Thermochemistry question

[dorjiako]

---

Members don't see this ad.

Ethanol metabolism in yeast consists of the conversion of ethanol C2H5OH to ethanoic acid CH3COOH. What is the enthalpy of the reaction if 0.1 mmol of ethanol is metabolized?
Inputs will be appreciated.

 

[vayntraubinator]

Ethanol metabolism in yeast consists of the conversion of ethanol C2H5OH to ethanoic acid CH3COOH. What is the enthalpy of the reaction if 0.1 mmol of ethanol is metabolized?
Inputs will be appreciated.

Enthalpy is usually given in J/mol.

So, start by converting .1mmol to mol ==> 1x10^(-4)mol and multiply by enthalpy = (to get) Joules

So, if you rearrange this relationship, you'll see dividing joules by moles youll get the enthalpy of the reaction (J/mol).

There's some missing info here, tho

 

[dorjiako]

Enthalpy is usually given in J/mol.

So, start by converting .1mmol to mol ==> 1x10^(-4)mol and multiply by enthalpy = (to get) Joules

So, if you rearrange this relationship, you'll see dividing joules by moles youll get the enthalpy of the reaction (J/mol).

There's some missing info here, tho

The above question is exactly what was asked. I don't know if they expect you to use outside knowledge of Hess law. In that case, the preceding question that was the combustion of propane used the following dissociation energies. O=O is 497KJ/mol, c=O is 805kj/mol, O-H is 464kj/mol, c-c is 347, as you can see no c-H bond dissociation energy was given but it was determined to be 413 in that problem. Then, the next question was exactly what I wrote above. I tried solving it using same number and I was getting 0.041kj. The answer in the book is 0.0021kj. The only problem that I am thinking is that is that the bond energy that was given for c=o(805) was what I used for both c=o and c-o. I don't know what value to use since none was given.
Help still needed.

 

[FutureDoctor503]

I solved this question and I also got 0.041 kJ. For the bond energy of c-o, I didn't plug in any value becasue there is one c-o bond on each side of the reaction and you can cancel it out.

 

[dorjiako]

I solved this question and I also got 0.041 kJ. For the bond energy of c-o, I didn't plug in any value becasue there is one c-o bond on each side of the reaction and you can cancel it out.

You are truly awesome for trying to help out a sister here. Anyway, I got exactly the answer you got and then changed gears. I looked up the heat of formation of ethanol(-277), for the 2(O) =249X2=498, for H20=-241 and for ethanoic acid = 483.16.
Using the usual formular of delta heat of reaction = delta heat of formation of product -delta heat of formation of reactants.
(483-241)-(-277+498)=
242-221= 21.
therefore for 0.1x10^-3mol = 21x 10^-4 = 0.0021KJ.
It took me a long time to get get the right answer but I eventually did. However, I am still at loss at why the bond energy formular did not give us the same answer, I wonder if there are steps we were missing.

Have you tried the other thermochemistry question that I posted. I absolutely have no idea on how to tackle that one. All input will be greatly appreciated. Thanks a lot.

 

[FutureDoctor503]

That's nice. Usually when I encounter bond enthalpy questions, the question stem includes heat of formation reactions and the only thing you have to do is balance the number of moles and add or subtract accordingly. I was surprised that this question did not include those reactions.
I tried to solve the other question as well but couldn't come up with a solution. I'll try to work on it again.

 

[dorjiako]

That's nice. Usually when I encounter bond enthalpy questions, the question stem includes heat of formation reactions and the only thing you have to do is balance the number of moles and add or subtract accordingly. I was surprised that this question did not include those reactions.
I tried to solve the other question as well but couldn't come up with a solution. I'll try to work on it again.

Thanks. Have you tried my PH/PKA PROBLEM. Still lost for now.

 

[ctag912]

so i figured it out...since you're taking away 2 hydrogen bonds, you have to multiply 413 by 2... you get 826... then 826- 805 = 21 kj. 21kj * 1E^-4 = .0021kj

 

[WarDaddy]

The bonds broken in ethanol are 5 C-H bonds, 1 O-H bond, 1 C-C bond.
The bonds formed in acetic acid are 1 C=0, 1 O-H, 1 C-C, and 3 C-H bonds.

As ctag said, it doesn't matter that you don't know the bond dissociation energy of the O-H bond, because they cancel each other out. Moreover, continue cancelling everything out that you can on each side of the equation. This really simplifies the math for these problems types and will same you time on test day. The problem becomes . . .

(2 C-H bonds in the reactant) - ( 1 C=O bond in the product)
So, thats 826 - 805 = 21 kJ/mol
21X10^-3 = 0.0021 kJ