# Why at constant pressure and temperature Gibbs energy change of a process can be negative?

1. $dG < 0$, for a spontaneous reaction at constant Temperature $T$ and Pressure $P$, this means that at contant $T$ and $P$, $dG$ can be negative
2. $dG = VdP -SdT$, this means that at constant $P$ ($dP =0$) and constant $T$ ($dT = 0$), $dG = V\cdot 0-T\cdot 0 = 0$. At constant $P$ and $T$ according to this equation $dG$ is always zero.

Statement $1$ contradicts statement $2$. Why does this happen?

• Statement 2 is applicable when there is no reaction, otherwise it would be more complicated. – Ivan Neretin Dec 25 '15 at 21:30

The relation $dG = VdP - SdT$ implies that the Gibbs free energy of the system depends only on the two variables $T$ and $P$, e.g., as in the case of a single-component gas consisting of a fixed number of molecules. For such systems, the macrostate is completely determined if $T$ and $P$ are given; nothing can change further.
On the other hand, suppose that there is an additional parameter $X$ describing the system and that $X$ is not fixed. Then, according to the second law of thermodynamics, the system, being at constant $T$ and $P$, self-adjusts $X$ in such a way that $G$ is minimized.
In case there exist multiple species that can undergo a chemical reaction, $X$ could be a measure of how far the reaction has proceeded. The condition $\partial G/\partial X = 0$ gives the well-known law of chemical equilibrium.
The entropy change of the system in any process is given by $ΔS=\int \:\frac{dq_{rev}}{T}$, the calculation of which assumes a reversible process taking place between the initial and final states. In this revesible process, both the temperature and pressure can vary. Hence while calculating entropy change of the system, which is subsequently used in the calulation of $ΔG$ using $ΔG = ΔH-TΔS$, the assumption that temperature and pressure are constant, i.e. statement 2, becomes invalid.