Question

The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product stream is analyzed and found to contain 58.7 mol% C2H5Br and 18.3 mol% HBr. The feed to the reactor conains only ethylene and hydrogen bromide. Calculate the fractional conversion of the limiting reactant and the percentage by which the other reactant is in excess. If the molar flow rate of the feed stream is 305 mol/s, what is the extent of reaction

Answer 1

Answer:

Explanation:

The first thing to do is to write out the chemical equation showing the reaction between the two chemical compounds.

C2H4 + HBr ------------------------> C2H5Br.

The chemical species/compounds reacted in the ratio of 1 : 1 : 1. That is one mole of C2H4 reacted with one mole of HBr to give one mole of C2H5Br.

So, we are given; C2H5Br = 58.7 mol%, HBr = 18.3 mol%. Thus, for C2H4, we have [ 100 - (58.7 + 18.3)] = 23 mol%. Therefore, the limiting reagent is HBr.

So, if we have 100 mol/s ;

C2H5Br = 100 × 0.587 = 58.7 mol/s, HBr = 100 × 0.183 = 18.3 mol/s and C2H4 = 100 × 0.23 = 23 mol/s.

Therefore, the total value of the flow rate in the feed = (HBr in feed) mol/s + (C2H4 in feed) mol/s = (58.7 + 18.3)mol/s + (58.7 + 23) mol/s =( 77 + 81.7) mol/s = 158.7 mol/s. .

So, 100/158.7 × 30.5 = 192.2.

Also the feed composition in HBr and C2H4 is;

HBr = 77/158.7 = 0.4852, C2H4 = 81.7/158.7 = 0.515.

Therefore, the number of HBr in feed and C2H4 in feed = 160m/s and 200m/s respectively (approx.)

Therefore, the fraction conversion is;