Mass Action Vs Gibbs Free Energy?

[SyrianHero]

I'm having a problem understanding why a K= 0 indicates deltaG = 0, K > 1 indicates delataG < 0, and K < 1 indicates deltaG > 0. Can anyone explain why this is the case?

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[Altius Premier Tutor]

Mathematically, the formula helps get the relationships clear and then we can talk more about why: deltaG0 = -RTln(Keq).

You NEED to know for the MCAT that the ln(n<1) = negative; ln(1.0) = zero; ln(n>1) = positive. You also need to know that negative deltaG = spontaneous; positive deltaG = non-spontaneous.

Now, you'd also have to understand Keq as a ratio of products over reactants at equilibrium. If there are more products, Keq > 1; if there are more reactants than products Keq < 1; if products = reactants, Keq = 1.0. Without me making the connection for you, can you see how this predicts the exact sign of deltaGo for any of those three circumstances above?

Now, THINK about what Keq really means...its just a number, a ratio, which represents the condition of a reaction--specifically, where the reaction stops moving in either direction in a net, measurable way. This is where entropy is at a max, you could say the system is at its maximum "happiness" if you like. If it had more products than it has at that exact ratio, it would NOT be as happy, so it would move back TOWARD that ratio of product to reactant. If it had less, it would NOT be ideal, so it would move toward Keq. Thus, the Keq is the place where the reaction will end up on its own given standard conditions. If this is a place where there are a lot of products relative to reactants, products must be relatively more stable than having more reactants would be. Any time this is true, the reaction will run spontaneously (without added energy input), as long as it is at the same conditions that were used to calculate the Keq. Put even more simply, you can "get products" from such a reaction without doing anything--this reaction "wants" to form products. However, if that happy place (Keq) is a ratio of more reactants than products, then we know that having more reactants must be relatively more stable than having less reactant and more product. Things NEVER spontaneously move toward something less stable. Such a reaction DOES require you doing something if you want to get products--we call this non-spontaneous. You are going to have to essentially "force" the reaction to make products by adding energy. Finally, if Keq = 1, then reactant and product concentrations at that sweet spot (Keq) are the SAME. No one is favored. You are going to have to add energy to make it go either way.

One of the very helpful conceptual things to remember about applying such topics to the MCAT is that we NEVER start reactions at Keq ratios of product/reactant. We usually start with 100% reactants.

This is a very high value MCAT-2015 topic, so make sure you get this one nailed down in your head. My little synopsis won't be enough. Good luck!

 

[SyrianHero]

Mathematically, the formula helps get the relationships clear and then we can talk more about why: deltaG0 = -RTln(Keq).

You NEED to know for the MCAT that the ln(n<1) = negative; ln(1.0) = zero; ln(n>1) = positive. You also need to know that negative deltaG = spontaneous; positive deltaG = non-spontaneous.

Now, you'd also have to understand Keq as a ratio of products over reactants at equilibrium. If there are more products, Keq > 1; if there are more reactants than products Keq < 1; if products = reactants, Keq = 1.0. Without me making the connection for you, can you see how this predicts the exact sign of deltaGo for any of those three circumstances above?

Now, THINK about what Keq really means...its just a number, a ratio, which represents the condition of a reaction--specifically, where the reaction stops moving in either direction in a net, measurable way. This is where entropy is at a max, you could say the system is at its maximum "happiness" if you like. If it had more products than it has at that exact ratio, it would NOT be as happy, so it would move back TOWARD that ratio of product to reactant. If it had less, it would NOT be ideal, so it would move toward Keq. Thus, the Keq is the place where the reaction will end up on its own given standard conditions. If this is a place where there are a lot of products relative to reactants, products must be relatively more stable than having more reactants would be. Any time this is true, the reaction will run spontaneously (without added energy input), as long as it is at the same conditions that were used to calculate the Keq. Put even more simply, you can "get products" from such a reaction without doing anything--this reaction "wants" to form products. However, if that happy place (Keq) is a ratio of more reactants than products, then we know that having more reactants must be relatively more stable than having less reactant and more product. Things NEVER spontaneously move toward something less stable. Such a reaction DOES require you doing something if you want to get products--we call this non-spontaneous. You are going to have to essentially "force" the reaction to make products by adding energy. Finally, if Keq = 1, then reactant and product concentrations at that sweet spot (Keq) are the SAME. No one is favored. You are going to have to add energy to make it go either way.

One of the very helpful conceptual things to remember about applying such topics to the MCAT is that we NEVER start reactions at Keq ratios of product/reactant. We usually start with 100% reactants.

This is a very high value MCAT-2015 topic, so make sure you get this one nailed down in your head. My little synopsis won't be enough. Good luck!

This was very helpful. Thanks a lot for taking some of your valuable time to help me with this!