July 10, 2020, 04:49:13 AM
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Topic: Question regarding ΔG° and ΔG  (Read 141 times)

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Offline asheys666

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Question regarding ΔG° and ΔG
« on: March 31, 2020, 03:07:28 PM »
I know that standard conditions for ΔG° requires all reactants and products to be present in the beginning at 1M or 1atm. However, this only applies to solutions and gases.

What if we had a reaction that was something like YZ₂ (g) -> Y (s) + Z₂(g). We start with only 1 atm of YZ₂ (g) in the beginning of the reaction and we are told ΔG°> 0.

I know that initially, the reaction will proceed forward and that because ΔG° is greater than 0, it will be reactant favored which means that the Z₂(g) product will be less than 1atm.

However, what about about Y(s)? Would I say that Y(s) produced will be less than 1mol or should I just not even regard the solid product?

Thank you and sorry if my wording is a little confusing!

PS: I think the main reason I was confused with this question was that it was a scenario my teacher never introduced.

She only introduced two scenarios: when liquid turns into gas and when a solid turns into a solution. Let’s say the same scenario is applied to these reactions, that we start with only reactants and our Delta Gnot is greater than 0.

For the liquid to gas reaction, the vapor pressure would be less than 1atm and for the solid to aqueous reaction, the solution will be less than 1M.

However for the reaction in the question it’s going from a gas to a solid and a gas, so I don’t know how I should describe it? The partial product of the gaseous product is less than 1atm and the solid product will be less than 1mol? Or is there another way to describe this?

Offline Enthalpy

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Re: Question regarding ΔG° and ΔG
« Reply #1 on: April 01, 2020, 07:13:24 AM »
The deep material of a solid, even as a fine powder precipitate, does not participate in the reaction (to make things simpler). So the amount of solid doesn't change an equilibrium. That's why it's not accounted like a gas nor solute in an equilibrium equation.

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