Question: The gas-phase reaction between methanol and acetic acid to form methyl acetate and water CH 3 OH + CH 3 COOH = CH 3 COOCH
The gas-phase reaction between methanol and acetic acid to form methyl acetate and water
CH3OH + CH3COOH = CH3COOCH3 + H2O
(A) (B) (C) (D)
Takes place in a batch reactor and proceeds to equilibrium. When the reaction mixture conies to equilibrium, the mole fractions of the four reactive species satisfy the relation yCyD/ yA yB = 4.87
(a) Suppose the feed to the reactor consists of nA0, nB0, nC0, nD0, and gram-moles of A. B. C, D, and an inert gas, I. respectively. Let (mol) be the extent of reaction. Write expressions for the gram-moles of each reactive species in the final product. nA(ζ). nB(ζ). nC(ζ), and nD(ζ). Then use these expressions and the given equilibrium relation to derive an equation for, the equilibrium extent of reaction, in terms of nA0, . . . n10.
(b) If the feed to the reactor contains equimolar quantities of methanol and acetic acid and no other species calculate the equilibrium fractional conversion.
(c) It is desired to produce 70 mol of methyl acetate starting with 80 mol of acetic acid. If the reaction proceeds to equilibrium, how much methanol must be fed? What is the composition of the final product?
(d) If you wanted to carry out the process of part (b) or (c) commercially, what would you need to know besides the equilibrium composition to determine whether the process would be profitable? (List several things)
Step by Step Solution
3.46 Rating (159 Votes )
There are 3 Steps involved in it
a ABCD n n4 5 nB nBo nc nco nDDo 5 n1 n10 Total nr Y n SnT YB B 5n Yc nc 5n YD nD 5nT b C At equil... View full answer
Get step-by-step solutions from verified subject matter experts
Document Format (2 attachments)
13-E-C-E-C-P (145).pdf
180 KBs PDF File
13-E-C-E-C-P (145).docx
120 KBs Word File
