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Topic: Thermodynamics: finding ΔG° from pH  (Read 7192 times)

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Offline beakin

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Thermodynamics: finding ΔG° from pH
« on: December 02, 2008, 07:45:12 PM »
1. The problem statement, all variables and given/known data
The pH of pure water is temperature dependent, for example it is 6.55 at 50 C. Determine ΔG° for the ionization of water at 50C


2. Relevant equations
ΔG=ΔG°+RTln(Q)

at eqm ΔG=0
q=Keq
pure water:
pH 7.00 at 25C


3. The attempt at a solution
Not sure if the process is at eqm at 50C

but if it was you could just use
ΔG°=-RTln(Keq)
and use the pH to get the keq and you end up with 40.5KJ/mol which doesn't seem right

what is the difference between ΔG° and ΔGf°

thanks

Offline opuktun

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Re: Thermodynamics: finding ΔG° from pH
« Reply #1 on: December 25, 2008, 01:43:33 PM »
Your idea seems correct. Here's my calculation:

H2O <=> H+ + OH-

At eqm,  :delta: G = :delta: G⁰ + RT ln Q = 0
:delta: G = - RT ln K with K=Kw=10-14 at 298 K
:delta: G = - 8.31451 x 298 ln (10-14) = 79.9 kJ mol-1
:delta: G = :delta: G⁰ + RT ln Q = 79.9 + 8.31451 x 298 ln [10-6.55x2] / 1000= -1.14 kJ mol-1

Offline Hunt

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Re: Thermodynamics: finding ΔG° from pH
« Reply #2 on: December 28, 2008, 06:53:26 PM »
1. The problem statement, all variables and given/known data
The pH of pure water is temperature dependent, for example it is 6.55 at 50 C. Determine ΔG° for the ionization of water at 50C


2. Relevant equations
ΔG=ΔG°+RTln(Q)

at eqm ΔG=0
q=Keq
pure water:
pH 7.00 at 25C


3. The attempt at a solution
Not sure if the process is at eqm at 50C

but if it was you could just use
ΔG°=-RTln(Keq)
and use the pH to get the keq and you end up with 40.5KJ/mol which doesn't seem right

what is the difference between ΔG° and ΔGf°

thanks

The process should be at thermal and chemical equilibria at 50 C.
How did u calculate Kw ? Remember that in this case Kw = 10-2pH because [ OH- ] = [ H+ ]

Offline Loyal

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Re: Thermodynamics: finding ΔG° from pH
« Reply #3 on: December 29, 2008, 01:23:36 AM »
Unless there are strict corrections (Which I do not see from the problem listed) he got the equilibrium constant from the usual equation.

pH = -log([H30+])
So  [H30+] = 10-pH

then

K = [H30+][OH-
with [OH-]  = [H30+]

K(25 C) = x2 = (10-7.00)2 = 10-14
Chemistry Student(Senior) at WSU

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