## Value of K in Gibbs free energy equation

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Katherine Grillo 1B
Posts: 69
Joined: Fri Sep 28, 2018 12:28 am

### Value of K in Gibbs free energy equation

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.

Vincent Li 4L
Posts: 48
Joined: Fri Sep 28, 2018 12:19 am

### Re: Value of K in Gibbs free energy equation

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, $\Delta G = \Delta G^{\circ} + RTlnK$ , 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.

Miya Lopez 1I
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Joined: Tue Nov 14, 2017 3:00 am
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### Re: Value of K in Gibbs free energy equation

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?

Jayasuriya Senthilvelan 4I
Posts: 31
Joined: Thu Jan 10, 2019 12:17 am

### Re: Value of K in Gibbs free energy equation

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).