## Equilibrium Value K

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

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

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

RyanTran2F
Posts: 44
Joined: Thu Jul 27, 2017 3:00 am

### Equilibrium Value K

At equilibrium, K can be 1, >1, or <1. If K>1 then products are favored and ΔG is negative (the reaction favors the products). We also learned that a reaction proceeds spontaneously towards equilibrium (which is when ΔG = 0) so then does that mean, based on the equation learned in class, that at equilibrium K is supposed to be 1? I am pretty sure this is not the correct assumption but I do not know why since I thought we said that at equilibrium ΔG = 0. Please help? Thanks :)

Edit: Does it just mean that a reaction tries to reach ΔG = 0 and in the process releases energy that we can use for work (but it never actually reaches this state since usually K is >1 or <1 so one side is favored more)?
Last edited by RyanTran2F on Sun Feb 04, 2018 6:13 pm, edited 1 time in total.

Leah Thomas 2E
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am

### Re: Equilibrium Value K

I'm pretty sure Dr. Lavelle mentioned that at most times reaction will not have a K=1, or be at equilibrium. However, if K=1 then delta G will be equal to 0 because from the equation delta G=-RTln(K) taking the ln(1)=0 so the overall delta G will also equal 0.

Cyianna 2F
Posts: 34
Joined: Wed Nov 16, 2016 3:04 am

### Re: Equilibrium Value K

At equilibrium, K=1; this would represent a reaction at equilibrium, since both the reactants nor products are favored. $\Delta G$ is zero at equilibrium due to the fact that in the equation: $\Delta H-T*\Delta S$ if both enthalpy and entropy equal zero, then there is no change, and $\Delta G$ would be 0.

804899546
Posts: 51
Joined: Sat Jul 22, 2017 3:00 am

### Re: Equilibrium Value K

Actually I think K can be any value, its just a ratio of products to reactants at equilibrium. When K>1 its favored to have more products than reactants at equilibrium, and visa versa when K<1. K=1 is really rare because you'd have to have exactly the same concentration of products and reactants.