Question

Ethane is chlorinated in a continuous reactor: C2H6 + Cl2 --> C2H5Cl + HCl (Rxn 1) Some of the product monochloroethane is further chlorinated in an undesired side reaction: C2H5Cl + Cl2 --> C2H4Cl2 + HCl (Rxn 2) The reactor is designed to yield a 15% conversion of ethane and a selectivity of 15 mol C2H5Cl/mol C2H4Cl2, with a negligible amount of chlorine in the product gas. Calculate the fractional yield of monochloroethane. Use a basis of 100 mol C2H5Cl produced. Report your answer with three significant figures. Note: Reactor feed is ethane and chlorine. Identify the limiting reactant to answer the question.

Answer 1

conversion of ethane = 15

(Moles of C2H6 in - moles of C2H6 out) x 100 / (Moles of C2H6 in) = 15

moles of C2H6 out = 0.85 x Moles of C2H6 in ............ Eq1

selectivity = (mol of C2H5Cl out) / (mol of C2H4Cl2 out)

15 = 100 / (mol of C2H4Cl2 out)

mol of C2H4Cl2 out = 100/15 = 6.66 mol

Carbon balance

2 x C2H6 in = 2 x C2H6 out + 2 x 100 + 2 x C2H4Cl2 out

2 x C2H6 in = 2 x C2H6 out + 2 x 100 + 2 x 6.66

2 x C2H6 in = 2 x C2H6 out + 213.32

From eq1

2 x C2H6 in = 2 x 0.85 x C2H6 in + 213.32

Moles of C2H6 in = 711.06 mol

moles of C2H6 out = 0.85 x 711.06 mol = 604.41 mol

Hydrogen balance

6 x C2H6 in = 6 x C2H6 out + 5 x 100 + 4 x 6.66 + HCl out

6 x 711.06 = 6 x 604.41 + 5 x 100 + 4 x 6.66 + HCl out

Moles of HCl out = 113.26 mol

Cl balance

2 x Cl2 in = 100 + 2 x C2H4Cl2 out + Moles of HCl out

2 x Cl2 in = 100 + 2 x 6.66 + 113.26

Moles of Cl2 in = 113.29 mol

Moles of Cl2 in = moles of HCl out

Limiting reactant = Cl2

fractional yield of C2H5Cl

= (moles of C2H5Cl out) / (maximum C2H5Cl can be produced from limiting reactant)

= 100 / Moles of Cl2 in

= 100 / 113.29

= 0.883