# Why is dG = dH − TdS?

I get that $$G=H-TS$$ because then: \begin{align}\mathrm dG&=\mathrm dH-T\,\mathrm dS-S\mathrm dT\\&=T\,\mathrm dS+V\,\mathrm dp-T\,\mathrm dS-S\,\mathrm dT\end{align} Therefore, by cancelling: $$\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$$ which is the equations for $$\mathrm dG$$. However, I can’t get this result from using $$\mathrm dG=\mathrm dH-T\,\mathrm dS$$.

$$\mathrm dH=S\,\mathrm dT+V\,\mathrm dp$$ $$\mathrm dG=\left(S\,\mathrm dT+V\,\mathrm dp\right)-T\,\mathrm dS$$

This does not give me the right equation. Where am I going wrong?

$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S - S\,\mathrm{d}T$$
• If temperature is the same but not constant (initial and final state have same temperature) can we say $ΔG=ΔH-TΔS$? Commented Mar 10, 2021 at 17:47
If you write $\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S$, you drop off one term of the differentiation of $G$ which is $S\mathrm{d}T$. By definition $G = H - TS$. So, $$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S-S\,\mathrm{d}T$$ Then, you should write $\mathrm{d}H = T\,\mathrm{d}S + V\,\mathrm{d}p$ (not as you have written $\mathrm{d}H = S\,\mathrm{d}T + V\,\mathrm{d}p$ ).
Finally, $$\mathrm{d}G = T\,\mathrm{d}S + V\,\mathrm{d}p - T\,\mathrm{d}S - S\,\mathrm{d}T$$ $$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T$$