The Rate Law background: For most chemical reactions, changing the concentrations of the reactants will change the rate of the reaction. The rate of a reaction is the change of concentration taking place per change in time. For the generic reaction: 2 A + B C we would measure the rate by measuring either the increase in the concentration of the product, [C], with a change in time, t, or by measuring the decrease in the concentration of one of the reactants. In this example, there are two moles of substance A reacting with one mole of substance B. To account for the different number of moles of A and B reacting, we would use [A] to show the decrease in 2 the concentration of substance A, and [B] to show the decrease in the concentration of substance B. So, the rate of our generic reaction could be shown in any one of the following equivalent forms: rate = [A] = [B] = [C]. 2 t t t Rate has the units: moles per liter per time. It might seem intuitively obvious to you that if we increase the amount of substance A or substance B present, the reaction will go at a higher rate. That is to say, the more of substance A or substance B that is present, the faster substance C gets produced. Indeed, this is usually the case. The quantitative relationship between the concentrations of substances in a reaction, and the rate of a reaction, is called the rate law for a reaction. If the generic reaction used here had a typical and simple rate law, it would be: rate = k[a] x [B] y k is called the rate constant. Its value depends on the nature of the substances and the temperature. Each concentration is raised to a power. The value of the power is called the order of the reaction for each reactant. For example, if x and y were found to each equal 1, then the rate law for the reaction would be: rate = k[a][b]. Both A and B would be first order reactants. The overall rate order of the reaction would be 2 (the sum of the reactant orders). The order of a reaction can only be determined by experiment. A balanced equation for a reaction gives no clues to the rate law for the reaction. You will study the reaction (net ionic reaction shown): 6 I (aq) + Br (aq) + 6 H + (aq) 3 I 2 (aq) + Br (aq) + 3 H 2 O You will determine the rate law for the reaction, and in a following experiment, you will study the effect of temperature on the rate constant. Look carefully at the reaction. You see that iodide ions on the left are oxidized to iodine molecules on the right. Also notice that the bromine in Br, the bromate ion, is reduced to Br during the reaction. This reaction could be described in the following way: Bromine atoms in a plus five oxidation state pick electrons off of six iodide ions in an acidic solution. The bromines end up with a minus 1 state, the oxygens that were attached to them attach to hydrogen ions, ending up in water molecules, and the iodines end up as aqueous iodine molecules. How do you suppose this all takes place? Do 13 particles (six I - s, one Br, six H + s) all somehow collide, hold on to each other, and react? Or does the reaction take place in a series of collisions, each one involving two or three particles, each one resulting in part of the reaction taking place? This is one of the purposes behind the study of rate laws establishing the sequence of events occurring during the transformation of reactants into products. This sequence of events is called a reaction mechanism. The reaction you are studying has the rate law: rate = k[i ] x [Br ] y [H + ] z. (eqn 1) You will experimentally determine the values of x, y, and z. Some reactions may be zero order with respect to a particular reactant, which means that the reactant will not appear in the rate law expression. Prior experimental work has already been done to present the rate law in this form. The initial rate method will be used to determine the order of each reactant (the values of x, y, and z). In this method, the reaction is clocked as it proceeds for only a small portion of the total time it would take to reach completion. You will repeatedly time the reaction during the first 5% or so of the reactants being used up. Each time you run the reaction, you will double one, then another, then the third reactant s concentration compared to the original set of concentrations. 1
If you divide the rate of the reaction with the original set of concentrations into the rate of the reaction with the concentration of one of the reactants doubled, you can calculate how the change in concentration of that one reactant affected the rate of the reaction. That is to say, you can determine the order for that one reactant. If, for example, you ran the reaction twice, and everything was the same in the two runs except that you doubled the concentration of I, and the rate doubled after you doubled the [I ], then you could conclude that the value of x in equation 1 is 1. If the rate quadrupled after you doubled the [I ], then the value of x is 2. And if the rate went up eight-fold, the value of x is 3 (2 1 = 2, 2 2 = 4, 2 3 = 8). Orders do not have to be whole numbers, although in the reaction you are running, they are small, whole numbers. The timing method you will use is based on the properties of one of the products of the reaction, the I 2 molecule. Two reactions that the I 2 molecule can undergo in this experiment are: A fast reaction: 1) I 2 (aq) + starch deep blue complex; And a very fast reaction: 2) I 2 (aq) + 2 (aq) 2I (aq) + S 4 O 6 (aq) Given the quantity of O 3 used (0.000200M), equation 1 on page 1 becomes: 3.33x10 5 M = k[i ] x [BrO t 3 ] y [H + ] z (eqn 2) The change in the concentration of O 3 is a minus value, since it goes down from a 0.000200M to 0M. That is why the 3.33 x 10-5 is a positive rather than a negative value. The first run you do will be compared to each of the next three runs, so take extra care on run one. After the first four runs, you will do a fifth, but only after calculating the values of x, y, and z, and then calculating the average value of k for the reaction. The fifth run will test your measuring and calculating abilities. For the quantities of materials mixed, you are asked to predict the time the reaction should take. The closeness of your prediction and the actual time of reaction may result in bonus points added to this lab grade, depending on the instructor. The experiment will be run at room temperature, which we will assume stays constant. Record the temperature at the beginning and at the end of the experiment to see how good an assumption this is. If there is any (thiosulfate ion) present, I 2 will react with it rather than react with starch. Every time you run the reaction in this experiment, you will add some starch and enough Na 2 solution to give a [ ] of 0.000200M. The presence of starch and of does not interfere with any of the reactants used, only with the I 2. The rate of the chemical reaction you are studying could be given in any of the following equivalent ways, by measuring either the rate of disappearance of reactants or the rate of appearance of products: or rate = [I ] 6 t ] = [Br t rate = [I ] 2 = [Br ] 3 t t = [H + ] 6 t = [H 2 O] 3 t Reaction rate experiments depend upon the ability to measure concentrations. The measuring technique employed in this experiment depends on iodine, so: rate = [I 2 ] 3 t. Because the iodine will be reacting with ions: rate = [S O 2 3 6 t 2 ] 2
experiment Work in pairs. Supplies: 5-10 ml graduated cylinders (3 from cart) 1 stopwatch (from cart) 5 transfer pipets 1-125ml, 1-250 ml flask 1 thermometer 80 ml of 0.0100M KI, 70 ml of 0.00100M Na 2, 80 ml of 0.0400M KBr 85 ml of 0.100M HCl (Use the smaller beakers from the lab drawers. Estimate volumes from the markings on the beakers. Do not waste reagents.) Bring the reagents to your work area in clean, labeled containers. Put about 50 ml of deionized water in another beaker. One of the graduated cylinders and one of the transfer pipets will be used for each reagent. Label these as well. Make sure they are clean and dry to start. If necessary, rinse the cylinder and pipet with a small amount of its reagent. The amount of reagent the directions give is enough for an extra run, if necessary, as well as a few extra milliliters for rinsing. General mixing information: You will run the reaction five times, four at the beginning of the experiment and one after calculations. You will use 10.0 ml of the Na 2 solution for each run. Use the quantities of each of the other reagents given in the table below. Take care in measuring the volumes. After pouring each solution into the graduated cylinder, use the transfer pipet to adjust the bottom of the meniscus right to the line. Place the KI, the Na 2, and any water into the 250 ml flask. Place the KBr, the HCl, and 3 to 4 drops of starch solution into the 125 ml flask. Pour the contents of the 125 ml flask into the 250 ml flask, starting the stopwatch at the moment of transfer. Swirl the mixture for 10 seconds or so. Set the flask down on a piece of white paper and watch for the first appearance of a blue color. When the color first appears, stop the stopwatch and record the time. Pour the mix into the sink. Rinse the flasks well with deionized water, allow them to drip-drain, then reload them for the next run. Do not dry the inside of Initial Room Temperature R eaction 250 ml flask contains 10.0 ml Na 2, plus: 125 ml flask ( + 3-4 drops starch) Mixture 0.0100M K I (ml) H 2 O (ml) 0.0400M K B r (ml) 0.100M H C l (ml) 1 10.0 10.0 10.0 10.0 2 20.0 0 10.0 10.0 3 10.0 0 20.0 10.0 4 10.0 0 10.0 20.0 and after calculations are finished, 5 9.0 10.0 6.0 15.0 Room Temperature at End of Experiment Each mix has a total volume of 50.0 ml (including the Na 2 ). This is purposely done to ease the calculations. When the reagents are poured together, each one dilutes the other. The concentrations given at the head of each column are the pre-mix concentrations. You will use the post-mix concentrations in the calculations. To calculate concentrations after a dilution, use the following equation: original volume final conc = original conc x final volume In the table on the following page, record the time (in seconds) for each reaction and calculate the concentrations of each substance in each mix. Also, calculate the rate for each reaction, by dividing the time into 3.33 x 10 5 M (see equation 2 on the previous page). Write out the rates using scientific notation. If you write each rate with the same exponent, it will be easier to visualize what is happening. That is, if one rate is 3.50 x 10 7, and some other one shows up on your calculator as 1.45 x 10-6, re-write the second number as 14.5 x 10-7, so the two numbers have the same index. It s easier to notice the 1:4 ratio of these numbers in this form. 3
data and calculations (3.33 x 10-5 M/ t) C on cen tr ation s after M ixin g R eaction M ix T ime ( t) (sec) R ate (M sec -1 ) [I ] [B r ] + [H ] 1 M M M 2 M M M 3 M M M 4 M M M For the next mix, just calculate the concentrations. Do not mix until later. 5 M M M Equation 2 on page 34 gives: rate = 3.33 x 10 5 M = k[i ] x [BrO t 3 ] y [H + ] z. You will use the data in runs 1 and 2 to calculate x, the data in runs 1 and 3 to calculate y, and the data in runs 1 and 4 to calculate z. Calculation of x: Write out equation 2 using the numbers in the above data table for runs 2 and 1: rate = k [I ] x [BrO - 3 ] y [H + ] z 2) = k Calculation 1) = k Here is how to do it: Divide, term by term, the data from run 1 into the data from run 2. Notice how all the units cancel, how the k s cancel, and how the [BrO 3 ] and [H + ] terms cancel. x 0.00400M The [I ] terms should look like this: 0.00200M, which works out to (2.00) x. Example: Suppose the value obtained when rate 1 is divided into rate 2 is 2.25. You would have the following equation to solve: 2. 25 = ( 2. 00 )x.this can be rewritten as: log2.25 = x log2.00. To solve, you would take: log2. 25 x = log2.00, which gives x = 1.17. This reaction has whole number orders, so you would round x to 1. Calculation of y: Write out equation 2 using the numbers in the above data table for runs 3 and 1: Your value for x:, rounded = rate = k [I ] x [Br - ] y [H + ] z 3) = k Calculation 1) = k Your value for y:, rounded = Calculation of z: Write out equation 2 using the numbers in the above data table for runs 4 and 1: rate = k [I ] x [Br - ] y [H + ] z 4) = k Calculation 1) = k Your value for z:, rounded = 4
Calculation of the value of k: Use the values of x, y, z, and the concentrations and rates from the table on page 36 to calculate the value of k for each run. rate k = rate = k[i ] x [BrO ] y [H + ] z, so [I ] x [BrO ] y [H + ] z. Remember to substitute the actual rounded off values of x, y, and z. Show your setup for each calculation, then give the value for each k. Also calculate the average value of k. If one value of k is out of line with the other three, check with the instructor. Reaction 1 setup: k 1 = Reaction 2 setup: k 2 = Reaction 3 setup: k 3 = Reaction 4 setup: k 4 = -1 Average value of k = L 3 mol -3 sec Run number 5: Now, finally, you can run the reaction with mix number 5. First, however, calculate how many seconds the reaction should take. Use rate = k[i ] x [BrO ] y [H + ] z. Use the average value of k and the concentrations for the reactants in mix 5 from the table on page 36. Apply the values of x, y, and z correctly. Solve the equation for rate. Since rate = 3.33 x 10 5 M t, you can now solve for t. Show the instructor the calculated value for t before you begin. Call the instructor to your place to observe the color change and the time of the reaction. If the reaction takes place within 5 seconds of your predicted time, you may receive bonus points. Calculation: Predicted value for t sec Observed value for t sec Instructor s comment Care in all measurements is important in this experiment. However, the uncertainties in measuring require a bit of good luck to come within one or two seconds of the predicted time. So if you are a good experimenter, but you missed out on any bonus points in this experiment, it might be disappointing, but it is not unexpected. 5
Name Grade Date Questions 1. How much did the temperature change during the course of the experiment? Increasing temperature speeds up a reaction. Decreasing temperature slows down a reaction. If there were a temperature change of over 2 degrees, try to interpret if variations in your data are consistent with the temperature change. 2. On page 33, three lines from the bottom of the left-hand column, the rate law for the generic reaction could be verbally described as: The reaction is first order in substance A, first order in substance B, and second order overall. Use this style to describe the rate law for the reaction you used in this experiment. 3. If the order of a reaction is the same as its stoichiometry, the order of each reactant is the same number as the coefficient of that reactant in the balanced equation. When this is the case, it is possible (but not necessary) that the reaction takes place in one step. If the order of a reaction is different than its stoichiometry, then the reaction must take place in steps. For the reaction used in this experiment, show how the rate law (equation 1 on page 33) would look if the reaction took place in one step: 4. Since the observed rate law is different than the one in question 3, the reaction must take place in steps. Remember, particles are in constant motion, and at room temperature, they reach fairly high velocities. Think of a series of collisions, each one resulting in a partial change for the colliding species. For example, the first step might be the collision of a Br with an I to form BrO 2 and IO. The reaction in each step of a series of steps is called an elementary process. Since neither BrO 2 nor IO show up as final products in the reaction, they would be called intermediate products. They must be consumed in subsequent steps in the series of reactions that make up the reaction mechanism. Working out reaction mechanisms is an important part of chemical kinetics. Working out reaction mechanisms in biochemical processes is at the heart of modern medical research. Studies on AIDs and other diseases operate using this technique. They look to see what molecules a disease organism uses as it attacks a host, and how these molecules fit in with the host molecules. If a researcher discovers all the molecular steps involved in a disease, what do you think her next step is? 6