# Gibb's Free Energy and Keq

Discussion in 'MCAT Study Question Q&A' started by MedPR, Jan 14, 2012.

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1. ### MedPRBanned

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Dec 1, 2011
MDApps:
This is a 2 question thread, if you don't mind

Question 1: Can someone give me a simple, but thorough D) explanation on how deltaG and Keq are related? The specific example that triggered the question is related to transesterification. Explained here by TBR:

"A C-O and a O-H bond were both broken in the reactants and formed in the products. This implies that the enthalpy for the reaction is roughly 0. Going from an ester and non-cyclic alcohol to an ester and non-cyclic alcohol generates a change in entropy of roughly 0, because the reaction starts and finishes with roughly the same degrees of freedom. This implies that the change in free energy for transesterification is around 0. This means the equilibrium constant for transesterification is approximately 1."

I understand the reasoning behind deltaH = 0, deltaS = 0 and deltaG=0, but I don't understand how you can determine that Keq is 1. I know that Keq = 1 indicates that there is equilibrium between products and reactants, since Keq is [P]/[R], but just because Gibbs is 0 doesn't mean the reaction is in equilibrium does it? I know that negative delta G means the reaction is spontaneous, but doesn't the activation energy still need to be reached? Meaning, not every reaction with negative deltaG occurs constantly.

Question 2: The next line says "Transesterification can be used for shuttles in the cell membrane"

Are there any examples of this that we should know for the MCAT?

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3. ### Ssina 2+ Year Member

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Oct 10, 2011
with question one, use &#916;Go=-RT lnKeq .... so if &#916;G is zero, then Keq will have to be 1.
for the second one... no idea I guess more specific stuff should come in a passage for me to know

4. ### milski1K member 5+ Year Member

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Dec 30, 2009
Where the rain grows
deltaS, deltaH and deltaG being zero mean that there is no real difference (from thermodynamic point of view) between the two states. In other words, the molecules are equally happy to be in either product or reactant state. Since both are equally stable, it is logical that the equilibrium will lie in the middle, where you have the same amount of each, making Keq=1.

Activation energy has no influence on the final result of the equilibrium, only on how fast it will be reached. Having high activation energy might mean that it will take a really long time but if thermodynamics say that Keq is 1, eventually equilibrium will be achieved.

Here is somewhat made up example. Let's say that you have a few hundred coins. They are equally happy to lay on the ground as both heads or tails. Based on that thermodynamics says that if you have them lying around eventually you'll end up with same amount of heads and tails. It does not say how often that will happen.

Now let's talk about kinetics and activation energy. In normal conditions, it's pretty hard to flip a coin sitting on the floor. That's the equivalent of very high activation energy and means that it will take a really long time to achieve that equilibrium. You'll need some unusual spikes of energy to get there - earthquakes, strong winds, whatever.

Imagine that the coins are sitting on a floor which is experiencing strong vibrations and is making the coins bounce in the air all the time. That would be the equivalent of the molecules being in a high energy/temperature environment. Since you have plenty of energy the equilibrium will be achieved much faster. Note that it is still the same equilibrium.

If you wanted an example of a coin with lower activation energy, you can think of spherical or close to spherical coins - they would be much easier to flip and would get to equilibrium faster even on a normal non-vibrating floor.

5. ### MedPRBanned

18,606
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Dec 1, 2011
MDApps:
Thanks, I completely forgot about that equation .

Thanks for the explanation. I wasn't look at it the right way initially.

6. ### chiddler 5+ Year Member

Apr 6, 2010
"Transesterification can be used for shuttles in the cell membrane"

Curious if anybody has a good response for this.

What i'm thinking is that phospholipids are part of the membrane, yes? They contain fatty acids, which are connected to the glycerol component of the phospholipid. Maybe R-OH attacks the carbonyl of the fatty acid and replaces it, THEN the phospholipid flip flops its position with the help of an enzyme (because it is non-spontaneous), then another alcohol attacks the phospholipid where the R-OH was, allowing it inside. Transport!

So brilliant, it just might work!

7. ### MedPRBanned

18,606
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Dec 1, 2011
MDApps:

So you're saying that the R in ROH is much longer/more branched than R' in R"COOR'. So when transesterification occurs, the new ester is more nonpolar due to the larger alkyl group? That sounds like it would work to me

8. ### jaybird12

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Jul 24, 2014
So let me make sure I understand the relationship between ΔG˚ and Keq.
if K=1 then ΔG˚ = 0
if K>1 then ΔG˚ < 0
if K<1 then ΔG˚ > 0

ΔG˚ shows which reaction direction is favored thermodynamically at standard state. So if ΔG˚ is negative for the forward direction, then the products are favored over the reactants and Keq is greater than 1. But if ΔG˚ is positive for the forward direction, then that direction is unfavorable thermodynamically speaking, and the reactants are favored over the products and Keq is less than 1.
Conversely, if one side is favored over the other, say products over reactants, then the forward reaction is favored because it gives more products, and so the forward reaction is spontaneous and ΔG˚ is negative.
Is that right?

9. ### Teleologist 2+ Year Member

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Jul 7, 2013
on your 6
The forward reaction is thermodynamically favored.

10. ### Teleologist 2+ Year Member

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Jul 7, 2013
on your 6
You mean concentration.

Epic example. Kudos dude for successfully simplifying statistical mechanics and the Boltzmann distribution regarding kinetic energy of a collection of molecules.

11. ### type12 2+ Year Member

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Jun 12, 2012
Why did you revive a 2-year old thread?

Besides that, you're 100% golden. You're correct in your reasoning.

12. ### justadream 5+ Year Member

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Apr 29, 2011
@type12

I found an example in TBR that seems to contradict jaybird's conclusion:

TBR GC Book II page 137 Example 8.3

“A reaction that is spontaneous in the forward direction corresponds with which of the following features??

Correct: It has a ratio of Q(of reaction) to Keq that is less than 1.

Incorrect: It has an equilibrium constant that is greater than 1.0.

TBR says "Favorability in the forward direction implies there is an excess of reactants and a shortage of products relative to equilibrium. This tell us nothing about the value of the equilibrium constant. The value of Keq may or may not be greater than 1.0"

13. ### type12 2+ Year Member

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Jun 12, 2012
There's no contradiction. Q is always trying to reach Keq. A Q/Keq less than one means that Q needs to increase to go towards Keq; How do you increase Q, which is [Products]/[Reactants]?

Last edited: Aug 19, 2014
14. ### Teleologist 2+ Year Member

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Jul 7, 2013
on your 6
Tricky. Only at std. state conditions does Keq > 1 mean that a reaction is spontaneous.

15. ### justadream 5+ Year Member

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Apr 29, 2011
Okay so let's say I'm at non-standard state conditions.

Can I still make conclusions about spontaneity based on the relative values of Q and K (namely, Q<K means spontaneous and vice versa)?

16. ### type12 2+ Year Member

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Jun 12, 2012
You can make conclusions, but you're looking at Q and K strangely.

Q is where you currently are, the ratio of products over reactants. That's all it says.

K is the same ratio, but it's not where you are, but where the reaction will have the lowest free energy: where your reaction will be happy (at equilibrium). Thermodynamics tells us the reactions wants to rid itself of as much free energy as possible.

So, Q (current state) will try to become the same value as K (happy, low free energy state), standard conditions or otherwise.

17. ### justadream 5+ Year Member

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Apr 29, 2011
So when people talk about "spontaneous" reactions, they are talking about reactions spontaneous at standard conditions, correct?

If so, could you make a reaction typically called "nonspontaneous" become spontaneous by varying the conditions?

18. ### type12 2+ Year Member

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Jun 12, 2012
No to the first; yes to the second. I'm getting the sense you are confusing terms possibly.

You have a spontaneous direction in both standard and nonstandard conditions, and that refers to the relationship between Q and K. If Q = K, that means your reaction is at equilibrium and you will not see an increase in either products or reactions. If they are not, the spontaneous direction is the direction that will make Q equal to the value of K.

And yes, you can make a direction nonspontaneous become spontaneous by varying the conditions. For the MCAT, you need to know you can do this change by changing temperature, pressure, or concentrations.

Last edited: Aug 20, 2014
19. ### justadream 5+ Year Member

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Apr 29, 2011
@type12

What I meant by my first statement was:

Like when people define a reaction as "spontaneous", don't they need to specify the conditions they are talking about (e.g., standard conditions)?

Because technically, every reaction could be spontaneous/nonspontaneous if you change the conditions?

20. ### type12 2+ Year Member

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Jun 12, 2012
Yes and yes. This is correct.