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

The following two statements appear to contradict each other:

1. For a spontaneous reaction at constant temperature $$T$$ and pressure $$p$$ the change in Gibbs free energy can be negative $$(\mathrm dG < 0).$$
2. Since $$\mathrm dG = V\,\mathrm dp - S\,\mathrm dT,\tag{1}$$ at constant pressure $$(\mathrm dp = 0)$$ and constant temperature $$(\mathrm dT = 0)$$ the change in Gibbs free energy is always zero: $$\mathrm dG = V\cdot 0 - T\cdot 0 = 0.\tag{2}$$

Why does this happen?

• Statement 2 is applicable when there is no reaction, otherwise it would be more complicated. Dec 25, 2015 at 21:30

The relation

$$\mathrm dG = V\,\mathrm dp - S\,\mathrm dT\tag{1}$$

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 $$\frac{\partial G}{\partial X} = 0\tag{2}$$

gives the well-known law of chemical equilibrium.

This is a 6 yr old query. But I will try to answer this question. Equation of dG for electrochemistry (for galvanic cell for example) is dG = -SdT + VdP + EdQ. Where E = electric potential of anode/cathode of the galvanic cell and dQ is the amount of charge reduced from anode or added to the cathode. So for galvanic cell, dG < 0 since it is spontaneous (no required power source to force a reaction). Negative charge like electrons flow from lower potential to higher potential with no power source required. For anode, dG_A = +E_AdQ (since it reduces electrons or "adds positive charge"). For cathode, dG_C = -E_CdQ (since it adds electrons or "adds negative charge"). Adding these two equations: dG_total = dG_A + dG_C = (E_A - E_C)dQ Since E_C > E_A, dG_total is negative or dG_total < 0 (spontaneous process).

Another equation of dG is dG = -SdT + VdP + (mu)dN. Where mu = chemical potential and dN is the number of molecules added/reduced to/from the system. Analogous to the electric potential and charge. Hope this clarifies even though this is an old query.

• Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown. Also, I'm not sure how this is related to the question. Mar 26 at 11:32

The entropy change of the system in any process is given by

$$ΔS=\int\frac{\mathrm dq_\mathrm{rev}}{T},\tag{1}$$

the calculation of which assumes a reversible process taking place between the initial and final states. In this reversible process, both the temperature and pressure can vary. Hence while calculating entropy change of the system, which is subsequently used in the calculation of $$ΔG$$ using $$ΔG = ΔH - TΔS,\tag{2}$$ the assumption that temperature and pressure are constant, i.e. statement 2, becomes invalid.