Chemical Kinetics lab
Summary
"Iodine Clock" refers to a group of reactions which involve the mixing of two colorless solutions to produce a solution which remains colorless for a precise amount of time, then suddenly changes to a deep purple-blue color. The time is controlled by the temperature and/or the concentrations of the reactants.
The reactions involve the oxidation of iodide ion (I-) to dissolved iodine (I2) or tri- iodide ion (I3-). Either of these two species will combine with a starch indicator to produce the characteristic purple-blue color. This color is visible to the naked-eye when the concentration of iodine or tri-iodide ion exceeds 10-5 moles/liter.
The equation for the reaction being studied in this experiment is:
(#1) 2 H+(aq) + 2 I-(aq)+ H2O2(aq) → I2 + 2 H2O(l)
The rate of the reaction is dependent on the concentrations of the iodide ion and hydrogen peroxide. Therefore, we can write a generic equation for the rate law of such reaction, as follows:
(#2) Rate = k [I- ]a * [H2O2]b
Starch is added to the reaction mixture in order to track (measure) the time in which the reaction occurs. This will tell you when the iodine is formed, but you have no idea as to the amount of hydrogen peroxide that has reacted in a given amount of time, i.e., the rate of reaction.
NOTE #2: While performing experiments, we often have to rely on physical changes (like color changes) in order to identify when chemical changes are occurring (or have occurred). Starch forms a complex with iodine which has a very characteristic and strong navy-black color, hence why it is used in this experiment. It must be noted that starch does not act as a catalyst in this reaction. It is simply used as a "chemical trick" used in the laboratory to track reaction progress.
Alternatively, the thiosulfate ion, S2O3-2, is also added to the reaction mixture so that the following reaction will occur simultaneously with reaction (1).
(#3) I2 (aq)+ 2 S2O3-2(aq) → 2 I-(aq) + S4O6 -2(aq)
Thus, the iodine that is formed in reaction (#1) is immediately transformed into iodide ion and we do not see the blue-black color of the starch-iodide complex until all the thiosulfate ion has reacted with I2 (aq). When this occurs, we will then know the amount of hydrogen peroxide that has reacted and the time it took to react.
For example, if you start with 5.0 x 10-5 M of Na2S2O3 in the reaction flask and you see a color change after 20 seconds, then you can say that 2.5 x 10-5 M of I2 (aq) has been formed in 20 seconds and that 2.5 x 10-5 M of hydrogen peroxide has reacted in 20 seconds (Remember for every 2 moles of thiosulfate ion reacting, only 1 mole of iodine and, therefore, 1 mole of hydrogen peroxide are reacting. This is the stoichiometric ratio from the two reaction equations, (#3) and (#1)). The rate of disappearance of hydrogen peroxide is 2.5 x 10 -5 M divided by 20 seconds. This gives a value of 1.25 x 10 -6 M/s.
CollapseLearning Goals
Goals of this lab:
Part I:
- To determine orders of reaction and rate constant, for the reaction of iodide with hydrogen peroxide to produce iodine, and
- Using MATLAB to find order of the reactions, rate constant and the rate law.
Part II:
- To demonstrate the applications of Arrhenius Equation, and
- To determine the Activation Energy (Ea) using MATLAB and graphical analysis of results from Part II.
Context for Use
This is a lab activity for undergraduate students who are studying General Chemistry 2 course. This lab is about 3 hours. Half of the class possess technical skills and experience with MATLAB. Students needs to be skillful in using chemical equipment, data collection and working in a team. This activity is during third week of the course. Some students found easy and other students found it very hard.
Description and Teaching Materials
Experiment and Calculations Explained:
Part I:
In all the reactions the amount of thiosulfate ion used will be kept small relative to the amount of hydrogen peroxide. Therefore, the concentration of hydrogen peroxide will change very little, and you will essentially be working with initial concentrations of reactants and initial rates of reaction. You will do a total of three reactions at room temperature, one reaction at cold temperature (about 5 °C), and one reaction at hot temperature (about 40 °C, using a hot water bath).
The three reactions at room temperature are labeled #1, #2 , and #3. These make up Part I of the experiment. For reaction #1 and #2 the concentration of iodide is the same but the concentration of the other reactant, hydrogen peroxide, does vary. For reaction #1 and #3 the concentration of hydrogen peroxide is the same, but the concentration of the other reagent involved in the rate expression, the iodide ion, does vary.
The amounts, and therefore the concentrations, of the starch, sulfuric acid, and sodium thiosulfate are constant for each reaction combination. In the reaction where the concentration of hydrogen peroxide (a molecular compound) is varied, the difference in the total volume of solution is compensated by the equivalent amount of another molecular compound, water. In the reaction where the concentration of iodide ion is varied, the difference in total volume is compensated by the equivalent amount of another ion of the same charge, the chloride ion. Effectively the total volume of the reaction, and the total ionic strength of the substances present is maintained at a constant value.
After you have determined the reaction rate for each of the three reactions at room temperature you will substitute the values of the reaction rate, [I -], and [H2O2] for reaction #1 into equation (#2). Then divide this entire equation by substituting the values for the reaction rate, [I -], and [H2O2] for reaction #2 into equation (#2). This will give you a ratio of the reaction rates for reactions #1 and #2 on the left side; and a ratio of the products of k, [I -]a , and [H2O2]b for reactions #1 and #2 on the right side. The values of k, and [I -]a will cancel on the right side since they are the same for both reaction #1 and #2. This gives you a number on the left equal to
`([H_2 O_2]^b#)/([H_2O_2]^b##)`on the right. The ratio on the right can be rewritten as `(([H_2 O_2]#)/([H_2 O_2]##))^b`, which is a single number raised to the power b. The value of b can now be calculated.
Now repeat the calculations for the two reactions where [H2O2] is kept constant and determine the value for a. You can now calculate the value of k for each of the reactions since you have the values of a , b , [I -], [H2O2], and the reaction rate. The average of these three values is the value of k at room temperature.
Part II
The results of the three reactions run at different temperatures (cold, room temp and hot) make up Part II of the experiment. The mathematical relation, the Arrhenius equation, relates the rate constant, k, and the temperature in the following equation:
(#4) `ln(k)= -(E_a/R)(1/T) + ln (A)`
where Ea is the activation energy for the reaction, T is the temperature in K, k is the rate constant at T, and R is the ideal gas constant in units of J/mol·K and having a numerical value of 8.314 J/mol·K. The equation is comparable to that of a straight line: y = mx + b. If "ln k" is plotted against "1/T", then the slope of the resulting line will equal -Ea/R. The activation energy can be obtained by this method. See figure 1 below.
Kinetics lab (Microsoft Word 2007 (.docx) 104kB Oct12 23)
Teaching Notes and Tips
I am planning to use MATLAB to find order of this chemical reactions, rate constant and rate law. I am planning to learn from this workshop and integrate in this lab activity.
Assessment
Chlorine dioxide, ClO2, is a reddish-yellow gas that is soluble in water. In basic solution, it gives ClO3- and ClO2- ions:
2ClO2(aq) + 2 OH- (aq)→ ClO3- (aq) + ClO2-(aq) + H2O(l)
To obtain the rate law for this reaction, the following experiments were run and, for each, the initial rate of reaction of ClO2 was determined. Obtain the rate law and the value of the rate constant.
Initial Concentrations (M) Initial Rate (mol/L·s)
Experiment # ClO2 OH -
Exp. 1 0.060 0.030 0.0248
Exp. 2 0.020 0.030 0.00276
Exp. 3 0.020 0.090 0.00828
Nitrogen monoxide, NO, reacts with hydrogen to give nitrous oxide, N2O, and water.
[code math
]2 NO(g) + H2 (g) → N2O(g) + H2O(g)[end code]
If the reaction is first order towards NO and zeroth order towards H2,
Write the rate equation.
What would be effect of doubling the concentration of NO on the rate of the reaction?
What would be effect of doubling the concentration of H2 on the rate of the reaction?
If the reaction is second order towards NO and first order towards H2, what would be the effect of doubling the concentration of both reactants on the rate of the reaction?
References and Resources
https://lonestar.yuja.com/V/Video?v=5180109&node=23577654&a=492779562&autoplay=1
This is a lab prepared by LSC faculties