Chemistry 112 Final Exam, Part II February 16, 2005

Transcription

1 Name KEY. (35 points) Consider the reaction A + B + C + D + E + F Æ P, which has a rate law of the following form: d[p]/dt = k[a]a[b]b[c]c[d]d[e]e[f]f The data sets given or displayed below were obtained through various types of studies. a. (30) What are the exponents a, b, c, d, e and f? (The units of d[p]/dt are moles/liter hr and the concentrations are in terms of moles/liter.) b. (5) Consider a different reaction than that described above, but with : stoichiometry in all reactants and the product. If the order of the reaction with respect to reactant X and the overall order were both determined to be 2.5, what would be the units of the rate constant for this reaction? () When initial rate experiments were done, the following table of initial rates, corresponding to various starting concentrations of reactants, was generated. In each of these experiments, the initial concentrations of E and F were set at the same values for all experimental runs. (d[p]/dt) 0 [A] 0 [B] 0 [C] 0 [D] (2) When all the reactants, except B, were maintained at the same initial values, the data plotted on the graph on the left below was obtained (d[p]/dt) 0 /[E] [B] t(hr) (3) When all of the reactants, except E, were maintained in great excess the data plotted on the right hand graph above was measured. (4) When the reaction of F was studied in solution with all of the other reactants present in large excess, then the half life observed when [F] 0 = mole/liter was 0 seconds, while the half life found when [F] 0 = 0.5 mole/liter was 50 seconds.

2 Name KEY Solution: (a) From the Table, we see when that [A] 0 and [B] 0 are both doubled simultaneously (data sets and 2) and all other initial concentrations were maintained at constant values, the initial rate of reaction does not change. From this observation, we can state that a + b = 0. Now, looking at data sets 2 and 3, we see that decreasing [C] 0 by a factor of two while simultaneously holding all other initial concentrations constant results in a 6 fold increase in the initial rate of reaction. Letting R 32 = (d[p]/dt) 03 /(d[p]/dt) 02, we find R 32 = {[C] 03 /[C] 02 } c = 6 = (0.5) c. Taking the logs of both sides, log 6 =.204 = c log 0.5 = -0.30c. Therefore, c = -4. Examination of data sets and 4 allows d to be evaluated. For this case, in which all initial concentrations, except that of D, are held constant results in an equation stating that R 4 = {[D] 04 /[D] 0 } d = 5.66 = 2 d. Taking the logs of both sides, log 5.66 = = d log 2 = 0.3d. Thus, d = 2.5. Based on the work in the third conference, we know that the equation log (d[p]/dt) 0 = b log [B] 0 + constant describes the graph above on the left. The slope of the log-log plot of this equation will be equal to the exponent b. Picking two points on this graph, we can calculate the slope as m = (log 0.7 log.0)/(log 2- log ) = -0.54/0.3 = b = - /2. Since, from above, we know that a + b = 0, we can immediately determine that a = /2. Examination of the plot of /[E] vs. t, all other components present in great excess, immediately indicates that the reaction second order in [E]. The half-life data for F indicates that the half-life drops by a factor of two as the initial concentration halves in value. For zero order reactions, the half-life equation is t /2 = [A] 0 /2k. This equation indicates that the half-life will fall by a factor of two when the initial concentration drops by a factor of two. Hence the rate law for disappearance of F is zero order in [F] and f = 0. (b) Given that the reaction is 2.5 order overall, we can write d[x]/dt =-k[x] -2.5, where X represents the reactant X. To obtain the units of k, we form the ratio (d[x]/dt)/[x] This yield units of (moles/l t)/(moles -2.5 /l -2.5 ) = moles 3.5 /l 3.5 t).

3 Name KEY 2. (35 points) Consider the reaction sequence immediately below, which occurs in solution: A B k k - k 2 B C M + B C + M This reaction can be shown to have the steady state rate law that follows: d[c] dt = k 3 k k [A] 2 k + k + - k3 2 [M] + k k 3 [A][M] k + k + - k3 2 [M] When k - >> (k 2 + k 3 [M]), then this rate law simplifies to the form d[c] dt = k 2 k [A] k - + k k 3 [A][M] k The dependences of the ratio of rate constants k to k - (k /k - ) and the rate constant ratios of k 2 and k 3 to temperature, namely k 2 /T and k 3 /T, on /T are illustrated in Figures a., b. and c. respectively on the next page. (a) (4) Is the above reaction a catalyzed reaction? Why or why not? (b) (4) What are the values of H 2 and S 3 for reaction 2 and reaction 3 respectively? Note the equation for the least squares fit above the graph of log k 3 /T vs. /T. (c) () What is the value of H - H -? (d) (7) Consider an analogous reaction, involving closely related reactant D, intermediate E and product F, for which the same mechanism as described above holds and for which the same form of rate law as described in equation () is true. In this case, it was found that H - H - = -30 kj, H 2 = 70 kj, H 3 = 35 kj, S - S - = 25 J/K, S 2 = J/K and S 3 = -25 J/K. At what temperature will the rate of the reaction along the first order pathway in equation () above (in which D, E and F have replaced A, B and C) have the same rate as the reaction along the pathway involving M when [M] = M? Solution: (a) This is a catalyzed reaction. The species M enters into step 3, but is not consumed in the reaction. (b) The slope of graph b. can be calculated as being Thus, H 3 = Rm = J, The y-intercept of graph c is = (log 2x + S 3/2.3R); thus S 3 = -20 J/K. (c) The slope of the line in graph a. is 67; since H - H - = -2303Rm, H - H - = -20,400 J. (d) Since the reaction rate along the two pathways is equal, we can write k 2 k [A]/k - = k 3 k [A][M]/k -. Canceling out terms that appear on both sides, we find k 2 = k 3 [M] or k 3 [M]/k 2 =, which can be set equal to the ratio given below. (Since [M] =, it does not appear in the equation.) - ()

4 Name KEY k 3 k 2 = 2x T S 3-3 e R e HT R S 2-2 2x T e R e HT R Taking the natural logs of both sides, we obtain ln = 0 = ( S 3 - S 2)/R {( H 3- H 2)/RT}. Since R appears in both terms on the right, it can be cancelled out. Substituting in for the values given in the statement of the problem, the expression can then be written as 35 +(35000/T) = 0. Solving for T, we find that a temperature of 00 K would be required for the two reaction rates to be equal.

5 University of California, School of Pharmacy Chemistry 2 Final Examination, Problem 2. Graphs displaying how (a) k /k -, (b) k 2 /T and (c) k 3 /T vary with /T. a k /k /T, K - b. 00 c. y = x k 2 /T 0 log (k 3 /T) /T, K - /T, K -

6 Name KEY 3. (35 points) Consider the pair of reactions shown below: Z A Z B k Z A + Z B A + B D () Z A Z C k 2 Z A + Z C A + C E (2) When the rate constant for each reaction was studied individually as a function of I /2, the pair of graphs shown below resulted. Separate experiments indicated that Z B was negative. (You may assume that the ionic reactants and products make a negligible contribution to the ionic strength of the reaction solution.) In answering the questions below, do not make any extrapolations outside of the graph shown below. a. (8) What are the values of the products Z A Z B and Z A Z C? b. (9) What are the values of Z A, Z B and Z C? c. (7) What is the value that the ratio k 2 /k would have at an ionic strength of zero? d. (7) What ionic strength would be required to make k 500 times larger than k 2 if the ratio of k 20 /k was 0.? e. (4) What concentration of magnesium sulfate is needed to make I =? 5 4 log k 3 log k log k I /2 Solution: a. According to the Bronsted-Bjerrum (BB) equation (log k = log k 0 +.8Z x Z y I /2 ), we can obtain the product of the charges entering the transition state of a bimolecular reaction by measuring the slope of a plot of log k vs. I /2. This is given for both reactions by the graph above. The slope of the plot of log k vs. I /2 can be determined to be 3., which leads to Z A Z B = 3, while the slope of the graph of log k 2 vs. I /2 is.02, which yields Z A Z C = -. b. By examining the two products above, we must conclude that Z A has a magnitude of and Z B a magnitude of 3. (If Z A were to have a magnitude of 3, then Z C would have to

7 Name KEY take a non-integer value of magnitude /3.) Now, it is known that Z B is negative in sign; therefore Z B = -3. Since Z A Z B = 3, Z A =. Since Z A Z C = -, Z C = +. c. Subtracting the BB equation describing k from that describing k 2, we obtain the following: log (k 2 /k ) = log (k 20 /k ) +.08Z A (Z C -Z B )I /2. Picking values of k 2 and k off of the graph at a particular I /2, we can calculate the value of log (k 20 /k ) at that I /2. For example, picking points at I /2 = 0., we find log k 2 =.90 and log k = 2.6. Since log (k 2 /k ) = log k 2 log k, we find 0.7 = log (k 20 /k ) +.08(-)(4)(0.) = log (k 20 /k ) or log (k 20 /k ) = This yields (k 20 /k ) = 0.5. d. Using the equation from above, namely log (k 2 /k ) = log (k 20 /k ) +.08Z A (Z C -Z B )I /2, we substitute in values as follows: log = = (-)(4)I /2. Solving, we find I /2 = -.69/ = Therefore, we find I = e. I = = 0.5{[Mg +2 ](2) 2 + [SO 4-2 ](-2) 2 }. However, in MgSO 4, the concentration of magnesium and sulfate ions is the same. Thus, we can write that = 0.5(8[Mg +2 ]). Solving for [Mg +2 ], we find that [Mg +2 ] = 0.25 M. This is also the concentration of MgSO 4 required.

8 Name KEY 4. (40 points) The following two graphs represent plots of data obtained from enzyme kinetic studies of two proteins, Pro (on the left) and Pro2 (on the right). These two proteins, each of which was isolated in different laboratory, metabolize an antiparasitic drug used in the treatment of a mosquito-borne disease. The least squares fit of the data in each case is given above the corresponding graph. (Note, that in all of the studies represented below, the units of [S] is mm and v 0 is in terms of µm/sec.) 2.5 y = 0.096x y = -3.09x [S]/v 0.5 v [S] v 0 /[S] When the kinetics of degradation of the drug by a third protein Pro3, discovered by yet a third group, was studied, the data shown in the graph below was obtained (data indicated by /v 0 ). The kinetics of drug degradation by this third protein was also studied in the presence of two different inhibitors, I and I ; in each study, the inhibitor concentration was mm. The resulting data is shown by the plots labeled /v 0 and /v 0. For each study, the least squares fit of the data is above the graph..5 y = 0.302x (/v 0 ) y = 0.754x (/v 0 ') y = 0.999x (/v 0 ") /v /[S] /v 0 /v 0 ' /v 0 "

9 Name KEY a. (2) Do the enzyme kinetic data suggest that the three enzymes are likely to be the same protein, that two are likely the same protein and one is different or that the three enzymes correspond to three different proteins? Identify any proteins that are the same. Back up your answer with quantitative data and discussion of the reason(s) for your conclusions. b. (5) In the case of the third protein, on which the effect of inhibitors was studied, what kind of inhibition does I exert? What about I? c. (5) Evaluate the value of K I for the third protein. d. (5) Evaluate the value of K I for the third protein. e. (4) Studies with an uncompetitive inhibitor of this protein, namely I 3, indicated that the initial rate of reaction was three times as fast when [S] = mm and inhibitor was absent as when [S] = mm and [I 3 ] = mm. What is the value of K I for I 3? Solution: (a) If the K M and V max values describing the reaction of two enzymes with the same substrate S are quite close to one another, than it is likely that they are the same enzyme. If not, then they are likely to be different enzymes. We therefore analyze the graphical representations of the data for the three enzymes to obtain values of K M and V max. The plot in the top graph on the left is a Hanes plot. For this type of plot, m = /V max = and b = K M /V max. From this information, we find V max =.4 µm sec and K M = 3.27 mm. The top plot on the right is an Eadie-Hofstee plot. In this case, b = V max =.9 µm/sec and m = -K M = The bottom plot is a Lineweaver-Burk plot, for which b = /V max = or V max =.. Similarly, m = K M /V max = or K M = Examination of the K M indicates that there is about a 3% spread in values for the three enzymes, while there is about a 7% spread for the V max values. It is thus quite likely that the three enzymes are one and the same. The variation in V max and K M values likely comes about because of differing weighting of data points. (Indeed, the same set of data was used in constructing each of the three plots.) (b) I corresponds to a competitive inhibitor, while I is a noncompetitive inhibitor. (c) For competitive inhibition, m = (K M /V max ){ + (/K I )}. = 0.302( + /K I ) = Solving for K I, we find a value of 6.67 mm. (d) For noncompetitive inhibition, b = = (/Vmax){ + (/K I )} = /.9){ + (/K I )}. Solving for K I, we find a value of 4.86 mm. (e) For uncompetitive inhibition, we can write the following expression; v o /v o (inhibition) = [K M + [S]{ + ([I]/K I3 )}]/(K M + [S]) = 3 Substituting in values of [S] = [I] = mm and the values of K M and V max characteristic of the enzyme and solving for K I3, we obtain K I3 = 3.83 mm.