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khkwang
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Thermodynamics: Gibbs free energy from this "battery" reaction?
I'm reading my textbook 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?
dG = dH + TdS
dS = Q/T
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
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OR
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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?
Homework Statement
I'm reading my textbook 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?