28 Processes at solid surfaces

Transcription

1 28 Processes at solid surfaces Solutions to exercises E28.b E28.2b E28.3b Discussion questions The motion of one section of a crystal past another a dislocation results in steps and terraces. See Figures 28.2 and 28.3 of the text. A special kind of dislocation is the screw dislocation shown in Fig Imagine a cut in the crystal, with the atoms to the left of the cut pushed up through a distance of one unit cell. The surface defect formed by a screw dislocation is a step, possibly with kinks, where growth can occur. The incoming particles lie in ranks on the ramp, and successive ranks reform the step at an angle to its initial position. As deposition continues the step rotates around the screw axis, and is not eliminated. Growth may therefore continue indefinitely. Several layers of deposition may occur, and the edges of the spirals might be cliffs several atoms high Fig Propagating spiral edges can also give rise to flat terraces Fig Terraces are formed if growth occurs simultaneously at neighbouring left- and right-handed screw dislocations Fig Successive tables of atoms may form as counter-rotating defects collide on successive circuits, and the terraces formed may then fill up by further deposition at their edges to give flat crystal planes. Consult the appropriate sections of the textbook listed below for the advantages and limitations of each technique. AFM: 28.2h and Box 28.; FIM: 28.5c; LEED: 28.2g; MBRS: 28.6c; MBS: 28.2i; SAM: 28.2e; SEM: 28.2h; and STM: 28.2h. In the Langmuir Hinshelwood mechanism of surface catalysed reactions, the reaction takes place by encounters between molecular fragments and atoms already adsorbed on the surface. We therefore expect the rate law to be second-order in the extent of surface coverage: A + B P ν = kθ A θ B Insertion of the appropriate isotherms for A and B then gives the reaction rate in terms of the partial pressures of the reactants. For example, if A and B follow Langmuir isotherms eqn 28.5, and adsorb without dissociation, then it follows that the rate law is ν = kk A K B p A p B + K A p A + K B p B 2 The parameters K in the isotherms and the rate constant k are all temperature dependent, so the overall temperature dependence of the rate may be strongly non-arrhenius in the sense that the reaction rate is unlikely to be proportional to exp E a /. In the Eley-Rideal mechanism ER mechanism of a surface-catalysed reaction, a gas-phase molecule collides with another molecule already adsorbed on the surface. The rate of formation of product is expected to be proportional to the partial pressure, p B of the non-adsorbed gas B and the extent of surface coverage, θ A, of the adsorbed gas A. It follows that the rate law should be A + B P ν = kp A θ B The rate constant, k, might be much larger than for the uncatalysed gas-phase reaction because the reaction on the surface has a low activation energy and the adsorption itself is often not activated. If we know the adsorption isotherm for A, we can express the rate law in terms of its partial pressure, p A. For example, if the adsorption of A follows a Langmuir isotherm in the pressure range

2 PROCESSES AT SOLID SURFACES 453 of interest, then the rate law would be ν = k Ap B + A. E28.4b E28.5b If A were a diatomic molecule that adsorbed as atoms, we would substitute the isotherm given in eqn 28.8 instead. According to eqn 28.24, when the partial pressure of A is high in the sense A l, there is almost complete surface coverage, and the rate is equal to kp B. Now the rate-determining step is the collision of B with the adsorbed fragments. When the pressure of A is low A, perhaps because of its reaction, the rate is equal to k A p B ; and now the extent of surface coverage is important in the determination of the rate. In the Mars van Krevelen mechanism of catalytic oxidation, for example in the partial oxidation of propene to propenal, the first stage is the adsorption of the propene molecule with loss of a hydrogen to form the allyl radical, CH 2 =CHCH 2. An O atom in the surface can now transfer to this radical, leading to the formation of acrolein propenal, CH 2 =CHCHO and its desorption from the surface. The H atom also escapes with a surface O atom, and goes on to form H 2 O, which leaves the surface. The surface is left with vacancies and metal ions in lower oxidation states. These vacancies are attacked by O 2 molecules in the overlying gas, which then chemisorb as O 2 ions, so reforming the catalyst. This sequence of events involves great upheavals of the surface, and some materials break up under the stress. Zeolites are microporous aluminosilictes, in which the surface effectively extends deep inside the solid. M n+ cations and H 2 O molecules can bind inside the cavities, or pores, of the Al O Si framework see Fig of the text. Small neutral molecules, such as CO 2,NH 3, and hydrocarbons including aromatic compounds, can also adsorb to the internal surfaces and this partially accounts for the utility of zeolites as catalysts. Like enzymes, a zeolite catalyst with a specific composition and structure is very selective toward certain reactants and products because only molecules of certain sizes can enter and exit the pores in which catalysis occurs. It is also possible that zeolites derive their selectivity from the ability to bind to stabilize only transition states that fit properly in the pores. Numerical exercises The number collisions of gas molecules per unit surface area is Z W = a For N 2 N A p 2πM / mol 0.0Pa i Z W = 2π kg mol J K mol 298 K /2 = m 2 s = cm 2 s ii Z W = mol Torr Pa/760 Torr 2π kg mol J K mol 298 K /2 = m 2 s = cm 2 s

3 454 INSTRUCTOR S MANUAL b For methane mol 0.0Pa i Z W = 2π kg mol J K mol 298 K /2 = m 2 s = cm 2 s ii Z W = mol Torr Pa/760 Torr 2π kg mol J K mol 298 K /2 = m 2 s = cm 2 s E28.6b The number of collisions of gas molecules per unit surface area is Z W = p = N A p 2πM /2 so p = Z WA2πM /2 N A A s mol π / m 2 2π kg mol J mol K 525 K /2 = Pa E28.7b The number of collisions of gas molecules per unit surface area is Z W = N A p 2πM /2 so the rate of collision per Fe atom will be Z W A where A is the area per Fe atom. The exposed surface consists of faces of the bcc unit cell, with one atom per face. So the area per Fe is A = c 2 and rate = Z W A = where c is the length of the unit cell. So rate = N A pc 2 2πM / mol 24 Pa m 2 2π kg mol J K mol 00 K /2 = s E28.8b The number of CO molecules adsorbed on the catalyst is N = nn A = pv N A =.00 atm L mol L atm K mol 273 K = The area of the surface must be the same as that of the molecules spread into a monolayer, namely, the number of molecules times each one s effective area A = Na = m 2 = 8.8m 2

4 PROCESSES AT SOLID SURFACES 455 E28.9b If the adsorption follows the Langmuir isotherm, then θ = + so K = θ p θ = V/V mon p V/V mon Setting this expression at one pressure equal to that at another pressure allows solution for V mon V /V mon p V /V mon = V 2 /V mon p 2 V 2 /V mon V mon = so p V mon V V = p 2V mon V 2 V 2 p p kpa = = 9.7cm3 p /V p 2 /V /.60 04/2.73 kpa cm 3 E28.0b The mean lifetime of a chemisorbed molecule is comparable to its half life: Ed t /2 = τ 0 exp J mol s exp J K mol = 200 s 500 K E28.b The desorption rate constant is related to the mean lifetime half-life by t = ln 2/k d so k d = ln 2/t The desorption rate constant is related to its Arrhenius parameters by E d k d = A exp so ln k d = ln A E d and E d = ln k ln k 2 R T2 T = ln.35 ln J K mol 600 K 000 K E d = J mol E28.2b The Langmuir isotherm is θ = + so p = θ K θ 0.20 a p = kpa = 0.32 kpa b p = kpa 0.75 = 3.9kPa E28.3b The Langmuir isotherm is θ = + We are looking for θ, so we must first find K or m mon θ K = p θ = m/m mon p m/m mon Setting this expression at one pressure equal to that at another pressure allows solution for m mon m /m mon p m /m mon = m 2 /m mon p 2 m 2 /m mon so p m mon m m = p 2m mon m 2 m 2 p p kpa m mon = = = 0.84 mg p /m p 2 /m / /0.2 kpa mg So θ = 0.63/0.84 = 0.75 and θ 2 = 0.2/0.84 = 0.25

5 456 INSTRUCTOR S MANUAL E28.4b The mean lifetime of a chemisorbed molecule is comparable to its half-life E d t /2 = τ 0 exp a At 400 K : t /2 = J mol s exp J K mol 400 K At 800 K : b At 400 K : At 800 K : = s t /2 = J mol s exp J K mol 800 K = s t /2 = J mol s exp J K mol 400 K = s t /2 = J mol s exp J K mol 800 K =.4s E28.5b The Langmuir isotherm is θ = + so p = θ K θ For constant fractional adsorption K pk = constant so p K = p 2 K 2 and p 2 = p K 2 ad H But K exp so K ad H = exp K 2 R T T 2 ad H p 2 = p exp R T T J mol = 8.86 kpa exp J K mol 298 K = 6.50 kpa 38 K E28.6b The Langmuir isotherm would be a θ = + b θ = /2 + /2 c θ = /3 + /3

6 PROCESSES AT SOLID SURFACES 457 A plot of θ versus p at low pressures where the denominator is approximately would show progressively weaker dependence on p for dissociation into two or three fragments. E28.7b The Langmuir isotherm is θ = + so p = θ K θ For constant fractional adsorption pk = constant so p K = p 2 K 2 and ad H But K exp and ad H = R T T 2 so p 2 ad H = exp p R ln p p 2, p 2 p = K K 2 T T 2 ad H = J K mol 80 K 350 kpa ln 240 K kpa = J mol = 6.40 kj mol E28.8b The time required for a given quantity of gas to desorb is related to the activation energy for desorption by E d t exp so and E d = R ln t T T 2 t 2 t E d = exp t 2 R T T 2 E d = J K mol 873 K ln 856 s 02 K 8.44 s = J mol a The same desorption at 298 K would take J mol t = 856 s exp J K mol b The same desorption at 500 K would take J mol t = 8.44 s exp J K mol 298 K = s 873 K 500 K 02 K = s

7 458 INSTRUCTOR S MANUAL Solutions to problems Solutions to numerical problems P28.2 Refer to Fig Figure 28. The 00 and 0 faces each expose two atoms, and the face exposes four. The areas of the faces of each cell are a 352 pm 2 = cm 2, b pm 2 = cm 2, and c pm 2 = cm 2. The numbers of atoms exposed per square centimetre are therefore a b c cm 2 =.6 05 cm cm 2 =.4 05 cm cm 2 = cm 2 For the collision frequencies calculated in Exercise 28.5a, the frequency of collision per atom is calculated by dividing the values given there by the number densities just calculated. We can therefore draw up the following table Hydrogen Propane Z/atom s 00 Pa 0 7 Torr 00 Pa 0 7 Torr P28.4 V cz = V mon z{ cz} This rearranges to [28.0, BET isotherm,z= pp ] z zv = c z + cv mon cv mon Therefore a plot of the left-hand side against z should result in a straight line if the data obeys the

8 PROCESSES AT SOLID SURFACES 459 BET isotherm. We draw up the following tables a 0 C,p = 3222 Torr p/torr z z zv /cm b 8 C,p = 648 Torr p/torr z z zv /cm The points are plotted in Fig. 28.2, but we analyse the data by a least-squares procedure. The intercepts are at a and b Hence cv mon = a cm 3, b cm 3 The slopes of the lines are a 76.0 and b Hence c cv mon = a cm 3, b cm 3 Solving the equations gives c = a 63.3, b and hence c = a 64, b 264 V mon = a 3.cm 3, b 2.5cm 3 20 a b Figure 28.2

9 460 INSTRUCTOR S MANUAL P28.7 We assume that the Langmuir isotherm applies. θ = + [28.5] and θ = + For a strongly adsorbed species, and θ =. Since the reaction rate is proportional to the pressure of ammonia and the fraction of sites left uncovered by the strongly adsorbed hydrogen product, we can write dp NH3 dt = k c p NH3 θ k cp NH3 H2 To solve the rate law, we write p H2 = 3 2 {p 0NH 3 p NH3 } [ NH3 2 N H 2] from which it follows that, with p = p NH3 dp dt = kp p 0 p, k = 2k c 3K This equation integrates as follows p p t 0 dp = k dt p 0 p 0 or p p 0 t = k + p 0 t ln p p 0 We write F = p 0 t ln p p 0,G= p p 0 t and obtain G = k + F = p 0 F Hence, a plot of G against F should give a straight line with intercept k at F = 0. Alternatively, the difference G F should be a constant, k. We draw up the following table t/s p/torr G/Torr s F /Torr s G F /Torr s Thus, the data fit the rate law, and we find k = 0.02 Torr s. P28.9 Taking the log of the isotherm gives ln c ads = ln K + ln c sol /n so a plot of ln c ads versus ln c sol would have a slope of /n and a y-intercept of ln K. The transformed data and plot are shown in Fig

10 PROCESSES AT SOLID SURFACES Figure 28.3 c sol /mg g c ads /mg g ln c sol ln c ads K = e.9838 mg g = 0.38 mg g and n = /.7 = 0.58 In order to express this information in terms of fractional coverage, the amount of adsorbate corresponding to monolayer coverage must be known. This saturation point, however, has no special significance in the Freundlich isotherm i.e. it does not correspond to any limiting case. P28. The Langmuir isotherm is θ = + = n n so n + = n and p n = p + n Kn So a plot of p/n against p should be a straight line with slope /n and y-intercept /Kn. The transformed data and plot Fig follow p/kpa n/mol kg p/n kpa mol kg Figure 28.4

11 462 INSTRUCTOR S MANUAL n = mol = 5.78 mol kg kg The y-intercept is b = so K = = Kn bn kpa mol kg 5.78 mol kg K = kpa = 7.02 Pa P28.2 For the Langmuir adsorption isotherm we must alter eqn 5 so that it describes adsorption from solution. This can be done with the transforms p concentration,c V amount adsorbed per gram adsorbent,s Langmuir isotherm and regression analysis c s = c + s Ks = 0.63 g mmol, standard deviation = 0.07 g mmol s = 35.6 mmol L g mmol, standard deviation = 5.9 mmol L g mmol Ks R Langmuir = K = 0.63 g mmol 35.6 mmol L g mmol = L mmol Freundlich isotherm and regression analysis s = c c /c 2 c = 0.39, standard deviation = 0.02 = 0.539, c 2 standard deviation = R Freundlich = Temkin isotherm and regression analysis s = c lnc 2 c c =.08, standard deviation = 0.4 c 2 = 0.074, standard deviation = R Temkin = The correlation coefficients and standard deviations indicate that the Freundlich isotherm provides the best fit of the data.

12 PROCESSES AT SOLID SURFACES 463 Solutions to theoretical problems P28.7 θ = +, θ = V V p = θ K θ = V KV V dp dv = KV V + dµ = σ V KV V 2 = V dlnp = pσ V dp KV V = V σ V V dv = σ V V Therefore, we can adopt any of several forms, dµ σ V σ = dv = dv = V V θ V KV V 2 V KV V 2 V σ θ dv dθ = V σ dln θ P28.8 For the Langmuir and BET isotherm tests we draw up the following table using p = 200 kpa = 500 Torr [Examples 28. and 28.3] p V p/torr p / Torr cm V 0 3 z z zv /cm 3 is plotted against p in Fig. 28.5a, and z zv is plotted against z in Fig. 28.5b Figure 28.5a

13 Figure 28.5b 464 INSTRUCTOR S MANUAL We see that the BET isotherm is a much better representation of the data than the Langmuir isotherm. The intercept in Fig. 28.5b is at , and so = cm 3. The cv mon slope of the graph is 9.93, and so c cv mon = cm 3 Therefore, c = 2.98, and hence c = 3.98, V mon = 75.4cm P28.22 a Kunit: g R L [g R = mass grams of rubber] /n K F unit: mg gr L /n K L unit: mg L M unit: mg g R b Linear sorption isotherm q = Kc eq K = q c eq so K is best determined as an average of all q/c eq data pairs. K av = 0.26g R L, standard deviation = 0.04g R L 95 per cent confidence limit: g R L If this is done as a linear regression, the result is significantly different. K linear = 0.083g R L, R linear = standard deviation = g R L Freundlich sorption isotherm: q = K F c /n eq, using a power regression analysis, we find that K F = 0.64, standard deviation = 0.37

14 PROCESSES AT SOLID SURFACES 465 c = 0.877, standard deviation = 0.3; n =.4 n R Freundlich = Langmuir sorption isotherm q = K LMc eq + K L c eq q = + K L M c eq M K L M = g R L, standard deviation =.03; K L = g R L M = g R mg, R Langmuir = standard deviation = 0.985; M = 233 mg g R All regression fits have nearly the same correlation coefficient so that cannot be used to determine which is the best fit. However, the Langmuir isotherm give a negative value for K L. If K L is to represent an equilibrium constant, which must be positive, the Langmuir description must be rejected. The standard deviation of the slope of the Freundlich isotherm is twice as large as the slope itself. This would seem to be unfavourable. Thus, the linear description seems to be the best, but not excellent choice. However, the Freundlich isotherm is usually preferred for this kind of system, even though that choice is not supported by the data in this case. q rubber 0.64 c.4 eq = q charcoal c.6 eq = 0.64 c 0.46 eq The sorption efficiency of ground rubber is much less than that of activated charcoal and drops significantly with increasing concentration. The only advantage of the ground rubber is its exceedingly low cost relative to activated charcoal, which might convert to a lower cost per gram of contaminant adsorbed.