0
$\begingroup$

What I know: under constant pressure and assuming all work done is expansion work, enthalpy is equivalent to heat using the formula for its definition.

However, non-expansion work done during electrochemical processes alter the enthalpy value according to the equation ΔH-TΔS=-nFE, where the voltage provides the non-expansion work if I'm correct.

Under this condition, is the change in enthalpy still equivalent to the amount of heat transferred? If not, what is the physical significance of ΔH in this scenario?

$\endgroup$
2
  • 1
    $\begingroup$ Replace isobaric condition by isochoric and enthalpy by internal energy. Change of internal energy is equal to accepted heat only with zero non-volume work. Similarly for enthalpy and isobaric process. $\endgroup$
    – Poutnik
    Commented Feb 16, 2023 at 14:35
  • $\begingroup$ Yes, you are right. When there is non-pV work the equivalence fails to apply. $\endgroup$
    – Buck Thorn
    Commented Feb 17, 2023 at 6:37

1 Answer 1

0
$\begingroup$

Heat is the transfer of thermal energy. Enthalpy is a property of a sample or system. So they are never equivalent.

The change in enthalpy is equal to the heat transfer if you have a closed system at constant pressure, and the only work done or done to the system is pressure-volume work. This follows, with a couple of steps of derivation, from the conservation of energy.

$\endgroup$
1
  • $\begingroup$ Heat content is the sum of the internal KE of the molecules+ any latent heats. It is not the process of heat transfer. The internal energy U +PV is a state function tht has been given the name enthalpy H. Its virtue is that if the only work done is PV work Q, the change in heat content, now is a state function. This allows calorimetric measurements at constant pressure to be used in calculating other state functions. $\endgroup$
    – jimchmst
    Commented Feb 16, 2023 at 20:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.