Timeline for Enthalpy change defined at constant pressure only?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2020 at 12:32 | comment | added | Chet Miller | @Buraian This derivation is in every thermo book. | |
| Sep 10, 2020 at 10:11 | comment | added | Brian | how did you derive the terms in square brackets? | |
| May 11, 2020 at 15:50 | comment | added | Chet Miller | I thought about this some more, and I have an additional answer for you. The equation you wrote describes the change in enthalpy between two closely neighboring (differentially separated) thermodynamic equilibrium states. For a chemical reaction at constant temperature and pressure, the summation represents the change in Gibbs free energy, which is zero at equilibrium for all differential variations in the number of moles of the various species. So, under these circumstances, you are left with dH=TdS. | |
| May 8, 2020 at 20:44 | comment | added | Chet Miller | For a closed system at constant pressure with no mass entering or leaving, irrespective of whether there is chemical reaction occurring, the 1st law tells us the $Q=\Delta H$. Good luck with integrating the equation that you have written, given that s is a function of the chemical composition also. | |
| May 8, 2020 at 19:06 | comment | added | Antonios Sarikas | Why we say that for chemical reactions at constant pressure $dH=dq$ at constant pressure? It is true that $$dH=Tds+Vdp+\sum_i μ_i dn_i$$ Only the middle term cancels out so $dH \neq dq$. | |
| Apr 15, 2016 at 18:08 | vote | accept | Utkarsh Barsaiyan | ||
| Apr 15, 2016 at 16:47 | history | answered | Chet Miller | CC BY-SA 3.0 |