Respuesta :
Answer:
Fractional conversion=0.749
Percentage by which the other reactant is in excess=16.56%
extent = 56.23 mol/s
Explanation:
reaction
[tex]C_{2}H_{4} + HBr - - -> C_{2}H_{5}Br[/tex]
molar composition of the product stream:
[tex]C_{2}H_{5}Br[/tex] = 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.
[tex]C_{2}H_{4}[/tex] = 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:
[tex]mol_{i}=MF_{T}*molar-percentaje[/tex]
MF= total molar flow.
[tex]C_{2}H_{5}Br = 100 \frac{m}{s}*0.517=51.7 mol/s[/tex]
HBr =17.3 mol/s
[tex]C_{2}H_{4}[/tex] = 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 [tex]C_{2}H_{5}Br[/tex] we needed 51.7 moles of HBr and [tex]C_{2}H_{4}[/tex] respectively.
[tex]mol_{in}=mol_{used}+mol_{out}[/tex]
[tex]HBr = 51.7 mol + 17.3 mol =69 mol [/tex]
[tex]C_{2}H_{4}[/tex] = 82.7 mol
with this we can see that the limit reagent in this reaction is HBr.
82.7 mol of [tex]C_{2}H_{4}[/tex] need the same number of mol of HBr but there is just 69 moles of this.
The molar composition in the inlet is:
HBr =[tex]\frac{mol-HBr}{total-mol}*100=\frac{69}{69+82.7}*100[/tex] = 45.4%
[tex]C_{2}H_{4}[/tex] = 100% - 45.4% = 54.51%
Fractional conversion of HBr (FC) = [tex]\frac{mol_{used}}{mol_{in}}[/tex]
[tex]=\frac{69 mol-17.3mol}{69 mol}= 0.749[/tex]
percentage of [tex]C_{2}H_{4}[/tex] in excess =[tex]\frac{mol_{in}-mol_{needed}}{mol_{in}}*100=\frac{69 mol}{82.7}*100[/tex] =16.56%
b. The extent of reaction (e) is defined as:
[tex]e=\frac{n_{out}-n_{in}}{(+-)v}[/tex]
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
[tex]FC_{HBr}=\frac{mol_{in}-mol_{out}}{mol_{in}}[/tex]
mol HBr_{out}=[tex]mol_{in}-mol_{in}*FC=75.04-75.04*0.749=18.81 mol[/tex]
v=-1
[tex]e=\frac{18.81mol-75.04mol}{-1}=56.23 mol[/tex]
The extent of the given reaction in the continuous reactor is; 56.25 mol/s
What is the extent of the reaction?
We are told that ethylene reacts with hydrogen bromide to form ethyl bromide. This reaction is;
C₂H₄ + HBr = C₂H₅Br
We are given composition as;
Molar composition of C₂H₅Br = 51.7%
HBr Molar composition = 17.3%
Thus, molar composition of ethylene = 100% - (51.7% + 17.3%) = 31%
For the reaction side, we have;
x mole of C₂H₄ and (1 - x) mole of HBr
For the product stream side, we have;
0.31 mole of C₂H₄
0.173 mol of HBr
0.517 mol of C₂H₅Br
Since we are told that the molar flow rate of the feed stream is 165mol/s;
For C balance we have; 165 * x * 2 = n'(0.31 * 2) + n'(0.517 * 2)
⇒ 330x = 1.654n' ------(eq 1)
For Br Balance, we have;
165(1 - x)*1 = n'(0.173 * 1)
⇒ 165 - 165x = 0.173n' ----(eq 2)
Solving eq 1 and 2 simultaneously gives;
n' = 108.77 mol/s; x = 0.545 mol of C₂H₄/mol and (1 - x) mol of HBr is 0.455 mol of HBr/mol
The feed ratio will now be 0.545:0.455 and since it is greater than the stoichiometric ratio of 1, then HBr is the limiting agent.
(n'_HBr)_fed = 165 * 0.455 = 75.08 mol HBr
Fractional conversion of HBr = [[75.08 - (0.173 * 108.8)]/75.08] * 100% = 0.749 HBr react per mol fed
(n'_C₂H₄)_stoich = 75.08 mol C₂H₄
(n'_C₂H₄)_fed = 165 * 0.545 = 89.93 mol C₂H₄
%Excess of C₂H₄ = (89.93 - 75.08)/75.08 = 19.8%
Thus;
Extent of reaction = 108.8 * 0.517 = 56.25 mol/s
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