The following pictures represent equilibrium mixtures for the interconversion of A molecules (red) and X, Y, or Z molecules (blue):
What is the sign of ∆G° for each of the three reactions?

Question

The following pictures represent equilibrium mixtures for the interconversion of A molecules (red) and X, Y, or Z molecules (blue):

Three cases showing equilibrium mixtures of A (green) with X, Y, Z (yellow) for Gibbs Free Energy analysis.

What is the sign of ∆G° for each of the three reactions?

Accepted Answer

Hello everyone today. We have the following problem shown below our system representations of three equilibrium mixtures for the inter conversions of the molecules to X. B molecules to Y. And the molecules to see the molecules of B are shown as yellow spheres while molecules of X, Y or Z are green determine a sign of our gibbs. Free energy for each case. So first we need to let the number of molecules, I need to let the number of molecules be proportional to our concentration. So for our case number one we have to be molecules. So for case number one we have to be molecules For every four x molecules. For case # two we have three B molecules for every three Y molecules. And lastly for case number three we have For the molecules for every two Z molecules. And so we need to let our gives free energy reaction relate to the gibbs free energy. In the following way we have our negative of our gas constant, r times r temperature times our natural log of K. Or constant. And Kay is going to be found by having the concentration of our products divided by the concentration of our reactant. We need to do this for each case. So for case number one, that is simply going to be our concentration of X over our concentration of B That's going to be four divided by two or simply to. So without plugging any math variables in we know that our K value is greater than one and if our K value is greater than one then our natural log of K is going to be positive and therefore our Gibbs free energy is going to be negative For case # two. So we'll just to note that this is case number one and this is case number two. So for case number two we're going to do the same process or concentration of why over our concentration of B For why we have three and 4 R. B. We have three, meaning we have one. So in this situation RK is going to be equal to one and so our natural log of K is going to be equal to zero. And because of this, our Gibbs free energy is also going to be equal to zero. And lastly we will do case number three down here, that's going to be our concentration of Z divided by the concentration of B. In other words, that's going to be 2/4, which simplifies to one half. In our third case R K is equal to one half, meaning that our natural log of K is going to be negative because it is less than one. And so if it is negative, then our gifts for energy is as you can probably guess the opposite which is going to be positive. And so we have our answers for each of our cases here negative for case, 10 For case two and positive for case three. And with that we have our answers overall, I hope this helped and until next time.

The following pictures represent equilibrium mixtures for the interconversion of A molecules (red) and X, Y, or Z molecules (blue):
What is the sign of ∆G° for each of the three reactions?

Verified Solution

Key Concepts

Gibbs Free Energy (∆G)

Equilibrium Constant (K)

Reaction Quotient (Q)

The following pictures represent equilibrium mixtures for the interconversion of A molecules (red) and X, Y, or Z molecules (blue): What is the sign of ∆G° for each of the three reactions?

Verified Solution

Key Concepts

Gibbs Free Energy (∆G)

Equilibrium Constant (K)

Reaction Quotient (Q)

The following pictures represent equilibrium mixtures for the interconversion of A molecules (red) and X, Y, or Z molecules (blue):
What is the sign of ∆G° for each of the three reactions?