Question

1. Ethylene is produced by the dehydrogenation of ethane and the following reactions took place

C₂H₆ --> C₂H₄ + H₂

C₂H₆ + H₂ --> 2CH₄

Using molecular balance. Calculate % Yield of C₂H₄, if the % conversion of C₂H₆ is 80 % and the selectivity, mole C₂H₄ per mole CH₄, is 1.75. Calculate also the % composition by mole of the product gases.

Answer 1

Let 100 mol C2H6 is fed.

C2H6 conversion = 80%

C2H6 converted = 0.8 × 100 mol = 80 mol

Unreacted C2H6 = (100-80) mol = 20 mol

Let y be the moles of C2H6 reacted through reaction 1 and (80-y) reacted through reaction 2.

From reaction stoichiometry, y mol of C2H4 and 2(80-y) moles of CH4 is produced.

Given, y/[2(80-y)] = 1.75

y/(80-y) = 3.5

y = 280 - 3.5y

4.5 y = 280

y = 62.22

C2H4 produced = 62.22 mol

CH4 produced = 2× (80-62.22) mol = 35.56 mol

H2 is produced in reaction 1 and consumed in reaction 2.

H2 produced = y-(80-y) = 2y-80 = 44.44 mol

% yield of C2H4 = {(moles of C2H4 produced)/(moles of C2H4 that would have produced for the complete conversion of C2H6 through reaction 1 only)} × 100

=(62.22/100)× 100 = 62.22%

Composition of product gas:

C2H6 = 20 mol

C2H4= 62.22 mol

CH4 = 35.56 mol

H2 = 44.44 mol

Total moles in product stream = 162.22 mol

Molar % composition of product stream is:

C2H6 = (20/162.22) × 100 =12.33%

C2H4 = (62.22/162.22)× 100= 38.35%

CH4 = (35.56/162.22) × 100= 21.92%

H2 = (44.44/162.22) × 100= 27.39%