- #1

- 739

- 3

## Main Question or Discussion Point

There are 2 types of reactions 1) Being fully irreversible and 2) reversible

So for 1) the graph of Gibbs energy against the progress of reaction would look like this: http://postimg.org/image/phycri0vr/

So the difference between 100% product and 100% reactant is the ΔG of the reaction which makes sense.

But for 2) the graph of Gibbs energy against the progress of reaction would look like this: http://postimg.org/image/grmh7oknz/

So now the ΔG° would be for 1 mole of reactant to form 1 mole of product and also, the actual difference is just the G value of the products minus the G value of the reactants. However, during equilibrium the ΔG is now 0. However, if we just take the difference between the G value of the equilibrium point, E and the 1 mole of reactant, that value is non-zero so why is it always known for the ΔG at equilibrium to be 0?

Lastly, I also read that no reaction is fully irreversible so in fact all reactions has the graph on 2). However, in that case if all reaction has an eventual ΔG of 0 then how can any work be done? And we always say that using the formula ΔG°=ΔH°-TΔS° would give us the maximum work being done. But since all reactions would have an actual ΔG of 0, how can there any work be done?

So for 1) the graph of Gibbs energy against the progress of reaction would look like this: http://postimg.org/image/phycri0vr/

So the difference between 100% product and 100% reactant is the ΔG of the reaction which makes sense.

But for 2) the graph of Gibbs energy against the progress of reaction would look like this: http://postimg.org/image/grmh7oknz/

So now the ΔG° would be for 1 mole of reactant to form 1 mole of product and also, the actual difference is just the G value of the products minus the G value of the reactants. However, during equilibrium the ΔG is now 0. However, if we just take the difference between the G value of the equilibrium point, E and the 1 mole of reactant, that value is non-zero so why is it always known for the ΔG at equilibrium to be 0?

Lastly, I also read that no reaction is fully irreversible so in fact all reactions has the graph on 2). However, in that case if all reaction has an eventual ΔG of 0 then how can any work be done? And we always say that using the formula ΔG°=ΔH°-TΔS° would give us the maximum work being done. But since all reactions would have an actual ΔG of 0, how can there any work be done?