Engineering, 16.10.2019 05:00 jnsoccerboy3121

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 51.7 mole% c2h5br and 17.3% hbr. the feed to the reactor contains 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 165mol/s, what is the extent of reaction?

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Answer from: josemartinez2025

Fractional conversion=0.749

Percentage by which the other reactant is in excess=16.56%

extent = 56.23 mol/s

Explanation:

reaction

$C_{2}H_{4} + HBr - - - C_{2}H_{5}Br$

molar composition of the product stream:

$C_{2}H_{5}Br$ = 51.7%

HBr = 17.3 %

when you add these two compositions you don't get 100% (69%) so there is also ethylene in the product stream.

$C_{2}H_{4}$ = 100% - 69% = 31%

We are going to suppose a flow rate in the product stream of 100 mol/s. Although you have a flow rate for the feed stream since you don't know the molar composition in the entrance is easier suppose a molar flow in the outlet.

with this information we can estimate the number of moles in the product stream as:

$mol_{i}=MF_{T}*molar-percentaje$

MF= total molar flow.

$C_{2}H_{5}Br = 100 \frac{m}{s}*0.517=51.7 mol/s$

HBr =17.3 mol/s

$C_{2}H_{4}$ = 31 mol/s

The stoichiometry for this reaction is 1: 1. With this information we can estimate the moles of each reagent in the feed stream. To produce 51.7 mol of $C_{2}H_{5}Br$ we needed 51.7 moles of HBr and $C_{2}H_{4}$ respectively.

$mol_{in}=mol_{used}+mol_{out}$

HBr = 51.7 mol + 17.3 mol =69 mol

$C_{2}H_{4}$ = 82.7 mol

with this we can see that the limit reagent in this reaction is HBr.

82.7 mol of $C_{2}H_{4}$ need the same number of mol of HBr but there is just 69 moles of this.

The molar composition in the inlet is:

HBr = $\frac{mol-HBr}{total-mol}*100=\frac{69}{69+82.7}*100$ = 45.4%

$C_{2}H_{4}$ = 100% - 45.4% = 54.51%

Fractional conversion of HBr (FC) = $\frac{mol_{used}}{mol_{in}}$

$=\frac{69 mol-17.3mol}{69 mol}= 0.749$

percentage of $C_{2}H_{4}$ in excess = $\frac{mol_{in}-mol_{needed}}{mol_{in}}*100=\frac{69 mol}{82.7}*100$ =16.56%

b. The extent of reaction (e) is defined as:

$e=\frac{n_{out}-n_{in}}{(+-)v}$

where n indicate the moles of a compound in the inlet (in) and in outlet (out) respectively and v is the stoichiometric coefficient for that compound in the reaction. It can be negative if it's a reagent or positive if it's a product.

In this case we have 165 mol/s a flow rate in the inlet. Assuming that the composition in the inlet and the fractional conversion are the same we have.

mol HBr (inlet) = 165 mol/s * 45.4% = 75.04 mol

$FC_{HBr}=\frac{mol_{in}-mol_{out}}{mol_{in}}$

mol HBr_{out}= $mol_{in}-mol_{in}*FC=75.04-75.04*0.749=18.81 mol$

v=-1

$e=\frac{18.81mol-75.04mol}{-1}=56.23 mol$

Answer from: Quest

explanation:

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Answer from: Quest

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