# Why is change in Gibbs free energy not always zero? ($q = T \Delta S$) [duplicate]

We know that ,

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

We know that $$\Delta H = q = T \Delta S$$

so,

$$\Delta G = 0$$

Where is my mistake here?

• the formulas are for the change in Gibbs free energy yet the title talks about the value of the Gibbs free energy - which do you mean? – user245141 May 11 at 11:04
• yeah made the question more accurate – DDD4C4U May 11 at 11:20
• Essentially answered here: physics.stackexchange.com/questions/218068/… – ratsalad May 12 at 14:21
• It is close but the thing is that in that answer he explains why it doesn't work in that case. I am trying to find the reason why it would work in the general case compared to that case. – DDD4C4U May 12 at 19:33

## 1 Answer

The change in Gibbs free energy is zero for a reversible process (as long as no non-PV work is being done). The equivalence that you provide, $$\Delta H = q = T \Delta S$$, is only valid for reversible processes at constant $$T$$ and $$P$$. If the process is irreversible then $$T \Delta S \neq q$$, and thus $$\Delta G \neq 0$$.