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We know the basic Thermodynamic equations are: $$ dU=-PdV+ TdS + \mu dN$$ $$ dH=dU+d(PV)= TdS+ \mu dN + VdP$$ we typically assume constant Pressure for a reaction carried out in air and neglect the dP term $$dG=dH-d(TS)= \mu dN - SdT+ VdP$$ once again we neglect the dT and dP terms but for heat of reaction we use the enthalpy not just the TdS term. When we have a battery though the electrical energy is the Gibbs free energy with the heat of reaction being TdS and other energy losses also contributing to the heat. When the reaction is not separated (not an electrochemical cell) then the Gibbs is part of the heat.

For a typical reaction where the chemicals are in contact why do they use the enthalpy to determine the heat of the reaction produced?

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  • $\begingroup$ fixed the equation. Interesting how a non spontaneous reaction can become spontaneous with temperature $\endgroup$
    – ChemEng
    Commented Oct 19, 2020 at 0:10
  • $\begingroup$ I'm not entirely clear on what your question is, but the reason for using $\Delta H$ in a lot of equations rather than $q$ is because enthalpy is a state function, whereas heat is not. $\endgroup$
    – Andrew
    Commented Oct 19, 2020 at 14:10

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When a reaction occurs at constant pressure and no work other than expansion work (such as electrochemical) is performed, then it can be shown, from the definition of the enthalpy and the first law of thermodynamics applied to a closed system, that the change in enthalpy is equal to the heat:

$$\begin{align}dH&=dU+d(pV)\\&=dw_{pV}+dw_{other}+dq+pdV+Vdp\\&=-p_{ext}dV+dw_{other}+dq+pdV+Vdp\\&=dq+dw_{other} \end{align}$$

The last equality follows when the pressure is constant and equal to the external pressure. When only pV work is done then clearly heat and enthalpy change are equal. Readily measured heats of reaction provide a very convenient way to determine values of the enthalpy state function which can be tabulated and employed in further thermodynamic computations.

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    $\begingroup$ Yes this is a great explanation. Once again at constant pressure and temperature whether a reaction occurs in the forward or reverse is still dependent on the temperature and pressure which means the chemical potential is a function of both $$ \mu= RTlna $$ where the activity (effective concentration) is also a function of both $\endgroup$
    – ChemEng
    Commented Oct 19, 2020 at 13:15
  • $\begingroup$ @ChemEng It's worh recalling that $(\partial \mu/\partial T)_p=\bar S$ $\endgroup$
    – Buck Thorn ♦
    Commented Oct 19, 2020 at 13:25
  • $\begingroup$ from the multivariable chain rule we know $$ \mu=\frac{\partial{G}}{\partial{N}}$$ and $$S=\frac{\partial{G}}{\partial{T}} $$ so the second partials are equal $\endgroup$
    – ChemEng
    Commented Oct 19, 2020 at 18:08
  • $\begingroup$ Sorry, I miss-typed. Should have written $(\partial \mu_i/\partial T)_p=-\bar S_i$. This is analogous to $(\partial G/\partial T)_p= -S$ except true for the partial molar quantities (indicated by the bar). You can integrate this relation to obtain the potential at some other T. $\endgroup$
    – Buck Thorn ♦
    Commented Oct 19, 2020 at 18:16
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    $\begingroup$ yes this would be an interesting way to measure entropy as gibbs is the free energy or -nFE for electrochemistry and enthalpy-reversibleheat = heat of reaction so we can measure those $\endgroup$
    – ChemEng
    Commented Nov 3, 2020 at 21:03

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