Gibbs free energy can be defined as: $$dG=VdP-SdT+\sum_i\mu_idn_i$$ where $\mu=(\frac{\partial G}{\partial n})_{P,T}$. Last term, $\sum_i\mu_idn_i$ allows for open systems to be considered (where $dn$ does not equal zero). However, this can also be used for state functions like enthalpy too: $$dH=TdS+VdP+\sum_i\mu_idn_i$$ But my question is why can this be used for enthalpy, considering that $\mu=(\frac{\partial G}{\partial n})_{P,T}$ (i.e it's a partial derivative of $G$ not $H$). Perhaps it's clearer for me to write the total derivative out like this: $$dH=(\frac{\partial H}{\partial S})_{P,n}dS+(\frac{\partial H}{\partial P})_{S,n}dP+(\frac{\partial G}{\partial n})_{P,T}dn$$ where I have considered a single-component open system (the component may enter or leave the system)

  • $\begingroup$ I am sorry .. I am afraid my errors in our comment discussion on your other post led you to ask this question (which reflects an incorrect assumption that I fear you got from me). It is in fact perfectly ok to define chemical potential as a partial derivative of other thermodynamic potentials with respect to composition. Those definitions don't have much practical use (since they require constant entropy), so I had forgotten about them, but they are still completely valid. $\endgroup$ Commented Jan 9, 2015 at 23:15
  • $\begingroup$ Thanks, Why do they require constant entropy? $\endgroup$
    – RobChem
    Commented Jan 9, 2015 at 23:30

1 Answer 1


Actually, there is a completely valid definition of chemical potential in terms of enthalpy (I realize your confusion on this point may be my fault, since I initially told you the opposite on another post.)

The correct form for chemical potential as a partial molar enthalpy is (as I think you already suspected):

$$ \mu_i=(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}} $$

So the total exact differential is expressible as you had originally expected:

$$ dH=TdS+VdP+\sum_i\mu_idn_i=(\frac {\partial H}{\partial S})_{p,n_i}dS+(\frac {\partial H}{\partial p})_{S,n_i}dS+\sum_i(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}}dn_i $$ The definition of chemical potential as a partial molar Gibbs free energy is only valid at constant temperature and pressure, so it wouldn't make any sense to write it that way in the expression for the exact differential of enthalpy.

Sorry for any confusion my initial errors may have caused you. Hope this clears it up.

  • $\begingroup$ Thank you very much. I think I understand - the chemical potential $\mu$ is a quantity that defines how a state variable changes with composition of the system but the state variable that it describes is dependent on the variables that you hold constant. Do you agree? $\endgroup$
    – RobChem
    Commented Jan 9, 2015 at 23:37
  • $\begingroup$ Please see above - comment edited. $\endgroup$
    – RobChem
    Commented Jan 9, 2015 at 23:45
  • 1
    $\begingroup$ Well, I'd phrase it a little differently, but I think you have it essentially right. Chemical potential can be defined as the partial molar form of any thermodynamic potential, and the variables held constant determine the relevant thermodynamic potential. So if you hold entropy and pressure constant, chemical potential is partial molar enthalpy, and so on. The Gibbs free energy form (constant T,p) is the most commonly used since those are the most relevant conditions for a wide range of chemical experiments. $\endgroup$ Commented Jan 10, 2015 at 0:09
  • $\begingroup$ Great, that is what I meant. It makes perfect sense now. Thanks very much. $\endgroup$
    – RobChem
    Commented Jan 10, 2015 at 0:10

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