# How to get the enthalpy of a reaction from the enthalpy of formations with Hess' law?

Let's consider a generic reaction: $$\sum_{i=0}^{i=I}n_{i}A_{i} \rightarrow \sum_{j=0}^{j=J}m_{j}B_{j}$$ where $$n$$ and $$m$$ are the stoichiometric coefficients, while $$A$$ and $$B$$ are reagents and products. In order to compute the enthalpy of this reaction we use the enthalpy of formation of each product and each reagent. We know that for the reagents: $$\sum_{s=0}^{s=S}w_{s}^{i}C_{s}^{i} \rightarrow A_{i} \qquad \Delta H^{A_i}_{f}$$ By multiplying for $$n_{i}$$: $$n_{i} \sum_{s=0}^{s=S}w_{s}^{i}C_{s}^{i} \rightarrow n_{i} A_{i} \qquad \Delta H^{A_i}_{f}n_{i}$$ Summing over $$i$$ you get: $$\sum_{i=0}^{i=I}n_{i} \sum_{s=0}^{s=S}w_{s}^{i}C_{s}^{i} \rightarrow \sum_{i=0}^{i=I}n_{i} A_{i} \qquad \sum_{i=0}^{i=I}\Delta H^{A_i}_{f}n_{i}$$ Note that all these passages are guaranteed by Hess law. Let's pose for simplicity: $$\sum_{i=0}^{i=I}n_{i} \sum_{s=0}^{s=S}w_{s}^{i}C_{s}^{i}=K_{1}$$ We can do exactly the same process for products obtaining: $$\sum_{j=0}^{j=J}m_{i} \sum_{g=0}^{g=G}z_{g}^{j}D_{g}^{j} \rightarrow \sum_{j=0}^{j=J}m_{j} B_{j} \qquad \sum_{j=0}^{j=J}\Delta H^{B_J}_{f}m_{j}$$ Let's pose for simplicity: $$\sum_{j=0}^{j=J}m_{i} \sum_{g=0}^{g=G}z_{g}^{j}D_{g}^{j}=K_{2}$$ With an easy passage we finally get: $$\sum_{i=0}^{i=I}n_{i}A_{i}+K_{2} \rightarrow \sum_{j=0}^{j=J}m_{j}B_{j} +K_{1} \qquad \Delta H=\sum_{j=0}^{j=J}\Delta H^{B_J}_{f}m_{j}- \sum_{i=0}^{i=I}\Delta H^{A_i}_{f}n_{i}$$ Now, in my book it is written that the last enthalpy we got, is the reaction enthalpy of my initial reaction, but how you can see, it is the reaction enthalpy of the last equation, that it is not the same (there are $$K_{1}$$ and $$K_2$$ which do not cancel each other). Someone could explain this?

• I am pretty sure that $K_1$ and $K_2$ have to be the same, otherwise your reaction will not balance. Unfortunately I don't have the time to prove this. – Martin - マーチン Nov 16 '19 at 15:37