Chemical Kinetics FR
Paper requirement for Chem 26.1...
CHEMICAL KINETICS: THE IODINE CLOCK REACTION J. FERNANDEZ INSTIUTE OF BIOLOGY, COLLEGE OF SCIENCE UNIVERSITY OF THE PHILIPPINES, DILIMAN, QUEZON CITY 1101, PHILIPPINES DATE SUBMITTED: DECEMBER 13, 2013 DATE PERFORMED: DECEMBER 6, 2013 ABSTRACT Chemical kinetics determines the rate of a certain reaction. Rate of reaction can be expressed as the ratio of the change in concentration of a reactant/product over the change in time. There are many factors that affect the rate of reaction and three of these factors were investigated in this paper. Different runs of the used mixture were subjected to varying parameters. The effect of concentration was first examined. Based on the results, the higher the concentration, the immediate is the reaction. The effect of temperature was also observed and it was very evident in the results that rate and temperature are directly related. The presence of catalyst was the last factor to be studied wherein the change in rate was very evident. The presence of catalyst resulted to very fast reaction. The effect of these factors can be attributed to the changes that happen in the collision of reacting particles. The catalyst worked by lowering the activation energy which is the minimum energy required for a reaction to take place.
INTRODUCTION Chemical kinetics is the branch of chemistry that deals with how fast a reaction occurs. Unlike thermodynamics that just tells us if a reaction can occur spontaneously, chemical kinetics determines the reaction rates. The rate of reaction reflects the rate of change in the concentrations of the reactants and products of a reaction. For a hypothetical reaction below,
the rate of reaction can be written as:
The process by which a reaction occurs, called the reaction’s mechanism, can often be deduced by measuring how fast the reaction occurs. This is possible because a reaction’s rate depends on the rate of effective collisions between reacting species. Two theories account for how chemical reactions occur, the Collision Theory and the Transition State Theory. Based on the Collision Theory, three basic events must happen for a reaction to occur. The atoms, molecules, or ions involved must (1) collide, (2) collide with enough energy to break and form bonds, and (3) collide with proper orientation. The transition state theory, meanwhile, postulates that reactants form a high energy intermediate, the transition state, which then falls apart into products. For a reaction to
occur, reactants must acquire sufficient energy (activation energy, Ea) to form the transition state. Generally, the rate of reaction can be affected by factors such as nature of reactants, concentration of reactants, temperature of reaction, and the presence of a catalyst. In this specific experiment called the iodine clock reaction, only the last three factors were studied. Also, the study of chemical kinetics is of major importance in chemical research. Knowledge about it helps in fields like pyrotechnics and medicine especially in treatment of diabetes and cardiovascular disease. METHODOLOGY The experiment was divided into three parts. Part A studied the effect of changes in concentrations on the reaction rate, Part B the changes in temperature, and Part C the presence of a catalyst. For Part A, solutions of Beaker A (KI and KCl) and Beaker B (K2S2O8, K2SO4, Na2S2O3 and fresh starch) with varying concentrations were mixed together representing different runs. The reactions were timed until the mixture turned blue. For Part B, 2 sets of “Run 2” were prepared. One set was put into a hot water bath while the other in an ice bath. After reaching desired temperatures, contents of Beakers A and B were mixed. The reactions were timed until the mixture turned blue. For Part C, another set of “Run 2” was prepared. Content of Beaker A was mixed with B. This time, 4 drops of 1M CuSO4 was added to the mixture. The reaction was timed until the mixture turned blue. RESULTS AND DISCUSSION For the experiment, the balanced equation is
so the rate can be known from
2−¿ 2−¿ S2 O¿8 S O¿4 ¿ ¿ = = −∆ ¿ ∆¿ Rate=¿ ¿
∆[ I 2] ∆t
To study the effect of concentration, 5 different runs with varying volumes and molarity of reactants were prepared. Initial concentrations were computed based on the given data. Upon computation, similar values of initial concentrations were obtained for some of the set-ups. The rates were also estimated using 1/t. The values of t were obtained experimentally. Using all the data on the datasheet (See Appendix) and the computed initial rates, order of reaction with respect to each reactant was calculated. For thiosulfate, the ratio of Run 1 and Run 2 was used. Run 2 and Run 5 were used to get the order with respect to iodide. After substituting the values, it was determined that the reaction was 2nd order with respect to thiosulfate and 1st order with respect to iodide. This gives us an overall kinetic order of 3. The rate constant which is 75 was known through substitution of values. Therefore, the rate law is expressed as: Rate = 75 [S2O82-]2 [I-] Unfortunately though, the computed orders and rate law were a bit different from the supposed values. Theoretically, the overall kinetic order would only be 1. The reaction’s order with respect to thiosulfate should have been equal to 1 and to iodide equal to 0. The next factor that was considered was temperature. Temperature is a measure of heat, which is the average kinetic energy of an object’s molecules. An increase in the temperature would excite the molecules, making them move faster and hit each other more frequently. In line with this, introducing heat to a reaction will increase its reaction rate, while doing the reverse would slow it down. Just by observing the data in the datasheet, it can be observed that rate was fast at higher temperature and
slow at lower temperature. This relationship can be expressed through the Arrhenius equation:
implying the direct relationship of rate and temperature. Since the reaction is theoretically 1st order, it is safe to assume that 1/t is equal to the rate constant. Getting the natural logarithm of the equation, we get:
solution. Naturally, this reaction proceeds forward very slowly but due to the presence of the catalyst, the reaction went fast. Basically, the presence of CuSO4 produced a different path with a lower activation energy thus speeding up the reaction. Difference in the activation energies can be observed through the 2 graphical representations.
which could be compared to the line equation y=mx+b where the slope, m, is equivalent to –Ea/R. After plotting the values, a negative slope of the line would be evident. Generally, the graph of the relationship of temperature to the rate of reaction would look like this:
Figure 2. Uncatalyzed reaction
Figure 1. Arrhenius plot The equation of the line was calculated to derive the activation energy and the Arrhenius constant. The activation energy is the minimum energy required for a reaction to take place. Its sign is always positive since it is energy absorbed by the system. Part 3 of the experiment studied the effect of the presence of catalyst on the reaction rate. CuSO4 was the catalyst used. The reaction was done in a solution with starch where the iodine part reacts to form a blue
Figure 3. Catalyzed reaction Cu2+ was used as the metal catalyst and starch as the indicator. Though a catalyst may contribute to the rate of reaction, it is important to note that it is not used up
in the reaction, it will have the same initial and final concentration even after the reaction is completed.
SUMMARY AND CONCLUSIONS With all the data and information obtained, it is safe to assume that reaction rates depend on many factors. As observed in the experiment, rate of reaction is affected by the concentration of reactants, changes in temperature and the presence of catalyst. Generally, the higher the concentration, the immediate the reaction is. The higher the temperature, the faster the reaction occurs. The presence of catalyst significantly increases the rate of the reaction.
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0.02 ¿ x ¿ 0.02 ¿ x ¿ 0.08 ¿ y ¿ ¿ k¿ k¿ Rate 1 2.4 x 10−3 = =¿ Rate 2 1.6 x 10−3
0.02 ¿ ¿ 0.04 ¿ x ¿ y 0.04 ¿ ¿ 0.04 ¿ y ¿ ¿ k¿ k¿ Rate 2 1.6 x 1 0−3 = =¿ Rate 5 3.4 0 x 10−3 log 0.34 = x log 0.5 + y log 1
log 1.5= x log1 + y log 2
x = log 0.34/ log 0.5
y= log 1.5/ log 2
x = 1.55 ~ 2
y= 0.6 ~ 1
rate = k [S2O82-]2 [I-] k = rate/ [S2O82-]2 [I-] k = 2.4x10-3/ (0.02)2 (0.08) k = 75
Sample Calculations Run 1 [S2O82-] = 0.1M x 5 mL / 25 mL = 0.02 M [I-] = 0.2 M x 10 mL / 25 mL = 0.08 M
Rate = 1/t Y = A + Bx Y = 5.75 – 3657.1 x Ea = m x –R Ea = (-3657.1) x (8.314) Ea = 30.4 kJ/mol A = e5.75 A= 314.2