How Do Certain Factors Affect the Rate of a Chemical Reaction?

Transcription

1 EXPERIMENT 7 How Do Certain Factors Affect the Rate of a Chemical Reaction? INTRODUCTION Two important questions may be asked about a chemical reaction. () How completely do the reactants combine to give products? and (2) How fast is the reaction? How completely? is a question of chemical equilibrium, the realm of chemical thermodynamics. How fast? is the realm of chemical kinetics, the subject of this experiment. These two realms are connected for elementary reversible reactions, where the equilibrium constant is the ratio of the rate constants for the forward and reverse reactions. However, a study of the factors affecting the rate often yields information about the mechanism of more complex reactions, information not obtainable from thermodynamics. This experiment will be concerned with measuring the rate of a chemical reaction and with finding an equation, called the experimental rate law, which expresses the rate as a function of the concentrations of the various reactants. The rate of a chemical reaction is expressed as the rate of change of a reactant or product concentration. Thus for a reaction 2 A + B P (Eq. ) the rate can be described in terms of the time derivatives of any of the reactant or product concentrations. Rate ¼ d½pš dt ¼ 2 d½aš dt ¼ d½bš dt (Eq. 2) The rate is taken as a positive number with units mol L - s - or mol L - min -. The rate of change of product concentration will be positive (product concentration increases with time), but the rate of change of reactant concentrations will be negative (reactant concentrations decrease with time) so that negative signs must be inserted to make the definition consistent. Since the stoichiometry requires that two moles of A be consumed for each mole of B consumed or of P formed, the rate of change of A will be twice as big as that of B. Again, to make the definition consistent, the factor of /2 must be inserted. The rate of a chemical reaction may depend on many variables, but almost always the rate depends on the concentrations of the reacting species and on the temperature. Rates of gas-phase reactions depend on pressure as well. Photochemical reactions, involving the absorption of a photon, have rates dependent on the wavelength and intensity of incident light. Reaction rates often depend on other less obvious variables, including the concentrations of species that are not involved in the stoichiometric reaction. These species are called catalysts if they increase the rate and inhibitors if they decrease the rate. This experiment will focus on three questions.. How does the concentration of each of the reacting substances affect the rate? For a homogeneous chemical reaction, the rate is usually proportional to the concentrations of the reactants, each concentration being raised to a power called the order of the reaction with respect to the reactant. For example, suppose that A homogeneous reaction is one occurring uniformly throughout one phase (e.g., a solution). A heterogeneous reaction occurs at the interface of two phases, e.g., a liquid and a gas or a liquid and a solid. The rate of a heterogeneous reaction depends on the surface area of the interfacial region.

2 a reaction has the overall stoichiometry of Equation. In general, we expect the rate law to be of the form Rate =k [A] a [B] b (Eq. 3) where the proportionality constant k is called the rate constant 2. The bracketed symbols [A] and [B] refer to the molar concentrations of A and B, as usual. If the conversion of reactants to products involved a single elementary reaction 3, in which all reacting molecules come together to form all the products, the orders a and b would correspond simply to the stoichiometric coefficients of A and B in the overall reaction. 2 and, respectively. However, in general the orders cannot be predicted from the overall stoichiometry and must be determined by experiment. This is because reaction mechanisms usually include several elementary reaction steps which sum to the stoichiometric equation, and often one of these steps will be rate limiting. The rate law then corresponds to the rate of the slowest step in the overall reaction. To illustrate how the rate law depends on the reaction mechanism, Table shows three different hypothetical mechanisms, each having the overall stoichiometry of Equation. The predicted rate law for each mechanism is shown, together with the orders with respect to A and B, and the overall order, a + b. For each mechanism, the sum of the elementary steps is the stoichiometric equation. Note that the first mechanism with a single elementary step gives a predicted rate law with orders equal to stoichiometric coefficients. Table Rate expressions for different reaction paths for the reaction of Equation. A and B are reactants, P is the product, and C, D, and E are intermediates in these reaction mechanisms. Mechanism Rate Law Order Order Overall in A in B Order 2A+B P rate =k[a] 2 [B] A C (slow) B+C P (fast) A + B D + E (slow) A + D E (fast) 2 E P (fast) B C (slow) A + C D (fast) A + D P (fast) rate =k[a] rate =k[a][b] 2 rate =k[b] 0 2. How does the temperature at which a reaction occurs affect the rate? Most ordinary chemical reactions in the gas phase and in solution depend on temperature according to the Arrhenius equation 4, k =A e -Ea /RT (Eq. 4) where A is called the frequency factor and Ea is the activation energy. 3. How does a catalyst affect a reaction? A catalyst is a substance that alters a reaction pathway in such a way as to increase the rate of the reaction. One way a catalyst can do this is to decrease the activation energy required for the reaction. A lower activation energy for the catalyzed reaction means that at a given 2 The term rate constant is really a misnomer since k depends on temperature and any other variables not included in Equation 3. 3 The reaction mechanism of a chemical reaction is composed of a set of molecular reactions called elementary reactions or elementary steps. Be careful to distinguish elementary reactions, which purport to describe what is actually happening at a molecular level, from stoichiometric chemical reactions, which describe the overall reactants and products without suggesting the means of getting from one to the other. In the (rare) case that a reaction occurs in a single step, the stoichiometric reaction will be an elementary step and the reaction orders will be the same as the stoichiometric coefficients. 4 The Arrhenius equation is basically an empirical expression that has been found to accurately describe the temperature-dependence of many chemical reaction rates. The rates of combustion or explosion reactions often increase smoothly up to some critical temperature where there is a discontinuity, indicating a change in mechanism. Many enzyme-catalyzed reactions show Arrhenius behavior at lower temperatures, but decrease in rate at higher temperatures as the conformation of the enzyme catalyst is altered and/or enzyme denaturation occurs. 2

3 temperature, a larger fraction of the molecules will possess enough energy to react, and thus the catalyst increases the rates of both the forward and backward reactions without affecting the equilibrium constant for the reaction. Although the catalyst is not consumed in the overall reaction, it may participate in the formation of intermediates and then be re-formed in subsequent steps to produce the final products. To answer these three questions, this experiment studies the rate of the reaction H 2 O 2 (aq) + 2 I - (aq) + 2 H + (aq) I 2 (aq) + 2 H 2 O(l) (Eq. 5) The product I 2 will react stoichiometrically and rapidly with a known, limiting amount of thiosulfate ion, according to the following reaction. 2 S 2 O 3 2- (aq) + I 2 (aq) 2 I - (aq) + S 4 O 6 - (aq) (Eq. 6) Here the time it takes to consume all the thiosulfate is measured, i.e., the time to produce an equivalent amount of I 2 (one mole of I 2 consuming 2 moles of S 2 O 3 2- ). As soon as the thiosulfate is consumed and I 2 begins to accumulate, it forms a bright blue complex with starch. The rate of the reaction will be the number of moles of I 2 that needed to be formed in order to react with all the thiosulfate, divided by the observed time at which I 2 begins to accumulate and the blue complex becomes visible. In Equation 6, the thiosulfate rapidly regenerates the iodide, keeping its concentration constant so that the reaction under study (Equation 5) does not proceed in reverse. The reaction order in each reactant will be calculated from observations of how changes in the concentrations of H 2 O 2, I -, and H + affect the reaction time. The final volume of each reaction mixture is 0 ml. Thus, if the volumes of reagents used were recorded, the initial concentration of each reactant can be calculated. Equation 2 refers to an instantaneous reaction rate, whereas the average rate over a finite interval is measured here. This is a good approximation if the reactant concentrations change less than about 3%. The effect of a catalyst is also evaluated. By definition a catalyst increases the rate without being consumed in the net reaction. Measuring the uncatalyzed and catalyzed rate confirms that the catalyzed reaction has a lower Ea. PROCEDURE (WORK SINGLY) Safety Precautions The iodine solutions can stain skin or clothing. Wash thoroughly if any of the reagents come into contact with the skin. Wear safety goggles. Material General Equipment Special Equipment Reagents 25 ml Erlenmeyer flask 50 ml buret M potassium iodide (KI) 6 5 ml beakers 0 ml buret Acetate buffer containing: Hot plate/stirrer Stopwatch M acetic acid (CH 3 CO 2 H) 50 ml beaker Automatic pipet tip M sodium acetate (CH 3 CO 2 Na) Automatic pipet Thermometer M sodium thiosulfate (Na 2 S 2 O 3 ) Stir bar 2 ml graduated pipet 0.05% starch 5 ml graduated pipet ml graduated pipet >.000 M hydrogen peroxide (H 2 O 2 ) containing 0.00 M sulfuric acid (H 2 SO 4 ) as a stablizer 0.30 M acetic acid (CH 3 CO 2 H) 0.00 M ammonium heptamolybdate ((NH 4 ) 6 Mo 7 O 24 4H 2 O). Dirty glassware guarantees bad data. Clean all the necessary glassware and plastic beakers with soap 3

4 and water until water drains freely. The glassware must be rinsed with deionized water and allowed to drain thoroughly, but it doesn t need to be dried. Water does not drain very efficiently from the surface of the plastic beakers; tapping the beakers firmly on a paper towel on the benchtop can help shake some of the water droplets out. Rinse the burets and glass pipets with a small volume of reagent before filling them. It is not necessary to clean the disposable plastic automatic pipet tip. 2. Set up the 50 ml buret to deliver distilled water and the 0 ml buret to deliver buffer/thiosulfate/starch solution. Be sure to label the burets. Refer to the Introduction chapter posted on Canvas for instructions on filling a buret, if necessary. Use a 5 ml graduated pipet for the acetic acid, a 2 ml graduated pipet for the potassium iodide solution and a ml graduated pipet for the molybdenum catalyst. Following the instructions given in the introduction module posted on Canvas, set the automatic pipet for 0.5 ml. To calibrate the automatic pipet, tare an empty beaker and dispense a volume of distilled water into it with the automatic pipet. Reweigh the beaker and calculate the mass of the water. Record this mass as the exact volume for which the pipet is set. Please repeat this calibration procedure for the.0 ml automatic pipet. Remember to keep the automatic pipet vertical to prevent the solutions from entering the inside of the pipet. All seven kinetic runs follow the same general procedure, but with the following variations. Mixtures a and b. Do two kinetic runs of this same composition in order to assess the reproducibility of measurement of the reaction time. The reaction time should be in the neighborhood of one to two minutes and the times for the two runs should agree within a few seconds. If they do not agree, repeat the run until agreement is obtained. The correct time for Δt is important because all of the calculations for the other runs depend on it. Mixtures 2 4 and 6. The temperatures for these runs should be within 0.5 o C of that for mixture. Be sure to use a cool stirring motor for mixture 6, not one that is still hot from reaction 5. Mixture 5. Solution 5 has the same composition as solution and 4, but the temperature should be 2 5 o C above that of solution. If you perform experiments on mixtures -5 first and use your partner's cold plate, you need to calibrate the auto pipet to 0.5 ml for the reactions -5, and then calibrate to.0 ml for the reaction 6 If you perform all room temperature experiments first (-4, 6) you need to calibrate micropipettes each time when you add different amount of H 2 O Label and prepare the reaction mixtures as follows, with all the reagents except the H 2 O 2, which is added last to start the reaction. Be especially careful to measure reagents accurately. Record the exact volumes of each reagent added and the exact concentrations of the stock solutions supplied. Inaccuracies most often arise when mistakes are made in measuring out the reactant (or forgetting to add a reactant), or when the solution is inadequately stirred while the hydrogen peroxide is being added. 4. When the vessels contain all the reactants except H 2 O 2, add a small magnetic stirring bar to the first mixture and stir the reaction mixture vigorously enough to create a vortex. Clamp a thermometer toward the side of the beaker so as not to interfere with the stir bar. 4

5 Composition of the reaction mixtures Reaction a b Mixture Container 5 B B B B B E B Temperature RT RT RT RT RT RT+5 C RT ml H 2 O ml buffer mixture ml 0.05 M KI ml 0.3 M CH 3 COOH ml 0.00 M Mo(VI) ml.0 M H 2 O Added last, to start the reaction To start each room temperature reaction, use the calibrated automatic pipet to measure the H 2 O 2 solution (0.50 or.00 ml, as needed) directly into the reaction mixture while it is being stirred. Start the timer when the hydrogen peroxide is added; stop the timer as soon as the solution starts to turn faintly blue. Be sure to check beforehand whether the stopwatch reads in seconds, minutes and seconds, or hundredths of a minute. Record the temperature at the end of the reaction. Using a larger stir bar, retrieve the stir bar from the mixture. Rinse it and the thermometer sensor thoroughly with distilled water and dry with a KimwipeTM between runs to avoid contaminating mixtures or diluting them by an unknown factor. Rinse the stir bar and thermometer especially well after the run with the catalyst (mixture 4). 6. For the heated run (5), clamp the Erlenmeyer flask so that the contents are below water level in a 50 ml beaker of hot (~40 C) tap water set on top of a room temperature stirrer with the heat turned off. The magnetic field from the stirrer will be strong enough to turn the stirring bar in the Erlenmeyer flask. Do not stir the water bath, as this tends to make the temperature rise too quickly. Place the Erlenmeyer flasks at the bottom of the beaker and make sure the contents are below the water level. Clamp the thermometer in the flask and measure the temperature of the reaction mixture; it should reach a steady temperature 2 5 C above room temperature. If the tap water is not warm enough, turn the heat on to a low setting briefly and monitor the temperature in the water bath. Turn the heat off when the water temperature begins to rise; it will continue to rise and stay warm long enough for the experiment. Wait for the temperature of the warmed mixture to stabilize, changing less than about a degree in a minute. Then add 0.50 ml of H 2 O 2 solution and start the timer. Record the initial temperature in the reaction mixture just after the addition of the H 2 O 2. When the solution turns blue, stop the timer and again note and record the temperature. The average of the two readings is taken as the temperature of the reaction mixture. 7. Before cleaning up, check your data. Recall the way concentration and temperature can influence the rate of reaction. Repeat any questionable measurements. 8. Turn off the thermometer and pour all the solutions into a waste beaker. Be careful not to lose the stir bar. Wash the glassware and plastic beakers. 5 B = 5 ml plastic beaker, E = 25 ml Erlenmeyer flask 5

6 Collect all used reaction mixtures and unused solutions in a beaker and transfer them to the waste container in the hood at the end of the experiment. Dispose of the auto pipet tips in the medical waste container not the broken glass container. INTERPRETATION OF THE DATA. Calculating the Reaction Orders. Because the same amount of thiosulfate ion is in each reaction mixture, the time of the endpoint represents formation of the same number of moles of I 2. Therefore the ratio of the reaction rates for any two reaction mixtures will be equal to the inverse of the ratio of the reaction times. This is merely a restatement of the equation distance = rate x time, where the distance is the amount of iodine formed in Equation 5. To get the reaction order with respect to H 2 O 2, note that the reaction rate is inversely proportional to the reaction time, Δt, as shown by Equation 7. Rate 6 Rate = "t "t 6 (Eq.7) Expressions for the rates of reaction in reaction mixtures and 6 can be derived from Equation 5. For reaction mixture For reaction mixture 6 Rate = k[h2o2] a [I - ] b [H + ] c Rate6 = k[h2o2] 6 a [I - ] b [H + ] c (Eq.8) (Eq.9) Dividing Equation 9 by Equation 8 yields an expression for the isothermal ratio of rates. [ ] 6 [ ] Rate 6 = "t # = H O 2 2 Rate "t % 6 $ H 2 O 2 & ( ' a Eq.0) Note that all of the concentrations that have been held constant in the two reaction mixtures cancel out, leaving only the ratio of the concentrations of H 2 O 2, raised to the power of the reaction order in peroxide, a. In addition, if the temperature is constant, the rate constant k appearing in Equations 8 and 9 will also cancel out. Since the reaction times and the concentrations of H 2 O 2 in reaction mixtures and 6 are known, order in H 2 O 2 can be calculated. The nearest integer value of the reaction order can usually be determined by inspection. If necessary, a can be calculated by taking the logarithm of both sides of Equation 0 and rearranging. # log "t & % ( $ "t a = 6 ' # log H 2O 2 % $ H 2 O 2 [ ] 6 [ ] & ( ' (Eq. ) 6

7 The same method is used to obtain the reaction order with respect to I - (using the data for reaction mixtures 2 and ) and the reaction order with respect to H + (using the data for reaction mixtures 3 and and the known pka of acetic acid). K a ¼ ð½hþ Š½CH 3 CO 2 ŠÞ ¼ 0 4:7 ½CH 3 COOHŠ (Eq. 2) 2. Effect of Temperature. In reaction mixtures 5 and the concentrations of all the reactants are the same, but the temperature of mixture 5 is greater than that of mixture. As before, the ratio of the rates is given by the inverse ratio of the reaction times. Because all of the concentration terms are the same, they cancel out, but the rate constants do not cancel because the rate constant is temperature-dependent. Therefore Rate 5 Rate ¼ Dt Dt 5 ¼ k T 5 k T (Eq. 3) where k T5 and k T are the rate constants at the temperatures T 5 and T of reaction mixtures 5 and, respectively. Once the ratio of rate constants at two temperatures is known, the activation energy E a can be calculated from the Arrhenius equation. Taking the natural log and assuming that the constant A is independent of temperature, ln k ¼ ln A E a RT (Eq. 4) A plot of ln k against (/T) therefore is a straight line 6 with slope of -E a /R, but with only two points the data need not be plotted. Instead, E a can be calculated algebraically using Equation 5. Be sure to use absolute temperatures (Kelvins) and the correct units for R. ln k 5 ¼ ln k 5 ln k ¼ D ln k ¼ E a k R T 5 T ¼ E a R D T (Eq. 5) ð 3. Rate Constants. Once the orders a, b, and c are established, the rate constant can be calculated by solving Equation 8 for k. It is necessary to calculate the concentrations of the reactants in the actual reaction mixture (final volume 0 ml or 0.00 L) and the rate. All concentrations should be expressed as mol L, so that the final dimensions of the rate constant will be in L, mol, and s. The value of the rate constant will depend, of course, upon the accuracy of the time measurement, and hence will be a little different for the different solutions even if their temperatures are identical. Changing the temperature changes k. 4. Effect of a Catalyst. Using the measured reaction times, we can calculate the ratio of the reaction rates from the reaction times for mixtures 4 and. Rate 4 ðcatalyzedþ Rate ðuncatalyzedþ ¼ Dt 4 ¼ k ðcatþ Dt4 k ðuncatþ! (Eq. 6) 6 Measuring the rate at more than two temperatures would permit verification of the linearity of the plot and also give a sense of the precision of the rate measurements obtained. The rate is measured at only two temperatures in this experiment due to time constraints. 7