Equilibrium Value K

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Equilibrium Value K

Postby RyanTran2F » Sun Feb 04, 2018 11:47 am

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
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Joined: Fri Sep 29, 2017 6:06 am

Re: Equilibrium Value K

Postby Leah Thomas 2E » Sun Feb 04, 2018 1:00 pm

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
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Joined: Wed Nov 16, 2016 3:04 am

Re: Equilibrium Value K

Postby Cyianna 2F » Sun Feb 04, 2018 1:03 pm

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

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Re: Equilibrium Value K

Postby 804899546 » Sun Feb 04, 2018 1:32 pm

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.

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