\(\displaystyle |\nu _1| A_1 +|\nu _2| A_2+\cdots \rightleftharpoons |\nu _3| A_3 +|\nu _4| A_4 +\cdots \)

At standard conditions \(\displaystyle (P_0,T_0)=(1 \) bar \(\displaystyle , 298.15 \) K \(\displaystyle ) \) we calculate the enthalpy of reaction via tabulated values of enthalpy of formation:

\(\displaystyle \Delta H_r(P_0,T_0)=\sum\limits _i \nu _i H_{f,i}(P_0,T_0) \)

Until now that's ok! If I want to evaluate the change in enthalpy of reaction at another condition, say \(\displaystyle (P,T) \)

**most textbooks come with the following expression**:

\(\displaystyle \Delta H_r(P,T)=\Delta H_r(P_0,T_0)+\int\limits _{T_0}^{T} \Delta C_P d\tau \)

where

\(\displaystyle \Delta C_P\equiv \sum\limits _i\nu _i C_{P,i} \)

For me that's all right. I understand the path (changing the reactants from \(\displaystyle (P,T) \) to \(\displaystyle (P_0,T_0) \), proceeding with the reaction and than taking the products from \(\displaystyle (P_0,T_0) \) to \(\displaystyle (P,T) \)).

What I can't understand is:

**why the pressure change is not considered?**

I mean, the expression is correct only for ideal gases, which \(\displaystyle H=H(T) \). Shouldn't I put the residual terms in the expression? Like these:

\(\displaystyle \Delta H_r(P,T)=\Delta H_r(P_0,T_0) + \int\limits _{T_0}^{T} \Delta C_P d\tau + H^R_{reac.}(\overline{y}_{reac.},P_0,T_0) - H^R_{reac}(\overline{y}_{reac.},P,T)+H^R_{prod.}( \overline{y}_{prod.},P,T)-H^R_{prod.}(\overline{y}_{prod.},P_0,T_0) \)

and then use some EOS to compute the residuals...

I would be grateful if someone help me... Thanks!