∆G = ∆Gº + RT lnQ






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Ashley Wang 4G
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∆G = ∆Gº + RT lnQ

Postby Ashley Wang 4G » Sun Feb 16, 2020 11:10 pm

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
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Joined: Thu Jul 25, 2019 12:17 am

Re: ∆G = ∆Gº + RT lnQ

Postby Leyna Dang 2H » Sun Feb 16, 2020 11:16 pm

∆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
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Re: ∆G = ∆Gº + RT lnQ

Postby kausalya_1k » Mon Feb 17, 2020 11:11 pm

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

Gerald Bernal1I
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Re: ∆G = ∆Gº + RT lnQ

Postby Gerald Bernal1I » Mon Feb 17, 2020 11:21 pm

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
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Re: ∆G = ∆Gº + RT lnQ

Postby Matthew Tsai 2H » Tue Feb 18, 2020 12:02 am

∆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
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Re: ∆G = ∆Gº + RT lnQ

Postby sarahforman_Dis2I » Tue Feb 18, 2020 8:46 am

Ashley Wang 4G wrote: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
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Re: ∆G = ∆Gº + RT lnQ

Postby Marni Kahn 1A » Tue Feb 18, 2020 9:08 am

∆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
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Re: ∆G = ∆Gº + RT lnQ

Postby Renee Grange 1I » Tue Feb 18, 2020 9:39 am

∆Gº is the delta G at standard conditions

205405339
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Re: ∆G = ∆Gº + RT lnQ

Postby 205405339 » Tue Feb 18, 2020 9:40 am

∆Gº is basically just ∆G but under standard conditions

Brian_Ho_2B
Posts: 221
Joined: Fri Aug 09, 2019 12:16 am

Re: ∆G = ∆Gº + RT lnQ

Postby Brian_Ho_2B » Tue Feb 18, 2020 10:17 am

Ashley Wang 4G wrote: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
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Re: ∆G = ∆Gº + RT lnQ

Postby Katie Bart 1I » Tue Feb 18, 2020 12:02 pm

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

Brian_Ho_2B
Posts: 221
Joined: Fri Aug 09, 2019 12:16 am

Re: ∆G = ∆Gº + RT lnQ

Postby Brian_Ho_2B » Tue Feb 18, 2020 12:49 pm

Katie Bart 1I wrote: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
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Re: ∆G = ∆Gº + RT lnQ

Postby Sally Qiu 2E » Tue Feb 18, 2020 1:51 pm

delta G is the gibbs free energy under standard conditions


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