chemical reaction engineering 3ed edition作者 octave Levensp(2)

?rA??dPVdPA13.6623.66?(?A)?PA?(CART)2

VRTdtRTdtRTRT22 ?(3.66RT)CA?kCA So we can get that the value of

k?3.66RT?3.66?0.08205?400?120.1

2.9 The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol.

How much faster the decomposition at 650℃ than at 500℃? Solution:

Lnr2kE11300kJ/mol11?Ln2?(?)??(?)?7.586 3r1k1RT1T28.314kJ/(10mol?K)173K923K?r2?1970.7 r1

2.11 In the mid-nineteenth century the entomologist Henri Fabre noted that French ants (garden variety) busily bustled about their business on hot days but were rather sluggish on cool days. Checking his results with Oregon ants, I find

Running speed, m/hr Temperature, ℃

150 13

160 16

230 22

295 24

370 28

What activation energy represents this change in bustliness? Solution:

r?k0e?ERTf(concentration)let?f(concentration)?constant??ak0e1E TR1, T?ERT?ke'?ERT

?LnrA?Lnk'?Suppose y?LnrA,x?so slope??

E, intercept?Lnk' R5

r?1150 160 230 295 370

A/(m?h)

Lnr-3.1780 -3.1135 -2.7506 -2.5017 -2.2752 A

T/oC 13 16 22 24 28 13.4947

3.4584

3.3881

3.3653

3.3206

T?10?3 4-y = 5417.9x - 15.686R2 = 0.973r nL2-100.00330.003350.00340.003450.00351/T

-y = -5147.9 x + 15.686

Also slope??E??5147.9K, intercept ?Lnk'R= 15.686 , E??5147.9K?8.3145J/(mol?K)?42.80kJ/mol

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Chapter 3 Interpretation of Batch Reactor Data

3.1 If -rA = - (dCA/dt) =0.2 mol/liter·sec when CA = 1 mol/liter, what is the rate of

reaction when CA = 10 mol/liter? Note: the order of reaction is not known.

Solution: Information is not enough, so we can’t answer this kind of question.

3.2 Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A is

converted in a 5-minute run. How much longer would it take to reach 75% conversion? Solution: Because the decomposition of A is a 1st-order reaction, so we can express the rate equation as:

?rA?kCA

We know that for 1st-order reaction, LnCAo?kt, CALnCAoC?kt1, LnAo?kt2 CA1CA2CA1?0.5CAo, CA2?0.25CAo

So t2?t1?CC111(LnAo?LnAo)?(Ln4?Ln2)?Ln2 equ(1) kCA2CA1kkt1?C11Ln(Ao)?Ln2?5min equ(2) kCA1kSo t2?t1?t1?5min

3.3 Repeat the previous problem for second-order kinetics. Solution: We know that for 2nd-order reaction, So we have two equations as follow:

11??kt, CACA011211?????kt1?k5min, equ(1) CA1CA0CAoCAoCAo 7

11411????3()?3kt1?kt2, equ(2) CA2CAoCAoCAoCAoSo t2?3t1?15min, t2?t1?10min

3.4 A 10-minute experimental run shows that 75% of liquid reactant is converted to product

1by a -order rate. What would be the fraction converted in a half-hour run?

2Solution: In a

dCA10.5?order reaction: ?rA???kCA, 2dtkt0.50.5?CAo?CA1, 2So we have two equations as follow:

10.50.50.50.50.5CAo?CA?C?(CAo)?0.5CAo?kt1?k(10min), equ(1) 1Ao4After integration, we can get:

0.50.5CAo?CA2?kt2?k(30min), equ(2)

0.50.5Combining these two equations, we can get:1.5CAo?kt2, but this means CA2?0, which is

impossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted XA?1.

3.5 In a hmogeneous isothermal liquid polymerization, 20% of the monomer disappears in 34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rate equation represents the disappearance of the monomer?

Solution: The rate of reactant is independent of the initial concentration of monomers, so we know the order of reaction is first-order,

?rmonomer?kCmonomer And LnCo?(34min)k 0.8Cok?0.00657min?1

?rmonomer?(0.00657min?1)Cmonomer

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3.6 After 8 minutes in a batch reactor, reactant (CA0 = 1 mol/liter) is 80% converted; after 18 minutes, conversion is 90%. Find a rate equation to represent this reaction. Solution:

11Lntk1?XA2Ln1??1.43, dissatisfied. In 1st order reaction, 2?11t1Ln5Lnk1?XA111111(?)?kCA2CAo0.1CAoCAo9/CAo9tIn 2nd order reaction, 2????, satisfied.

11111t14/CAo4(?)?kCA1CAo0.2CAoCAoAccording to the information, the reaction is a 2nd-order reaction.

3.7 nake-Eyes Magoo is a man of habit. For instance, his Friday evenings are all alike—into the joint with his week’s salary of ＄180, steady gambling at ―2-up‖ for two hours, then home to his family leaving ＄45 behind. Snake Eyes’s betting pattern is predictable. He always bets in amounts proportional to his cash at hand, and his losses are also predictable—at a rate proportional to his cash at hand. This week Snake-Eyes received a raise in salary, so he played for three hours, but as usual went home with ＄135. How much was his raise? Solution:

nAo?180, nA?13, t?2h,

'nA?135, t;?3h, ?rA?knA

'nAonAoLn()Ln(')nAnA ??kt, 'ttSo we obtain LnnAonA'nAo180LnLn13?135, n'?28

A233.9 The first-order reversible liquid reaction

A ? R , CA0 = 0.5 mol/liter, CR0=0

takes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while equilibrium conversion is 66.7%. Find the equation for the this reaction.

Solution: Liquid reaction, which belongs to constant volume system,

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