dGº=-TRlnK

[MedGrl@2022]

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Does anyone know where I can find a good explanation for the equation dGº=-TRlnK? It seems counter-intuitive that as K increases that the standard state would become more spontaneous? Or does K always =1 since we are at equilibrium and therefore the only variable is T and as T increases then so does the spontaneity?

If anyone could help me understand and/or show me were I can go to further dissolve this equation in my brain that would be GREAT!

Thank you for all your help fellow SNDers!!!! We will succeed on the MCAT and through med school! 🙂

 

[Bru]

K is the rate constant so it's not really going to change except for temperature.

Think of it this way: if K is large then the reaction strongly goes from reactants to products. This corresponds well mentally to a large deltaG which indicates a reaction that is (more) spontaneous.

If anyone has a better explanation I'd love to hear that.

 

[WSHRocks]

K is not the rate constant (little k) here but the equilibrium constant (big K). For a given reaction, Keq will be constant and is the ratio of products to reactants at equilibrium, where Q is any deviation from equilibrium. The larger K is, the more products are favored in the reaction over reactants. This increases the absolute value of delta G, where it is always negative so long as K>1, meaning spontaneous (whether or not it happen in a short or long period of time).

 

[advair250]

Gen chem 2.

 

[Endure]

Does anyone know where I can find a good explanation for the equation dGº=-TRlnK? It seems counter-intuitive that as K increases that the standard state would become more spontaneous? Or does K always =1 since we are at equilibrium and therefore the only variable is T and as T increases then so does the spontaneity?

If anyone could help me understand and/or show me were I can go to further dissolve this equation in my brain that would be GREAT!

Thank you for all your help fellow SNDers!!!! We will succeed on the MCAT and through med school! 🙂

If the equilibrium obtained by a reaction is energetically favorable, this will be reflected in the ΔG value as the reaction goes from standard state conditions (i.e., when Q = 1) to equilibrium.

Whether or not increasing T "increases spontaneity" is difficulty to judge without more information. In other words, an exothermic reaction could be offset by a decrease in entropy when the temperature of the system increases.

 

[MedGrl@2022]

If the equilibrium obtained by a reaction is energetically favorable, this will be reflected in the ΔG value as the reaction goes from standard state conditions (i.e., when Q = 1) to equilibrium.

Whether or not increasing T "increases spontaneity" is difficulty to judge without more information. In other words, an exothermic reaction could be offset by a decrease in entropy when the temperature of the system increases.

I thought that is K>1 then there are more products and the reverse equation has a higher probability (more spontaneity).

 

[Endure]

I thought that is K>1 then there are more products and the reverse equation has a higher probability (more spontaneity).

I think you're mixing up the reaction quotient, Q, with the equilibrium constant, K.

&#916;G° is a property that is intrinsic to the reaction being investigated, much like the value of K (assuming external conditions remain the same). So if I have a reaction with K > 1, then log(K) > 0 and &#916;G° < 0.

Imagine the reaction A + B &#8594; C, where K at 25°C is 10. If I get a beaker and add reactants and products such that A, B and C are present at 1.0 M concentration, then my reaction mixture is at standard state and Q = 1 (since [A] = = [C]). If I allow the reaction to proceed to equilibrium, more products will be formed to achieve K. This process happens because it is energetically favorable and is reflected by a &#916;G value < 0. In other words, progression from standard state to equilibrium is spontaneous.

Hope that helps 🙂