Q2 [4.42 of the textbook]

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 % [tex]\text{CH}_5\text{Br}[/tex] and 17.3% [tex]\text{HBr}[/tex]. 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 165 mol/s, what is the extent of reaction? (Give its numerical value and its units.)

\[ \text{Reaction: } \text{C}_2\text{H}_4 + \text{HBr} \rightarrow \text{C}_2\text{H}_5\text{Br} \]

To calculate the fractional conversion of the limiting reactant and the percentage by which the other reactant is in excess, we need to first identify the limiting reactant.

Given:
- Product stream contains 51.7 mole % CH3Br and 17.3% HBr.
- Feed stream contains only ethylene (C2H4) and hydrogen bromide (HBr).
- Molar flow rate of the feed stream is 165 mol/s.

To identify the limiting reactant, we can compare the mole ratios of the reactants in the product stream to the mole ratios of the reactants in the balanced chemical equation for the reaction:

C2H4 + HBr -> C2H5Br

From the product stream analysis, we can determine the mole ratio of CH3Br to HBr as 51.7/17.3 = 2.99/1.

From the balanced chemical equation, the mole ratio of CH3Br to HBr is 1/1.

Since the mole ratio of CH3Br to HBr in the product stream is greater than 1/1, it means that HBr is the limiting reactant.

Now, let's calculate the fractional conversion of the limiting reactant (HBr). We can use the mole fraction of HBr in the feed stream and the product stream:

Mole fraction of HBr in the feed stream = (17.3 mol %) / 100 = 0.173

Mole fraction of HBr in the product stream = (17.3 mol %) / 100 = 0.173

The fractional conversion of HBr is given by the difference in the mole fractions:

Fractional conversion of HBr = (Mole fraction of HBr in feed stream) - (Mole fraction of HBr in product stream)
= 0.173 - 0.173
= 0

The fractional conversion of HBr is 0, which means that all the HBr in the feed stream is consumed in the reaction.

To calculate the percentage by which the other reactant (ethylene, C2H4) is in excess, we need to calculate the mole fraction of C2H4 in the feed stream and the product stream:

Mole fraction of C2H4 in the feed stream = 1 - Mole fraction of HBr in the feed stream
= 1 - 0.173
= 0.827

Mole fraction of C2H4 in the product stream = 1 - Mole fraction of CH3Br in the product stream
= 1 - (51.7 mol %) / 100
= 1 - 0.517
= 0.483

The percentage by which C2H4 is in excess is given by the difference in the mole fractions:

Percentage by which C2H4 is in excess = (Mole fraction of C2H4 in feed stream) - (Mole fraction of C2H4 in product stream)
= 0.827 - 0.483
= 0.344

The percentage by which C2H4 is in excess is 34.4%.

Finally, to calculate the extent of reaction, we can use the stoichiometry of the reaction:

1 mol of HBr reacts with 1 mol of C2H4 to form 1 mol of CH3Br.

Since the fractional conversion of HBr is 0, it means that none of the HBr is consumed in the reaction. Therefore, the extent of reaction is 0 mol/s.

In summary:
- The fractional conversion of the limiting reactant (HBr) is 0.
- The percentage by which the other reactant (C2H4) is in excess is 34.4%.
- The extent of reaction is 0 mol/s.

To know more about reaction:

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