Lab Report on Kinetics of an Ionic Reaction

The lab report below was submitted as part of the coursework for CM2132 Physical Chemistry. Please do not plagiarise from it as plagiarism might land you into trouble with your university. Do note that my report is well-circulated online and many of my juniors have received soft copies of it. Hence, please exercise prudence while referring to it and, if necessary, cite this webpage.

1.      Aim
This experiment aims to (i) understand the order of reaction with respect to [Br-], [BrO3-] and [H+], (ii) evaluate the rate constant at 25 oC and (iii) observe the qualitative effects of varied ionic strength on the rate of reaction.
2.      Introduction
In an acidic solution, Br- and BrO- react to form Br2 according to the equation below;
5Br- + BrO3- + 6H+ 3H2O + 3Br2
The rate of this reaction is influenced by the ionic strength of the reaction medium and the concentration of the three ionic reactants.
In this experiment, the concentrations of the reactants are varied and their respective reaction rates, measured. The initial rate law is as follows,
rate = k[Br-]0a [BrO3-]0b [H+]0c
where k is the rate constant, while a, b, and c are the orders of reaction with respect to [Br-], [BrO3-] and [H+] respectively. The components of the rate law must be found by experimental means for it cannot be predicted from stoichiometric ratios of a balanced overall equation.
Initial rate method is used to determine the order of reaction with respect to the concentration of these ions while maintaining the ionic strength of reaction medium. The initial rate, R0 is found to follow the equation R0 = k [Br-]1[BrO3-]1[H+]2. The rate constant is then determined from the rate law. The effect of changing ionic strength on the reaction rate is investigated by comparing the rate between two reaction mediums of different ionic strength.
3.       Experimental procedures
Chemicals: 1.0M KBr, 0.2M KBrO3, 1.0M HNO3, 0.1M NaHCO3, 1M KI, 0.02174M Na2S2O3
Apparatus: Stoppered bottle or flask, pipette, burette
In this experiment, 9 sets of titrations are conducted. With the exception of solution 2, each set of solution are filled with the requisite amount of KBr, KBrO3 and NaNO3 (refer to Table 1) in a 50 ml volumetric flask before being diluted with water. Another solution containing 1M HNO3 diluted to 50 ml is also made.
Both solutions are added in a stoppered bottle and 10 ml of aliquots are withdrawn at 2, 4 and 7 min intervals after the start of reaction. To quench the reaction, these aliquots are added to 18 ml of solution containing 13 ml of NaHCO3 and 5 ml of KI. They are then titrated with Na2S2O3 with starch as indicator.
To determine the order of reaction with respect to the reactants, different volumes of the reactants are used while maintaining constant ionic strength. From solution 1 to 9, except solution 2, different volumes of KBr, KBrO3, HNO3 and NaNO3 are used and NaNO3 is replaced by Na2SO4 in solution 2. For solution 2, Na2SO4 is used instead of NaNO3. Since SO42- has a higher charge than NO3-, different volume of Na2SO4 has to be used in place of NaNO3 so that the ionic strength of reacting mixture remains the same.
The ionic strength for solution 1 is calculated to be
I = ½ Ʃ (K+ + Br- + K+ + BrO3- + Na+ + NO3- + H+ + NO3-)                                                                                          = ½ x ( 5/100 x 1 x 12 + 5/100 x 1 x 12 + 5/100 x 0.2 x 12 + 5/100 x 0.2 x 12 + 5/100 x 1 x 12 + 5/100 x 1    x 12 + 7.5/100 x 1 x 12 + 7.5/100 x 1 x 12)                                                                                                            = 0.185 M
In order to maintain constant ionic strength of reaction solutions to 0.185M,
In solution 2 for example, let v ml be the volume of Na2SO4 needed,
I = ½ Ʃ (K+ + Br- + K+ + BrO3- + 2Na+ + SO42- + H+ + NO3-)                                                                               0.185 = ½ x (5/100 x 1 x 12 + 5/100 x 1 x 12 + 5/100 x 0.2 x 12 + 5/100 x 0.2 x 12 + 2v/100 x 1 x 12 + v/100 x 1 x 22 + .5/100 x 1 x 12 + 7.5/100 x 1 x 12)                                                                                             v =10/ 6 = 1.67 ml 1.70 ml
Similar calculations are done to solutions 3 to 8. The volumes are calculated as shown below.
Table 1:
 Solution KBr (ml) KBrO3 (ml) HNO3 (ml) NaNO3 (ml) Na2SO­4 (ml) 1 5.0 5.0 7.5 5.0 - 2 5.0 5.0 7.5 - 1.70 3 3.0 5.0 7.5 7.0 - 4 1.0 5.0 7.5 9.0 - 5 5.0 3.0 7.5 5.4 - 6 5.0 1.0 7.5 5.8 - 7 5.0 5.0 5.0 7.5 - 8 5.0 5.0 3.0 9.5 - 9 5.0 5.0 7.5 15.0 -
Using the table above, solutions 1, 3 and 4 are compared to find the reaction order with respect to [Br-] as only [Br-] is varied while keeping the rest constant. Solutions 1, 5 and 6 are compared to find the reaction order of [BrO3-] and solutions 1, 7 and 8 are compared to find reaction order of [H+] for the same reasoning.
In solution 9, the amount of NaNO3 added was doubled than the amount added in to solution 1. The ionic strength will then be higher in solution 9 and thus the effect on the rate of reaction can be studied with different ionic strength.
4.      Results and calculation
Given:
[KBr] = 1.000M   [HNO3] = 1.007M   [KBrO3] = 0.2000M   [Na2SO4] = 1.000M
[NaNO3] = 1.000M   [Na2S2O3] = 0.02174M
Solution 1
 Time 1 min 55 sec 3 min 55 sec 6 min 55 sec Initial volume of Na2S2O3 / ml 0.00 0.00 0.00 Final volume of Na2S2O3 / ml 2.70 4.70 6.70 Volume of Na2S2O3 used / ml 2.70 4.70 6.70
Solution 2
 Time 2 min 4 min 7 min Initial volume of Na2S2O3 / ml 0.00 0.00 0.00 Final volume of Na2S2O3 / ml 2.20 3.90 5.70 Volume of Na2S2O3 used / ml 2.20 3.90 5.70
Solution 3
 Time 2 min 4 min 7 min Initial volume of Na2S2O3 / ml 0.00 0.00 0.00 Final volume of Na2S2O3 / ml 1.80 3.10 4.70 Volume of Na2S2O3 used / ml 1.80 3.10 4.70
Solution 4
 Time 2 min 4 min 7 min Initial volume of Na2S2O3 / ml 0.00 0.00 0.00 Final volume of Na2S2O3 / ml 0.70 1.20 1.90 Volume of Na2S2O3 used / ml 0.70 1.20 1.90

Solution 5
 Time 2 min 4 min 7 min Initial volume of Na2S2O3 / ml 0.00 0.00 0.00 Final volume of Na2S2O3 / ml 1.90 3.30 4.70 Volume of Na2S2O3 used / ml 1.90 3.30 4.70

Solution 6
 Time 2 min 4 min 7 min Initial volume of Na2S2O3 / ml 0.00 0.00 0.00 Final volume of Na2S2O3 / ml 0.60 1.00 1.60 Volume of Na2S2O3 used / ml 0.60 1.00 1.60

Solution 7
 Time 2 min 4 min 7 min Initial volume of Na2S2O3 / ml 0.00 0.00 0.00 Final volume of Na2S2O3 / ml 1.30 2.40 3.60 Volume of Na2S2O3 used / ml 1.30 2.40 3.60

Solution 8
 Time 2 min 15 sec 4 min 7 min Initial volume of Na2S2O3 / ml 0.00 0.00 0.00 Final volume of Na2S2O3 / ml 0.70 1.00 1.60 Volume of Na2S2O3 used / ml 0.70 1.00 1.60

Solution 9
 Time 2 min 4 min 7 min Initial volume of Na2S2O3 / ml 0.00 0.00 0.00 Final volume of Na2S2O3 / ml 2.50 4.50 6.20 Volume of Na2S2O3 used / ml 2.50 4.50 6.20

Graph 1: Volume of Na2S2O3 used (ml) against time (s) in solution 1 From graph of volume of Na2S2O3 used (ml) against time (s) in solution 1 above,
V = -2 x 10-5 t2 + 0.025 t + 0.021
dV/dt = - 4 x 10-5 t + 0.025
When t=0, dV/dt = 0.025 ml/s
Br2 + 2I- à2Br- + I2
I2 + 2S2O32- à 2I- + S4O62-
2 mol of S2O32- = 1 mol of I2 produced = 1 mol of Br2 produced
Initial rate of production of S2O32- =0.025 x 10-3 x 0.02174 = 5.435 x 10-7 mol / s
Initial rate of production of Br2, R0 = 5.435 x 10-7 /2 / (10/1000) = 2.718 x 10-5 M/s
Similar calculation was done to get R0 of the other solutions using graph 2 to 9.
Varying Concentrations of Br-
In solution 1, [Br-]0 = 1.000 x 5 / 100 = 0.0500 M
In solutions 1, 3 and 4, concentration of Br- is varied while [BrO3-] and [H+] are kept constant and the following data was obtained:
Table 2:
 Solution [Br-]0 (M) log [Br- / M]0 R0 (M/s) log ( R0 / M s-1) 1 0.0500 -1.301 2.718 x 10-5 -4.566 3 0.0300 -1.523 1.631 x 10-5 -4.788 4 0.0100 -2.000 5.435 x 10-6 -5.265

Since [BrO3-] and [H+] are kept constant, initial rate can be written as, R0 = k [Br-]0a
log R0 = log k + a log [Br-]0
Plotting log R0 against log [Br-]0
From the graph of log (R0 / M s-1) against log [Br- / M]0 (graph 10),
y = x – 3.265
gradient of line = a = 1
Varying concentrations of BrO3-
In solution 1, [BrO3-]0 = 0.2000 x 5.0 / 100 = 0.0100 M
In solutions 1, 5 and 6, concentrations of BrO3- are varied and the following data is obtained:
Table 3
 Solution [BrO3-]0 (M) log [BrO3- / M]0 R0 (M/s) log ( R0 / M s-1) 1 0.0100 -2.000 2.718 x 10-5 -4.566 5 0.0060 -2.222 1.848 x 10-5 -4.733 6 0.0020 -2.699 4.348 x 10-6 -5.362
Initial rate, R0 = k[BrO3-]0b
log[R0] = log k + b log[BrO3-]0
From the graph of log R0 against log [BrO3-]0 (graph 11),  y = 1.167x – 2.193
Gradient of line = b = 1.167 1
Varying concentrations of H+
In solution 1, [H+]0 = 1.007 x 7.5 / 100 = 7.553 x 10-2 M
In solutions 1, 7 and 8, concentrations of H+ were varied and the following data was obtained:
Table 4:
 Solution [H+]0 (M) log [H+ / M]0 R0 (M/s) log ( R0 / M s-1) 1 0.07553 -1.122 2.718 x 10-5 -4.566 7 0.05035 -1.298 1.196 x 10-5 -4.922 8 0.03021 -1.520 5.435 x 10-6 -5.265
Initial rate, R0 = k[H+]0c
log[R0] = log k + c log[H+]0
From the graph of log R0 against log[H+]0 (graph 12),
Gradient of line = c = 1.747 2
Hence overall, the initial rate = k[Br-]1[BrO3-]1[H+]2
Order of reaction, n = 1 +1 + 2 = 4

Rate constant
Rate Constant, k = Initial Rate / ( [Br-]1[BrO3-]1[H+]2 )
In solution 1, k = 2.718 x 10-5 / ( 0.050 x 0.010 x 0.075532) =11.54
Considering only solutions 1 to 8, except 2, where ionic strength is constant, the following data was obtained.
Table 5:
 Solution Initial Rate, R0 ( M/s) [Br-]0 (M) [BrO3-]0 (M) [H+]0 (M) k (L3 mol-3 s-1) 1 2.718 x 10-5 0.050 0.010 0.07553 9.53 3 1.631 x 10-5 0.030 0.010 0.07553 9.53 4 5.435 x 10-6 0.010 0.010 0.07553 9.53 5 1.848 x 10-5 0.050 0.006 0.07553 10.8 6 4.348 x 10-6 0.050 0.002 0.07553 7.62 7 1.196 x 10-5 0.050 0.010 0.05035 9.43 8 5.435 x 10-6 0.050 0.010 0.03021 11.9

Average k= 9.76 L3 mol-3 s-1
Varying ionic strength
In solutions 1 and 9, concentrations of Br-, BrO3- and H+ are kept constant while the ionic strength is increased in solution 9 by using 15.0 ml of NaNO3. As the concentrations of the other ions remain constant in the solutions, we only consider the ionic strength of NaNO3.
Concentration of NaNO3 in solution 1 = 5.0/100 x 1 = 0.050M
Ionic strength of NaNO3 in solution 1 = ½ [(0.05 x 12) + (0.05 x 12)] = 0.050 M
Concentration of NaNO3 in solution 9 = 15.0/100 x 1 = 0.150M
Ionic strength of NaNO3 in solution 9 = ½ [(0.15 x 12) + (0.15 x 12)] = 0.150 M
From the data obtained earlier,
Table 6:
 Solution [NaNO3] (M) Ionic Strength (M) R0 (M/s) 1 0.050 0.050 2.718 x 10-5 9 0.150 0.150 2.500 x 10-6

This shows that the reaction proceeds at a slower rate in the solution of higher ionic strength.

5.      Discussion
Initial rate method
The initial rate method is employed to determine the order of reaction with respect to the reactants. This method involves measuring the reaction rate at very short times before any significant changes in concentration occur. By varying the concentration of one particular reactant only, the order of reaction with respect to that reactant can be determined. For example, only the concentration of Br- is varied in solutions 1, 3 and 4 (Table 1), while the concentrations of other reagents remain constant.
The effect of changing the concentration of Br- is correlated to the change in initial rate by the initial rate equation, R0 = k [Br-]0a.  The graph log R0 = log k + a log [Br-]0 is plotted and the gradient, a, is representative of the order of reaction with respect to [Br-].
Reaction mechanism
The reaction rate equation is experimentally determined to be = k [Br-] [BrO3-] [H+]2. Therefore, the overall order of reaction is 4. This suggests that there are 4 ions reacting directly in the rate determining step. However, this is unlikely for several reasons: for a reaction to occur, the particles must have sufficient energy and collide in a proper orientation at the same time.
It is improbable for the rate-determining step to be a reaction between 4 particles – this requires the simultaneous collision of four sufficiently energetic particles in the proper orientation. Chances of such a phenomenon are slim. Hence, a rate equation can only suggest the molecularity and mechanism of a rate-determining step, but not prove it absolutely. By this reasoning, the reaction must have been a multi-step mechanism with preliminary equilibria forming intermediates which then react.
A mechanism is hence suggested:
 H+ + Br- → HBr Step 1 - fast H+ + BrO3- → HBrO3 Step 2 - fast HBr + HBrO3 → HBrO + HBrO2 Step 3 - slow
Oppositely charged ions react together to form intermediates in Steps 1 and 2. The two intermediates (produced from 4 reacting particles) then further react in a bimolecular rate-determining step (Step 3). This accounts for why the rate equation has an overall order of 4.
 HBrO2 + HBr → HBrO Step 4 -fast HBrO + HBr → H2O + Br2 Step 5 – fast
Eventually, the HBrO and HBrO2, produced in Step 3, reacts with more HBr to form water and bromine.
Checking for consistency in the rate equation via the rate-determining step, also known as the elementary step,   This mechanism, like all other mechanisms, is not conclusively proven. It is a suggestion based on logical reasoning of empirical observations.
Effect of Ionic strength
The ionic strength of the reacting solution must be kept constant as it influences the rate of reaction. To ensure so, the volumes of KBr, KBrO3, HNO3 and NaNO3 used have been pre-calculated. To understand the effects of ionic strength on the rate of reaction, the reaction rates of solutions 1 and 9 are compared (Table 6). In solution 1, where there is a lower relative ionic strength, the initial rate of solution 1 is 2.718 x 10-5 M/s. This is greater than the initial rate in solution 9 of 2.500 x 10-5M/s.
This observation may be explained qualitatively in terms of interactions between the reactants and activated complex and the ionic atmospheres of oppositely charged ions which surround them in solution. Three cases may then be identified:
(a)   If the reactants share the same polarity, the activated complex will be more highly charged than the reactants. Increasing the ionic strength of the solution will therefore have a greater stabilizing effect upon the complex than on the reactants, and will thus increase the rate constant by lowering the effective activation energy.
(b)   If the reactants are oppositely charged, the charge on the activated complex will be lower than the charges on the reactants, and the rate constant will decrease with ionic strength.
(c)    If one of the reactants is uncharged, there will be no change in the rate constant with ionic strength.

Effect of temperature
According to the Arrehenius equation, , the rate constant, k, is affected by temperature, where A is the collision frequency factor, and Ea the activation energy of the reaction. Generally, increasing temperature increases average speed of particles and therefore their collision frequency.
However, in this experiment, temperature is assumed to be 25oC throughout even though there may be variations. Since the temperature changes are not taken into account, the discrepancy in values of k is expected.
To obtain more accurate results, the experiment should be conducted in an environment whereby temperature fluctuations are minimized.
Possible sources of error
The main source of error in this experiment is the time error. 10.0 ml aliquots have to be drawn at time 2, 4 and 7 minutes after the start of reaction and quenched. The reaction mixture is quenched using potassium iodide which reacts with bromine produced from the reaction of Br- and BrO3-. The formation of iodine is detected by the colour change of starch indicator from blue black to colourless. However, it is challenging to stop the reaction at exactly 120s, 240s and 420s from the start of reaction as the reaction does not stop in the pipette and the long time taken for liquid to be transfer from pipette to the conical flask containing KI. The variation in time will affect to the amount of Na2S2O3 needed and in turn the initial rate, hence deviation in rate constant, k.

Precautions
In this experiment, bromine is a product. Inhalation of bromine may cause breathing difficulties, headaches and dizziness. Any skin contact may cause irritation and chemical burns. As such, care must be taken to avoid direct contact with the chemicals.
Also, all used chemicals were transferred into plastic waste containers, instead of being washed down sinks.
6.      Conclusion
With the ionic strength of reaction solutions kept constant, the order of reaction with respect to [H+] is found to be 2, [Br-] to be 1 and [BrO3-] to be 1. From the rate law, R0 = k [Br-]1[BrO3-]1[H+]2, the rate constant k is calculated to be 9.76 L3 mol-3 s-1. Higher ionic strength is found to lead to a slower rate of reaction.
7.      References
Chemistry, Biology and Pharmacy Information Center. Article retrieved on 4 March 2012: <http://www.infochembio.ethz.ch/links/en/physchem_kinetik_lehr.html>
D.A. Skoog, D.M. West, F.J. Holler, S.R. Crouch, “Fundamentals of Analytical Chemistry”, 8th edition, Thomson Brooks/Cole, USA, 2004
M. C. Gupta, “Statistical Thermodynamics”, New Age International, 2007
P Atkins, T Overton, J Rourke, M Weller, F Armstrong, “Inorganic Chemistry” 4th edition, Oxford, 2006

8.      Exercises

1.    Why does changing the ionic strength of the solution affect reaction rate?

Ionic strength is a measure of the concentration and charge of the ions in the solution. Increasing the ionic strength of a solution will increase its ionic atmosphere, hence providing a greater stabilizing effect for species which are more charged.
If the reactants have the same charge, the activated complex will be more highly charged than the reactants. Increasing the ionic strength of the solution causes a greater stabilising affect on the complex than on the reactants. Thus, the rate constant increases by lowering the effective activation energy.
Conversely, in a solution with oppositely charged ions, cations are surrounded by anions and anions are surrounded by cations, therefore decreasing the activities of the reactants and the overall reaction rate. This effect can be observed from comparing experiments 1 and 9; in experiment 9, an increase in the concentration of NaNO3 means that higher amounts of Na+ and NO3- ions are present in the reaction mixture. The reacting ions’ activities will be lowered and, as a corollary, the rate of reaction is reduced as well.
2.    Consider the equilibrium: CH3COOH      CH3COO-   +   H+. Which substance must you add to produce more CH3COOH?
At pH = 5, what is the prevalent species in solution? What is prevalent at pH = 2?

According to Le Chatelier’s Principle (LCP), a reaction’s equilibrium position shifts to counteract a change imposed, thereby establishing a new equilibrium. To produce more CH3COOH, more H+ and/or CH3COO- ions must be added into the solution. When either aforementioned species are added, the equilibrium shifts to the left in order to counteract the imposed change in concentrations, thereby increasing the concentration of CH3COOH.

According to literature,
pKa of ethanoic acid = 4.75.

By Henderson-Hasselbalch equation, pH = pKa + lg At pH 5, the solution has a pH higher than that its pKa. By Henderson-Hasselbach equation, = 1.78 which suggests that CH3COO- dominates. By LCP, at a higher pH, the concentration of H+ is lower, the equilibrium therefore shifts right, causing more CH3COO- to be formed.

At pH 5, the solution has a pH higher than that its pKa. By Henderson-Hasselbach equation, = 0.00178 which suggests that CH3COOH dominates. At pH 2, the [H+] in the solution is high. In the presence of high concentration of H+, by LCP, the equilibrium will be shift to the left, causing more CH3COOH to be produced. Therefore, the prevalent species in the solution is CH3COOH.
9.      Graphs and Datasheet
Graph 2: Volume of Na2S2O3 used (ml) against time (s) in solution 2 Graph 3: Volume of Na2S2O3 used (ml) against time (s) in solution 3 Graph 4: Volume of Na2S2O3 used (ml) against time (s) in solution 4 Graph 5: Volume of Na2S2O3 used (ml) against time (s) in solution 5 Graph 6: Volume of Na2S2O3 used (ml) against time (s) in solution 6 Graph 7: Volume of Na2S2O3 used (ml) against time (s) in solution 7 Graph 8: Volume of Na2S2O3 used (ml) against time (s) in solution 8 Graph 9: Volume of Na2S2O3 used (ml) against time (s) in solution 9 Graph 10: log ( R0 / M s-1) against log [Br- / M]0 Graph 11: log ( R0 / M s-1) against log [BrO3- / M]0 Graph 12: log ( R0 / M s-1) against log [H+ / M]0 