C3H6 + C6H6 C9H12 C3H6 + C9H12 C12H18 A) Draw and label a flowchart of the proce

Question

C3H6 + C6H6 C9H12

C3H6 + C9H12 C12H18

A) 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.

B) 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.

C) 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).

Explanation / Answer

Number of variables are - 2

Number of independent atomic balances - 2

Number of nonreactive species - 0

Degree of freedom = 2-2 = 0

The conversion with respect to benzene is 20% which means that only 20% of the benzene in the feed is converted now the amount of propylene in the feed will have to be limiting as all the propylene is consumed such that we get cumene to DIPB in the ratio of 10:1

So propylene consumed is 10 for cume and 2 for DIPB

so for every 12 moles of propylene consumed 11 moles of benzene is consumed (10 for cumene and 1 for DIPB)

so if 100% benzene was to be consumed the ratio of propylene to benzene in the feed should have been 12 moles of propylene and 11 moles of benzene. But since 20% benzene is consumed we need to have the feed to benzene to be such that the 11 moles consumed is only 20% of the total in the feed.

So the feed should have 11moles /0.2 = 55 moles of benzene should be in the feed when 12 moles of propylene are present.

So feed ratio should be 55 moles C6H6/12 moles C3H6 = 4.58

Fractional yield of cumene will be 10 moles of cumene will be 10 moles x 120.20 g/mol = 1202 g cumene

1 mole of DIPB will be 162.27 g/mol = 162.3 g

1202g/(1202+162.3) = 0.881 is the fractional yield of cumene

We would need information on the heat required for the reaction, selectivity differences at different temperatures. The temeperature of the output feed.

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