Question

The chlorination of ethane proceeds as follows:

C₂H_6(g) + Cl_2(g) --> C₂H₅Cl_(g) + HCl_(g)

However, some of the formed chloroethane further reacts with excess chlorine in an undesired side reaction:

C₂H₅Cl_(g) + Cl_2(g) --> C₂H₄Cl_2(g) + HCl_(g)

If only C₂H₆ and Cl₂ are in the feed stream, determine their feed ratio (mol C₂H₆/mol Cl₂) and the % yield of C₂H₅Cl given that the fractional conversion of C₂H₆ is 0.350 and the selectivity is 12. All chlorine is consumed in the reaction (i.e., no Cl₂ in the product stream). Use the extent of reaction method for your solution.

Answer 1

The reactions occuring are

Selectivity of C₂H₅Cl over C₂H₄Cl = 12: 1

Fractional conversion of C₂H₆ = 0.350

Basis : 1 mole ethane

Ethane reacted = 1(0.350) = 0.350 mol

Selectivity = 12

Ratio of C₂H₅Cl : C₂H₄Cl = 12:1

But C₂H₅Cl is reactant for second reaction

So moles of C2H5Cl formed initially

=0.350 moles

According to stiochometry moles of Cl2 required = 0.350(1) = 0.350 moles

Out of 0.350 moles C2H5Cl some of them react in second reaction to form C2H4Cl2

Moles of C2H5Cl reacted in second reaction = (0.350) (1/13) = 0.02692 moles

Moles of C2H5Cl in product = 0.350-0.02692 = 0.32307 moles

Moles of C2H4Cl2 in product = 0.02692 moles

C2H5Cl : C2H4Cl2 = 0.32307/0.02692 = 12

Moles of Cl2 reacted in second reaction according to stiochometry = 0.02692(1) = 0.02692 moles

Total Cl2 reacted = 0.02692+ 0.350

= 0.3769 moles

All Cl2 is consumed so

Cl2 in feed = 0.3769 moles

Feed ratio= 1/(0.3769)

= 2.6532 mol ethane/mole Cl2

Yield of C2H5Cl

Yield = (actual moles/(theoretical moles) )(100)

Actual moles = 0.32307 moles

Theoretical moles = 0.350 moles

Yield = (0.32307/0.350) (100) = 92.3057%

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