Exam I Solutions Chem 6, 9 Section, Spring 2002

Transcription

1 1. (a) Two researchers at the University of Nebraska recently published a paper on the rate of the disappearance of World Wide Web links, a phenomenon called link rot. They asked the question, If I place on the web a page containing links to sites over which I have no control, how rapidly do those sites disappear? They studied web sites for three Nebraska courses that, together, had 515 unique links to other sites when they were first placed on the web. They found that links disappeared in a first-order fashion with a 55 month half life. The entire General Chemistry web site has 13 external links right now. How many can I expect to be invalid next year at this time? A 55 month half-life for a first-order reaction (which here is valid link invalid link) corresponds to a first-order rate constant k = ln()/(55 months) = month 1. If we start with 13 valid links now, then the number still valid in one year (1 months) is Remaining links = (Currently links) e kt = 13 e (0.016 month 1 )(1 months) = so that the number invalid should be about (13 11) =. (b) The mass-60 isotope of cobalt, 60 Co, is widely used in radiotherapy. Hospitals buy this isotope in a standard amount and thus a fixed initial radioactivity. The sample must be replaced when the radioactivity falls to 75% of its initial strength. How long does this take, given that 60 Co has a half-life of 5.6 years? When the radioactivity has fallen to 75% of its initial strength, the ratio of the initial amount of 60 Co to the final amount is 0.75, and the time this takes is given by solving 0.75 = e kt for t where k = ln()/t 1/ and t 1/ is the half-life. We find t = ln(0.75)/k = ln(0.75)(t 1/ )/ln() =.18 yr.. In class, we saw several examples of the catalytic decomposition of hydrogen peroxide, H O, according to the net reaction H O H O + O In acidic solution, bromide ion, Br, also acts as a catalyst, according to the following twostep mechanism: (1) Br + H + + H O Br + H O () Br + H O Br + H + + O (a) Assuming step (1) is much slower than step (), predict the net reaction rate law. Be sure to define any symbols you need. If step (1) is the slow step, then the rate of that step will determine the net reaction rate. Let k 1 be the rate constant for step (1) in the forward direction. Then we can write the rate law for that step as d[h O ] = k 1 [Br ] [H + ] [H O ]

2 Next, we note that a fast step () means that as soon as step (1) destroys one H O, step () immediately destroys another H O. This tells us that the net rate of destruction of H O must be twice as big as that for step (1) alone: Net rate of peroxide destruction = d[h O ] = k 1 [Br ] [H + ] [H O ] But the stoichiometry of the net reaction says that we should write the net reaction rate law as Rate of net reaction = 1 d[h O ] = 1 k 1 [Br ] [H + ] [H O ] = k 1 [Br ] [H + ] [H O ] so that our final answer does not contain a factor of. (b) Next, assume step () is much slower than step (1). What net reaction rate law does this assumption predict? If step () is the slow step, then we define k to be that step s forward rate constant and write d[h O ] = k [Br ] [H O ] Next, we assume step (1) reaches equilibrium and use its equilibrium constant K eq (expressed also in terms of the forward and reverse rate constants of step (1) in the usual way) to find an expression for the bromine concentration, [Br ]: K eq = k 1 k 1 = Substitution gives d[h O ] [Br ] [Br ] [H + ] [H O ] or [Br ] = K eq [Br ] [H + ] [H O ] = k [Br ] [H O ] = k K eq [Br ] [H + ] [H O ] but here, one peroxide can t be destroyed in step () until another is destroyed first in step (1). This makes the net rate of destruction of H O twice as big as the expression above suggests: Net rate of peroxide destruction = d[h O ] = k K eq [Br ] [H + ] [H O ] Finally, we go back to the net reaction and use its stoichiometry to write the rate of the net reaction, and once again, the factor of disappears, just as it did in part (a): Rate of net reaction = 1 d[h O ] = 1 k K eq [Br ] [H + ] [H O ] = k K eq [Br ] [H + ] [H O ] (c) Bromide ion is colorless, but Br in a water solution has a characteristic red-brown color. If a solution of bromide ion is added to a solution of H O, one observes that the solution quickly develops a red-brown color that lasts until the solution stops generating O gas, at which point the solution becomes colorless again. How does this observation let you decide which step in the mechanism really is the slower step? Step () must be slow to allow [Br ] to reach a great enough value to color the solution. If step () is fast, Br is destroyed as fast as it is produced, and we see no color.

3 3. At high temperatures, the gas phase reaction between NO and H to produce N and H O proceeds at a measurable rate according to the net reaction NO(g) + H (g) N (g) + H O(g) In three separate experiments, various initial amounts of NO and H were mixed at constant temperature in a container of fixed volume, and the initial rates of appearance of water were measured, as shown in the table below. Note that the amounts are expressed as partial pressures of each gas in atm units. Experiment P NO /atm P H /atm (dp HO/)/atm s (a) Fill in the blanks below: Net Rate = 1 dp NO = 1 dp H = + dp N = + 1 dp H O (b) Find the rate law and the rate constant for this reaction. Explain, in words or mathematics, how you arrived at your answer. Comparing experiments 1 and, we see that doubling P NO increases the water appearance rate by a factor of four, which tells us the reaction is second order in NO. Likewise, comparing experiments 1 and 3 tells us that the water appearance rate doubles when P H doubles, making the reaction first order in H. We can now write the net reaction rate law: Net rate = + 1 dp H O = k P NO P H We solve this expression for the rate constant k and use data from any of the three experiments to calculate its numerical value: k = 1 dp H O/ = atm s 1 P NO P H (c) In a fourth experiment, the inital partial pressure of H was P o H =.0 atm, but the inital partial pressure of NO was only P o NO = atm. How long would it take for the partial pressure of NO to fall to half this value? Note: Use the rate law and rate constant you derived in part (b) above. If you did not get an answer to part (b), you may make up any rate constant and rate law that allows you to answer this part for, at most, half credit. Here, the initial amount of H is much greater than the initial amount of NO, which tells us that P H is essentially constant as NO disappears. This allows us to define an effective rate constant

4 k eff = k P o H = atm s 1.0 atm = atm 1 s 1 and to write the rate law as Net rate = + 1 dp H O = k P NO P H = k eff P NO This is a simple second-order rate law, and the half-life expression for a second-order process of this type gives us our final answer: t 1/ = 1 = k eff P o 4 s NO (d) This reaction has a large equilibrium constant (i.e., it goes to completion ). What is the final, equilibrium total pressure of the second of the experiments tabulated at the start of this problem? In this net reaction, whenever four moles of gaseous reactants disappear, they are replaced by three moles of gaseous products. The total initial pressure of the reactants in the second experiment is the sum of the reactant partial pressures: 0.5 atm atm = 1.0 atm. When the reactants are all gone (at equilibrium), the final total pressure will thus be three-fourths of this initial pressure: 0.75 atm. 4. The experimental activation energy in a net reaction reflects the activation energies of all the elementary steps in the net reaction s mechanism in a way that depends on the mechanism. For example, we considered in class the simple net reaction A + B C in terms of an equally simple two-step mechanism involving a single intermediate I: (1) A + B I () I C k (a) Apply the steady-state approximation to the intermediate I to find an expression for the experimental rate constant k in terms of the three rate constants of the mechanism, k 1, k 1, and k. (This should be easy we did this in class!) Rate = d[c] = k [I] d[i] Steady state: = 0 = k 1 [A] [B] k 1 [I] k [I] k Thus: [I] = 1 [A] [B] k 1 + k Rate = d[c] k = 1 k [A] [B] = k [A] [B] k 1 + k k k = 1 k k 1 + k k 1 k 1

5 (b) Now assume step (1) has an activation energy E 1 in the forward direction and E 1 in the backward direction, and assume step () has an activation energy E that is much larger than either E 1 or E 1. Use your result from part (a) to find an expression for the observable activation energy of the experimental rate constant k. If E is much larger than either E 1 or E 1, we can say that, first, step () is the slow step (so that k is small) and that, second, the observable net reaction activation energy should be close to E because it dominates the activation energies of each step in the mechanism. Let s now prove this by writing an Arrhenius expression for each rate constant in the mechanism and introducing these expressions into the expression for k found in part (a). k k = 1 k k 1 k because k k 1 + k k is small (E is large) 1 k 1 = A 1 e E 1 /RT k 1 = A 1 e E 1 /RT k = A e E /RT so k = A 1 e E 1 /RT A e E /RT A 1 e E 1 /RT = A 1 A e A E 1 + E E 1 /RT 1 but also k = A exp e E a /RT where E a is the observable activation energy Comparing these two expressions for k shows that A exp = A 1 A and A 1 E a = E 1 + E E 1 E (because E is larger than E 1 or E 1 ) 5. (a) An elementary bimolecular gas-phase reaction with zero activation energy will have (circle one) the same rate a slower rate a faster rate a different rate in an unpredictable way if the temperature of the reaction is increased but the reactant concentrations are kept fixed. The bimolecular gas-phase rate constant expression depends on temperature in two ways: the activation energy factor and the relative collision speed factor ( 4RT/πM). If the activation energy is zero, then the temperature influences the rate constant only through the relative collision speed factor, which increases with increasing temperature, making the rate faster. (b) Is it likely that the net reaction in problem 3, NO + H N + H O, is an elementary reaction? Why or why not? (Do not use your answer to problem 3 to justify your answer here.) An elementary reaction represents a realizable, simultaneous, physical collision among the reactants (or, for a unimolecular process, a single reactant undergoing a spontaneous, isolated physical change). Here, four reactant molecules would have to collide at the same time, which is too improbable.

6 (c) An elementary reaction with a 55 kj mol 1 forward activation energy and a 30 kj mol 1 net internal energy change has a reverse activation energy of (circle one) 85 kj mol kj mol 1 5 kj mol 1 +5 kj mol kj mol 1 A 30 kj mol 1 internal energy change means that the products are 30 kj mol 1 lower in energy than the reactants. Thus, to reach the transition state, the products have to surmount an activation energy barrier that is 30 kj mol 1 higher than the reactants, for a total activation energy of ( ) kj mol 1 = 85 kj mol 1. (d) A catalytic enzyme that changes a reaction s activation energy from a value E 1 to a lower value E such that E 1 = E but does not change the rate constant s pre-exponential factor A changes the reaction rate constant by a factor of (circle one) e e E /RT e E /RT e E 1/RT e E 1/RT e E /RT Let k 1 be the rate constant without the catalyst and k be the rate constant with the catalyst. Then the factor we seek is k /k 1, and we can write k 1 = A e E 1 /RT = A e E /RT and k = A e E /RT so that the factor is k = A e E /RT k 1 A e E /RT = ee /RT (e) In the mechanism of the net reaction between H and Br (H + Br HBr), which is a chain reaction mechanism, there are two steps that are called chain propagation steps. Write the elementary reaction for either of these two propagation steps. A chain propagation step consumes one intermediate while generating a second intermediate that, in turn, can regenerate the first in another step in the chain. In this reaction, the intermediates are H atom and Br atom, and the two propagation steps are H + Br HBr + Br Br + H HBr + H