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 mol% [tex]C_2H_5Br[/tex] and 13.3 mol% [tex]HBr[/tex]. The feed to the reactor contains only ethylene and hydrogen bromide.

1. Calculate the fractional conversion of the limiting reactant.
2. Calculate the percentage by which the other reactant is in excess.
3. If the molar flow rate of the feed stream is 285 mol/s, what is the extent of reaction?

To calculate the fractional conversion of the limiting reactant, first, determine the moles of each reactant present in the product stream. Then, calculate the fractional conversion of the limiting reactant using the mole flow rates. To calculate the percentage by which the other reactant is in excess, subtract the mole flow rate of the limiting reactant in the product stream from the initial mole flow rate of the other reactant and divide it by the initial mole flow rate. The extent of reaction is equal to the fractional conversion of the limiting reactant.

Explanation:

Calculation of Fractional Conversion of Limiting Reactant:

To calculate the fractional conversion of the limiting reactant, first, determine the moles of each reactant present in the product stream:
Ethyl Bromide (C2H5Br) = 51.7 mol% of 285 mol/s = 147.2055 mol/s
Hydrogen Bromide (HBr) = 13.3 mol% of 285 mol/s = 37.905 mol/s

The limiting reactant is the reactant that is completely consumed in the reaction. In this case, hydrogen bromide (HBr) is the limiting reactant because it has a lower mole flow rate compared to ethylene (C2H4).

Now, calculate the fractional conversion of the limiting reactant using the mole flow rates:
Fractional Conversion of HBr = (Initial Mole Flow Rate of HBr - Mole Flow Rate of HBr in Product Stream) / Initial Mole Flow Rate of HBr
Fractional Conversion of HBr = (37.905 mol/s - 13.3 mol/s) / 37.905 mol/s = 0.649

Calculation of Excess Reactant:

Now, let's calculate the percentage by which the other reactant, ethylene (C2H4), is in excess in the feed stream:
Mole Flow Rate of Ethylene (C2H4) = Initial Mole Flow Rate - Mole Flow Rate of Limiting Reactant in Product Stream
Mole Flow Rate of Ethylene (C2H4) = 285 mol/s - 147.2055 mol/s = 137.7945 mol/s

Percentage Excess of Ethylene (C2H4) = (Mole Flow Rate of Excess Reactant / Initial Mole Flow Rate) x 100
Percentage Excess of Ethylene (C2H4) = (137.7945 mol/s / 285 mol/s) x 100 ≈ 48.3%

Calculation of Extent of Reaction:

The extent of reaction represents the fraction of the limiting reactant that has been converted into product. In this case, since the limiting reactant is hydrogen bromide (HBr), the extent of reaction is equal to the fractional conversion of HBr:
Extent of Reaction = Fractional Conversion of HBr = 0.649

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