# Keq from Gibb's Free Energy

by kq6up
Tags: energy, free, gibb
 P: 209 Just for kicks. I tried calculating Keq from Gibb's free energy. This should be a straight forward calculation, but the answer is no where near close. Here is my calculations in SAGE: sage: R=8.314; T=298; G=237000 ; Keq=var('Keq'); f=G+R*T*log(Keq) sage: f.solve(Keq) [Keq == e^(-59250000/619393)] sage: float(e^(-59250000/619393)) 2.8588096844432612e-42 Note the positive sign before R, that is because when I solve for Keq sage assumes f=0. I assume using the -(G) should give the self dissociation constant for water. That is 10^(-14) Thanks, Chris Maness
 P: 209 Yes, I see that it is poorly formatted. I have students today, and I just through it up there expecting a quick answer. The reaction is the self ionization of water $${ H }_{ 2 }O(l)\Leftrightarrow { H }^{ + }(aq)+{ OH }^{ - }(aq)$$ However, in my haste I over looked something. I used the Gibbs free energy thinking that the products would be elemental. I need to use the heat of formation for Hydroxide and Hydronium and try it again. And yes, I do agree it should be in the chemistry topic. Thanks, Chris Maness
 P: 209 Keq from Gibb's Free Energy I have it now. Looked up Gibbs. Not sure why H+ Standrd Gibbs of Formation is 0. All the other ions have a value. I imagine it has to do with $${ 2H }^{ + }+2{ e }^{ - }\rightarrow { H }_{ 2 }$$ has a half cell potential of Zero. Chris