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8. A company has a large ethane (C2H6) stream available and has demands for both ethylene

(C2H4) and acetylene (C2H2). The relative amounts of these two chemicals required varies from

time to time, and the company proposes to build a single plant operating at atmospheric pressure

to produce either material.

(a) Using the data below, calculate the maximum temperature at which the reactor may operate

and still produce essentially ethylene (not more than 0.1% acetylene).

(b) Calculate the minimum temperature at which the reactor can operate and produce essentially

acetylene (not more than 0.1% ethylene).

(c) At what temperature will one produce equimolal quantities of acetylene and

ethylene?

Data and Notes: Assume that only the following reactions occur.

C2H6C2H4+ H2

C2H6C2H2+ 2H2

Neglect the effect of temperature on H0.

Answer :

Data from Appendix A An Introduction To Chemical Engineering Kinetics and Reactor

Designby Charles G Hill p.570 and 571

1. Ethylene (C2H4)

Gf0

= 16.282 kcal/gmoleHf

0 = 12.496 kcal/gmole

2. Acetylene (C2H2)

Gf0 = 50 kcal/gmole

Hf0 = 54.194 kcal/gmole

3. Ethane (C2H6)

Gf0 = -7.860 kcal/gmole

Hf0 = -20.236 kcal/gmole

C2H6C2H4+ H2

G10 = 1Gf

0C2H4-1Gf

0C2H6

= 1(16.282 kcal/gmole)1(-7.860 kcal/gmole)

= 24.142 kcal/gmole

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H10 = 1Hf

0C2H4-1Hf

0C2H6

= 1(12.496 kcal/gmole)1(-20.236 kcal/gmole)

= 32.732 kcal/gmole

C2H6C2H2+ 2H2

G20 = 1Gf

0C2H2-1Gf

0C2H6

= 1(50 kcal/gmole)1(-7.860 kcal/gmole)

= 57.860 kcal/gmole

H20 = 1Hf

0C2H2-1Hf

0C2H6

= 1(54.194 kcal/gmole)1(-20.236 kcal/gmole)

= 74.43 kcal/gmole

a. G10/(T2-T1) = H10/T.............................................................................persamaan 2.2.8 page 8

Neglect the effect of temperature on H0.

24.142 kcal/gmole / (T2-298.15K) = 32.732 kcal/gmole

32.732 T29759.0458 = 24.142

32.732 T2 = 9783.1878

T2 = 298.888 K

b. G20/(T2-T1) = H2

0/T.............................................................................persamaan 2.2.8 page 8

Neglect the effect of temperature on H0.

57.860 kcal/gmole / (T1-298.15K) = 74.43 kcal/gmole

74.43 T22,2191.3045 = 57.860

74.43 T2 = 2,2249.1645

T2 = 298.927 K