The Gas Phase Reaction Between Methanol A And Acetic Acid B To Form Methyl Acetate C And Water D Ch3oh Ch3cooh 1 (56.96 KiB) Viewed 44 times
The gas-phase reaction between methanol (A) and acetic acid (B) to form methyl acetate (C) and water (D) CH3OH + CH3COOH CH3OOCCH3 + H₂O takes place in a batch reactor and proceeds to equilibrium. When the reactor mixture comes to equilibrium, the mole fractions of the four reactive species are related by a reaction equilibrium constant = Keq = 4.87 усур УА УВ The feed to the reactor consists of nÃO, no, no, no and nio gram moles of A, B, C, D and an inert gas I respectively. a. If nÃO = 1.400 ngo, and all other input quantities are 0.0, what is the equilibrium fractional conversion of A? i b. It is desired to produce 65.0 moles of methyl acetate starting with 72.0 moles of methanol, an unknown amount of acetic acid, and no other components. If the reaction proceeds to equilibrium, how much acetic acid must be fed? Required acetic acid: i mol
What is the composition of the final product? Yacetic acid: Ymethylacetate: Ymethanol: Ywater: i Extent: i i i c. Suppose it were important to reduce the concentration of methanol by making its conversion at equilibrium 99.5%. Again, assuming the feed to the reactor contains only methanol and acetic acid and that it is desired to produce 65.0 moles of methyl acetate, determine the extent of reaction and moles of methanol and acetic acid that must be fed to the reactor. Required methanol: i i Required acetic acid: i mol mol
Hint The first part may require using the quadratic equation.
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