Nov 6, 2019
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  1. Pre-Medical
Here is a problem from Kaplan:

At 25 degrees C, the delta G naught for a certain reaction A < - -> B + 2C is 0. If the concentration of A, B, and C in the cell at 25 degrees C are all 10 mM, how does the delta G compare to the measurement taken at 1 M concentrations?
A. Delta G is greater than delta G naught, thus the reaction is spontaneous
B. Delta G is less than delta G naught, thus the reaction is spontaneous
C. Delta G is greater than delta G naught, thus the reaction is nonspontaneous
D. Delta G is less than delta G naught, thus the reaction is nonspontaneous

The answer is B and you can plug in the numbers into the equation, delta G = delta G naught + RT lnQ to get a negative delta G, which means it's spontaneous.

My question is two parts:
Part 1) from a conceptual perspective, why is delta G negative? If I change the concentration of both the reactants and products by the same amount shouldn't the reaction favor the products, since there is more products than reactants (3 products, which are B and 2C versus 1 reactant, A). I understand that this would be true for Keq<Q, which does not apply here, since we aren't given Keq (i.e. you don't know where equilibrium is). It might be helpful to see how someone goes through this problem conceptually without any equation and plugging in numbers.
Part 2) How do we account for the change in delta G if we were to use 20 M (a non-decimal number) instead of 10 mM?
 
D

deleted936470

When considering the "value" of delta G, you can ask yourself, which variable overpowers the system? Delta G naught or RT ln Q? Since both consider concentration (entropy as the sum of products minus reactants for delta G naught, and that Q = concentration quotient), you want to consider which variable is more greatly affected? Hint: [C]^2 is important here when comparing large differences in concentration magnitude (mM vs M); due to the fact that you're taking a natural log, I recommend you plug and chug for the value of ln Q before trying to conceptualize.

1) Why is delta G negative? It is equal to RT ln Q when delta free naught is 0. By changing the reactants and products by the "same amount", you are greatly changing the system being that product C has a superscript^2 when calculating ln Q.

delta G naught in initial conditions is 0 under STP with concentrations of each species = 10 mM, then after plug and chug delta G ~ - 2.3e^4 and spontaneous.

2) At some point, concentration overpowers thermodynamics, i.e., increasing the concentration of everything will ultimately lead to a large [C]^2 and delta G becomes positive.

Under STP with concentrations of each species = 20M, then delta G ~ 1.5e4 and non-spontaneous.
 
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