1. All the reactants were placed in mixture A in the order in which they appeared in Table 2 except the H2O2 in a 250 cm3 conical flask.
2. The temperature of the solution mixture was read and recorded.
3. Specific amount of standardized H2O2 solution (according to mixture 1 in Table 2) was pipetted to the conical flask and stir continuously.
4. The stop-watch was started when half of the H2O2 solution had been drained from the pipette.
5. The conical flask was swirled to mix the contents well.
6. Carefully watched the solution mixture for the sudden appearance of the blue color, and the timer was stopped when blue color appears.
7. The time was recorded for the reaction in Table 3. The solution was discarded and the conical flask was rinsed.
Results
Table 1: Standardization of H2O2 solution.
Average titre: 34.43cm3
˚
Tabble3: Reaction rate measurements
Question
1(a) Calculate the order of the reaction with respect to hydrogen peroxide, H2O2.
Ans:
Since
2S2O32-(aq) + I2(aq) → 2I-(aq) + S4O62-(aq)
The average titre of S2O32- is 34.43cm3
The no. of mole of S2O32- = 1/2 no. of mol of I2
The no. of mole of S2O32- = 0.05 mol dm-3 x (34.43/1000) dm-3
= 0.001722 mol
The no. of mol of I2 = 1/2 x 0.00172 mol
= 8.61x10-4 mol
Since
H2O2 (aq) + 2 I-(aq) + 2H+(aq)→I2 (aq) + 2H2O(l)
The no. of mol of I2 = The no. of mol of H2O2
= 8.61x10-4 mol
The conc. of H2O2 in 25cm3 = 8.61x10-4 mol / (25/1000) dm3
= 0.0344 mol dm-3
Let the exact conc. of H2O2 is M2
Since M1V1 = M2V2
0.0344 x 25 = M2 x 1
M2 = 0.0344 x 25 / 1
=0.861 mol dm-3
the concentration of H2O2 in the reaction mixture in A,
= 0.861 mol dm-3 x (10/1000) dm3 / [75+30+25+5+5+10)/1000] dm3
= 0.0574 mol dm-3
In B, the concentration of H2O2 in the reaction mixture
=0.861 mol dm-3 x (5/1000) dm3 / [80+30+25+5+5+5)/1000] dm3
=0.0287 mol dm-3
The order of reaction respect to hydrogen peroxide:
Since x = log(∆t2/ ∆t1) / log([H2O2]1/[H2O2]2)
x = log(355/ 172)/log[(0.0574)/(0.0287)]
x = 1.05 = 1
It is a first order respect to hydrogen peroxide, H2O2.
(b) Calculate the order of the reaction with respect to iodide, I--
Ans:
the concentration of I- in the reaction mixture in A
= 0.05 mol dm-3 x (25/1000) dm3 / [75+30+25+5+5+10)/1000] dm3
= 8.33x10-3mol dm-3
the concentration of I in the reaction mixture in C
=0.05 mol dm-3 x (50/1000) dm3 / [50+30+50+5+5+10)/1000] dm3
=1.67x10-2mol dm-3
The order of reaction respect to iodide:
Since x = log(∆t2/∆t1) / log([I-]1/[I-]2)
x = log(91/ 172)/log[(8.33x10-3)/(1.67x10-2)]
x = 0.915 = 1
It is first order respect to iodide, I-.
(c) Calculate the order of the reaction with respect to H+. Taking Ka for acetic acid as 1.8 ×10-5 M, calculate the H+ concentration from the below equation using the known concentrations of acetic acid and sodium acetate contained in reaction mixtures A and D. (In mixture D be sure that you account for the acetic acid contained in both the buffer and the added 0.3 M acetic acid.). Show your calculations of the H+ concentration.
Ka = [H+] [CH3COONa] / [CH3COOH]
Ans: In A, the concentration of acetic acid in the reaction mixture
=0.05 mol dm-3 x (30/1000) dm3 / [(75+30+25+5+5+10) /1000] dm3
=0.01 mol dm-3
In A, the concentration of sodium acetate in the reaction mixture
=0.05 mol dm-3 x (30/1000) dm3 / [(75+30+25+5+5+10) /1000] dm3
=0.01 mol dm-3
Since Ka = [H+] [CH3COONa] / [CH3COOH]
1.8x10-5 = [H+] (0.01) / (0.01)
[H+] = 1.8x10-5mol dm-3
In D, the concentration of acetic acid in the reaction mixture
No. of mol of acetic acid = 0.05 mol dm-3 x (30/1000) dm3
= 0.0015 mol
The no. of mol of acetic acid added = 0.3 mol dm-3 x (45/1000) dm3
= 0.0135mol
The concentration of acetic acid in the reaction mixture
= (0.0015 + 0.0135) mol / [(30+30+45+25+5+5+10) /1000 dm3]
= 0.1 mol dm-3
In D, the concentration of sodium acetate in the reaction mixture
= 0.05 mol dm-3 x (30/1000) dm3 / [(30+30+45+25+5+5+10) /1000 dm3]
= 0.01 mol dm-3
Since Ka = [H+] [CH3COONa] / [CH3COOH]
1.8x10-5 = [H+] (0.01) / (0.1)
[H+] = 1.8x10-4mol dm-3
The order of reaction respect to iodide:
Since x = log(∆t2/ ∆t1) / log([H+]1/[H+]2)
x = log(169/172)/log[(1.8x10-5)/( 1.8x10-4)]
x = 0.0076 = 0
It is zero order reaction respect to H+.
2(a) From your measurements, what are the reaction orders with respect to each of the three reactants H2O2, I- and H+?
Ans: It is zero order respect to H+, first order respect to iodide, I- , and first order respect to hydrogen peroxide, H2O2.
(b) What is the rate expression (rate law) for the uncatalyzed reaction? What is the total reaction order?
Ans:
The rate law is:
Rate = k[H2O2][I-] where k is rate constant
Therefore it is second order respect to the reaction.
(c) Is the rate law consistent with the idea that the mechanism of the reaction is
H2O2 (aq) + 2 I-(aq) + 2H+(aq)→ I2 (aq) + 2H2O(l) (all in one step)?
Explain why or why not.
Ans:
If the mechanism involves only one step only, then it means that this step is the rate determining step. The rate law is Rate = k[H2O2][I-]. The rate determining step should include only one hydrogen peroxide and one iodide molecule only, but the above reaction involve one hydrogen peroxide , two iodide and two hydrogen ion molecule.
Also, the hydrogen ion should not be present in the rate determining step as it is zero order and the rate law does not include it.
Actually, according to collision theory, it is not favour that 3 molecules collides together and proceed a reaction. Collision of two or one molecule is more favourable.
The more ideal mechanism should be:
H2O2(aq) + I-(aq) → intermedate1 Rate = k1[H2O2][I-]
I-(aq) + intermedate1→ intermedate2 + I2(aq) Rate = k2[I-][intermedate1]
Intermedate2 + H+(aq) → intermedate3 + H2O(aq) Rate = k3[intermedate2][H+]
Intermedate3 + H+(aq) → H2O(aq) Rate = k4[intermedate3][H+]
(d) The following reaction pathway has been proposed for the pH range 3.5 to 7:
H2O2 (aq) + I-(aq)-->OH-(aq) + HOI (aq)
H+(aq)+ OH-(aq) -->H2O(l)
HOI (aq) + I-(aq) + H+(aq)→I2 (aq) + H2O(l)
Add the equations above. Do they give the overall reaction? According to your results,
which of the steps would be the rate-determining step?
Ans: They can give the overall reaction. Since,
H2O2 (aq) + I-(aq)-->OH-(aq) + HOI (aq)
H+(aq)+ OH-(aq) -->H2O(l)
HOI (aq) + I-(aq) + H+(aq)→I2 (aq) + H2O(l)
Finally, It gives
H2O2 (aq) + 2 I-(aq) + 2H+(aq)→ I2 (aq) + 2 H2O(l)
According to my results, the rate law is Rate = k[H2O2][I-]
From the above mechanism, the rate determining step is:
H2O2 (aq) + I-(aq) -->OH-(aq) + HOI (aq)
Since the above step only involve one hydrogen peroxide and one iodide molecule which obey the rate law.
- Calculate the reaction rate ratio (rate E / rate A) for reaction mixtures E and A. When the temperature is increased, does the reaction rate decrease or increase? Explain your results briefly.
Ans: The reaction rate ratio (rate E / rate A) for reaction mixtures E and A is:
Rate = 1/ time
Rate E = 1/77 = 0.0130 Rate A = 1/172 = 0.00581
The reaction rate ratio (rate E / rate A) = 0.0130 : 0.00581 = 2.24 : 1
When the temperature increases, the reaction rate increases. As the more energy provide to the molecules, the speed of the particles increase, the kinetic energy of the particles increase, the more frequent and more violent for the molecules collision. and increase the chance of the effective collision. The rate is double when the temperature increases by 10.
4. Calculate the reaction rate ratio (rate F / rate A) for reaction mixtures F and A. When there is an addition of catalyst, does the reaction rate decrease or increase? Explain your results briefly.
Ans: In A, the concentration of Mo(VI) in the reaction mixture is zero. Since no catalyst is added in the reaction mixture.
In F, the concentration of Mo(VI) in the reaction mixture is:
= 0.01 mol dm-3 x (5/1000) dm3 / [(70+30+25+5+5+5+10)/1000 dm3]
= 3.33x10-4 mol dm-3
The reaction rate ratio (rate F/ rate A) for reaction mixtures F and A is:
Rate = 1/ time
Rate F= 1/56= 0.0179 Rate A = 1/172 = 0.00581
The reaction rate ratio (rate E / rate A) = 0.0179 : 0.00581 = 3 : 1
When there is an addition of catalyst, the reaction rate increases.
The catalyst we have used is homogenous catalyst as the whole reaction usually occurs in a single phase only. Catalyst provide another path of lower activation energy. The number of particles with kinetic energy is greater or equal to activation energy increase. The number of effective collisions increase, then the rate increase.
5. What is the use of Na2S2O3(aq) in Part I?
Ans:
The Na2S2O3 (aq) in Part I is used for determination the concentration of H2O2. Since
2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)
I2 is titrated with known volume and concentration of S2O32-. Then the no. of mole of I2 can be calculated by comparing the mole ratio.
Since
H2O2(aq) + 2I-(aq) + 2H+(aq) → I2(aq) + 2H2O(I )
The concentration of H2O2 can be calculated by comparing the mole ratio with I2. While the volume of H2O2 is known, then the concentration of H2O2 can be found.
6. Na2S2O3 (aq) in Part II take a very important role in this experiment, what is the use of this chemical?[Hints: 2S2O32-(aq) + I2(aq) → 2I-(aq) + S4O62-(aq) ; the use of Na2S2O3(aq) in Part II is related to a name called “clock reaction”
Ans: Na2S2O3 has two functions in part II.
Firstly, the thiosulphate reacts with the iodine as in the following equation:
2S2O32-(aq) + I2(aq) → 2I-(aq) + S4O62-(aq)
At the instant that all the thiosulphate has reacted, free iodine is produce in the solution and its presence is shown by appearance of blue black color of the iodine-starch complex. The thiosulphate ions therefore act as a “monitor” indicating the point at which a certain fixed amount concentration of iodine has been formed. The initial rates of these series of reactions are given by rate = concentration / time. As long as the quantity of thiosulphate ions added for each run is the same the formation of iodine is being monitored at a constant concentration. So long as concentration is constant, it needs not necessarily be quantified, and 1/time can be used as a measure of initial rate of reaction. For this reason the reaction is often referred to as an iodine ‘clock’ reaction, for which rate of reaction is proportional to 1/ time where time is taken to reach a specified stage.
Secondly, adding small but fixed amount of S2O42- is used to delay the time for the appearance of blue color. If no adding small but fixed amount of S2O42-, then the time for the appearance of blue color is so fast that we cannot start the stopwatch at the time that for the appearance of blue color.
Discussion
- In part I, it should have a large different and trial 2 is larger then trial 1.Since
2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)
The volume of S2O32- increase in trial 2, and the no. of mole of S2O32- increase, the no. of mole of iodine is also increase, then the concentration of iodine increase. The concentration of iodine increase is due to the iodide can be oxidized by oxygen which is promoted by acids.
2H2O2(aq) → 2H2O(I) + O2(g)
Hydrogen peroxide is unstable, then it decomposes to water and oxygen by time.
H2O2(aq) + 2I-(aq) + 2H+(aq) → I2(aq) + 2H2O(I )
In this reaction, there are acid, iodide and the oxygen which is decomposed by hydrogen peroxide. Then the concentration of iodine increase.
4I-(aq) + O2(g) + 4H+(aq) → 2I2(aq) + 2H2O(aq)
Therefore, the volume of sodium thiosulphate in trial 2 is larger then trial 1.
However, our results in trial 1 and trial 2 are nearly the same. Since the H2O2(aq) is taken from reagent in both trial 1 and trial 2. So, there is not enough time for H2O2(aq) oxidized.
The second factor is that the sodium thiosulphate is decomposed by acidic medium.
S2O32-(aq) + 2H+(aq) → S(S) + SO2(g) + H2O(I).
Therefore, the volume of sodium thiosulphate in trial 2 is larger then trial 1.
- A buffer is a solution which can resist the change in pH by dilution, addition of small amount of strong acid or strong base, the pH of a buffer does not change easily. A buffer can be prepared by mixing the following solutions with comparable concentrations:
weak acid + conjugate base or salt of the weak acid (acidic buffer),
eg. a solution containing1M CH3COOH & 1M CH3COONa, pH < 7.
weak base + conjugate or salt of the weak base (alkaline buffer),
eg. a solution containing 1M NH3 & 1M NH4CI, pH > 7.
In this experiment, the buffer used is acidic buffer by adding 0.05M CH3COOH and 0.05 CH3COONa.
- Since the color change from iodine to iodide cannot be used to accurate the end point (the change in color brown to yellow to colourless is difficult to observe), starch is used as the indicator.
Starch + iodine → blue complex
- The rate equation for a reaction may help to deduce the mechanism of the reaction. The order of a reaction is an important implication of the rate determining step. From the order, we can deduce the mechanism and the rate determining step. The orders are experimentally found, they are not necessarily equal to the coefficient of the reactant in the equation unless the reaction is a single reaction.
- A catalyst is a substance which alters the rate of a particular chemical reaction, but itself is chemically unchanged at the end of the reaction. A catalyst can change the rate of a reaction by providing an alternative pathway for the reaction, usually with a pathway of lower activation energy than that of the uncatalyzed reaction.
There are some improvements in this experiment.
First, hydrogen peroxide is unstable, it decomposes to water and oxygen by time. Therefore do the titration as quick as possible.
2H2O2(aq) → 2H2O(I) + O2(g)
Second, the concentration of iodine increase is due to the iodide can be oxidized by oxygen which promoted by acids. Therefore do the titration as quick as possible.
4I-(aq) + O2(g) + 4H+(aq) → 2I2(aq) + 2H2O(aq)
Third, as for the human error, the problem can be minimized by performing the titration by the same person. So, the reading can be taken by the same person and the color change can be observed by the same person.
Conclusion
In the experiment, the reaction was found to be zero order respect to (H+), it is first order respect to iodide, (I-) , it is first order respect to hydrogen peroxide, (H2O2). Hence the rate law is Rate = k[H2O2][I-]. The rate of reaction is increase when the temperature is increase and the rate is increase when a positive catalyst is added to the reaction.