Bromination of ethylene (C 2 H 4 ) is carried out in a reactor according to the

Question

Bromination of ethylene (C2H4) is carried out in a reactor according to the following reaction
C2H4 + HBr => C2H5Br

Feed stream containing both reactants (C2H4 and HBr) is fed to the reactor at a total molar flow rate of 100 mol/s. The mole fraction of ethylene in the feed stream is 0.6. The fractional conversion of ethylene is 0.5.

Use the atomic balance approach to calculate the composition, in mole fractions of each component, of the product stream.

The following abbreviations apply:

E: ethylene (C2H4)
B: hydrogen Bromide (HBr)

R: ethyl bromide (C2H5Br)

Explanation / Answer

100 mol/s is the total molar rate in which the mole fraction of ethylene in the feed stream is 0.6.so .4 will be HBr

C2H4 + HBr => C2H5Br

In this equation 1 mol of C2H4 will react with 1 mol of HBr to form 1 mol of C2H5Br.

so for unit time 40 mol of HBr react with 40 mol of C2H4 and form 40 mol of C2H5Br.

so ater reaction

mol of C2H4=20mol , mol of HBr=0 mol, mol of C2H5Br=40 mol

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