Keq from Gibb's Free Energy

In summary: ManessIn summary, Chris Maness tried to calculate Keq from Gibb's free energy, but the answer was no where near close. He incorrectly used the -(G) to get the self dissociation constant for water, and then he had to look up Gibbs to figure out why the H+ standard Gibbs of Formation is zero.
  • #1
kq6up
368
13
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
 
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  • #2
kq6up said:
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
First of all, this is probably more appropriate for the Chemistry Board.

Your post is a model of how not to explain a problem. You have not explained what reaction you are you dealing with. You have not explained how you get your formula for f or what f is. You have not explained the theory behind what it is you are trying to do. You have not explained the mathematics either. If you want someone to help you, you have to do a much better job of presenting your question.

AM
 
  • #3
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 [tex] { H }_{ 2 }O(l)\Leftrightarrow { H }^{ + }(aq)+{ OH }^{ - }(aq) [/tex]

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
 
  • #4
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 [tex] { 2H }^{ + }+2{ e }^{ - }\rightarrow { H }_{ 2 } [/tex] has a half cell potential of Zero.

Chris
 
  • #5


I would first commend you for your efforts in attempting to calculate Keq from Gibb's free energy. It is a complex calculation and it is not surprising that your result is not close to the expected value. There may be a few factors contributing to this discrepancy.

Firstly, it is important to make sure that all your units are consistent in the calculation. In your calculation, you have used the value of R in Joules/mol*K, but G in kJ/mol. This could lead to a significant difference in the final result.

Additionally, there may be some errors in the mathematical operations or in the values used. It is always a good practice to double check the values and equations used in a calculation to ensure accuracy.

Lastly, it is possible that your calculation is correct, but the result is not close to the expected value due to experimental limitations. There may be other factors at play that affect the equilibrium constant, such as temperature, pressure, and the presence of catalysts or inhibitors.

In any case, it is important to carefully review your calculation and consider all possible factors that may have contributed to the discrepancy. If you are still unable to obtain a reasonable result, it may be helpful to consult with a colleague or seek assistance from a mentor or expert in the field. Remember, science is a continuous learning process and it is perfectly normal to encounter challenges and make mistakes along the way. Keep up the good work and continue to explore and learn from your experiences.
 

What is Keq from Gibb's Free Energy?

Keq from Gibb's Free Energy is a mathematical equation used in thermodynamics to determine the equilibrium constant, or the ratio of products to reactants, for a chemical reaction.

How is Keq from Gibb's Free Energy calculated?

Keq from Gibb's Free Energy is calculated using the equation Keq = e-ΔG/RT, where ΔG is the change in free energy, R is the gas constant, and T is the temperature in Kelvin.

What does a high Keq value indicate?

A high Keq value indicates that the reaction strongly favors the formation of products at equilibrium. This means that the reaction is efficient and proceeds in the forward direction.

How is Keq from Gibb's Free Energy related to spontaneity?

Keq from Gibb's Free Energy is related to spontaneity through the equation ΔG = -RTln(Keq). If ΔG is negative, the reaction is spontaneous and the Keq value will be greater than 1. If ΔG is positive, the reaction is non-spontaneous and the Keq value will be less than 1.

Can Keq from Gibb's Free Energy be used to predict the direction of a reaction?

Yes, Keq from Gibb's Free Energy can be used to predict the direction of a reaction. If the calculated Keq value is greater than 1, the reaction will proceed in the forward direction. If the Keq value is less than 1, the reaction will proceed in the reverse direction. If the Keq value is equal to 1, the reaction is at equilibrium and no net change will occur.

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