Gibbs Free Energy Question

[tncekm]

Okay, so here is the question:

What is the standard Gibbs free energy of formation of water vapor at 25°C, if, for the reation shown below under standard conditions, ΔH = -484 kJ/mol, and ΔS = -89 J/K-mol

2H₂(g) + O₂(g) --> 2H₂O(g)

A) -457 kJ/mol
B) -395 kJ/mol
C) -229 kJ/mol
D) Water vapor does not form at 25°C

I used ΔG = ΔH - TΔS and a number that most closely corresponded to (A), however, (C) was given as the correct answer. Am I missing something here?

I noticed that if I cut ΔH and ΔS in half I get the right answer, so were they just referring ot the formation of a single mole of H₂O? If so, do you find the question as poorly worded as I do?

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[Foghorn]

The balanced chemical equation results in 2 mols of water vapor.

 

[tncekm]

Thanks for the response. However, I noticed the the equation is balanced as given and yields two moles of water. I just figured the question was asking for the change in gibbs free energy for the entire balanced the reaction.

Anyway, I'm pretty sure that I know the answer to my problem, and its just that I needed to adjust the given ΔH and ΔS values to yield just one mole of product.

Given that, would an AAMC test give a question like the one above? And, if so, should I just always assume to reduce the ΔH to reflect just one mole of product?

 

[iA-MD2013]

I'm not sure if I agree with the answer. Since ΔG is given in terms of kJ/mol, it seems like the answer should be what you said (A). If the answer were just in kJ, then that would require you to take into account the 2 mols of H2O(g) produced. I'm guessing it's something about the wording of the question...
Where is this question from?...I want to see their justification behind their answer.

 

[TheBoondocks]

I'm not sure if I agree with the answer. Since ΔG is given in terms of kJ/mol, it seems like the answer should be what you said (A). If the answer were just in kJ, then that would require you to take into account the 2 mols of H2O(g) produced. I'm guessing it's something about the wording of the question...
Where is this question from?...I want to see their justification behind their answer.

I agree. It's a bad question. The units leave you in kj/mole. If it were kj then it would make sense.

 

[tncekm]

I'm not sure if I agree with the answer. Since ΔG is given in terms of kJ/mol, it seems like the answer should be what you said (A). If the answer were just in kJ, then that would require you to take into account the 2 mols of H2O(g) produced. I'm guessing it's something about the wording of the question...
Where is this question from?...I want to see their justification behind their answer.

EK1001 G.Chem.

Justification (verbatim):
C is correct. Use the equation: ΔG°=ΔH°-TΔS° and don't forget to convert joules to kilojoules for entropy.

I'm just glad to see other people don't like the wording.

 

[iA-MD2013]

EK1001 G.Chem.

Justification (verbatim):
C is correct. Use the equation: ΔG°=ΔH°-TΔS° and don't forget to convert joules to kilojoules for entropy.

I'm just glad to see other people don't like the wording.

Yeah, I think its really crappy wording. Ummm...I guess the thing to learn from this is keep an eye out for whether they give you the values in J/mol or J.

 

[tncekm]

Yeah, I think its really crappy wording. Ummm...I guess the thing to learn from this is keep an eye out for whether they give you the values in J/mol or J.

Even then, they gave the H and S values in J/mole and the reaction shown had 2 moles of water forming, which would have meant we needed to multiply by 2, not divide by 2... Hmm...

 

[iA-MD2013]

Even then, they gave the H and S values in J/mole and the reaction shown had 2 moles of water forming, which would have meant we needed to multiply by 2, not divide by 2... Hmm...

EXACTLY!!! I was going to mention that...but then I thought I was completely wrong. But, yes, the math doesnt work out in either way (I think)...
Awesome...books make mistakes too...I hope this is one of them

 

[Vihsadas]

Okay, so here is the question:

What is the standard Gibbs free energy of formation of water vapor at 25°C, if, for the reation shown below under standard conditions, ΔH = -484 kJ/mol, and ΔS = -89 J/K-mol

2H₂(g) + O₂(g) --> 2H₂O(g)

A) -457 kJ/mol
B) -395 kJ/mol
C) -229 kJ/mol
D) Water vapor does not form at 25°C

I used ΔG = ΔH - TΔS and a number that most closely corresponded to (A), however, (C) was given as the correct answer. Am I missing something here?

I noticed that if I cut ΔH and ΔS in half I get the right answer, so were they just referring ot the formation of a single mole of H₂O? If so, do you find the question as poorly worded as I do?

Yeah I think you guys are right. They made one of two errors:
1) They meant for ΔS to be = -890, not -89
or
2) They meant for the two quantities to be given as kJ and J/K respectively.

 

[iA-MD2013]

Thank you Vihsadas! That makes me feel better

 

[TheBoondocks]

Thank you Vihsadas! That makes me feel better

indeed.

 

[tncekm]

Yeah I think you guys are right. They made one of two errors:
1) They meant for ΔS to be = -890, not -89
or
2) They meant for the two quantities to be given as kJ and J/K respectively.

Well, if it were option #1 the value comes out to be exactly 218.78, which isn't any of the listed values and we know that answer (d) is wrong.

And, if it were option #2 that doesn't go well with most enthalpy values I've seen. ΔH° values are usually given as energy/mole which makes sense because enthalpy is an extensive property and depends on amount.

I'm starting to think either their answer is just wrong or they left out info.

Either way, the fact that everyone else is confused has me feeling better