semibatch reaction:
C6H6(l) +Cl2(g) ==> C6H5Cl(l) + HCl(g)
C6H5(l) + Cl2(g) ==> C6H4Cl2(l) + HCl(g)
C6H4Cl2(l) + Cl2(g) ==> C6H3Cl3(l) + HCl(g)
\reactor contained 11.0 mole of benzene and a product mixture in which the amt of mono-chlorobenzene was 3x the amt of dichlorobenzene, and the amount of the latter was 2x the amount of tri-chlorobenzene. The total amount of chlorine fed during the operation was 41.0 mol Cl2.
Find:
a) The fractional conversion of benzene.
b) The amt of each liquid component besides benzene in the reactor
c) the amount of HCl produced
d) the fractional conversion of chlorine
To solve this problem, we can use material balance equations to determine the amount of each component involved in the reaction.
Let's denote:
x1 = moles of mono-chlorobenzene (C6H5Cl)
x2 = moles of dichlorobenzene (C6H4Cl2)
x3 = moles of trichlorobenzene (C6H3Cl3)
x4 = moles of benzene (C6H6)
x5 = moles of HCl
We are given the following information:
- Initial moles of benzene (x4) = 11.0 mol
- The amount of mono-chlorobenzene (x1) is three times the amount of dichlorobenzene (x2) -> x1 = 3x2
- The amount of dichlorobenzene (x2) is two times the amount of trichlorobenzene (x3) -> x2 = 2x3
- Total amount of chlorine fed (Cl2) = 41.0 mol
Now, let's solve the problem step by step:
a) Fractional conversion of benzene:
The fractional conversion of benzene (x) can be calculated by dividing the moles of benzene reacted (initial moles - remaining moles) by the initial moles of benzene.
Initial moles of benzene = 11.0 mol
Remaining moles of benzene = x4
Fractional conversion of benzene (x) = (Initial moles of benzene - Remaining moles of benzene) / Initial moles of benzene
b) Amount of each liquid component besides benzene in the reactor:
From the given information, we know that x1 = 3x2 and x2 = 2x3. By substituting these relationships, we can calculate the moles of each component.
x1 = 3x2 = 3(2x3) = 6x3
x2 = 2x3
x4 = 11.0 mol (initial moles of benzene)
c) Amount of HCl produced:
The amount of HCl produced is equal to the moles of mono-chlorobenzene (x1) since one mole of mono-chlorobenzene produces one mole of HCl.
Amount of HCl produced = x1
d) Fractional conversion of chlorine:
The fractional conversion of chlorine (y) can be calculated by dividing the moles of chlorine reacted (initial moles - remaining moles) by the initial moles of chlorine.
Initial moles of chlorine = Total amount of chlorine fed (Cl2) = 41.0 mol
Remaining moles of chlorine = ?
Fractional conversion of chlorine (y) = (Initial moles of chlorine - Remaining moles of chlorine) / Initial moles of chlorine
By solving for x, substituting the relationships for b, and calculating c and d, you should be able to find the answers to each part of the problem.