If ΔG° for a reaction is less than zero, then

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)  If ΔG° for a reaction is less than zero, then __________. If the reation is equal to zero, than ________

A) K = 0

B) K = 1

C) K > 1

D) K < 1

E) More information is needed.

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Explanation & Answer

Thank you for the opportunity to help you with your question!

First, ΔG° (standard Gibbs free energy) and K (the equilibrium constant) are both related to equilibrium.

We know that if ΔG°<0, we have a spontaneous reaction, which means it should move toward the products (although we do not know the RATE at which this happens. A spontaneous reaction can still take a long time, but that does not matter for this question) does not , and that if ΔG°>0, it is a non-spontaneous reaction.

K, the equilibrium constant, relates the concentration of products to the concentration of reactants. It's generally [products]/[reactants].

ΔG° and K are related to one another through the equation:

ΔG° = -RT*ln(K)

Where R is the gas constant, T is the absolute temperature (Kelvin), and K is the equilibrium constant.

So, if we know ΔG° < 0, then we have a negative number on the right hand side of the equation. Since both R and T must be positive, we just have to make sure the right hand side of the equation also ends up negative. We can use the mathematical properties of the natural log function to accomplish this.

If the argument of the natural log function, which in this case is K, is less than 1, the natural log will be negative. If the argument of the natural log function is greater than 1, the natural log of the number will be positive. You can see this quickly if you plot ln(x). It will cross the x-axis from negative values to positive values at 1.

What does this mean for us?

We know we have, as signs on our numbers:

(-) = -*(+)(+)*ln(K), which simplifies down to: (-) = -ln(K)

Since we need both sides of the equation to be negative in this case, we need the ln(K) to be positive. We know that K must be greater than 1 then, due to the properties of the natural log function.

So, for the first blank, if ΔG° < 0, then K > 1.

For the second blank, if ΔG° = 0, then that means we are at equilibrium, and so K must be 1 (for the second blank).
We can reason through this the same way as we did for the first blank mathematically.

0 = -(+)(+)*ln(K), which simplifies to 0 = ln(K). This occurs at K = 1.

Please let me know if you need any clarification or have any questions. I'm always happy to answer or clarify anything. Hopefully I was able to help!

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