Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: beakin 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
-
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
-
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+ ]
-
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