Question

The reactions:

           C₂H₆ → C₂H₄ + H₂

           C₂H₆ + H₂ → 2CH₄

take place in a gas phase continuous reactor operated at steady state. The feed contains 87 mole% ethane and the balance inerts, I. The conversion of ethane is 57% and the yield of ethylene is 51%. Yield, in this case, is defined as the moles of desired product formed/moles that would have been formed if there were no side reactions and the limiting reactant were completely consumed. Calculate the:

      a) mole percentage of ethane in the product

      b) mole percentage of ethylene in the product

      c) mole percentage of hydrogen in the product

      d) selectivity of ethylene to methane

Answer 1

Given reactions:

C2H6 --> C2H4 + H2 .....(1)

C2H6 + H2 ---> 2CH4........ (2)

gas phase continuous steady state reactor and feed contains

C2H6 = 87 mol%

Inert = 13 mol%

basis: 100 mol of feed to the reactor.

C2H6 = 87 mol

Inert = 13 mol

Let us assume extent of reaction e1 for rx(1) and e2 for rx(2).

At reactor outlet products containing following compositions,

C2H6 = 87 - e1 - e2

C2H4 = e1

H2 = e1 - e2

CH4 = 2e2

Given conversion of ethane X= 0.57

Yield of ethylene , Y = 0.51

at reactor outlet,

C2H6 = 87* (1-X)

87 - e1 - e2 = 87*(1-0.57)

87 - e1 - e2 = 37.41

e1 + e2 = 49.59 mol ......(I)

For ethylene,  

C2H4 = e1

87*Y*X = e1

87*0.51*0.57 = e1

e1 = 25.29 mol

From above eq(I)

e1 + e2 = 49.59

e2 = 49.59 - 25.29

e2 = 24.3 mol

so here extent of reaction both reactions,

e1 = 25.29 mol

e2 = 24.3 mol

products :

C2H6 = 37.41 mol

C2H4 = 25.29 mol

H2 = 0.99 mol

CH4 = 2*24.3 = 48.6 mol

a) mol percentage of ethane in products,

total moles of products, P = 37.41 + 25.29 + 0.99 + 48.6= 112.29 mol

Mole fraction of ethane = ethane mol/total product mol = 37.41/112.29= 0.333

mole percentage of ethane = 0.333*100 = 33.3%

ans: 33.3%

b) mole percentage of ethylene in the product,

mole fraction of ethylene = ethylene mol/P = 25.29/112.29 = 0. 225

Mole percentage of ethylene = 0.225*100 = 22.5%

ans: 22.5 %

c)  mole percentage of Hydrogen

H2 mole fraction = 0.99/112.29 = 0.0088

mole percentage of H2 = 0.0088*100 = 0.88%

ans: 0.88%

d) selectivity of ethylene to methane: It is the ratio of desired product to undesired product,

here selectivity = ethylene mol/methane mol = 25.29/48.6 = 0.52

ans: 0.52