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Is change in enthalpy defined only at constant pressure?

I know that $Q-W=\Delta U $, and $Q$ at constant pressure equals $\Delta H$ (enthalpy change).

Can $Q$ (heat given to the system) be used interchangeably with $\Delta H$ in the first equation?

Is $ \Delta H =nC_p\Delta T$? How?

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2 Answers 2

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Is change in enthalpy defined only at constant pressure?

No. The definition of $\Delta H$ is $\Delta H = \Delta U + \Delta (PV)$ Simple as that.

I know Q-W=del(U) And Q at constant pressure equals del(H) (enthalpy change).

Can Q (heat given to the system) be used interchangeably with del(H) in the first equation?

No. Only if the applied pressure is held constant during the change.

Is del(H) =nCp del(T)? How?

No. Only for an ideal gas. For real gases, this equation is not correct outside the limit of ideal gas behavior. For an ideal gas, $\Delta U=nCv\Delta T$ and $\Delta (PV)=nR\Delta T$, so $\Delta H=n(C_V+R)\Delta T$. And then $C_p=(\partial H/\partial T)_P=C_v+R$, so $\Delta H=nC_P\Delta T$.

For a real gas, liquid, or solid, $$dH=nC_PdT+\left[V-T\left(\frac{\partial V}{dT}\right)_P\right]dP$$ Note that the term in brackets is zero for an ideal gas.

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  • $\begingroup$ 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 $dH \neq dq$. $\endgroup$
    – ado sar
    Commented May 8, 2020 at 19:06
  • $\begingroup$ 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. $\endgroup$ Commented May 8, 2020 at 20:44
  • $\begingroup$ 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. $\endgroup$ Commented May 11, 2020 at 15:50
  • $\begingroup$ how did you derive the terms in square brackets? $\endgroup$
    – Babu
    Commented Sep 10, 2020 at 10:11
  • $\begingroup$ @Buraian This derivation is in every thermo book. $\endgroup$ Commented Sep 10, 2020 at 12:32
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Is change in enthalpy defined only at constant pressure?

No, enthalpy-change is not meant only for isobar processes.

Can Q (heat given to the system) be used interchangeably with del(H) in the first equation?

No, not always. Only when the non-compression work done by the system is zero, then only $\delta Q$ be interchanged with $\mathrm dH\;.$ See my answer to this ChemSE post.

Is del(H) =nCp del(T)?

Yes, indeed.

How?

Check my answer to this ChemSE post.

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  • $\begingroup$ As you said $dH=\delta H$ only when non-compression work done is zero. Does this mean that the formula $$dG_{T,P}=-TdS_{total}$$ is only valid when the non-compression work done is zero? $\endgroup$
    – ado sar
    Commented Mar 29, 2020 at 16:03

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