## Go=0

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

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

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

Moderators: Chem_Mod, Chem_Admin

EMurphy_2L
Posts: 128
Joined: Sat Sep 07, 2019 12:16 am

### Go=0

when does G*=0

Wendy 1E
Posts: 111
Joined: Sat Aug 17, 2019 12:17 am
Been upvoted: 2 times

### Re: Go=0

In lecture today we learned that ΔG is zero at a reaction's boiling point.

JesseAuLec1Dis1G
Posts: 52
Joined: Mon Jun 17, 2019 7:23 am

### Re: Go=0

Gibb's free energy is at zero when the reaction is at equilibrium.

Abigail Sanders 1E
Posts: 112
Joined: Wed Sep 11, 2019 12:16 am

### Re: Go=0

To clarify the last two responses, Gibbs Free energy is 0 both at the boiling point and when the reaction is at equilibrium.

alex_4l
Posts: 109
Joined: Thu Sep 26, 2019 12:18 am

### Re: Go=0

G = 0 when the rate of forward reaction equals the rate of the reverse reaction (equilibrium)

preyasikumar_2L
Posts: 101
Joined: Fri Aug 09, 2019 12:17 am

### Re: Go=0

ΔG = 0 means that the system is at equilibrium - the forward and reverse reactions are occurring at the same rate at the same time.

Eva Zhao 4I
Posts: 101
Joined: Sun Sep 29, 2019 12:16 am

### Re: Go=0

∆G˚=0 when products and reactants are in their standard state and K=1. Note that this is not common.

Justin Quan 4I
Posts: 104
Joined: Sat Sep 14, 2019 12:17 am

### Re: Go=0

To add on, you can figure out when ∆G = 0 by using the mathematical relationships between ∆G˚= - RTlnK and ∆G = ∆G˚ + RTlnQ. We know that at equilibrium Q=K. If we substitute ∆G˚out, then ∆G = RTlnQ - RTlnK, thus ∆G = RTlnQ - RTlnK = RTlnK - RTlnK = 0 when the reaction is at equilibrium.

Return to “Gibbs Free Energy Concepts and Calculations”

### Who is online

Users browsing this forum: No registered users and 2 guests