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 01-14-2012, 07:08 PM #1 Banned   Status: Pre-Podiatry MDApps: View Profile Join Date: Dec 2011 Posts: 18,777 Gibb's Free Energy and Keq SDN Members don't see this ad. (About Ads) This is a 2 question thread, if you don't mind Question 1: Can someone give me a simple, but thorough () 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?
 01-14-2012, 07:17 PM #2 Senior Member   Status: Pre-Medical Join Date: Oct 2011 Posts: 410 with question one, use ΔGo=-RT lnKeq .... so if ΔG 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
 01-14-2012, 07:48 PM #3 1K Member     Status: Pre-Medical Join Date: Dec 2009 Location: Where the rain grows Posts: 1,853 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.
01-14-2012, 07:57 PM   #4
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 Originally Posted by Ssina with question one, use ΔGo=-RT lnKeq .... so if ΔG 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
Thanks, I completely forgot about that equation .

Quote:
 Originally Posted by milski 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.
Thanks for the explanation. I wasn't look at it the right way initially.

 01-14-2012, 09:02 PM #5 2K Member   Join Date: Apr 2010 Posts: 2,389 "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!
01-14-2012, 10:10 PM   #6
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 Originally Posted by chiddler "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!

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

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