
Hydrogen peroxide is corrosive to the eyes and the skin. The vapour irritates the respiratory tract. Ingestion of hydrogen peroxide may produce oxygen bubbles in the blood, resulting in shock.
Sulphuric acid is corrosive and it is very corrosive to the eyes the skin and the respiratory tract. Inhalation of an aerosol of sulphuric acid may cause lung oedema.
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