Mechanisms and Rate Laws
7.4 PROBLEMS FOR CHAPTER 7
Similarly, from 7.3-9 and -13,
CP, = kcM “P2 CM(kCMT)2 1 + kCMT = (1 + kCMT)2
Proceeding in this way, from 7.3-10, we obtain in general:
CP, =
CM( kc&-l
(1 + kCMT)‘-l = cM[l + (kcM+l]l-’
(7.3-14)
(7.3-15)
Thus, the product distribution (distribution of polymer species P,) leaving the CSTR can be calculated, if cMO, k, and T are known.
For a BR or a PFR in steady-state operation, corresponding differential equations can be established to obtain the product distribution (problem 7-15).
7.4 Problems for Chapter 7 1 7 1 7-4 The oxidation of NO to NOz, which is an important step in the manufacture of nitric acid by the ammonia-oxidation process, is an unusual reaction in having an observed third-order rate constant (k~o in (-mo) = k~&oco,) which decreases with increase in temperature. Show how the order and sign of temperature dependence could be accounted for by a simple mech- anism which involves the formation of (NO)* in a rapidly established equilibrium, followed by a relatively slow bimolecular reaction of (NO)2 with 02 to form NOz.
7-5 (a) Verify the rate law obtained in Example 7-3, equation 7.1-3.
(b) For the HZ + Brz reaction in Example 7-3, if the initiation and termination steps involve a third body (M), Br2 + M -+ 2Bf + M, and 2Br + M -+ Br;! + M, respectively, what effect does this have on the rate law in equation 7.1-3? (The other steps remain as in Example 7-3.)
7-6 The rate of decomposition of ethylene oxide, C&40(A), to CI& and CO, has been studied by Crocco et al. (1959) at 900-1200 K in a flow reactor. They found the rate constant to be given by
kA = 10” exp(-21,00O/T)
in s-l (with Tin K). They proposed a free-radical chain mechanism which involves the initial decomposition of C&O into radicals (C2HsO’ and HO), and propagation steps which involve the radicals CzHaO* and CHj (but not HO) in addition to the reactant and products; termination involves recombination of the chain carriers to form products that can be ignored.
(a) Write the following:
(i) an equation for the overall stoichiometry;
(ii) the initiation step in the mechanism;
(iii) the propagation steps;
(iv) the termination step.
(b) Derive the rate law from the steps of the mechanism, and state whether the form agrees with that observed. Clearly state any assumption(s) made.
(c) Estimate the activation energy (EAT) for the initiation step, if the sum of the activation energies for the propagation steps is 126,000 J mol-‘, and E,J for the termination step is 0.
7-7 Suppose the mechanism for the thermal decomposition of dimethyl ether to methane and formaldehyde
CH30CH3 + CHq + HCHO (4
is a chain reaction as follows:
CH30CH3 2 CH; + OCH; El
CH; + CH30CH3 2 CHq + CH20CH; E2
CH20CH; 2 CH; + HCHO E3
CH; + CH20CHj 2 CH3CH20CH3 E4
(a) Show how the mechanism is consistent with the stoichiometry for (A).
(b) Identify any apparent deficiencies in the mechanism, and how these are allowed for by the result in (a).
(c) Derive the rate law from the mechanism, clearly justifying any assumption(s) made to simplify it.
(d) Relate the activation energy, EA. of the reaction (A) to the activation energies of the indi- vidual steps.
7-8 A possible free-radical chain mechanism for the thermal decomposition of acetaldehyde (to CH4 and CO) is the Rice-Herzfeld mechanism (Laidler and Liu, 1967):
CH3CH0 2 CHj + CHO*
CHO’ 2 CO + H’
H’ + CHsCHO 2 CH3CO’ + H2 CHj + CHsCHO 2 CH4 + CH&O*
CH3CO’ 5 CHj + CO 2CH; + CZHeks (a) Which species are the chain carriers?
(b) Classify each step in the mechanism.
(c) Derive the rate law from the mechanism for CHsCHO + CH4 + CO, and state the order of reaction predicted. Assume Hz and CzH6 are minor species.
7-9 From the mechanism given in problem 7-8 for the decomposition of acetaldehyde, derive a rate law or set of independent rate laws, as appropriate, if Hz and CzH6 are major products (in addition to CH4 and CO).
7-10 From the mechanism given in Section 6.1.2 for the dehydrogenation of CzH6, obtain the rate law for CzH6 + Cz& + Hz (assign rate constants ki, . . , kg to the five steps in the order given, and assume C& is a minor product).
7-11 Repeat problem 7-10 for a rate law or set of independent rate laws, as appropriate, if CH4 is a major product.
7-12 (a) For the C& oxidative-coupling mechanism described in Section 7.2, verify the rate laws given in equations 7.2- 1 and -2, and show that 7.2-3 is consistent with these two equations.
(b) Show (i) that equation 7.2-1 reduces to 7.2-3 if CO2 is not formed; and (ii) that 7.2-2 reduces to 7.2-3 if CzH6 is not formed.
v0“OF
(c) From the rate laws in (a), derive an expression for the instantaneous fractional yield (se- lectivity) of C2H6 (with respect to CH4).(d) Does the selectivity in (c) increase or decrease with increase in coo?
7-13 In a certain radical-addition polymerization reaction, based on the mechanism in Example 7-4, in which an initiator, 1, is used, suppose measured values of the rate, (-rM), at which monomer, M, is used up at various concentrations of monomer, CM, and initiator, ct, are as follows (Hill, 1977, p. 125):
c&m01 mm3 9.04 8.63 7.19 6.13 4.96 4.75 4.22 4.17 3.26 2.07
ct/mol m-3 (- &/mol mm3 s-l
0.235 0.193
0.206 0.170
0.255 0.165
0.228 0.129
0.313 0.122
0.192 0.0937
0.230 0.0867
0.581 0.130
0.245 0.0715
0.211 0.0415
(a) Determine the values of k and n in the rate law (-TM) = kci’2c&.
7.4 Problems for Chapter 7 173 (b) What is the order of the dependence of the efficiency (f) of radical conversion to Pl
on CM?
7-14 In the comparison of organic peroxides as free-radical polymerization initiators, one of the measures used is the temperature (T) required for the half-life (ti,z) to be 10 h. If it is desired to have a lower T, would ri/2 be greater or smaller than 10 h? Explain briefly.
7-15 Starting from equations 7.3-3 to -6 applied to a constant-volume RR, for polymerization rep- resented by the step-change mechanism in Section 7.3.2, show that the product distribution can be calculated by sequentially solving the differential equations:
+ 2kcM* + k2c; = 0 dt
dcp,- + k%m,
dt = kcM+-, ; r = 2,3, . . (7.3-17)
7-16 This problem is an extension of problems 7-10 and 7-11 on the dehydrogenation of ethane to produce ethylene. It can be treated as an open-ended, more realistic exercise in reaction mech- anism investigation. The choice of reaction steps to include, and many aspects of elementary gas-phase reactions discussed in Chapter 6 (including energy transfer) are significant to this important industrial reaction. Solution of the problem requires access to a computer software package which can handle a moderately stiff set of simultaneous differential equations. E-Z Solve may be used for this purpose.
(a) Use the mechanism in Section 6.1.2 and the following values of the rate constants (units of mol, L, s, J, K):
( 1 ) C2H6 ---) 2CH; ; kl = 5 X 1014exp(-334000/RZ’) (2) CHj + C2H,j + C2H; + CH4 ; k2 = 4 X 1013 exp(-70300/RT) (3) C2H; + C2& + H* ; k3 = 5.7 X 10” exp(-133000/RT) (4) H’ + C2H6 --z C2H; + Hz ; k4 = 7.4 X 1014exp(-52800/RT) (5) H’ + C2H; --f C2H6 ; k5 = 3.2 X 1013
(i) Solve for the concentration of CsHs radicals using the SSH, and obtain an expression for the rate of ethylene production.
(ii) Obtain a rate expression for methane production as well as an expression for the reaction chain length.
(iii) Integrate these rate expressions to obtain ethane conversion and product distribution for a residence time (t) of 1 s at 700°C (1 bar, pure C2H6). Assume an isothermal, constant-volume batch reactor, although the industrial reaction occurs in a flow sys- tem with temperature change and pressure drop along the reactor.
(iv) From initial rates, what is the reaction order with respect to ethane?
(v) What is the overall activation energy?
(b) Integrate the full set of differential equations.
(i) Compare the conversion and integral selectivities in this calculation with those in pati (4.
(ii) Compare the ethyl radical concentrations calculated in the simulation with those predicted by the SSH.
(iii) Approximately how long does it take for the ethyl radicals to reach their pseudo- steady-state values in this calculation?
(iv) Run two different simulations with different ethane pressures and take the initial rates (evaluated at 100 ms) to obtain a reaction order. Compare with part (a).
(v) Run two different simulations with two different temperatures: take the initial rates (evaluated at 3% conversion) and calculate the activation energy. Compare with the answer from part (a).
(c) At temperatures near 700°C and pressures near 1 bar, the overall reaction rate is observed to be first-order in ethane pressure with a rate constant k = 1.1 x 1015 exp(-306000/RT).
How well does this model reproduce these results?
(d) Now improve the model and test the importance of other reactions by including them in the computer model and examining the results. Use the following cases.
(dl) Reversible reaction steps.
(i) Include the reverse of step (3) in the mechanism and rerun the simulation-does it affect the calculated rates?
(6) C21& + H’ + C2H; ; ks = 1013
(ii) How else might one estimate the significance of this reaction without running the simulation again?
(d2) Steps involving energy transfer.
How many of the reactions in this mechanism might be influenced by the rate of energy transfer? One of them is the termination step, which can be thought of as a three-step process (reactions (7) to (9) below). As described in Section 6.4.3, there
are possible further complications, since two other product channels are possible (reactions (10) and (11)).
(7) (8) (9) (10) (11)
CzHs + H* + C2H; ; k7 = 6 x 1013 CzH; + M + CzHs + M ; kg = 3 x 1013 C2H; --f C2H; + H* ; k9 = 2 x 1013
&Hi + 2CH; ; klo = 3 x 101 2 C2H; --z C2H4 + H2 ; kll = 3 x 101 2
Include these reactions in the original model in place of the original reaction (5).
(You can assume that M is an extra species at the initial ethane concentration for this simulation.) Use the values of the rate constants indicated, and run the model simulation. What influence does this chemistry have on the conversion and selec- tivity? How would you estimate the rate constants for these reactions?
(d3) The initiation step.
The initiation step also requires energy input.
(12) C& + M + C2H;i + M ; kn = 2 X 1013 exp(-340,00O/RT) The other reactions, (8) and (lo), have already been included.
At 1 bar and 700°C is this reaction limited by energy transfer (12) or by decompo- sition (lo)?
(d4) Termination steps.
Termination steps involving two ethyl radicals are also ignored in the original mech- anism. Include the following reaction:
(13) 2 C2H; + C2H.4 + C2H6 ; k13 = 6 x 10”
Does this make a significant difference? Could you have predicted this result from the initial model calculation?
(d5) Higher molecular-weight products.
Higher molecular-weight products also are made. While this is a complex pro- cess, estimate the importance of the following reaction to the formation of higher hydrocarbons by including it in the model and calculating the C4Hg product
7.4 Problems for Chapter 7 175 selectivity.
(14) C&, + C2H; --f C4Hs + Ho ; k14 = 2 x 1011
Plot the selectivity to C4Hs as a function of ethane conversion. Does it behave like a secondary or primary product? Consult the paper by Dean (1990), and describe additional reactions which lead to molecular weight growth in hydrocarbon pyroly- sis systems. While some higher molecular weight products are valuable, the heavier tars are detrimental to the process economics.
Much of the investigation you have been doing was described originally by Wojciechowski and Laidler (1960), and by Laidler and Wojciechowski (1961). Compare your findings with theirs.