k 3 ) and Κ3 /Κ 2 at 37 C? (d) (4) What will be the ratio of [D]/[C] after 25 min of reaction at 37 C? 1.0E E+07 k 1 /T 1.

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1 1. (35 points) Compound A reacts to form compounds B, C and D via parallel unimolecular pathways, as shown immediately below. A k 1 B (1) A k 2 C (2) A k 3 The plot on the graph shown below displays the temperature dependence of the ratio of rate constant to T (k 1 /T) for reaction 1. From data taken in a separate study, it was determined that the quantity (ΔH 3 -ΔH2 ) = 10 kj/mole as well as that (ΔS3 - ΔS2 ) = 50 J/mole K. (a) (15) Calculate ΔH and ΔS for reaction (1). (b) (10) At what temperature would the third reaction be equal in rate to the second reaction? (c) (6) What are the values for (ΔG 3 -ΔG2 ) and Κ3 /Κ 2 at 37 C? (d) (4) What will be the ratio of [D]/[C] after 25 min of reaction at 37 C? D 1.0E+08 (3) k 1 /T 1.0E E E E E E E-03 (a) We can evaluate ΔH and ΔS from the slope and y-intercept of the graph above. Picking two data points on the graph {(0.0026, ln 4x10 6) and (0.0033, ln 2x10 5 )}, we find the slope to have a value of m = Since the value of the slope is equal to - ΔH /R, we find that ΔH = J/mole = 35.6 kj/mole. Using the relation b = y mx in conjunction with the data point (0.0026, ln 4x10 6 ), we find the y-intercept to be Using the equation b = ln (2x10 10 ) + ΔS /R = ΔS /R, we obtain ΔS = 20.1 J/K mole. (b) For parallel reactions, we can write ln (k 3 /k 2 ) = (ΔS 3 - ΔS2 )/R - (ΔH3 -ΔH2 )/RT. When the two reactions are equal in rate, k 3 /k 2 = 1 and ln k 3 /k 2 = 0. Then, solving for T, we find T = (ΔH 3 -ΔH2 )/(ΔS3 - ΔS2 ) = 10,000J/mole/50 J/mole K = 200 K = T. 1/T

2 (c) We can calculate(δg 3 -ΔG2 ) by knowing that this quantity if given by (ΔH3 -ΔH2 ) - T(ΔS 3 - ΔS2 ) = 10,000 J/mole - 310K(50 J/mole K) = -5,500 J. We can then calculate Κ 3 /Κ 2 by using the fact that the equation ln (Κ3 /Κ 2 ) = -(ΔG3 -ΔG2 )/RT = Thus, Κ 3 /Κ 2 = (d) For 1 st order parallel reactions, we know that k 3 /k 2 = [D]/[C] at all times. Using the equation in part (b), we can evaluate ln (k 3 /k 2 ) = (50/8.314) 10,000/8.314(310) = = Taking the antilog of both sides, we obtain k 3 /k 2 = 8.4 = [D]/[C] after 25 min of reaction.

3 2. (35 points) Consider the reaction A + B + C + D + E + H + P, which has a rate law of the following form: d[p]/dt = k[a]a[b]b[c]c[d]d[e]e[h + ] h The data sets given or displayed below were obtained through various types of studies. What are the exponents a, b, c, d, e and h? (The units of d[p]/dt are moles/liter hr and the concentrations are in terms of moles/liter.) (1) 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 H + 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 H +, were either present in great excess or maintained at the same initial values, and the initial value of hydrogen ion concentration was varied, the data plotted on the 0 graph on the left below was obtained. [A] 0 (3) When the half-lives for decomposition of A were measured at various initial starting concentrations (denoted by [A] 0 ) under conditions where the concentrations of all reactants, except that of A, were either in great excess or maintained at a constant value, then the data displayed in the graph on the right side of the page were obtained. (4) When the reaction of E was studied in solution with all of the other reactants present in large excess or maintained at constant value, then the half-life observed when [E] 0 = 1 mole/liter was 100 seconds, while the half-life found when [E] 0 = 0.5 mole/liter was also 100 seconds. Examination of the first two rows of the table indicates that when [A] 0 and [B] 0, each at the same initial concentration, are both doubled while all other reactants are held constant, that the initial rate goes down by a factor of two. Abbreviating the initial rate as R, we can write R 2 /R 1 = [A] 02a [B] 02b /[A] 01a [B] 01b = ([A] 02 /[A] 01 ) a ([B] 02 /[B] 01 ) b. Substituting in actual concentrations, we obtain R 2 /R 1 = (2) a (2) b = 2 a+b = 0.5. Solving, we find that a + b = -1. Next, we note that when the concentrations of all components of the reaction mixture except C are held constant and [C] is doubled (rows 3 and 2 in the table), then the value of R 2 /R 3 = 4 = ([C] 02 /[C] 03 ) c = ((2/1) c = 2 c. We thus find that c = 2. In the case of the reactant D, we see that all components are held at constant concentration, except D, in rows 1 and 4. We can there fore write R 4 /R 1 = =

4 ([D] 04 /[D] 01 ) d = (1/3) d. Taking the ln of both sides, we obtain ln = d ln (1/3) or 3.84 = d. Solving, we find d = Next we note that when the graph on the right hand side represents half-life for loss of A when [A] was set at various initial values and all other reactant concentrations were maintained at constant values (e.g. by having them present in great excess. The appropriate relationship describing the half-lives for reactions having rate laws of the general form d[a]/dt = -k[a] n is t 1/2 = (2 n-1-1)/(n-1)k[a] 0 n-1 ). This can be written for our special case as ln t 1/2 = C + (1-a)ln[A] 0, where C is a constant. This equation indicates that the slope of a plot of ln t 1/2 vs. [A] 0 will have a slope of (1-a). Evaluating the slope of the graph gives m = (ln 0.3 ln 0.001)/(ln 1- ln 0.1) = 5.70/2.30 = Rounding this to 2.5 and setting this quantity equal to 1-a, we obtain a = Since a + b = -1, we can write b = -1 or b = 0.5. Next, we can determine the value of h by looking at the graph given on the left hand side above. Analogous to the experiments done in the conference/lab, we can write R = ln Constant + h ln [H + ]. Thus, the slope of the line in the graph yields h. Evaluating the slope, we obtain h = m = (ln 2 ln 1)/(ln 4 ln 1) = 0.5. Thus, h = 0.5. Finally, since the half-life is constant without regard to starting concentration of E, we can state that the order with respect to E is one or e = 1.

5 3. (35 points) Solve the following two problems. A. (19 points) Consider the following reversible bimolecular reactions, namely k 1 +4 W + X Y + Z k -1 When the ionic strength is very near zero, the equilibrium constant for the reaction has a value of 1 and k -1 has a value of 1 M -1 sec -1. When the ionic strength is 0.5, the rate constant k 1 has a value of 10. (Be sure to include appropriate units with your answers.) (1) (6 points) What is the value of the rate constant k 1 when the ionic strength is very low? (2) (7 points)what is the value of the equilibrium constant K when I = 0.5? (3) (6 points)what concentration of the strong electrolyte Li 3 PO 4 would be required to give a solution of this ionic strength? (The lithium cation bears a charge of +1.) (1) Since K = k 1 /k -1, we can evaluate k 1 = 1 M -1 sec -1. (2) Using the Brønsted-Bjerrum equation (log k = log k Z A Z B I 1/2 ), we calculate the values that k 1 and k -1 have when I = 0.5 M. Since Z Z = 0, we know that k -1 = k -10 = 1 M -1 sec -1. For k 1, we write log k = (3)(1)(0.5) 1/2 = 3.054(0.707) = 2.16 and k 1 = M -1 sec -1. Thus, K = k 1 /k -1 = (The answer 10 was also accepted, as k 1 was given, in an erroneous holdover from a previous incarnation of this problem, as 10 at I = 0.5.) (3) Ionic strength can be calculated from the relation I = 1/2 [C i ]Z i2. Thus, for lithium phosphate, we write I = 1/2{[Li +1 ](+1) 2 + [PO -3 4 ](-3) 2 } = 1/2{3[PO -3 4 ](+1) 2 + [PO -3 4 ](-3) 2 } = 6[PO -3 4 ]. Thus [PO -3 4 ] = M or M is the concentration that the solution should be made in Li 3 PO 4. B. (16 points) Consider another set of reversible bimolecular reactions, shown below: A + B C + D Z A Z B Z C Z D k 2 k-2 When the effects of ionic strength on the rate of the forward and reverse reactions were studied, the two graphs shown below were obtained. (k -20 refers to the value of the rate constant k -2 at zero ionic strength.) Other studies indicated that the charge on C was positive. (1) (4 points) What is the value of the product Z A Z B? (2) (4 points) What is the value of the product Z C Z D? (3) (8 points) What are the possible values of Z A and Z B and of Z C and Z D? (1) Applying the Brønsted-Bjerrum equation (log k 2 = log k Z A Z B I 1/2 ) to the first graph below, we note that the slope gives the quantity 1.018Z A Z B. Using sample points off the graph, we find m = (log 5x10 8 log 1x10 8 )/( ) = 8.22 or Z A Z B = Since this is very close to 8, we surmise that Z A Z B = 8. (2) Applying the alternate form of the Brønsted-Bjerrum equation, namely log (k -2 /k -20 ) = 1.018Z C Z D I 1/2 ), to the second graph below, we observe that the slope is given by 1.018Z C Z D. Determination of the slope of the graph yields m = 9.21 or Z C Z D = 9.05; this should be rounded to Z C Z D = 9.

6 (3) There are four sets of values for Z A and Z B that satisfy the relationship Z A Z B = 8: Z A = -1, Z B = -8; Z A = -8, Z B = -1; Z A = 8, Z B = 1; Z A = 1, Z B = 8; Z A = 4, Z B = 2; Z A = 2, Z B = 4; Z A = -2, Z B = -4; Z A = 4, Z B = -2. Similarly, there are four sets of values that satisfy the relationship Z C Z D = 9: Z C = 9, Z D = 1; Z C = -9, Z D = -1; Z C = 3, Z D = 3; Z C = -3, Z D = -3. However, the fact that C is known to be positive rules out the first two sets of values, leaving two possibilities: Z C = 9, Z D = 1 and Z C = 3, Z D = 3. Applying the law of conservation of charge, we know that Z A + Z B = Z C + Z D. Applying this relationship rules out all possibilities except Z C = 3, Z D = 3 and Z A = 4, Z B = 2 or Z A = 2, Z B = 4. The data do not allow us to definitively assign charges to A and B.

7 1E+09 Name KEY k 2 1E I 1/2 10 k -2 /k I 1/2

8 4. (35 points) Answer the following: A. (19 points) When the inhibition kinetics of an alcohol dehydrogenase, isolated from the intestinal wall of a parasitic form dwelling in an eel, was studied in the presence of various concentrations of ethanol (S) and in the presence of an inhibitor I, the following data represented on the graph immediately below were obtained. Note that concentrations of substrate are µm and rates are in terms of µm per second. Using information on this graph, determine: (1) (4 points) What kind of inhibition is acting on the enzyme? (2) (10 points) Evaluate the values of K m and V max. (3) (5 points) Evaluate the value of K I. 15 [I] = 1µM 10 1/v 0 5 [I] = 0 µm /[S] (1) This Lineweaver-Burk plot displays the behavior characteristic of noncompetitive inhibition. (2) From the data when [I] = 0, we can evaluate K m and V max. In particular, m = K m /V max and b = 1/V max, while the x-intercept = -1/K m. Since the x-intercept = -0.5, this implies K m = 2. We can evaluate the slope by m = (4-2)/( ) = 2/2 = 1 = Km/ V max. = 2/ V max or V max = 2 µm/sec. (3) In the presence of I we write for noncompetitive inhibition b = (1/ V max )(1 + [I]/K I ). From the plot, the y-intercept b in the presence of inhibit can be read to have a value of 1. Thus, 1 = 1/2(1 + 1/[K I ]). Thus, we find that K I = 1 µm.

9 B. (8 points) In another kinetic study on a different enzyme and using a different inhibitor, it was found that the value of K m was 5 µm, the value of K I was 2.5 µm and V max was 10 µm/sec. The substrate concentration was 5µM. (1) What is the value of the percent inhibition of this enzyme when I = 20 µm if the inhibition was competitive? (2) What is the value of the percent inhibition of this enzyme when I = 20 µm if the inhibition is uncompetitive? The definition of percent inhibition is given by % = 100(1-(v 0 ) i /v 0 ). (1) For competitive inhibition, the Enzyme Kinetics Handout gives the useful expression v 0 /(v 0 ) i = 1 + (K m [I]/([S]K I + K m K I )). Substituting in, we obtain: v 0 /(v 0 ) i = 1 + {(5)(20)/((5(2.5) + (5(2.5))} = 1 + (100/25) = 5. From this expression, we calculate (v 0 ) i /v 0 ) = 0.2 and % = 100(1-0.2) = 80%. (2) For uncompetitive inhibition, the analogous expression is given by: v 0 /(v 0 ) i = {K m + [S](1 + [I]/K I )}/(K m + [S]) Substituting in the appropriate values, we obtain: v 0 /(v 0 ) i = 5 + 5(1 + (20/2.5))/(5 + 5) = 5 From this value, we obtain (v 0 ) i /v 0 ) = 0.20 and then evaluate % as 100(1 0.20) = 80% C. (8 points) The rate of reaction induced by a third enzyme was studied and the kinetic data obtained is plotted in the graph shown below. What are the values of K m and V max? The concentrations of substrate are µm and rates are in terms of µm per second [S]/v [S] Solution: This is a Hanes plot; from lecture, the slope of a Hanes plot is m = 1/V max, the y- intercept is K m /V max and the x-intercept is K m. From the graph, we can determine that K m = -10 or K m = 10. Since the y-intercept = 2, we can then determine V max = K m /2 = 5 µm/sec.