plsfoldthx

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This is from my Kaplan book... I can't for the life of me figure out why A is the correct answer. I'm thinking it's a mistake

Methanol reacts with acetic acid to form methyl acetate and water as showin in the following table in the presence of an acid catalyst.

CH3OH (l) + CH3COOH (aq) -> CH3COOCH3 (aq) + H2O (l)

Bond dissociation energies (kj/mol)
C-C is 348 kj/mol
C-H is 413
C=O is 805
O-H is 464
C-O is 360

What is the heat of formation of methyl acetate in kJ/mol
A) -464 kJ/mol
B) +464 kJ/mol
C) -1,288 kJ/mol
D) +1,288 kJ/mol

I would have thought the heat of formation of methyl acetate would be the bond FORMATION energy of the C-O bond, hence -360 kJ/mol...

and the overall enthalpy change of the reaction is 0 right?
 

Geekchick921

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This is from my Kaplan book... I can't for the life of me figure out why A is the correct answer. I'm thinking it's a mistake

Methanol reacts with acetic acid to form methyl acetate and water as showin in the following table in the presence of an acid catalyst.

CH3OH (l) + CH3COOH (aq) -> CH3COOCH3 (aq) + H2O (l)

Bond dissociation energies (kj/mol)
C-C is 348 kj/mol
C-H is 413
C=O is 805
O-H is 464
C-O is 360

What is the heat of formation of methyl acetate in kJ/mol
A) -464 kJ/mol
B) +464 kJ/mol
C) -1,288 kJ/mol
D) +1,288 kJ/mol

I would have thought the heat of formation of methyl acetate would be the bond FORMATION energy of the C-O bond, hence -360 kJ/mol...

and the overall enthalpy change of the reaction is 0 right?
Okay, here's how I'm looking at it, I feel like the answer should be +464, not -464.

Two bonds need to be broken in the reactants.
H3C-OH <- That C-O bond, which requires +360 kJ/mol.
CH3COO-H <- That O-H bond, which requires +464 kJ/mol.

And we're only talking about methyl acetate for this problem, so we're only looking at the formation of one bond.
CH3COO-CH3 <- That C-O bond, which releases 360 kJ/mol.

Heat of formation = Bonds broken - bonds formed
= (464+360) - 360
= 464 kJ/mol

And yeah, since the formation of water requires another O-H bond, form, releasing another 464 kJ/mol of energy, the enthalpy change of the reaction is 0.
 

BerkReviewTeach

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Okay, here's how I'm looking at it, I feel like the answer should be +464, not -464.

Two bonds need to be broken in the reactants.
H3C-OH <- That C-O bond, which requires +360 kJ/mol.
CH3COO-H <- That O-H bond, which requires +464 kJ/mol.

And we're only talking about methyl acetate for this problem, so we're only looking at the formation of one bond.
CH3COO-CH3 <- That C-O bond, which releases 360 kJ/mol.

Heat of formation = Bonds broken - bonds formed
= (464+360) - 360
= 464 kJ/mol

And yeah, since the formation of water requires another O-H bond, form, releasing another 464 kJ/mol of energy, the enthalpy change of the reaction is 0.
If the OP typed the question correctly, they are looking for the heat of formation of methyl acetate and you have solved for the heat of reaction for the transesterification reaction shown.

With the information given, you can't solve for the heat of formation. You either need bond energies for C-C, H-H, and O-O to do it by the bond method or you need a table of heats of formation to do it using Hess' law.
 

Geekchick921

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If the OP typed the question correctly, they are looking for the heat of formation of methyl acetate and you have solved for the heat of reaction for the transesterification reaction shown.

With the information given, you can't solve for the heat of formation. You either need bond energies for C-C, H-H, and O-O to do it by the bond method or you need a table of heats of formation to do it using Hess' law.
OMG. Don't look at me! *Puts bag over head* :smack:

Buuuuut, since the heat of formation can't be found with the information given, and that it can't be found isn't one of the answers, I hope the OP comes back and clarifies.
 

plsfoldthx

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Hey guys.

I did type the question correctly and I noticed that they were looking for the heat of formation as well. I had thought that the heat of formation would be found just by adding all the bond energies together, but that would give me a value much higher than any of the answer choices so I assumed I was wrong.
 

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got the same answer.you can solve either using bond energies or heats of formation and we were given the bond energies. you do not need bond energies for h-h as there are no such bonds in product or reactant. however i got 464 too and would like to know why it is -464 if anyone can help a brother out.
 

dingyibvs

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Okay, here's how I'm looking at it, I feel like the answer should be +464, not -464.

Two bonds need to be broken in the reactants.
H3C-OH <- That C-O bond, which requires +360 kJ/mol.
CH3COO-H <- That O-H bond, which requires +464 kJ/mol.

And we're only talking about methyl acetate for this problem, so we're only looking at the formation of one bond.
CH3COO-CH3 <- That C-O bond, which releases 360 kJ/mol.

Heat of formation = Bonds broken - bonds formed
= (464+360) - 360
= 464 kJ/mol

And yeah, since the formation of water requires another O-H bond, form, releasing another 464 kJ/mol of energy, the enthalpy change of the reaction is 0.
This question can't be answered with the information given, you'll need to know the O-O and H-H bond energies. Your analysis is also incorrect for two other reasons:

1)The C-O bond broken is the one in CH3COOH, and the O-H bond broken is in CH3OH, not the other way around.

2)You neglected the O-H bond formed in H2O of the product.