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408 Chapter 7

m T =491.6 +38.45 X A

Now, evaluate the mole fraction, yj, for each component from Equation
7.14.4. Take one kgmol of ethyl benzene as the basis for the calculation. The
kgmol of steam per kgmol of incoming styrene is 11.78. Therefore, for the frac-
tional conversion of ethyl benzene, X A,

Ethyl Benzene 1 - XA
Steam 11.78
Styrene X A
Hydrogen X A

Total 12.78 + X A

y E = (l-x A )/(12.78+ X A)

= X A / (12.78 + X A)
y s
y w =11.78 / (12.78+ X A )

y H = x A /(12.78+ X A)

After Substituting these equations in terms of the conversion into Equation
7.14.3, we find that

2
f PSPH 1 f ( l - X A ) P X A P 2 1
= k f(p ;) = k I p E - ——— I = k | —————— - —————————— I
r A
L K P J L 12.78+ X A K P ( 12.78 + x A) 2 J
where the subscripts, E = ethyl benzene, S = styrene, and H = hydrogen.

From the above equations,

dT/dx A = -19610/m T

and

m T = 491.6 +38.45 X A

Also, from the problem statement,

k= 12600 exp(-l 1000 /T)

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