# Heat of Reaction (Enthalpy)

#### tom89

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!

#### kelvin

when [p,t] is not constant ?