The gas-phase reaction between methanol (A) and acetic acid (B) to form methyl a

Question

The gas-phase reaction between methanol (A) and acetic acid (B) to form methyl acetate (C) and water (D) CH_3OH + CH_3COOH CH_3OOCCH_3 + H_2O 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 realated by a reaction equilibrium constant y_C y_D/y_A y_H = K_eq = 4.87 The feed to the reactor consists of n_A0, n_B0, n_C0, n_D0 and n_10 gram moles of A, B, C, D and an inert gas I respectively. a. If n_A0 = 1.200 n_D0, and all other input quantities are 0.0. what is the equilibrium fractional conversion of A? b. It is desired to produce 60.0 mol methyl acetate starting with 67.0 mol of methanol an unknown amount of acetic and and no other components. If the reaction proceeds to equilibrium, how much acetic acid must be fed? Requited acetic acid: mol What is the composition of the final product? y_acetic acid y_methylacetate: y_methanol: y_water: c. Suppose it were important to reduce the concentration of methanol by maxing its conversion at equilibrium 99.0%. Again, assuming the feed to the reactor contains only methanol and acetic acid and that is is desired to produce 60.0 mol of methyl acetate, determine the extent of reaction and mol of methanol and acetic acid that must be fed to the reactor. Extent: 60.606 Required methanol: 60.6 mol Required acetic acid: 1.5.6 mol

Explanation / Answer

a)

Let initial moles of B be 1 mole

since NA0 = 1.2NB0

initial moles of A will be 1.2 moles

Now,

CH3OH + CH3COOH = CH3COOCH3 + H2O

Initially 1.2 mols 1 mols 0 mols 0 moles

At equilibrium 1.2-x 1-x x x

We know that

[C] [D] / [A] [B] = 4.87

from above,

[x]*[x] / [1.2-x]*[1-x] = 4.87

x2 / (1.2-2.2x+x2) = 4.87

x2 = 5.844-10.714x+4.87x2

3.87x2 - 10.714x + 5.844 = 0

x = 2.02, 0.75

Since the equilibrium composition can not exceed initial composition, we take x = 0.75

Fractional conversion of A =  moles of A reacted/moles of A fed

Fractional conversion of A = 0.75 / 1.2

Fractional conversion = 0.625

b)

Let initially 67 moles of CH3OH and B moles of CH3COOH are fed to form 60 moles of CH3COOCH3 at equilibrium.

CH3OH + CH3COOH = CH3COOCH3 + H2O

Initially 67 mols B mols 0 mols 0 moles

At equilibrium 67-60 B-60 60 60

We know that

[C] [D] / [A] [B] = 4.87

from above,

[60]*[60] / [67-60] [B-60] = 4.87

[60]*[60] / [7] [B-60] = 4.87

B-60 = 105.60

B = 105.60+60 = 165.60 mols

Hence Initially 165.60 mols of CH3COOH must be fed.

Final product composition = (moles of component at equilibrium)/(sum of all components at equilibrium)

Yacetic acid = 105.60 / (105.60+7+60+60) = 105.60 / 232.6 = 0.45

Ymethanol = 7 / (105.60+7+60+60) = 7 / 232.6 = 0.03

Ymethyl acetate = 60 / (105.60+7+60+60) = 60 / 232.6 = 0.26

Ywater = 60 / (105.60+7+60+60) = 60 / 232.6 = 0.26

c)

It is desired to produce 60 mols of methyl acetate with 99% conversion

we have

CH3OH + CH3COOH = CH3COOCH3 + H2O

Initially A mols B mols 0 mols 0 moles

At equilibrium A-60 B-60 60 60

Fractional conversion = Amount of A reacted / amount of A fed

0.99 = (60) / A

0.99A = 60

A = 60.606 moles (Required methanol)

Extent of reaction = Initial moles fed - moles reacted

Extent of reaction = 60.606 - 60

Extent of reaction = 0.606

Extent of reaction = 60.6%

Fractional conversion = Amount of B reacted / amount of B fed

0.99 = (60) / B

0.99B = 60

B = 60.606 moles (Required acetic acid)

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