Iodine And Propanone Reaction Lab Report
TITLE OF EXPERIMENT: An investigation of the iodine/propanone reaction.
DATA COLLECTION AND PROCESSING :
Chemical Equation:
CH3COCH3 (aq) + I2 (aq) ( CH3COCH2I + H+ (aq) + I- (aq)
Rate of reaction:
R = V T
Concentration of reactant after dilution :
Volume of reactant x concentration of reactant
Total volume of mixture
Uncertainty for rate of reaction:
R =( V + t ) x R V t
Eg for Exp 1, run 1: R = ( 1 + 0.1 + 0.1 + 0.1 + 0.01 ) x 0.36 = +0.03cm3s-1 20 8 0 4 11.16
Experiment 1 (without starch solution) :
| …show more content…
|Run 1 |Run 2 |Run 3 |Run 4 |
|Volume of 2M HCl/cm3 (+1cm3) |20 |10 |20 |20 |
|Volume of 2M propanon/cm3 (+0.1cm3) |8 |8 |4 |8 |
|Volume of water/cm3 (+0.1cm3) |0 |10 |4 |2 |
|Volume of 0.01M iodine/cm3 (+0.1cm3) |4 |4 |4 |2 |
|Time for colour to disappear/s (+0.01s) |11.16 |26.63 |27.53 |6.45 |
|Rate(Vol. Of I2(aq)/Time)/cm3s-1 |0.36 (+0.03) |0.15 (+0.02) |0.15 (+0.02) |0.31 (+0.05) |
Table 5 shows the rate of reaction for each run without the presence of starch.
Experiment 2 (with starch solution):
| |Run 1 |Run 2 |Run 3 |Run 4 |
|Volume of 2M HCl/cm3 (+1cm3) |20 |10 |20 |20 |
|Volume of 2M propanone/cm3 (+0.1cm3) |8 |8 |4 |8 |
|Volume of water/cm3 (+0.1cm3) |0 |10 |4 |2 |
|Volume of 0.01M iodine/cm3 (+0.1cm3) |4 |4 |4 |2 |
|Volume of starch solution/cm3 (+0.1cm3) |5 |5 |5 |5 |
|Time for colour to disappear/s (+0.01s) |37.44 |50.53 |55.19 |14.56 |
|Rate(Vol.
Of I2(aq)/Time) /cm3s-1 |0.11 (+0.01) |0.08 (+0.01) |0.07 (+0.01) |0.14 (+0.03) |
Table 6 shows the rate of reaction for each run with the presence of starch.
Experiment 1 (without starch solution) :
| |Run 1 |Run 2 |Run 3 |Run 4 |
|Concentration of HCl/moldm-3 |1.25 |0.625 |1.25 |1.25 |
|Concentration of propanone/moldm-3 |0.5 |0.5 |0.25 |0.5 |
|Concentration of Iodine/moldm-3 |0.00125 |0.00125 |0.00125 |0.000625 |
Table 7 shows the concentration of reactant for experiment 1.
Experiment 2 (with starch solution) …show more content…
:
| |Run 1 |Run 2 |Run 3 |Run 4 |
|Concentration of HCl/moldm-3 |1.08 |0.54 |1.08 |1.08 |
|Concentration of propanone/moldm-3 |0.43 |0.43 |0.22 |0.43 |
|Concentration of Iodine/moldm-3 |0.001 |0.001 |0.001 |0.00054 |
Table 8 shows the concentration of reactant for experiment 2.
FROM EXPERIMENT 1:
Compare Run 1 & 2:
|Run 1 |Run 2 |
|H+ Concentration = 1.25M |H+ Concentration = 0.625M |
|Rate = 0.36cm3s-1 |Rate = 0.15cm3s-1 |
→When the concentration of H+ is halved, the rate of reaction also becomes halved which indicates that the reaction is first order with respect to H+.
Compare Run 1 & 3:
|Run 1 |Run 3 |
|Propanone Concentration = 0.5M |Propanone Concentration = 0.25M |
|Rate = 0.36 cm3s-1 |Rate = 0.15 cm3s-1 |
→When the concentration of propanone is halved, the rate of reaction also becomes halved which indicates that the reaction is first order with respect to CH3COCH3.
Compare Run 1 & 4:
|Run 1 |Run 4 |
|Iodine Concentration = 0.00125M |Iodine Concentration = 0.000625M |
|Rate = 0.36cm3s-1 |Rate = 0.31cm3s-1 |
→When the concentration of iodine is halved, the rate of reaction remained the same which indicates that the reaction is zero order with respect to I2.
So, the rate law is given by:
Rate= k[ H+][CH3COCH3]
Rate constant, k = Rate [ H+][CH3COCH3]
| |Rate of reaction / dm3s-1 |[H+] / mol dm-3 |[CH3COCH] / mol dm-3 |k/ dm9 mol-2 s-1 |
|Run 1 |3.6x10-4 |1.25 |0.50 |1.4x10-4 |
|Run 2 |1.5x10-4 |0.625 |0.50 |1.2x10-4 |
|Run 3 |1.5x10-4 |1.25 |0.25 |3.0x10-5 |
|Run 4 |3.1x10-4 |1.25 |0.50 |1.2x10-4 |
Table 5 shows the rate constant for each run in experiment 1.
Average rate constant, k = 1.4x10-4+1.2x10-4+3.0x10-5+1.2x10-4 = 1.0x10-4 dm9 mol-2 s-1 4
→The overall order of the reaction is (1+1) = second order.
FROM EXPERIMENT 2 :
Compare Run 1 & 2:
|Run 1 |Run 2 |
|H+ Concentration = 1.08M |H+ Concentration = 0.54M |
|Rate = 0.11cm3s-1 |Rate = 0.08cm3s-1 |
→When the concentration of H+ is halved, the rate of reaction also becomes halved which indicates that the reaction is first order with respect to H+.
Compare Run 1 & 3:
|Run 1 |Run 3 |
|Propanone Concentration = 0.43M |Propanone Concentration = 0.22M |
|Rate = 0.11 cm3s-1 |Rate = 0.07 cm3s-1 |
→When the concentration of propanone is halved, the rate of reaction also becomes halved which indicates that the reaction is first order with respect to CH3COCH3.
Compare Run 1 & 4:
|Run 1 |Run 4 |
|Iodine Concentration = 0.001M |Iodine Concentration = 0.00054M |
|Rate = 0.11cm3s-1 |Rate = 0.14cm3s-1 |
→When the concentration of iodine is halved, the rate of reaction remained the same which indicates that the reaction is zero order with respect to I2.
So, the rate law is given by:
Rate= k[ H+][CH3COCH3]
Rate constant, k = Rate [ H+][CH3COCH3]
| |Rate of reaction / dm3s-1 |[H+] / mol dm-3 |[CH3COCH] / mol dm-3 |k/ dm9 mol-2 s-1 |
|Run 1 |1.1x10-4 |1.08 |0.43 |4.4x10-5 |
|Run 2 |8.0x10-5 |0.54 |0.43 |6.4x10-5 |
|Run 3 |7.0x10-5 |1.08 |0.22 |1.4x10-5 |
|Run 4 |1.4x10-4 |1.08 |0.43 |5.6x10-5 |
Table 6 shows the rate constant for each run in experiment 2.
Average rate constant, k = 4.4x10-5+6.4x10-5+1.4x10-5+ 5.6x10-5 = 4.5x10-5 dm9 mol-2 s-1 4
→The overall order of the reaction is (1+1) = second order.
(ii) QUALITITATIVE
ANALYSIS
1. Iodine is yellowish-brown in colour. 2. Starch solution, hydrochloric acid and propanone are colourless. 3. Hydrochloric acid has a weak vinegar smell. 4. Iodine, starch solution and propanone are odourless. 5. In experiment 1, after iodine is added into the mixture of propanone and hydrochloric acid, the solution changes from colourless to yellowish-brown. 6. In the end of experiment 1, the yellowish-brown solution changes from yellowish-brown to colourless. 7. In experiment 2, after iodine is added into the mixture of propanone, hydrochloric acid and starch solution, the solution changes from colourless to dark blue. 8. In the end of experiment 2, the dark blue solution turns into light blue.
DISCUSSION:
The reaction between iodine and propanone follows the equation as shown below:
CH3COCH3 (aq) + I2 (aq) ( CH3COCH2I (aq) + H+(aq)+ I-(aq)
In both of the experiments, run 1 act as the guidance to calculate the order of reaction for iodine, propanone and H+. In both of the experiments, the rate of reaction is halved when the volume of HCl is halved provided that the volume of propanone and iodine are the same. This shows that the rate of reaction is first order with respect to H+. On the other hand, when the volume of propanone is halved provided that the volume of HCl and iodine are the same, the rate of reaction is also halved. This proved that the rate of reaction is first order with respect to propanone as well. Next, when the volume of iodine is halved provided that the volume of propanone and HCl are the same, the rate of reaction remains the same which proved that the rate of reaction is zero order with respect to iodine is zero order.
The experiment is a” bimolecular process”(Pearson Baccalaureate: Higher Level Chemistry pg226. Retrieved 4 March 2012). The rate of reaction depends on the concentration of both the propanone and hydrogen ion. The rate determining step for propanone and hydrogen ion is shown below:
[pic] DIAGRAM 1 : Reaction mechanism for iodine propanone reaction Cited from: (Nelson Thornes. Chemistry in Context. Pg24 Reaction rate. Retrived 4 March 2012).
Referring to diagram 1, step 1 is the rate-determining step which is the slowest step in the chemical reaction. This step is the step which affects the rate of reaction the most. From diagram 1 too, I notice that CH3C(OH+)CH3 and CH2=C(OH)CH3 are the two intermediate species because they are cancelled out in the reaction step. Hydrogen ion catalysed this reaction. Hydrogen ion is the catalyst because it exists both in the reactant for the first step and the product for the third step. This means that it does not involve in the chemical reaction which is a characteristic of a catalyst. Iodine functions as the indicator so that the observations are clearer. When iodine is added into the mixture of propanone and hydrochloric acid in experiment 1, the solution turns from colourless to yellowish-brown. This is because iodine is yellowish-brown in colour. On the other hand, when iodine is added into the mixture of propanone, starch solution and hydrochloric acid, the solution turns from colourless to dark blue. This is because of the presence of starch. When iodine is added to starch, dark blue precipitate will form.
In the presence of starch solution, the rate of reaction decreases (as shown in table 6) because starch solution dilutes the reactants thus the concentration of the reactants decreases. The rate expression of this experiment is as below:
Rate = k[ CH3COCH3][H+]
The overall order of reaction is the sum of the powers of concentration of HCl and propanone in the rate equation. In this experiment, the overall order of reaction is (1+1) = 2.
The unit of rate constant can be deduced using the rate law equation:
Rate = k [H+] [CH3COCH3]
` dm3 s-1 = k (mol dm-3) ( mol dm-3) k = dm3 s-1 (mol2 dm-6) = dm9 mol-2 s-1
The average rate constant for experiment 1 is 1.0x10-4 dm9 mol-2 s-1and the average rate constant for experiment 2 is 4.5x10-5 dm9 mol-2 s-1.
Throughout the experiment, there are a few limitations that I faced. These include systematic error and random error.
One of the systematic error in the experiment that my group members and I conducted is there are colour stains at the bottom of the beaker which cannot be removed thus it was hard to observe the colour change of iodine. At that moment, there were limited beakers for use. The suggestion for this error is to use a brand-new beaker which has a clear, clean base. The second systematic error is the amount of iodine, distilled water, propanone, hydrochloric acid and starch solution were not measured accurately as they were measured with different apparatus with different uncertainties. Thus, we should use the same apparatus of more sensitive scale such as the burette to measure all the solution mentioned to get a very accurate measurement.
The random error that I faced in this experiment is the swirling time for the mixture of hydrochloric acid, propanone, distilled water, iodine and starch solution for each run are always different thus the recorded time for iodine to change to colourless in experiment 1 and to light blue in experiment 2 are not very accurate. Therefore, it is better for the same person to swirl the mixture with the same force in every experiment conducted. Next, the second random error is the colour change for the second experiment which is from dark blue to light blue cannot be observed accurately as we have different view of how a light blue colour should look like. In order to overcome this random error, I should choose one of my group members to observe the colour change for every run. It is also better to have a sample of the light blue solution at the side as a guide to detect the colour change. The third random error is we started the stopwatch at different time for each run which will influence the time taken for iodine to change colour. Thus, we should start the stopwatch once iodine is added to the mixture.
CONCLUSION:
The reaction of iodine is zero order, the reaction for H+ and CH3COCH3 is first order. Thus, the overall order of reaction between iodine and propanone is second order. The average rate constant for experiment 1 and 2 are 1.0x10-4 dm9 mol-2 s-1 and 4.5x10-5 dm9 mol-2 s-1 respectively. The rate law in this experiment is k[ CH3COCH3][H+].
References:
1. Pearson Baccalaureate Chemistry Higher Level.
2. http://books.google.com.my/books?id=1rZDv93KFWoC&pg=PT387&lpg=PT387&dq=mechanism+step+for+iodine+propanone+reaction&source=bl&ots=a0AoCGMHQz&sig=dPDmnANj0KTZRc3ny7cGQHE7R9I&hl=en&sa=X&ei=9TxTT46xKcvrrQe_npDPDQ&ved=0CGoQ6AEwCQ#
3. http://www.pearsonschoolsandfecolleges.co.uk/AssetsLibrary/SECTORS/Secondary/PDFs/Science/EdexcelScience/ALevelRevisionGuides/EdexcelA2ChemistryRG_9781846905964_pg8-17_web.pdf
[pic]
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R = Rate of reaction
V = Volume of iodine used
T = Time taken for colour of iodine to disappear
R = Uncertainty for rate of reaction • V = Uncertainty for volume of reactant • t = Uncertainty for time R = Rate of reaction V =Volume of reactant t = Time taken for the colour to disappear
STEP 1: CH3COCH3 + H+ CH3C(OH+)CH3 (Slow-rate determining step)
STEP 2: CH3C(OH+)CH3 ( CH2=C(OH)CH3 + H+ (Fast)
STEP 3: I2 + CH2=C(OH)CH3 ( CH3COCH2I + H+ + I- (Fast)
DATA COLLECTION AND PROCESSING :
Chemical Equation:
CH3COCH3 (aq) + I2 (aq) ( CH3COCH2I + H+ (aq) + I- (aq)
Rate of reaction:
R = V T
Concentration of reactant after dilution :
Volume of reactant x concentration of reactant
Total volume of mixture
Uncertainty for rate of reaction:
R =( V + t ) x R V t
Eg for Exp 1, run 1: R = ( 1 + 0.1 + 0.1 + 0.1 + 0.01 ) x 0.36 = +0.03cm3s-1 20 8 0 4 11.16
Experiment 1 (without starch solution) :
| …show more content…
|Run 1 |Run 2 |Run 3 |Run 4 |
|Volume of 2M HCl/cm3 (+1cm3) |20 |10 |20 |20 |
|Volume of 2M propanon/cm3 (+0.1cm3) |8 |8 |4 |8 |
|Volume of water/cm3 (+0.1cm3) |0 |10 |4 |2 |
|Volume of 0.01M iodine/cm3 (+0.1cm3) |4 |4 |4 |2 |
|Time for colour to disappear/s (+0.01s) |11.16 |26.63 |27.53 |6.45 |
|Rate(Vol. Of I2(aq)/Time)/cm3s-1 |0.36 (+0.03) |0.15 (+0.02) |0.15 (+0.02) |0.31 (+0.05) |
Table 5 shows the rate of reaction for each run without the presence of starch.
Experiment 2 (with starch solution):
| |Run 1 |Run 2 |Run 3 |Run 4 |
|Volume of 2M HCl/cm3 (+1cm3) |20 |10 |20 |20 |
|Volume of 2M propanone/cm3 (+0.1cm3) |8 |8 |4 |8 |
|Volume of water/cm3 (+0.1cm3) |0 |10 |4 |2 |
|Volume of 0.01M iodine/cm3 (+0.1cm3) |4 |4 |4 |2 |
|Volume of starch solution/cm3 (+0.1cm3) |5 |5 |5 |5 |
|Time for colour to disappear/s (+0.01s) |37.44 |50.53 |55.19 |14.56 |
|Rate(Vol.
Of I2(aq)/Time) /cm3s-1 |0.11 (+0.01) |0.08 (+0.01) |0.07 (+0.01) |0.14 (+0.03) |
Table 6 shows the rate of reaction for each run with the presence of starch.
Experiment 1 (without starch solution) :
| |Run 1 |Run 2 |Run 3 |Run 4 |
|Concentration of HCl/moldm-3 |1.25 |0.625 |1.25 |1.25 |
|Concentration of propanone/moldm-3 |0.5 |0.5 |0.25 |0.5 |
|Concentration of Iodine/moldm-3 |0.00125 |0.00125 |0.00125 |0.000625 |
Table 7 shows the concentration of reactant for experiment 1.
Experiment 2 (with starch solution) …show more content…
:
| |Run 1 |Run 2 |Run 3 |Run 4 |
|Concentration of HCl/moldm-3 |1.08 |0.54 |1.08 |1.08 |
|Concentration of propanone/moldm-3 |0.43 |0.43 |0.22 |0.43 |
|Concentration of Iodine/moldm-3 |0.001 |0.001 |0.001 |0.00054 |
Table 8 shows the concentration of reactant for experiment 2.
FROM EXPERIMENT 1:
Compare Run 1 & 2:
|Run 1 |Run 2 |
|H+ Concentration = 1.25M |H+ Concentration = 0.625M |
|Rate = 0.36cm3s-1 |Rate = 0.15cm3s-1 |
→When the concentration of H+ is halved, the rate of reaction also becomes halved which indicates that the reaction is first order with respect to H+.
Compare Run 1 & 3:
|Run 1 |Run 3 |
|Propanone Concentration = 0.5M |Propanone Concentration = 0.25M |
|Rate = 0.36 cm3s-1 |Rate = 0.15 cm3s-1 |
→When the concentration of propanone is halved, the rate of reaction also becomes halved which indicates that the reaction is first order with respect to CH3COCH3.
Compare Run 1 & 4:
|Run 1 |Run 4 |
|Iodine Concentration = 0.00125M |Iodine Concentration = 0.000625M |
|Rate = 0.36cm3s-1 |Rate = 0.31cm3s-1 |
→When the concentration of iodine is halved, the rate of reaction remained the same which indicates that the reaction is zero order with respect to I2.
So, the rate law is given by:
Rate= k[ H+][CH3COCH3]
Rate constant, k = Rate [ H+][CH3COCH3]
| |Rate of reaction / dm3s-1 |[H+] / mol dm-3 |[CH3COCH] / mol dm-3 |k/ dm9 mol-2 s-1 |
|Run 1 |3.6x10-4 |1.25 |0.50 |1.4x10-4 |
|Run 2 |1.5x10-4 |0.625 |0.50 |1.2x10-4 |
|Run 3 |1.5x10-4 |1.25 |0.25 |3.0x10-5 |
|Run 4 |3.1x10-4 |1.25 |0.50 |1.2x10-4 |
Table 5 shows the rate constant for each run in experiment 1.
Average rate constant, k = 1.4x10-4+1.2x10-4+3.0x10-5+1.2x10-4 = 1.0x10-4 dm9 mol-2 s-1 4
→The overall order of the reaction is (1+1) = second order.
FROM EXPERIMENT 2 :
Compare Run 1 & 2:
|Run 1 |Run 2 |
|H+ Concentration = 1.08M |H+ Concentration = 0.54M |
|Rate = 0.11cm3s-1 |Rate = 0.08cm3s-1 |
→When the concentration of H+ is halved, the rate of reaction also becomes halved which indicates that the reaction is first order with respect to H+.
Compare Run 1 & 3:
|Run 1 |Run 3 |
|Propanone Concentration = 0.43M |Propanone Concentration = 0.22M |
|Rate = 0.11 cm3s-1 |Rate = 0.07 cm3s-1 |
→When the concentration of propanone is halved, the rate of reaction also becomes halved which indicates that the reaction is first order with respect to CH3COCH3.
Compare Run 1 & 4:
|Run 1 |Run 4 |
|Iodine Concentration = 0.001M |Iodine Concentration = 0.00054M |
|Rate = 0.11cm3s-1 |Rate = 0.14cm3s-1 |
→When the concentration of iodine is halved, the rate of reaction remained the same which indicates that the reaction is zero order with respect to I2.
So, the rate law is given by:
Rate= k[ H+][CH3COCH3]
Rate constant, k = Rate [ H+][CH3COCH3]
| |Rate of reaction / dm3s-1 |[H+] / mol dm-3 |[CH3COCH] / mol dm-3 |k/ dm9 mol-2 s-1 |
|Run 1 |1.1x10-4 |1.08 |0.43 |4.4x10-5 |
|Run 2 |8.0x10-5 |0.54 |0.43 |6.4x10-5 |
|Run 3 |7.0x10-5 |1.08 |0.22 |1.4x10-5 |
|Run 4 |1.4x10-4 |1.08 |0.43 |5.6x10-5 |
Table 6 shows the rate constant for each run in experiment 2.
Average rate constant, k = 4.4x10-5+6.4x10-5+1.4x10-5+ 5.6x10-5 = 4.5x10-5 dm9 mol-2 s-1 4
→The overall order of the reaction is (1+1) = second order.
(ii) QUALITITATIVE
ANALYSIS
1. Iodine is yellowish-brown in colour. 2. Starch solution, hydrochloric acid and propanone are colourless. 3. Hydrochloric acid has a weak vinegar smell. 4. Iodine, starch solution and propanone are odourless. 5. In experiment 1, after iodine is added into the mixture of propanone and hydrochloric acid, the solution changes from colourless to yellowish-brown. 6. In the end of experiment 1, the yellowish-brown solution changes from yellowish-brown to colourless. 7. In experiment 2, after iodine is added into the mixture of propanone, hydrochloric acid and starch solution, the solution changes from colourless to dark blue. 8. In the end of experiment 2, the dark blue solution turns into light blue.
DISCUSSION:
The reaction between iodine and propanone follows the equation as shown below:
CH3COCH3 (aq) + I2 (aq) ( CH3COCH2I (aq) + H+(aq)+ I-(aq)
In both of the experiments, run 1 act as the guidance to calculate the order of reaction for iodine, propanone and H+. In both of the experiments, the rate of reaction is halved when the volume of HCl is halved provided that the volume of propanone and iodine are the same. This shows that the rate of reaction is first order with respect to H+. On the other hand, when the volume of propanone is halved provided that the volume of HCl and iodine are the same, the rate of reaction is also halved. This proved that the rate of reaction is first order with respect to propanone as well. Next, when the volume of iodine is halved provided that the volume of propanone and HCl are the same, the rate of reaction remains the same which proved that the rate of reaction is zero order with respect to iodine is zero order.
The experiment is a” bimolecular process”(Pearson Baccalaureate: Higher Level Chemistry pg226. Retrieved 4 March 2012). The rate of reaction depends on the concentration of both the propanone and hydrogen ion. The rate determining step for propanone and hydrogen ion is shown below:
[pic] DIAGRAM 1 : Reaction mechanism for iodine propanone reaction Cited from: (Nelson Thornes. Chemistry in Context. Pg24 Reaction rate. Retrived 4 March 2012).
Referring to diagram 1, step 1 is the rate-determining step which is the slowest step in the chemical reaction. This step is the step which affects the rate of reaction the most. From diagram 1 too, I notice that CH3C(OH+)CH3 and CH2=C(OH)CH3 are the two intermediate species because they are cancelled out in the reaction step. Hydrogen ion catalysed this reaction. Hydrogen ion is the catalyst because it exists both in the reactant for the first step and the product for the third step. This means that it does not involve in the chemical reaction which is a characteristic of a catalyst. Iodine functions as the indicator so that the observations are clearer. When iodine is added into the mixture of propanone and hydrochloric acid in experiment 1, the solution turns from colourless to yellowish-brown. This is because iodine is yellowish-brown in colour. On the other hand, when iodine is added into the mixture of propanone, starch solution and hydrochloric acid, the solution turns from colourless to dark blue. This is because of the presence of starch. When iodine is added to starch, dark blue precipitate will form.
In the presence of starch solution, the rate of reaction decreases (as shown in table 6) because starch solution dilutes the reactants thus the concentration of the reactants decreases. The rate expression of this experiment is as below:
Rate = k[ CH3COCH3][H+]
The overall order of reaction is the sum of the powers of concentration of HCl and propanone in the rate equation. In this experiment, the overall order of reaction is (1+1) = 2.
The unit of rate constant can be deduced using the rate law equation:
Rate = k [H+] [CH3COCH3]
` dm3 s-1 = k (mol dm-3) ( mol dm-3) k = dm3 s-1 (mol2 dm-6) = dm9 mol-2 s-1
The average rate constant for experiment 1 is 1.0x10-4 dm9 mol-2 s-1and the average rate constant for experiment 2 is 4.5x10-5 dm9 mol-2 s-1.
Throughout the experiment, there are a few limitations that I faced. These include systematic error and random error.
One of the systematic error in the experiment that my group members and I conducted is there are colour stains at the bottom of the beaker which cannot be removed thus it was hard to observe the colour change of iodine. At that moment, there were limited beakers for use. The suggestion for this error is to use a brand-new beaker which has a clear, clean base. The second systematic error is the amount of iodine, distilled water, propanone, hydrochloric acid and starch solution were not measured accurately as they were measured with different apparatus with different uncertainties. Thus, we should use the same apparatus of more sensitive scale such as the burette to measure all the solution mentioned to get a very accurate measurement.
The random error that I faced in this experiment is the swirling time for the mixture of hydrochloric acid, propanone, distilled water, iodine and starch solution for each run are always different thus the recorded time for iodine to change to colourless in experiment 1 and to light blue in experiment 2 are not very accurate. Therefore, it is better for the same person to swirl the mixture with the same force in every experiment conducted. Next, the second random error is the colour change for the second experiment which is from dark blue to light blue cannot be observed accurately as we have different view of how a light blue colour should look like. In order to overcome this random error, I should choose one of my group members to observe the colour change for every run. It is also better to have a sample of the light blue solution at the side as a guide to detect the colour change. The third random error is we started the stopwatch at different time for each run which will influence the time taken for iodine to change colour. Thus, we should start the stopwatch once iodine is added to the mixture.
CONCLUSION:
The reaction of iodine is zero order, the reaction for H+ and CH3COCH3 is first order. Thus, the overall order of reaction between iodine and propanone is second order. The average rate constant for experiment 1 and 2 are 1.0x10-4 dm9 mol-2 s-1 and 4.5x10-5 dm9 mol-2 s-1 respectively. The rate law in this experiment is k[ CH3COCH3][H+].
References:
1. Pearson Baccalaureate Chemistry Higher Level.
2. http://books.google.com.my/books?id=1rZDv93KFWoC&pg=PT387&lpg=PT387&dq=mechanism+step+for+iodine+propanone+reaction&source=bl&ots=a0AoCGMHQz&sig=dPDmnANj0KTZRc3ny7cGQHE7R9I&hl=en&sa=X&ei=9TxTT46xKcvrrQe_npDPDQ&ved=0CGoQ6AEwCQ#
3. http://www.pearsonschoolsandfecolleges.co.uk/AssetsLibrary/SECTORS/Secondary/PDFs/Science/EdexcelScience/ALevelRevisionGuides/EdexcelA2ChemistryRG_9781846905964_pg8-17_web.pdf
[pic]
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R = Rate of reaction
V = Volume of iodine used
T = Time taken for colour of iodine to disappear
R = Uncertainty for rate of reaction • V = Uncertainty for volume of reactant • t = Uncertainty for time R = Rate of reaction V =Volume of reactant t = Time taken for the colour to disappear
STEP 1: CH3COCH3 + H+ CH3C(OH+)CH3 (Slow-rate determining step)
STEP 2: CH3C(OH+)CH3 ( CH2=C(OH)CH3 + H+ (Fast)
STEP 3: I2 + CH2=C(OH)CH3 ( CH3COCH2I + H+ + I- (Fast)