The gas-phase reaction between methanol (A) and acetic acid (B) to form methyl acetate (C) and water (D) CH3OH + CH3COOH=CH300CCH3 + 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 Ус Ур = Keg = 5.87 Ул Ув The feed to the reactor consists of nAo, NBo, nco, Npo and njo gram moles of A, B, C, D and an inert gas I respectively. a. If nao = 0.600 ngo, and all other input quantities are 0.0, what is the equilibrium fractional conversion of A? i 0.8945 b. It is desired to produce 65.0 moles of methyl acetate starting with 70.0 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: 208.95 mol
The gas-phase reaction between methanol (A) and acetic acid (B) to form methyl acetate (C) and water (D) CH3OH + CH3COOH=CH300CCH3 + 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 Ус Ур = Keg = 5.87 Ул Ув The feed to the reactor consists of nAo, NBo, nco, Npo and njo gram moles of A, B, C, D and an inert gas I respectively. a. If nao = 0.600 ngo, and all other input quantities are 0.0, what is the equilibrium fractional conversion of A? i 0.8945 b. It is desired to produce 65.0 moles of methyl acetate starting with 70.0 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: 208.95 mol
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![The gas-phase reaction between methanol (A) and acetic acid (B) to form methyl acetate (C) and water (D) is represented by the following chemical equation:
\[ \text{CH}_3\text{OH} + \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{OOCCH}_3 + \text{H}_2\text{O} \]
This reaction occurs in a batch reactor and proceeds to equilibrium. At equilibrium, the mole fractions of the four reactive species are related by a reaction equilibrium constant:
\[
\frac{y_C y_D}{y_A y_B} = K_{eq} = 5.87
\]
The feed to the reactor consists of \(n_{A0}, n_{B0}, n_{C0}, n_{D0}\), and \(n_{I0}\) gram moles of A, B, C, D, and an inert gas I, respectively.
a. If \(n_{A0} = 0.600 \, n_{B0}\) and all other input quantities are 0.0, the equilibrium fractional conversion of A is calculated as:
- Equilibrium fractional conversion of A: 0.8945
b. To produce 65.0 moles of methyl acetate starting with 70.0 moles of methanol, an unknown amount of acetic acid is needed, with no other components.
- Required acetic acid required, if the reaction proceeds to equilibrium, is:
\[ \text{Required acetic acid: } 208.95 \, \text{mol} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F73303aca-d795-4569-9e8f-a8984c4a41f6%2Fbb0b22ff-4163-4147-b5c7-87d91f420f69%2Fhpzqqyp_processed.png&w=3840&q=75)
Transcribed Image Text:The gas-phase reaction between methanol (A) and acetic acid (B) to form methyl acetate (C) and water (D) is represented by the following chemical equation:
\[ \text{CH}_3\text{OH} + \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{OOCCH}_3 + \text{H}_2\text{O} \]
This reaction occurs in a batch reactor and proceeds to equilibrium. At equilibrium, the mole fractions of the four reactive species are related by a reaction equilibrium constant:
\[
\frac{y_C y_D}{y_A y_B} = K_{eq} = 5.87
\]
The feed to the reactor consists of \(n_{A0}, n_{B0}, n_{C0}, n_{D0}\), and \(n_{I0}\) gram moles of A, B, C, D, and an inert gas I, respectively.
a. If \(n_{A0} = 0.600 \, n_{B0}\) and all other input quantities are 0.0, the equilibrium fractional conversion of A is calculated as:
- Equilibrium fractional conversion of A: 0.8945
b. To produce 65.0 moles of methyl acetate starting with 70.0 moles of methanol, an unknown amount of acetic acid is needed, with no other components.
- Required acetic acid required, if the reaction proceeds to equilibrium, is:
\[ \text{Required acetic acid: } 208.95 \, \text{mol} \]
Transcribed Image Text:## Composition of the Final Product
- **Acetic Acid ($y_{acetic \, acid}$):** 0.516
- **Methyl Acetate ($y_{methyl \, acetate}$):** 0.0179
- **Methanol ($y_{methanol}$):** 0.233
- **Water ($y_{water}$):** 0.233
---
## Reaction Optimization Scenario
Suppose it is important to reduce the concentration of methanol by increasing its conversion at equilibrium to 98.5%. Assuming the reactor feed contains only methanol and acetic acid, with the goal to produce 65.0 moles of methyl acetate, determine the following:
- **Extent of Reaction:** 65
- **Required Methanol:** 65.98 mol
- **Required Acetic Acid:** 1459.5 mol
This analysis aims to adjust reactant quantities to achieve desired conversion efficiency and product yield.
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