# Gibbs energy and Entropy

In Gibbs energy equation:

$$\Delta G=\Delta H-T\,\Delta S$$

At constant pressure,

$$Q=\Delta H\tag{1}$$

and we know that $T \Delta S=Q$ (reversible) so finally

$$T\,\Delta S=Q\tag 2$$

Putting values of Equation $\text{(1)}$ and $\text{(2)}$ in Gibbs energy equation , we should always get $\Delta G=0$ for a reversible reaction. ..

Why isn't it so?

I don't know how to search it on Google; if anyone could give me a link or otherwise explain it?

Reason 1:

while $T\,\mathrm dS=\mathrm dQ$ for a reversible process, this does not mean $T\Delta S = Q$

Reason 2:

The term "reversible reaction" is unrelated to "reversible process".

Lame Reason 3:

Because your "always" statement does not repeat the "constant pressure" requirement assumed for $Q=\Delta H$

• @Loong This answer does not address the misconception that heat is transferred between systems that have to be specified. Apr 13 '19 at 12:46

The problem is that you are equating "reversible reaction" with "reversible process."

What you are seeing when you plug equations 2 and 1 into the Gibbs Free Energy equation is that at equilibrium,

$\Delta G = 0$

The reason that this works is because a reversible process is one that progresses by very small movements away from equilibrium states - when we say a process is "reversible" what we really mean is that we are making an approximation that allows us to ignore entropy generation. When you make this approximation and apply it to the Gibbs Free Energy equation, you are forcing it to stay at equilibrium - hence, $\Delta G = 0$.

I think your confusion is coming from the similar names of "reversible reaction" and "reversible process." A reversible process is one in which the entropy generation term is zero, but a reversible reaction is one in which the forward and reverse rates of reaction result in a mixture of reactants and products at equilibrium.

In other words, a reversible reaction can be in an equilibrium ($\Delta G = 0$) or non-equilibrium state ($\Delta G \neq 0$), but a reversible process is always in an equlibrium state.

• Even at equilibrium, $\Delta_r H$ and $\Delta_r S$ are typically not zero, so the OP's statements are incorrect even at equilibrium. Apr 13 '19 at 12:50

You are missing the subscripts. Heat transfer only makes sense if you define a system and its surrounding. Your equations become:

$$\Delta G_\text{sys}=\Delta H_\text{sys}-T\,\Delta S_\text{sys}$$

At constant pressure,

$$q = \Delta H_\text{sys}\tag{1}$$

and we know that $$T \Delta S=q$$ (reversible) so finally

$$T\,\Delta S_\text{surr}=-q\tag 2$$

Putting values of Equation $$\text{(1)}$$ and $$\text{(2)}$$ in Gibbs energy equation , we get

$$\Delta G_\text{sys} = -T\,\Delta S_\text{surr}-T\,\Delta S_\text{sys} = -T\,\Delta S_\text{univ}$$

So the reaction will proceed if $$\Delta G_\text{sys}$$ is negative or $$\Delta S_\text{univ}$$ is positive. This is how the textbooks tell it.