Fluid Catalytic Cracking (FCC) is commonly used to crack long chain hydrocarbons

Question

Fluid Catalytic Cracking (FCC) is commonly used to crack long chain hydrocarbons into shorter molecules. The hydrocarbon feed is heated and vaporized in a furnace prior to enter the FCC unit, such that the feed entering the reactor is a gas. The reactions in the FCC reactor can be represented in a simplified form as:

A 5 R

In order to design a new FCC reactor, you have measured the concentration of the uncracked hydrocarbon (CA) at the exit of a lab-scale high-temperature CSTR operating at 1000 K at different feed rates:

You performed a second set of experiments changing the reactor temperature, keeping a constant inlet molar flow rate FA0 = 1000 millimol / h, and obtained the following results:

In all the experiments, the internal void volume of the reactor was V = 0.1 L, and CA0 = 100 millimol/L.

(a) Find the order of the reaction

(b) Determine the activation energy for this reaction

Explanation / Answer

For a CFSTR, T= V/Vo= VCAO/FAO= CAOXA/(-rA)

-rA= rate of reaction

-rA= XAFAO/V (1)

CA= CAO*(1-XA)/ (1+eXA)

Where e= fractional change in volume for gas phase reaction = (5-1)/1= 4

Hence CA/CAO= (1-XA)/ (1+4XA)

(1+4XA)*CA/CAO = (1-XA)

CA/CAO + 4XA*CA/CAO= 1-XA

1-CA/CAO= XA*(1+4CA/CAO)

XA= (1-CA/CAO)/(1+4CA/CAO)

For the given data, prepare a table of conversion for different flow rates 1st. Then calculate –rA

let the order be n.

The rate equation can be represented as –rA= KCAn

taking ln, the equation becomes ln(-rA)= lnK + n lnCA

So a plot of ln(-rA) Vs lnCA gives a straight line whose slope is n and intercept is ln K. The plot is shown below

The slope obtained is 1. The order of reaction is 1.

So the rate expression becomes –rA= KCA. ( since K is temperature dependent)

2nd plot need to be prepared for

lnK= lnKo-Ea/RT,

so a plot of lnK vs 1/T gives a straight line whose slope is –Ea/R.

For 1st order reaction T= CAOXA/KCA ,   VCAO/FAO= CAOXA/KCA

K= FAOXA/V

XA= (1-CA/CAO)/ (1+4XA)

1st calculate XA for each temperature and then calculate K at each temperature.

The data is shown below.

K= FAOXA/V

XA= (1-CA/CAO)/ (1+4XA)

1st calculate XA for each temperature and then calculate K at each temperature.

The data is shown below.

K= FAOXA/V

XA= (1-CA/CAO)/ (1+4XA)

1st calculate XA for each temperature and then calculate K at each temperature.

The data is shown below.

The slope is –Ea/R = -3237, Ea (activation energy)= 3237*8.314 J/mole=26912 J/mole.

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