Heat of Reaction (Enthalpy)
Well, suppose we have a reaction:
$\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!
