What is the ΔG° for the ionization of water at 50C?

[beneakin]

Homework Statement

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

Homework Equations

ΔG=ΔG°+RTln(Q)

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

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

 

[GCT]

How exactly did you get to your answer? From the pH find the hydronium ion concentration then deduce the Keq by writing its equation in both equation and reaction form and plugging in the hydronium ion value.

 

[beneakin]

[H+] = 10^[-6.55] = 2.8*10^-7 = [OH-] since it is pure water?

so keq= [2.8*10^-7]^2 = 7.94*10^-14

ΔG°=-(8.31450)(273+50)ln(7.94*10^-14)

= 81.00 KJ mol-

is that correct, are we sure that it is in an equilbrium at 50C?

thanks again

 

[GCT]

[H+] = 10^[-6.55] = 2.8*10^-7 = [OH-] since it is pure water?

so keq= [2.8*10^-7]^2 = 7.94*10^-14

ΔG°=-(8.31450)(273+50)ln(7.94*10^-14)

= 81.00 KJ mol-

is that correct, are we sure that it is in an equilbrium at 50C?

thanks again

We are assuming equilibrium , your work seems fine.