Consider again the dissociation reaction

Question

(e) What is the value of ∆G for the dissociation reaction when the system is at equilibrium?

Accepted Answer

Hello everyone today. We have the following problem. The dissociation reaction here has the following system representations of possible initial states 12 and three, and the equilibrium state determine the delta G. Or the gibbs. Free energy for the dissociation reaction when the system is in a state of equilibrium. So in a state of equilibrium the concentrations, our concentrations of our substances are always going to be constant. So the concentrations are constant in this representation above, we have four molecules of our X, sub two and four molecules of our X in our equilibrium state. So whatever the type of reaction may be, may it be a dissociation, as is the case for here or an association reaction or any other type. The change in the free energy is always going to be equal to zero because our concentration of substances are constant. And so with that we've answered the question overall, I hope this helped. And until next time.

Consider again the dissociation reaction (e) What is the value of ∆G for the dissociation reaction when the system is at equilibrium?

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Consider again the dissociation reaction
(e) What is the value of ∆G for the dissociation reaction when the system is at equilibrium?