Thermodynamics: Gibbs free energy from this battery reaction?

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Thermodynamics: Gibbs free energy from this "battery" reaction?

Homework Statement


I'm reading my text book and it gives an example.

The dH of the reaction is -316kJ/mol so that much energy is released by the reaction itself. Additionally, the entropy of the products are higher than the reactants, so TdS = 78kJ/mol.

So conclusively, the textbook states that 394kJ/mol of electrical work gets produced from this reaction.

I'm confused though, because it describes the system energy decreasing by 316kJ/mol (released as electrical work), and then 78kJ/mol of heat comes in from the system... which also gets converted to electrical work? How does this heat automatically get turned into electrical work?

Homework Equations


dG = dH + TdS
dS = Q/T

The Attempt at a Solution



So I'm thinking of it as either one of two ways, but both seem wrong.

1) The dH is actually -394kJ/mol, released as electrical work. Then the environment supplies 78kJ/mol of energy into the system. But this doesn't make sense because why would dH be -394kJ/mol when it explicitly states that the reaction only produced -316kJ/mol.
In other words the process of reaction of one mole:
i) dH = 0; dG = 0; TdS = 0
ii) dH = -394kJ; dG = -394kJ; TdS = 0
iii) dH = -316kJ; dG = -394kJ; TdS = 78kJ

--------------------------------------------------
OR
--------------------------------------------------

2) dH = -316kJ, released as electrical work. Then 78kJ goes into the system as heat. Then that 78kJ turns into electrical work somehow. In other words:
i) dH = 0; dG = 0; TdS = 0
ii) dH = -316kJ; dG = -316kJ; TdS = 0
iii) dH = -316kJ; dG = -394kJ; TdS = 78kJ

Both seem wrong to me... could someone please clarify?
 

Answers and Replies

  • #2
Mapes
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Both explanations are a little off. The reaction will be spontaneous as long as [itex]\Delta G<0[/itex]; thus, 394 kJ/mol of work can be extracted from the system and transferred to the surroundings. In the process, higher-entropy products will be created. The entropy has to come from somewhere, so the system would cool down if it were isolated. Because it's not isolated, 78 kJ/mol is transferred from the surroundings in the form of heat to keep the system at ambient temperature. Although it's tempting to infer that heat is being transferred to work, this is unwise. Heat and work are not nouns but rather verbs; they describe processes of energy transfer.

Does this answer your question?
 
  • #3
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Thank you for the response, but I'm still confused... you explained why it happens, but not how explicitly.

The energy out of the reaction itself is only 316kJ. And 78kJ enters the system by heat. But with what mechanism does it leave the system? I don't think it'd be chemical because the reaction had already occurred. Would the remaining 78kJ leave by perhaps mechanical work?

Then
Q + dU = W_electrical + W_mechanical
78kJ + 316kJ = 316kJ + 78kJ
?

If so, then why does my text say it all comes out as electrical work?
 
  • #4
Mapes
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The energy out of the reaction itself is only 316kJ.

No, the energy out of the reaction itself is 394 kJ/mol, and that's what you can extract via electrical work. [itex]\Delta G[/itex] is the amount of electrical energy you can extract at constant temperature and pressure.

Compare that to [itex]\Delta H[/itex], with magnitude 316 kJ/mol, which is the amount of thermal energy you can extract at constant temperature and pressure.
 
Last edited:
  • #5
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So, in a reaction where the temperature were not constant, would dH be -394kJ? Then the temperature would drop; which would have to be made up for by entropy entering the system, by heat. Thus bringing dH to -316kJ?

Also, what exactly do you mean by thermal energy?

I'm I'm still way off base, could you possibly give me an analogy of the process?
 
  • #6
Mapes
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So, in a reaction where the temperature were not constant, would dH be -394kJ? Then the temperature would drop; which would have to be made up for by entropy entering the system, by heat. Thus bringing dH to -316kJ?

[itex]\Delta H[/itex] is going to stay at approximately -316 kJ/mol for conditions near STP; this parameter isn't too sensitive to temperature changes.

Also, what exactly do you mean by thermal energy?

I mean the disorganized movement of atoms and molecules that characterizes high temperatures.
 
  • #7
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Ok... thanks for the answers. Still horribly confused though. Hopefully some day soon I can look back a this and understand.
 
  • #8
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There are some problems with the question as posed by khkwang.
First it is a pity that he did not present the chemical reaction equation involved.
This equation would explain why TdS is possitive.
Second, in the 'relevant equations' he writes dG=dH+TdS, which should be dG=dH-TdS, and he assumes dS=Q/T, but this holds only for the reversible process, which is not stipulated.
To me it is unlear what the data under i) and ii) mean ('attemted answer') since the data under iii) are the ones given, hence cannot be disputed.
The more usefull equation would be dH=Q+W(elec), using the sign convention that q and w are >0 when energy enters the system.
(see attachment to https://www.physicsforums.com/showthread.php?t=338573)
For the reversible case, dG=max(W(el)) and Q=TdS.
 

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