∆G = ∆Gº + RT lnQ

[Ashley Wang 4G]

Hi,

I'm having trouble understanding the difference between the ∆G and ∆Gº terms in the equation ∆G = ∆Gº + RT lnQ.
Is ∆Gº the standard Gibb's free energy difference between the reactants and products, and ∆G the Gibb's free energy difference between the initial state and equilibrium (since we derived ∆Gº = -RT lnK at equilibrium)? Is this just because we can't assume the reaction is occurring at standard state conditions?

Thank you!

[Leyna Dang 2H]

∆G= the change of Gibbs free energy for a system while ∆G°= the change of Gibbs free energy for a system under standard conditions

[kausalya_1k]

deltaG knot is for when the system is at its standard state

[Gerald Bernal1I]

This equation relates equilibrium constant with Gibbs free energy. The refers to finding the G value when the reactants and products are in their standard states.

[Matthew Tsai 2H]

∆Gº refers to change in Gibbs free energy under standard conditions (usually this value will be given)
∆G refers to change in Gibbs free energy for the system (this is usually what we will be calculating for)

[sarahforman_Dis2I]

Hi,

I'm having trouble understanding the difference between the ∆G and ∆Gº terms in the equation ∆G = ∆Gº + RT lnQ.
Is ∆Gº the standard Gibb's free energy difference between the reactants and products, and ∆G the Gibb's free energy difference between the initial state and equilibrium (since we derived ∆Gº = -RT lnK at equilibrium)? Is this just because we can't assume the reaction is occurring at standard state conditions?

Thank you!

An important application of this could be reactions in biological systems.∆Gº and RT will both be constant, but ∆G will not. Like other students have said, ∆Gº is the change in gibs free energy under standard conditions whereas ∆G is the change in gibs free energy for the reaction. Since ∆Gº and RT are constant in biological systems, the reaction quotient Q must be changed in order to make a the ∆G negative. I know this isn't directly relating to your question, but I think it could be a useful application

[Marni Kahn 1A]

∆G is the change of Gibbs (free) energy for a system and ∆G° is the Gibbs energy change for a system under standard conditions (1 atm, 298K)

[Renee Grange 1I]

∆Gº is the delta G at standard conditions

[205405339]

∆Gº is basically just ∆G but under standard conditions

[Brian_Ho_2B]

Hi,

I'm having trouble understanding the difference between the ∆G and ∆Gº terms in the equation ∆G = ∆Gº + RT lnQ.
Is ∆Gº the standard Gibb's free energy difference between the reactants and products, and ∆G the Gibb's free energy difference between the initial state and equilibrium (since we derived ∆Gº = -RT lnK at equilibrium)? Is this just because we can't assume the reaction is occurring at standard state conditions?

Thank you!

To add on to what others have said, this equation is very important in chemistry and biology because it allows us to relate the delta G at standard conditions with the delta G in general. If we know what the delta G(not) is at standard 1 atm and 298K, we can use the equation to calculate a value for delta G at any given temperature and reaction quotient (any ratio of products to reactants).

[Katie Bart 1I]

Does the system have to be at equilibrium in order to use this equation?

[Brian_Ho_2B]

Does the system have to be at equilibrium in order to use this equation?

The system does not have to be at equilibrium or standard conditions to use this equation. The purpose of the "+ RTlnQ" is to account for differences in temperature and concentrations of reactants and products (not at equilibrium). If it was at equilibrium, then Q = K and delta G equals zero, meaning that delta G (not) is equal to "-RTlnK".

[Sally Qiu 2E]

delta G is the gibbs free energy under standard conditions