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Value of K in Gibbs free energy equation

Posted: Mon Feb 25, 2019 7:55 pm
by Katherine Grillo 1B
In the equation delta G = -RT*ln(K), K is the equilibrium constant right? But it's stated that a reaction is at equilibrium when delta G equals zero or when K equals 1. There are many reactions that have an equilibrium constant that is not equal to one but is in equilibrium. So I'm confused with this discrepancy.

Re: Value of K in Gibbs free energy equation

Posted: Mon Feb 25, 2019 8:09 pm
by Vincent Li 4L
When a chemical reaction is at equilibrium, the Gibbs free energy is, indeed, equal to zero, but that doesn't mean that the Gibbs free energy under standard conditions is (the little circle to the top right is what you're most likely missing). As you can see from the original equation, , when equilibrium is reached, delta G on the left hand side goes to zero, and solving the resulting equation for delta G naught yields -RTlnK. Hope this helps.

Re: Value of K in Gibbs free energy equation

Posted: Mon Feb 25, 2019 9:08 pm
by Miya Lopez 1I
Vincent Li 4L wrote:When a chemical reaction is at equilibrium, the Gibbs free energy is, indeed, equal to zero, but that doesn't mean that the Gibbs free energy under standard conditions is (the little circle to the top right is what you're most likely missing)...


Can you explain more about what the standard conditions are for Gibbs free energy?

Re: Value of K in Gibbs free energy equation

Posted: Mon Feb 25, 2019 9:16 pm
by Jayasuriya Senthilvelan 4I
Standard conditions for Gibbs free energy are some fixed temperature (usually 25 degrees celsius), pure liquid/solid, gases are 1 atm/bar, aqueous are 1 M. Also all compounds are at their standard states (so oxygen wouldn't be O in these conditions, it would be O2).