It is a simple Delta G naught reaction problem

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Homework Statement


Calculate Delta G naught rxn and K for the reaction:

2 Fe 3+ +2I- <--> 2 Fe 2+ +I2

Homework Equations

The Attempt at a Solution


I came up with -.24 V for E naught and plugged it into ( -2*96.5*-.24) and got Delta G as 46.32 KJ but I don't know if this is right and I'm not sure if it is delta G naught rather than just delta G.

Then from that answer ( if it is right ) I put Delta G into ( Keq = e^-delta G naught/ RT) and got 8*10^-9

Please let me know if I am doing this right. It is due Monday and I need to make sure I am doing it right for final exam on Tuesday. Thank You
 
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Try asking this in the chemistry section ... I don't have any tables (and you ignore units except for 46 kJ, which should be kJ/mole)
... it's been a long time, but 46 kJ/mol sounds suspiciuosly high for an iron 2 - iron 3 reaction. what the heck is 2*96.5 ?
 
lightgrav said:
Try asking this in the chemistry section ... I don't have any tables (and you ignore units except for 46 kJ, which should be kJ/mole)
... it's been a long time, but 46 kJ/mol sounds suspiciuosly high for an iron 2 - iron 3 reaction. what the heck is 2*96.5 ?
-2*96.5*-.46 is nFE Moles of electrons *Faraday's constant * E rxn and I am prett sure Volts and Moles cancel leaving
KJ.
The Answer however does sound high to me as well though.

Is there a way to move this to chemistry thread ( I am new to this),
or should I just go there and start new?

(Mentor Note: Moved)
 
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ok, a Faraday is a mole of electrons, with charge 6E23e/mol*1.6E-19C/e = 96 kC/mol
... Energy = charge * voltage = 2e/molecule * 96 kC/mol *0.24V = 46 kJ/mole.
the R is 8.3 J/mole*Kelvin, or =0.0083 kJ/mol*K ... so the moles cancel in the exponent, as they should.
I get 9.5E-9 at 300 Kelvin, meaning that the "weak" ¼ volt drives the reaction essentially to completion
(equilibrium is Temperature-dependent, but would not approach 50/50 until 5,500 Kelvin).

I don't know how to move a thread, either. sorry ; best I can do.
 
From my CRC Handbook:
DeltaG0 in kcal/gmol:

Fe3+... -2.52

Fe2+... -20.3

I-..... -12.35

I2..... 4.63

Using these, I did not get your answer.
 
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lightgrav said:
I get 9.5E-9 at 300 Kelvin, meaning that the "weak" ¼ volt drives the reaction essentially to completion
Do you mean completion to the right in the reaction as stated? With Keq = 9.5E-9?
 
State the reaction conditions.
lightgrav said:
46 kJ/mol sounds suspiciuosly high
Deudoronomy said:
sound high to me as well though
Do NOT use intuition.
Deudoronomy said:
Delta G naught rxn and K
Reread the problem statement as frequently as is necessary to avoid sidetracking yourselves.