1. (35 points) The reaction of A and B to form products is thought to go according to the following mechanism: A + 2B 2C + D k -1 2C 2C 2C + M k 2 k 3 k 4 G H J + M (a) (5) Identify the products in this reaction. (b) (5) Identify any catalyst(s) in this reaction. Explain why any species you identify as such are catalysts. (c) (12) Derive a rate law for formation of H, assuming that the steady state approximation can be applied to intermediate(s) in the above mechanism. (d) (7) Are there conditions that the rate law you derived in part (c) would be third order overall? If so, what are they? (e) (6) Are there conditions under which the rate law you obtained in part (c) would be 2nd order overall? If so, what are they? Solution: (a) D, G, H and J are products in this reaction (b) M; it is both consumed and regenerated in stoichiometric amounts in the reaction corresponding to k 4. (c) First, we write the rate law for formation of H as d[h]/dt = k 3 [C] 2. Next, we write the steady state rate law for formation of C: d[c]/dt = 2 [A][B] 2 2k -1 [C] 2 [D] 2(k 2 )[C] 2 2k 4 [C] 2 [M] 0 Then, solving for [C] 2, we obtain [C] 2 = [A][B] 2 /{k -1 )} Substituting this quantity into the equation for formation of H, we find the required rate law to be: d[h]/dt = k 3 [A][B] 2 /{k -1 )} (d) If k -1 [M] << (k 2 ), then d[h]/dt = k 3 [A][B] 2 /(k 2 ); this rate law is third order overall. (e) If k -1 [D] >> {k 4 )}, then the rate law becomes d[h]/dt = k 3 [A][B] 2 /k -1 [D] = k [A][B] 2 /[D] (where k = k 3 /k -1 ). This rate law has an overall order of 2. Similarly, if k 4 [M]>> (k 2 ) + k -1 [D], the rate law will have an overall order of 2.
2. (25) (a) Consider the reaction A + 2B P. In one study, the initial concentrations of A and B were set at 1 M and 3 M respectively. After 10 min, the concentration of A was found to be 0.2 M. What will be the concentration of B after 20 min if: (a) (7) The reaction is zero order in both A and B and zero order overall? (b) (7) The reaction is 1 st order in A, zero order in B and 1 st order overall? (c) (7) The reaction is 2 nd order in A, zero order in B and 2 nd order overall? (d) (4) The reaction is 1 st order in A, 1 st order in B and 2 nd order overall? Solutions: (a) Since the reaction is zero order in both A and B, the rate law is (1/1)d[A]/dt = -(1/2)d[B]dt = k. The integrated rate law for A is [A] = [A] 0 kt. From the data given, we can write the expression 0.2M = 1M k(10 min). Thus, k = 0.08 M/min. After 20 min, we can write [A] = 1M (0.08M/min)(20 min) = 1M 1.6M = -0.6M. Since A and B both stop reacting when [A] = 0, the total amount of A consumed is 1M. Therefore, the concentration of B consumed will be 2M. The concentration of B after 20 min of reaction will thus be 1M. (b) For the first order process, ln A = ln [A] 0 kt. Thus, we can write ln (0.2) = ln (1) k(10) or, taking natural logs, -1.61 = 0 k(10min). Therefore, k = 0.161 min -1. At 20 min reaction time, we can write ln [A] = 0 20 min(0.161 min -1 ) = -0.322 or [A] = 0.04M. Thus 0.96M of A has been consumed. This implies that 1.92M of B has been consumed. Thus the final concentration of B is 1.08M. (c) For the second order reaction that is also second order in A and zero order in B, we can write 1/[A] = 1/[A] 0 kt. Substituting in numerical values, we obtain 1/0.2M = 1/1M + k(10 min). Solving for k, we obtain a value of 0.4 M -1 min -1. After 20 min, the resulting value of A will be 1/[A] = 1/1M + (0.4 M -1 min -1 )20 min = 9 M -1. Thus [A] = 0.111M and we have consumed 0.889M of A. Therefore, we will have consumed 1.778M of B and the final concentration of B will be 1.222M. (d) For the case where the reaction is 1 st order in A, 1 st order in B and 2 nd order overall, we can write ln ([A]/[B]) = ln ([A] 0 /[B] 0 ) + {b[a] 0 a[b] 0 }kt. After the reaction has proceeded for 10 min, we will lose 0.8M of A and 1.6M of B. This leaves [A] = 0.2M and [B] = 1.4M. Thus, we can write ln ((0.2)/(1.4)) = ln((1)/(3)) + {2(1) 1(3)}k(10 min). Solving for k, we obtain a value of 0.0845 M - 1 min -1. Substituting this value of k, along with 20 min for the time of reaction, leads to a value of ln ([A]/[B]) = -2.89 or [A]/[B] = 0.0556 or [A] = 0.0556[B]. From the stoichiometry, we know that at any particular time ([A] 0 [A]) = ([B] 0 [B]); for our specific case, this becomes ([A] 0 0.0556[B] = ([B] 0 [B]). Substituting in values for the various concentrations of A and B, we find that 1 0.0556[B] = (3-[B]) = 1.5 [B]. This equation becomes 0.444[B] = or [B] = 1.14M after 20 min. The concentration of A present after 20 min reaction will be 0.0630M.
3. (25 points) The problems below refer to the graphs on the following page. A. (8 points) The kinetic data taken for the reaction A B is shown in Graph 3A. (1) What is the half-life of this reaction when the initial concentration of A is 0.15 M? (2) If the initial concentration of A is 5 M, what time will be required for the concentration of A to become 0 M? B. (10 points) The kinetic data taken for the reaction C D is shown in Graph 3B. (1) What is the half-life for this reaction when the initial concentration of C is 3 M? (2) What time would be required for the concentration of C to become 0 M if the initial concentration is 3 M? (3) What was the initial concentration of C when the reaction was run under the conditions shown on the graph? C. (7 points) The kinetic data taken for the reaction E F is shown in Graph 3C. (1) What is the half-life for this reaction when the initial concentration of E is 0.02 M? (2) What was the initial concentration of E when the reaction was run under the conditions shown on the graph? DO NOT OBTAIN YOUR ANSWERS BY MAKING EXTRAPOLATIONS FROM THE VARIOUS GRAPHS. ALL ANSWERS ARE TO BE CALCULATED FROM THE DATA CONTAINED WITHIN THE GRAPHS. ZERO CREDIT WILL BE GIVEN FOR ANSWERS THAT DO NOT FOLLOW THIS RULE! A. (1) From the graph, we see that the reaction is zero order (plotting [A] vs. t gives a straight line with equation [A] = [A] 0 kt). Using the points on the graph at 30 min and 10 min, we find m = (0.4 0.8)/(30 10) = -0.02 M min-1. Since k = -m, the rate constant k has a value of 0.02 M min -1. For zero order processes, t 1/2 = [A] 0 /2k. For [A] 0 = 0.15 M, t 1/2 = 3.75 min. (2) Using [A] = [A] 0 kt, we can write 0 = 5 M 0.02(t) or t = 250 min. B. (1) The graph indicates that the experimental data obey first order kinetics, for which ln [C] = ln [C] 0 kt. Using the points on the graph at 30 min and 10 min, we can evaluate the slope as ((- ) 0.3)/(30 10 min) = -0.8/20 = -0.04 min -1. Since k = -m, the value of k is 0.04 min -1. The half-life of a 1 st order reaction is independent on initial concentration and is given by t 1/2 = 0.693/k = 17.3 min. (2) For first order processes, there is no defined time at which the concentration of C will become exactly zero. Note that the natural log of 0 is undefined. The value of [C] can only approach zero with increasing time. (3) To calculate [C] 0, we need to pick a concentration on the graph, along with its corresponding time. Picking 10 min, with a corresponding value of ln [C] = 0.3, we can write 0.3 = ln [C] 0-0.04 min -1 (10 min). Solving for ln[c] 0, we get a value of 0.7. Taking the antilog, we obtain [C] = 2.01 M. C. (1) Since a plot of 1/[E] vs. time gives a linear plot, we can state that the reaction is second order in E and that the equation of the straight line fitting the data is 1/[E] = 1/[E] 0 + kt. Picking time points at 10 min and 50 min, the slope of the line in the graph can be calculated as (2.5 0.9)/(50 10) = 16/40 = 0.04 M -1 min -1 = k. For 2 nd order process, 2 nd order in a single component, t 1/2 = 1/k[E] 0. Thus t 1/2 = 1/(0.04 M -1 min -1 )(0.02 M) = 1250 min. (2) Using a time and concentration corresponding to a time of 10 min, along with the k determined above, we can write [0.9] = 1/[E] 0 + (0.04 M -1 min -1 )(10 min). Evaluating, we find 1/[E] 0 = or [E] 0 = 2M.
[A] Problem 3A Graph 0.9 0.8 0.7 0.6 0.4 0.3 0.2 10 20 30 40 50 [A] Problem 3B Graph 0 ln [C] - -1 ln [C] -1.5 10 20 30 40 50 60 3 Problem 3C Graph 1/[E] 2.5 2 1.5 1 10 20 30 40 50 60 1/[E]
4. (15 points) Answer the following: (a) (5) The half -life of a first order reaction X Y is observed to be 200 minutes at 27 C and 100 minutes at 37 C. At what temperature would the half-life be 30 minutes? To solve this problem, use the relation ln(k 2 /) = (E a /R)(T 2 -T 1 /T 1 T 2 ). We can evaluate the rate constants at the two temperatures by using the formula k = 0.693/t 1/2. Using the half-lives given at the two temperatures, we find that k at 27 C = 300 K = 0.00347 min and k at 310 K = 0.00693. Substituting these values into the above relationship, and using R = 8.314 J/K mol, we find E a = 53.58 kj. For a reaction that has a half-life of 30 min, the corresponding k will be 0.693/30 min = 0.0231 min -1. Substituting this value into the above equation with the and using the data for 300 K, we obtain ln ((0.0231)/0.00347)) = (53580/8.314){T 2-300)/300T 2 }. Carrying out multiplications and divisions, we obtain 1.89 = 6436T 2 1.931x 10 6 )/300T 2 or 567T 2 = 6436T 2 1.931 x 10 6 or 5869.T 2 = 1.931 x 10 6. Solving, we find T 2 = 329 K. (b) (5) The reaction W Z has a rate constant of 0,69 min -1 at 400 K in a reaction that has an activation energy of 53.6 kj. What is the value of the pre-exponential factor A for this reaction? We can write k 400 = Ae -53600/8.314(400) = 0.69 min -1 = Ae -16.12 = A(9.98 x 10-8 ) or A = 6.91 x 10 6 min -1. (c) (5) The unimolecular reaction F G has an energy of reaction (ΔU) of 5 kj and an activation energy E a of 25 kj. What is the activation energy of the reaction G F? To solve this problem, we can draw the diagram below, which depicts the relationship between E a for the process F G, E a for the process G H and ΔU. In particular, we see that E(F G) = E(G F) + ΔU. Thus, we obtain the E a for G F to be E(F G) - ΔU or E(G F) = (25 5) kj = 20 kj. t.s. E(F G) F E(G F) G ΔU