where a + b = 2 (this is the general case) These all come from the fact that this is an overall second order reaction.

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Walter Blair
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1 Chapter 7 Problems Page of 6 // Hydrolysis of ethyl acetate is as follows: EtAc + OH - Ac - + EtOH. At 5 ºC, the disappearance of OH - is used to determine the extent of the reaction, leading to data in the table (not shown here). Only reactants alter the rate of the reaction. (a) What is the overall order of the reaction? (b) Calculate the rate constant, with units. (c) How long would it take for [OH - ] to reach M for each experiment? (d) Based on (a), what are the possible rate laws for the reaction? (e) Describe an experiment that would allow you to determine the correct rate law. a) Note that the half-life decreases twofold as both reactant concentrations double. First order reactions do not have half-lives that depend on the initial concentration of ratelimiting species. This is then a second or higher order reaction. Both of the reactants start at the same initial concentration, so this will be act as a class I reaction even if it is class II. Since the half-life decreases by ½ as concentration doubles, this is a secondorder reaction. b) From part a) we see that this second order reaction has a half-life of t / /k [MeAc] or k /t / [MeAc] /{(000 sec)(0.0 M)} 0. M - sec -. This could also be computed with [OH - ] or with the top line of data. c) This is either a class I or a class II reaction, but both A and B are at the same concentration at the start, so it behaves like a class I reaction (see p. 33). Thus, /[OH - ] - /[OH - ] 0 k t: /(0.005 M) /(0.005 M) (0. M - s - )t for the first condition, or t (00 M - )/(0. M - s - ) 000 sec d) Possible rate laws could be v k [OH - ] v k [OH - ] [MeAc] v k [MeAc] v k [OH - ] a [MeAc] b /(0.005 M) /(0.0 M) (0. M - s - )t for the second condition, or t (300 M - )/(0. M - s - ) 3000 sec where a + b (this is the general case) These all come from the fact that this is an overall second order reaction. e) If you do a reaction in which the initial concentrations of the two reactants are different from each other (like [OH - ]) 0.0 M and [MeAc] M), then the initial rate of the reaction will tell you which of the rate laws is effective. If the half-life changes from 000 sec to 500 sec, then the second law is it. If the half-life decreases by half, then the first law is it. If the half-life doesn t change, then the 3 rd law is it. If the half-life somewhere between 000 and 000 sec, but not 500 sec, then you can determine the values of a and b for the fourth law.

2 Chapter 7 Problems Page of 6 // The stoichiometric equation for a reaction is A + B C + D Data in a table show the initial rate of formation of C as a function of initial concentrations of A and B. (a) What is the order of the reaction with respect to A? (b) With respect to B? (c) Write a differential equation for the production of C based on the answers to a and b. (d) What is the rate constant? (e) Give a possible mechanism and discuss how to show the mechanisms is consistent with the experiment. a) From the first two lines of the table, with [B] constant, v depends on [A] like [A] (when [A] doubles, v goes up by 4. So this reaction is second order with respect to A. b) Comparing the first and third lines, with [A] constant, there is no dependence of v on [B], so the reaction is zero order with respect to B. c) dc k[a]. dt d) Pick a line of the table, say the first one. From that we see that [A].0 M and v (dc/dt) is 0-3 M/s. Plug this into the equation in c) and rearrange terms: k dc dt [ A] 3 3 ( 0 ) 0 M )( s (M ) M s e) The rate determining step depends on two As interacting, so A + A A with rate constant k is a possible first step. Then the A dimer might rapidly interact with B to make C and D: A + B C + D + A with rate constant k is a possible second step. Note that only one A net has been consumed. The rate of formation of C is essentially equal to the rate of formation of A (from the slow reaction). When A is formed, it rapidly reacts with B to make C, D and A.

3 Chapter 7 Problems Page 3 of 6 // The following data were obtained for c vs. t for a reaction. Values were measured at sec intervals from 0 to 0 sec: 0, 6.9, 4.98, 4.3, 3.55, 3.,.6,.5,.,.9,.8,.65,.5,.36,.4,.3,.,.3,.09,, 0.9. (a) Plot c vs. t, ln(c) vs. t and /c vs. t. (b) Determine the kinetic order from the plots. Calculate k and write the simplest mechanism. (c) Describe a method that can be used to measure such rapidly changing concentrations. a) The three plots are shown below. Both c vs. t and ln(c) vs. t are clearly nonlinear, while the /c vs. t plot is basically linear. 0 c vs. t.5 ln(c) vs. t 8.5 c 6 ln (c) t (sec) t (sec). /c vs. t 0.8 /c t (sec) b) The linear /c vs. t plot suggests that this reaction follows second order kinetics. Doing a least-squares linear fit to the /c vs. t data gives a slope of , the units of which would be M - s - (because the slope is Δ(/c) / Δt, which has units (M - )/(sec). For a second order reaction, /c /c 0 + kt so the slope of a /c vs. t plot is just k, the rate constant. So k M - s -. The simplest possible mechanism uses two As to make product: A + A B with a rate constant k. c) Using a rapid mixing system would be appropriate. Detection by absorption (if either A or B is colored) would be fast. Using some optical detection method is usually helpful, if some sort of optical signal (absorption, fluorescence, optical rotation) can be measured. From what we re given here, it s hard to say more than that.

4 Chapter 7 Problems Page 4 of 6 // There is evidence that a critical concentration of a trigger protein is needed for cell division. This unstable protein is continually being synthesized and degraded. The rate of synthesis controls how long it takes to build up to the concentration needed for the start of DNA synthesis to initiate cell division. The trigger protein U is synthesized via a zero-order mechanism with rate constant k 0. It is degraded by a first-order mechanism with rate constant k. (a) Write a differential equation consistent with this mechanism. k (c) If U is being synthesized at a (b) Show it has a solution of the form k t [ U ] 0 ( e ) k constant rate with k 0 nm/sec and its half-life for degradation is 0.5 hr, calculate the maximum concentration that U will reach. How long will it take to reach this value? Plot [U] vs. t. (d) If a concentration of U of μm is needed to trigger DNA synthesis and cell replication, how long will it take to reach this concentration? (e) If the synthesis rate is halved, how long will it take for U to reach a concentration of μm? (f) What is the smallest rate of U synthesis that will allow cell replication if k is fixed? a) The synthesis step will have v k 0 and the degradation step v -k [U], so the net effect on velocity (the rate equation) will be v d[u]/dt k 0 k [U]. k b) If k t [ U ] 0 ( e ), then d[ U ] k0 d k t k t ( e ) k e 0. Plug these into our rate equation: k dt k dt k t k k t k t k e k k 0 ( e ) k k k e , which shows that d[u]/dt k 0 k [U]. k c) Since the half-life for degradation is 0.5 hr 30 min 800 sec, we know the rate constant for this first-order process: ln k 3.85 s. U will reach a t/ 800s maximum when d[u]/dt 0, that is, when 0 k 0 k [U] max. This says that [U] max k 0 /k ( nm/s)/( /s).6 μm. A plot of [U] vs. time is an increasing exponential that saturates at.6 μm. d) Because we know how [U] depends on t (from part b), we can calculate this value of t: 3 0 μm / s ( s ) t (μ M ) e or ( s ) t.6μ M (.6μM ) e or s ln( ) t s 0. 35hr s ln( e) As in d, 5 0 μm / s ) ( s ) t (μ M ) e or t s. 06hr s s f) We can just achieve [U] μm when the maximum value of [U] just makes it μm. So from part c, we have 0 k 0 k [U] max, or 0 k 0 ( s - ) ( μm), or k 0 ( μm s - ). 6

5 Chapter 7 Problems Page 5 of 6 // A reaction occurs with the stoichiometry A + P AP. The concentration of A was measured vs. time after mixing as shown in the table. (a) There is no dependence of the rate on [P]. What is the order of the reaction? Calculate the rate constant. (b) Propose a mechanism consistent with the kinetics and stoichiometry of the reaction. Write the rate law differential equation. (c) The data shown are for T 0 ºC. When the reaction was studied at 0 ºC, the rate constant doubled. What is the activation energy? a) A plot of ln[a] vs. t is a straight line, so this is first order with respect to A and zero order with respect to P, thus it is first order overall. The slope of the ln[a] vs. t plot is s - -k, so k s ln([a]) time (s) b) Because P doesn t enter into the rate, it isn t a reactant in the rate-limiting step reaction. A B (slow) B + P AP (fast) d[b]/dt k s [A] k f [B][P] 0 since we assume that [B] is near steady state. This says that [B] k s [A]/k f [P]. From the second equation, d[ap]/dt k f [B][P] k f (k s [A]/k f [P])[P] k s [A], where k s is the value calculated in a). RT T k kj mol K K K c) E (8.34 / )(73 )(83 ) A ln ln 44.5kJ / mol T T k 83K 73K

6 Chapter 7 Problems Page 6 of 6 // In aqueous solution the reaction of A to form B has the following rate expression: d[ A] + k[ A]{ + k'[ H ]}. (a) Propose a mechanism that is consistent with this dt experimental rate. Relate the ks in the mechanism to k and k. (b) From the ph dependence of the reaction, k was found to be 0 5 M -. In a ph 4 buffer, it took 5 min for a 0.3 M solution of A to reaction to give 0.5 M B. Calculate the value of k and find its units. a) It looks like this is a first-order reaction if the concentration of A is modified by the hydrogen ion concentration, so let s assume A + H + AH + fast AH + B slow (k ) A B slow (k ) Then d[b]/dt k [A] + k [AH + ] from the nd and 3 rd reactions. Note that the stoichiometric net reaction says that d[b]/dt -d[a]/dt, so -d[a]/dt k [A] + k [AH + ]. If the fast reaction is considered to be at steady state or equilibrium, then K [AH + ]/[A][H + ], so [AH + ] [A][H + ]K. Plugging this into the rate equation for [A] gives -d[a]/dt k [A] + k [AH + ] k [A] + (k K)[A][H + ] k [A]{ + (k K/k )[H + ]}, so k k and k k K/k. b) The half-time is 5 min 300 sec. From our rate equation, at ph 4, we have a value for the net rate constant that multiplies [A]: k{ + k [H + ]} k{ + ( 0 5 M - )(0-4 M)} k The half-life for a st order reaction is t / ln()/k, so 300 sec (0.693)/(k). Thus, k (0.693)/( 300 sec) sec min hr -.

Chemistry 112 Midterm January 30, 2006

Chemistry 112 Midterm January 30, 2006 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