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Author Topic: Log_k of O2 formation from water  (Read 2614 times)

paznativ

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Log_k of O2 formation from water
« on: 30/10/18 20:39 »
Hello all
in the phreeqc.dat and sit.dat, the log_k for O2 formation is 86.08 and 85.98, respectively.
Can anyone help me with that number?

for O2, the reduction equation is:
O2 + 4H+ + 4e- = 2H2O, E0=1.23V

log(Keq)= n*F*E0/2.303*R*T = 4*96485*1.23/2.303*298*8.314=83.19

what am I missing here? why aren't  the values equal?

thank you in advance

Paz
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dlparkhurst

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Re: Log_k of O2 formation from water
« Reply #1 on: 30/10/18 20:46 »
I think it is the difference between O2(g) and O2(aq).

log K 86.08 applies to the equation


2 H2O(l) = O2(aq) + 4 H+ + 4 e-
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paznativ

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Re: Log_k of O2 formation from water
« Reply #2 on: 30/10/18 20:49 »
Dear Dr. Parkhurst
Thank you so much!

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John Mahoney

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Re: Log_k of O2 formation from water
« Reply #3 on: 31/10/18 17:44 »
The difference is because the Henry's Law constant is included in the reaction.   The original equation in most textbooks is based upon just the gas pressures so you get something around -83.19,  but in PHREEQC you actually are using dissolved oxygen not the gas pressure over the solution.  So you have to correct for that condition. 

If we look just at dissolving Oxygen into water the reaction in the PHASES block is:

O2(g)
   O2 = O2
   -log_k   -2.8983
   -analytic -7.5001 7.8981e-3 0.0 0.0 2.0027e5
   -T_c  154.6; -P_c   49.80; -Omega 0.021

So in this case the log_K is the Henry's law constant which is the same as your difference of -2.89
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