CHEMISTRY 206 Experiment 4: A KINETIC STUDY

CHEMISTRY 206 Experiment 4: A KINETIC STUDY Instructor s Informal Preamble Chemists are interested in figuring out how reactions happen (i.e., mechanisms), and how quickly they occur (i.e., rates). Both of these topics lie within the realm of chemical kinetics. In this experiment, we will focus on investigating reaction rates. In general, the rate at which a reactant disappears (or a product forms) during a reaction is a function of: (1) the concentrations of the reactants, and (2) the temperature at which the reaction occurs. The importance of reactant concentrations is fairly intuitive: the more reactant there is, the more often the molecules will collide with other reactants, and thus, the faster the reaction will be. [If a reaction is a unimolecular decomposition, we see the same effect: the probability that one molecule will spontaneously fall apart at any given moment is higher when there are more molecules present.] The dependence of a reaction s rate on reactant concentration is summarized using a differential rate law: Rate (mol L -1 sec -1 ) = k [reactant A ] p [reactant B ] q [reactant C ] r... where the superscripts (p, q, r ) denote the order of the reaction with respect to reactants A, B, C respectively, and k is the rate constant. You should recall that if a reaction s rate is linearly dependent on a reactant s concentration, then the reaction is described as first order with respect to that reactant; similarly, if the rate depends on the square of the concentration, it is called second order. Experimentally, the rate law for a reaction is usually determined using one of two methods: (1) by changing the reactant concentrations one at a time and seeing how the initial rate changes (the method of initial rates), which yields the differential rate law; (2) by monitoring each reactant s concentration over a long period of time, and plotting the data various ways to match it with a possible integrated rate law. In this experiment, you will be using a combination of these two methods to figure out the rate law for a particular reaction. This brings us to the second important factor to consider: temperature. The rate constant, k, accounts for the fact that not all collisions between reactants actually lead to reaction. Collisions must be energetic enough and occur in an orientation that encourages the appropriate bonds in the reactants to start breaking and new bonds to start forming. The Arrhenius equation summarizes these requirements: k = Ae (Ea/RT) where R is the gas constant, R = 8.31451 J mol -1 K -1 The constant A contains information about how common collisions with a reactioninducing orientation are, while the exponential term compares the energy required (activation energy, E a ) with the thermal energy available at that temperature. At higher temperatures, reactions proceed more quickly for two reasons: more collisions occur, and more of them have sufficient energy to lead to reaction. 4-1

CHEMISTRY 206 Experiment 4: A KINETIC STUDY: THE REACTION OF CRYSTAL VIOLET WITH SODIUM HYDROXIDE Introduction In this experiment, you will study the kinetics of the reaction between a dye called crystal violet and the hydroxide ion. The reaction is depicted below: The large molecule on the left is the crystal violet cation ( cv + ) we have left out the anion associated with it (e.g., Cl - or Br - ) for clarity. Notice that every carbon atom in the central part of the cv + molecule has a double bond to a neighbouring atom plus two other single bonds; this means that every one of these C atoms is sp 2 -hybridized. This allows the electrons in the π-bonds to be spread out over the p-orbitals of all the carbons in the rings of the molecule; we describe this as a large conjugated system. Conjugated molecules like this are typically intensely coloured (explanations as to why are beyond the scope of this course), and the cv + cation in particular is violet in colour. When an OH - group bonds to the central carbon (which is therefore no longer sp 2 -hybridized), the conjugation between the rings is disrupted, and the molecule becomes colorless. The reaction of crystal violet with hydroxide ion can be followed colorimetrically, i.e., by monitoring the intensity of the violet colour, which disappears as the reaction proceeds. To monitor this, we use a device called a colorimeter, which measures the amount of light that passes through a solution (i.e., the percent transmittance). This quantity can be converted mathematically into another number called the absorbance, which is useful 4-2

because absorbance is directly proportional to the concentration of the coloured species. With our experimental setup, where light passes through 1 cm of cv + solution, the concentration is calculated by dividing the absorbance by 264488 L mol -1. Thus, the change in the crystal violet concentration over time can be determined by monitoring colour. In this experiment, the reactants to be considered are the crystal violet cation (cv + ) and the hydroxide anion, so the differential rate law will have the form: Rate = k [cv + ] p [OH ] q In this experiment, you will determine the order of the reaction with respect to each of the reactants (i.e., the values of p and q), as well as the value of the rate constant, k. You will also investigate the temperature dependence of the reaction s rate constant, in order to determine the activation energy of the reaction. PART 1: Determination of the Order of the Reaction with Respect to Crystal Violet In Part 1, in order to determine the order of the reaction with respect to crystal violet, you will monitor the disappearance of the crystal violet as a function of time. The reaction will be done using a very large excess of hydroxide ion, because under these conditions the concentration of hydroxide, [OH - ], will change so little during the course of the reaction that it can be treated as a constant. Thus, any changes in reaction rate observed as the reaction proceeds will be caused solely by the fact that the cv + concentration will be continually decreasing, which will enable us to evaluate the reaction s dependence on [cv + ] alone. Because k [OH - ] q will be approximately constant, we will call this term k eff (effective rate constant). The rate law therefore becomes: Rate = k [OH ] q [cv + ] p = k eff [cv + ] p The constant, k eff, is called a pseudo rate constant, because it consists of a true rate constant combined with a concentration term that remains essentially constant under the reaction conditions used. Most reactions are either zeroth order (p = 0), first order (p = 1), or second order (p = 2) with respect to any particular reactant, and normally chemists use a trial-and-error approach to determine which one applies. Using calculus, the equation above can be integrated to give the following three potentially correct integrated rate laws: If p = 0 ("zeroth order"): [cv + ] t = - k eff t + [cv + ] o If p = 1 ("first order"): ln[cv + ] t = - k eff t + ln[cv + ] o If p = 2 ("second order"): 1/[cv + ] t = k eff t + 1/[cv + ] o These equations all define straight lines (y = mx + b). To analyze the data collected in Part 1, you will make the above three trial plots. Only one of the plots will appear as a reasonably straight line: the one corresponding to the reaction s actual order with respect to crystal violet, p. The slope of the line will provide a value for the pseudo rate constant, k eff. [Because you ll have determined this constant in Part 1, you will label it as k eff-1.] 4-3

PART 2: Determination of the Order of the Reaction with Respect to Hydroxide In Part 2, you will collect another set of concentration vs. time data, this time using double the hydroxide concentration, which will lead to a different value for the pseudo rate constant k eff, this time labeled k eff-2. To get k eff-2, you will analyze the data the same way as in Part 1, but after that, your analysis will resemble a Method of Initial Rates analysis (where you would change a reactant s concentration and see how the initial rate is affected except that here, you ll look at a rate constant directly). Because you will have two pseudo rate constants obtained under two sets of conditions (hydroxide concentration of [OH ] 1 vs. [OH ] 2 ), you will have a two equations-two unknowns situation, which you can resolve as follows (please work through it yourself to verify ): From Part 1: k eff-1 = k [OH q ] 1 From Part 2: k eff-2 = k [OH q ] 2 = k ( 2[OH ] 1 ) q = k 2 q [OH q ] 1 So: k eff-2 / k eff-1 = k 2 q [OH ] q 1 / k [OH q ] 1 = 2 q Or, using the rules for manipulating exponents and logarithms: log(k eff-2 / k eff-1 ) = q log(2) Once you have solved for q (which is the order of the reaction with respect to hydroxide), you will be able to write a complete differential rate law for the reaction between crystal violet and hydroxide: Rate = k [cv + ] p [OH ] q. Having done this, you will then be able to go back and solve for the value of the true rate constant, k. PART 3: Determination of the Reaction s Activation Energy In Part 3, you will investigate the variation of reaction rate with temperature. A general "rule-of-thumb" for most organic reactions is that the reaction s rate will be somewhere between doubled and tripled for every 10 o C rise in temperature (assuming the initial reactant concentrations are not changed). However, if we know the activation energy, E a, for a reaction, we can accurately predict the dependence of the reaction s rate constant on temperature, rather than simply estimating it. In this last part of the experiment, we will determine the E a for the reaction of crystal violet with hydroxide. The rate constant for a reaction is a function of the activation energy and the temperature according to the Arrhenius equation, which can be written in logarithmic form: ln(k) = ln(a) E a /(RT) Remember: A is a constant for any particular reaction, R is the gas constant (R = 8.31451 J mol -1 K -1 ) and T, the temperature, must be in Kelvin. A plot of ln(k) versus 1/T, called an Arrhenius plot, has the form of a straight line, with slope = -E a /R. [Note that the intercept, ln(a), is not particularly meaningful in this experiment.] Thus, by experimentally measuring the rate constant at several different temperatures and constructing an Arrhenius plot, we can calculate the reaction s activation energy. This would give us the ability to predict the rate constant for our reaction at other temperatures, and could also shed light on the mechanism of the reaction since it could provide information about what the transition state of the reaction looks like (which is beyond the scope of this course). 4-4

Experiment Summary Parts 1 & 2: Determining the Rate Law You will acquire kinetic data at room temperature, consisting of the percent transmittance of the solution as a function of time for two different concentrations of (excess) sodium hydroxide. Percent transmittance data will be converted into absorbance data, which will be converted into concentration data. The trial-and-error analysis of the first set of concentration vs. time data (Part 1) will allow you to determine the order of the reaction with respect to the crystal violet concentration. A comparison of the first and the second (Part 2) sets of data will allow you to determine the order of the reaction with respect to the hydroxide ion concentration. You will then be able to calculate the rate constant for the reaction at room temperature, and write a complete rate law for the reaction. Part 3: Determining the Activation Energy By measuring the kinetic data for one starting concentration at several temperatures, you will be able to construct an Arrhenius plot and obtain the activation energy. Goals: for reaction cv + + OH cv OH, determine rate law rate = k [cv + ] p [OH ] q, and E a PART 1: To find p, the order of reaction with respect to crystal violet, cv + : Choose large initial [OH ] 1 so that [OH ] 1 q constant rate = k eff-1 [cv + ] p Monitor [cv + ] over time by following disappearance of colour Graphically analyze data to test for the three common orders (see flowchart below) Use graph that matches order with respect to cv + to determine value of k eff-1 = k [OH ] 1 q Plot [cv + ] vs. t If linear Zero order in cv + (p=0) If curved Plot ln[cv + ] vs. t If linear First order in cv + (p=1) If curved Plot 1/[cv + ] vs. t If linear Second order in cv + (p=2) If curved Unusual reaction order (p? 0,1,2) PART 2: To find q, the order of reaction with respect to hydroxide, OH : Choose new large initial [OH ] 2 rate = k eff-2 [cv + ] p (but now p is known ) Monitor[cv + ] over time & graph (once ) to find value of k eff-2 = k [OH ] q 2 Compare k eff-1 & k eff-2 to determine value of q Sub p & q into rate law to yield the complete differential rate law Use data from Parts 1 & 2 to find average value of true rate constant k at room temperature PART 3: To find the reaction s activation energy, E a : Perform three more runs (at 40, 35, 30 C) using Part 2 s initial concentrations Use k eff values from 40, 35, 30 C to determine rate constant k at each temperature Make Arrhenius plot (lnk vs 1/T for T = 40, 35, 30 C and room temperature) Use graph to determine E a 4-5

Prelaboratory Assignment Read the procedure and answer the Prelaboratory Questions before coming to the lab. Your demonstrator will inspect and collect in your prelab before you are permitted to begin the experiment - keep a copy of it for yourself, and have the TA sign your receipt record. Materials Apparatus Computer running LABWORKS (equipped with a colorimeter) 2 cuvettes, one with a black tube inserted Kimwipes to clean the cuvettes [Do not use paper towel it scratches them.] wash bottle waste beaker 2 small test tubes hot plate, 250 ml beaker (water bath), stand, and 2 clamps thermometer 5 ml graduated pipette for crystal violet solution 2 ml graduated pipette for NaOH solution Reagents and Materials 0.1 M sodium hydroxide (prepared by technical staff) 0.05 M sodium hydroxide (prepared by technical staff) 2.25 10-5 M crystal violet (prepared by technical staff) 6 pieces of Parafilm, or a rubber stopper or cork Procedure The colorimeter should be connected to the interface and ready for use (plugged into the Input 2 and Output 2 positions on the Labworks box). It will have to be calibrated first. Then, the reaction will be monitored by mixing the reactants and placing them in the cuvette so that the percent transmittance can be measured. Starting the Program 1. Double-click the Labworks icon. 2. Click on DESIGN, EXPERIMENT BUILDER and OPEN AN EXISTING FILE then OK. 3. Open the crysviol.exp file. Do not modify it! 4-6

This program is designed to start the timer as soon as the START button is clicked (when the NaOH solution is added to the crystal violet solution and the reaction starts). After the reaction mixture has been mixed and a sample of it is placed in the colorimeter, switch W is pressed on the Labworks interface box: from this moment, percent transmittance readings are measured every few seconds until the STOP button is clicked. The readings are shown graphically as the data is acquired. Calibration of the Colorimeter 1. Click on CALIBRATE in the top menu. 2. Click on %Trans2 in the colorimeter area of the calibration window. If you are "asked" about the light source, make sure switch X on the interface is on (up) and click OK. 3. Leave the DAC voltage at 1000 millivolts, and click in the area where it asks you to "enter the value" for the first calibration measurement. Put the cuvette containing the black tubing into the colorimeter well, with the smooth sides facing the LEDs. Cover the top with the black canister in which the cuvettes are stored, to block out stray light. 4. Wait for the reading to stabilize to 3 significant figures. This will take several minutes. Then type in 0.00 for the value (i.e., 0% transmittance) and press Enter. 5. Click in the area where it asks for the second measurement. Replace the cuvette with the black tube with a cuvette 3 / 4 full of distilled water, and cover it as before. 6. Wait for the reading to stabilize again (it will be much quicker this time), then enter a value of 100.00 (i.e., 100% transmittance) and press the Enter key. 7. Click on OK to end the calibration procedure. 8. Empty and drain the cuvette, and dry it with a rolled up Kimwipe. Part 1: Determining the order of reaction with respect to crystal violet 1. In a clean dry test tube, place about 2 ml of 0.05 M NaOH stock solution. 2. In a second clean dry test tube, add exactly 4.5 ml of crystal violet solution measured with a pipette. 3. Click on ACQUIRE in the top menu of the Labworks screen. 4. Position the mouse cursor over the START button, but do not click it until the instant the sodium hydroxide solution is added to the crystal violet solution. 5. Pipette 0.50 ml of the sodium hydroxide solution from the test tube into the crystal violet solution and click START immediately. 6. As quickly as possible, mix the contents of the test tube by covering it with a piece of Parafilm or a rubber stopper and inverting it twice. Then fill a cuvette to 3 / 4 full with the mixture, wipe outside of the cuvette with a Kimwipe, place it in the colorimeter, and cover it with the canister. 4-7

7. Wait 10 seconds (to allow the light detecting circuit time to re-stabilize) and press switch W on the interface to start taking percent transmittance readings. 8. After about 1000 seconds, click on STOP to stop the data collection. 9. Click on SAVE DATA and save the data file under an appropriate name. 10. Dispose the contents of the cuvette into the labelled waste container, rinse it with distilled water, and dry it with a Kimwipe to prepare for the next run. Data Analysis for Part 1 1. Click on ANALYZE (and if necessary, OPEN AN EXISTING FILE, then OK, and then open your file.) 2. The screen should look like the following figure. The left part of the window contains a spreadsheet, with column A containing time values (headed "secs") and column B containing percent transmittance values (headed "%T"). [By scrolling the spreadsheet window to the right, more empty columns can be uncovered.] The right hand side shows a plot of percent transmittance vs. time in seconds. To determine the order of the reaction with respect to cv + and to determine the rate constant, k eff-1, some data treatment must be done using the spreadsheet. 4-8

3. Create some new data columns: Click on a column letter, and in the menu bar click on COLUMN SETUP. The program will ask for a label, a formula and the number of decimal places as seen in the window shown at the right: 4. Create the new columns C through I using 3 decimal places for each column according to the table below: Column C D E F G H I Label Time (min) Trans Abs Conc Conc ln(conc) 1/(Conc) Formula A/60 B/100 log(d) E/264488 F ln(f) 1/F 5. When the columns are made, the following graphs must be plotted (see p. 4-10): Conc vs. time (minutes) ln(conc) vs. time (minutes) 1/Conc vs. time (minutes) This will allow you to determine if the reaction is zero, first or second order with respect to the crystal violet concentration, respectively. The Laboratory Report questions will lead you through this analysis. 4-9

Preparation of graphs: a. Click on GRAPH SETUP in the menu bar. In the x-axis box, select column C (Mins) using the arrows. In the Y1-axis box, select column G ( Conc) and then click OK. You will see the graph displayed. Print this graph by clicking on FILE in the menu bar and PRINT GRAPH and OK (as in earlier experiments, change the line colour to black first). b. Similarly, use GRAPH SETUP to plot and print separate graphs with column H ( ln(conc)) and I (1/Conc) on the Y1-axis vs. C (Time in min) on X1. c. Examine the three graphs and visually identify the one closest to a straight line (even the best may have a slight curve). This one corresponds to the reaction s order with respect to crystal violet it is very important to carefully evaluate your data at this point. Redisplay this straightest graph using GRAPH SETUP. d. Click on CURVE FIT (a small window will appear). Click tab X RANGE and select SELECT RANGE. Use the buttons («, <, >,») to select data in about the first 20% of the graph. Vertical lines will appear on the graph to show the range selected. e. Click on tab FORMULA, then select FIRST ORDER. [Note: In this instance, first order has nothing to do with the reaction order it simply instructs the analysis program to fit the data in the selected range with a straight line.] Close the little window by clicking on the X in the top right hand corner. Your graph might look like the one shown here. f. The value of k eff-1, the pseudo rate constant, is obtained from the slope of this line (which is the coefficient of x printed at the top of the graph). Print the graph, label it, and note the value of k eff-1. g. When you have completed the data collection and analysis, click on FILE in the menu bar and select CLOSE. When you are asked if you want to save changes to the file, click YES. Interpretation of graphs: The most linear graph corresponds to the correct integrated rate law for the reaction and therefore reveals the reaction s order with respect to cv +. Refer to the Introduction, the Experimental Summary and the questions in the Laboratory Report section for details. 4-10

Part 2: Determining the order with respect to hydroxide ion. Recall that the k eff obtained using our experimental design is a pseudo rate constant, because its value includes the concentration of OH, i.e., k eff = k[oh ] q. In the second part of the experiment, a different initial concentration of hydroxide will be used, so the pseudo rate constant obtained in this part, k eff-2, will have a different value than in Part 1. In Part 1, 4.5 ml of crystal violet solution and 0.5 ml of 0.05 M sodium hydroxide were mixed together. For Part 2, follow the same procedure as in Part 1, but this time use 0.5 ml of 0.1 M sodium hydroxide in order to double its concentration in the reaction mixture and see the effect on k eff. Comparison of the data obtained in Parts 1 and 2 will allow the value of q, the reaction s order with respect to hydroxide, to be determined. Once the form of the differential rate law has been determined, you can use your two data sets to calculate two values for the reaction s true rate constant (k 1 and k 2 ) and its average value k. You should report the temperature along with your rate constant; for example, label it as k 23 if the value was determined at 23 C. Important: It is essential to record the temperature of the crystal violet solution (which will be room temperature, ~ 20-25 o C). The exact temperature is required for the data analysis in Part 3. Primary data analysis for Part 2: The data analysis procedure is similar to Part 1, but you only need to plot and print the graph that corresponds to the cv + order (i.e., the p value) you determined in Part 1. Be sure to label the graph appropriately. This graph s slope will yield a value for the pseudo rate constant k eff-2 (see the Laboratory Report section for details). Instructions for the remainder of the data analysis (i.e., determining the value of q) are given in the Laboratory Report section. Note: You do not need to create data columns nor plot the graphs for the reaction orders that were eliminated in Part 1 however, if you are unsure about your conclusion from Part 1, you can print out all three graphs just in case. In any case, it is advisable to verify with your TA the correctness of your conclusion about the reaction s order with respect to cv + to be sure you are using the correct type of graph for Parts 2 & 3. 4-11

Part 3: Determining the activation energy In this part of the experiment, you will see how the magnitude of the rate constant k changes with temperature. You will run the reaction at three higher temperatures (30, 35 and 40 o C) using 0.1 M NaOH as in Part 2, and then determine the pseudo rate constants k eff-30, k eff-35 and k eff-40. From these, you will calculate the corresponding values of the true rate constant at these temperatures: k 30, k 35 and k 40. Including the room temperature rate constant obtained in Part 2, you will have rate constants for four different temperatures, which will permit you to make an Arrhenius plot of ln(k) vs. 1/T and thereby determine a value for the reaction s activation energy. Note: If this type of experiment were being performed under ideal conditions, the colorimeter cell well would be thermostatted to match the desired reaction temperatures. The simplified Labworks equipment does not have this feature, so the strategy is to record the earlier transmittance measurements for each run at approximately the correct temperature by starting about 1 o C too high. It is essential that you work with the warm solutions quickly and place the cuvette into the colorimeter as quickly as possible, otherwise the solutions will cool to significantly below the desired temperature and your data will be quite scattered. 1. Heat 200 ml of water in a 250 ml beaker on a hot plate on medium. 2. Place ~4.0 ml of 0.1 M NaOH in a test tube and stand the tube in the water bath. 3. Into another test tube, pipette 4.50 ml of crystal violet solution and stand the tube in the water bath. 4. Let the water bath reach 45 o C, then take it off the hot plate. Click on ACQUIRE and align the cursor on the START button. 5. Keep stirring the water gently (by moving the test tubes) and when the water cools to 41 o C, pipette 0.50 ml from the test tube containing the sodium hydroxide solution into the test tube containing the crystal violet solution. Click on START to start the program, mix the solution in the test tube, put a portion into the cuvette, wait 10 seconds, press the W switch, and collect data as before. Only collect data for about 3 minutes, because the solution will cool quite quickly. Save the data in a file. 6. To work efficiently, while data is being collected at a particular temperature, measure out another 4.5 ml of crystal violet solution, and put it in the water bath, which will probably cool to the next desired temperature just in time. 7. Start the other two runs at 36 and 31 o C. Primary data analysis for Part 3: Analyze the data as in Part 2 (that is, make only the appropriate plots) to get values for k eff at 40, 35 and 30 o C. Print out and label your graphs appropriately. Instructions for the rest of the analysis (the Arrhenius plot) are given in the Laboratory Report section. 4-12

Name: Section: Date: CHEMISTRY 206 Experiment 4: A KINETIC STUDY: THE REACTION OF CRYSTAL VIOLET WITH SODIUM HYDROXIDE Prelaboratory Questions 1. (2 marks) Write a brief summary, in your own words, describing the purpose of this experiment and the steps you will follow. One page - maximum. 2. (1 mark) The positive charge on the crystal violet cation is not shown. On which atom should it be placed? Explain. 3. (2 marks) The following data were collected for a reaction of type: A Products. Complete the last two columns and answer the questions below: Time (s) [A] (M) ln[a] 1/[A] 0 1.00 10-3 5 6.07 10-4 10 3.68 10-4 15 2.23 10-4 20 1.35 10-4 Determine the order of the reaction with respect to A and the value of the rate constant. Your answer must include a graph, drawn neatly by hand on graph paper, as well as a brief explanation. 4-13

Name: Section: Date: CHEMISTRY 206 Experiment 4: A KINETIC STUDY: THE REACTION OF CRYSTAL VIOLET WITH SODIUM HYDROXIDE Laboratory Report Observations and Data 1. (5 marks) Hand in copies of your graphs (labelled in detail) with this report. Calculation and Interpretation of Results Part 1: Determination of the Order of Reaction with Respect to Crystal Violet 2. (1 mark) Which graph gave the most linear plot? What is your conclusion about the order (p) of the reaction with respect to crystal violet? 3. (1 mark) What is the value of k eff-1, the pseudo rate constant printed by the curve fitting program for the most linear plot from Part 1? Suppose you were not given the equation at the top of the graph what calculation would you have had to do in order to determine the value of k eff-1? Show this calculation, and annotate the graph to show the data you used. 4-14

Part 2: Determination of the Order of Reaction with Respect to Hydroxide ion 4. (2 marks) What is the computer-generated value of the pseudo rate constant k eff-2 found using the mixture prepared with the 0.1 M solution of sodium hydroxide? Determine the ratio of the two k eff values (k eff-1 and k eff-2 ) and deduce the order (q) of the reaction with respect to hydroxide ion. Show your calculations. Note: the value of q should be an integer (0, 1 or 2). If it is not, it is because of experimental error. What is the nearest integer value? 5. (1 mark) Using the k eff values from Part 1 and Part 2, calculate an average value for the true rate constant, k. Show your calculations. [Take care in each case to correctly calculate the initial concentration of each species after mixing the solutions.] Part 3: Determination of the Activation Energy for the Reaction of cv + with OH 6. (1 mark) Calculate the values of the true rate constant k for the reactions at 30, 35 and 40 o C, including units. Show your work. 4-15

7. (1 mark) Complete the table below (where k is the true rate constant): Temperature (K) 1/T (K -1 ) k (units?) ln(k) (unitless) * ** 303 ** 308 ** 318 * Data from Part 2. **Data from Part 3. (2 marks) On graph paper (by hand, not using a computer), make a plot of ln(k) vs. 1/temperature, and determine the value of E a. (R = 8.31451 J mol -1 K -1 ). Hand in your graph, and show your calculations in the space below. 8. (1 mark) Using the values of E a and A from your graph, calculate the expected value of the rate constant for this reaction at 10 C below your value of room temperature (from Part 2). Does it seem that the 2-3 change in rate per 10 C change in temperature rule of thumb would apply to this reaction? 4-16