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

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Discussion Overview

The discussion centers on determining the standard Gibbs free energy change (ΔG°) for the ionization of water at 50°C, with a focus on the temperature dependence of pH and equilibrium considerations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that the pH of pure water at 50°C is 6.55 and questions whether the ionization process is at equilibrium at this temperature.
  • Another participant suggests calculating ΔG° using the equation ΔG° = -RTln(Keq) and expresses uncertainty about the correctness of their result of 40.5 kJ/mol.
  • A participant calculates the hydronium ion concentration from the pH and derives Keq, concluding it to be 7.94 x 10^-14, leading to a ΔG° calculation of 81.00 kJ/mol, while questioning the equilibrium assumption.
  • There is a reiteration of the calculation for Keq and ΔG°, with a request for confirmation of the equilibrium assumption at 50°C.
  • One participant acknowledges the assumption of equilibrium and indicates that the calculations appear to be correct.

Areas of Agreement / Disagreement

Participants express uncertainty regarding whether the ionization of water is at equilibrium at 50°C. There is no consensus on the correctness of the calculated ΔG° values, and multiple interpretations of the equilibrium condition are present.

Contextual Notes

The discussion includes assumptions about equilibrium and temperature effects on pH, as well as the dependence of ΔG° on the calculated Keq values. The calculations rely on specific temperature conditions and may not account for all variables affecting the ionization of water.

beneakin
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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
 
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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.
 
[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
 
beneakin said:
[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.
 

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