Question

you have designed a reactor to produce cumene via the reaction of propylene and benzene:     C3H6 + C6H6 → C9H12

In an undesired side reaction, cumene can react with propylene to form p-diisopropylbenzene (p-DIPB):

C₃H₆ + C₉H₁₂ → C₁₂H₁₈

1.

Draw and label a flowchart of the process and do the degrees-of-freedom analysis based on atomic species balances, taking a basis of 100 mol C₉H₁₂ produced. Assume that the feed contains only propylene and benzene, and that all of the propylene is consumed in the reaction.

2.

The reactor is designed to yield a 20% conversion of benzene and a selectivity of 10 mol C₉H₁₂/mol C₁₂H₁₈, with a negligible amount of propylene in the product gas. Calculate the feed ratio (mol C₆H₆/mol C₃H₆) and the fractional yield of cumene.

3.

Suppose that a heat exchanger was used to heat the feed mixture prior to entering the reactor. What information would you need to determine the required heat input? (Hint: start with the energy balance for an open system, and eliminate terms).

Answer 1

(there is some qty of DIBP is formed so actually not known exact qty) so , we have to use slight excess of reactants propylene like 1:1.1 or more) Benzene:propylene

2. the reactor is designed to yield 20% conversion of benzene and selectivity is 10:1,

for 1 mol of 20% = 0.20 mol conversion of benzene,

for 10 mol=1 mol selectivity therefore, ratio is 10

for 0.20 mol/10=.02 mol of DIBP, THEN WE WILL SUBSTRACT THE .02 FROM .20 MOL =0.18 MOL OF CUMENE IS FORMED.

FEED Ratio of propylene to benzene is should be 1( 1 mol benzene+ 1 mol propylene)

Yield of cumene=0.18 mol*120.19=21.63 gms actual gms

% yield =actual gms/therotical yield*100

=21.63/120.19=0.18*100=18%

therotical yield = moles of reactant*mol wt. of product/mol wt of reactant

=42.08 (1 mol)*120.19/42.08=120.19

3. to keep a track on how much energy needed for reaction we have to kept its accounting