CHE 441 Exam 2 Practice (Ch 1-4 Fogler)

(Adapted from previous exams)

Problem 1

(for the solution, click here)
The irreversible gas phase dimerization of trifluorochloroethylene ocurs as follows:

2 CF2CFCl --- 1 cyclo-(CF2CFCl)2


The following data were taken in a constant volume isothermal batch reactor at 440'C. Only CF2CFCl is present initially. Determine the order of the reaction and the reaction rate constant. The reaction order is either 1st or 2nd Order.

Time, s

Total Pressure, kPa

0

82.7

100

71.1

200

64.0

300

60.4

400

56.7

500

54.8

Problem 2

A reversible gas-phase reaction:

A === 2 B


occurs over a heterogeneous catalyst. The following mechanism is used to describe the reaction (external mass transfer and internal diffusion inside catalyst pores can be neglected):

S refers to vacant catalytic active sites. A series of experiments is performed where the initial rate of reaction is measured as a function of the initial partial pressure of A, PA0. At very low partial pressures, the initial rate of reaction is proportional to PA0 raised to the 0.5 power, while at high partial pressures, the initial rate is proportional to PA0 raised to the -0.5 power. Find the rate-limiting step of the mechanism. (solution)

CHE 441 Practice Exam 1 Solutions

Problem 1 Solution

Design Equation: -r_A = k PA^q = - dPA/dt, where -r_A is the rate of reaction, PA is the partial pressure of reactant ("A"), and q the order of the reaction.

Will use Integral Method to determine rate law.

If q = 1, then:

ln (PA/PA0) = - k t


If q = 2, then:

(1/PA) - (1/PA0) = k t


Thus, we need to plot ln PA vs t and (1/PA) vs. t.
But first, we need to convert P vs t data into PA vs t.

From Stoichiometric Table:

NA = NA0 (1 - X); implies PA = PA0 (1 - X); implies X = 1 - (PA/PA0)
[Recall that V and T are constant for this reactor.]

NT = NA0 (1 - 0.5 X); implies P = PA0 (1 - 0.5 X); implies X = 2 (1 - {P/PA0} )

Setting these two equations equal to each other: PA0 - PA = 2 (PA0 - P)
PA = - PA0 + 2P

From this relationship, the necessary values for graphing can be calculated:

Time, s

Total Pressure, kPa

PA, kPa

ln PA

1/PA, 1/kPa

0

82.7

82.7

4.415

0.0121

100

71.1

59.5

4.086

0.0168

200

64.0

45.3

3.813

0.0221

300

60.4

38.1

3.640

0.0262

400

56.7

30.7

3.424

0.0326

500

54.8

26.9

3.292

0.0371

The Plots show that 2nd order yields the best correlation, with slope = k = 5.04 e-5 / kPa

Alternate: PA = 2 P - PA0; - dPA/dt = - 2 dP/dt; sub into design equation:

- 2 dP/dt = k (2P - PA0)^q

If q = 2, then: -2 dP/dt = k (2P - PA0)^2; (2P - PA0)^{-2} dP = - 0.5 k t;

k t = (1/{2P - PA0}) - (1/PA0); plot of t vs. (1/{2P - PA0}) yields a straight line...

Problem 2 Solution

R Ads = k1 {Pa Cv - (1/KA) Ca.s}
R Surf = k3 {Ca.s Cv - (1/KS) Cb.s Cb.s}
R Desorpt = k5 {Cb.s - KD Pb Cv}

Desorption is the rate-determining step

RATE = k5 {Cb.s - (1/KD) Pb Cv}

(R Ads)/k1 = 0; implies Ca.s = KA Pa Cv

(R Surf)/k3 = 0; implies Cb.s Cb.s = KS Ca.s Cv = KS KA Pa Cv Cv
Thus, Cb.s = SQRT{KS KA Pa} Cv

Site Balance: Ct = Cv + Ca.s + Cb.s = Cv (1 + KA Pa + SQRT{KS KA Pa} )

Substitution into rate law:

RATE = k5 Ct [SQRT{KA KS Pa} - (1/KD) Pb} / [1 + KA Pa + SQRT{KS KA Pa} ]

RATE = k [ SQRT{Pa} - K1 Pb ] / [1 + K2 Pa + K3 SQRT{Pa} ]

Init RATE = k SQRT{Pao} / [ 1 + K2 Pao + K3 SQRT{Pao} ]

This rate law is consistent with the given data.


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November 6, 1996