# Why does enthalpy change differ in a chemical reaction because of temperature?

If enthalpy is a state function, shouldn't the change in enthalpy be the same between a given initial and final state, regardless of how the change took place? Kirchoff's equation describes this variation as $$\Delta H_{T_2} = \Delta H_{T_1} + \Delta C_p (T_2 - T_1).$$ I'm guessing that this change in enthalpy change is only because of the variation in $$C_p$$, but I'm not too sure about it. Am I completely wrong about this?

• Yes, enthalpy is a state function, and the rest is also true. Now what is your question? – Ivan Neretin Dec 20 '20 at 17:59
• You are right there. IF the total heat capacities of reagents and products were by chance equal, the reaction enthalpy would be temperature independent. But as these total heat capacities are generally different, the reaction enthalpy is temperature dependent. As the net heat for changing temperature of reactants and changing it back for products is nonzero. – Poutnik Dec 20 '20 at 18:09
• @OneEyedMushroom Yes, the change in enthalpy is independent of path, depending only on the initial and final states. But ΔH_T2 is going to be different from ΔH_T1. That's because the initial and final states for a reaction at T1 are different from a reaction at T2. Specifically, ΔH_T1 is the change in enthlapy in going from reactants at T1 to products at T1. ΔH_T2 is the change in enthalpy is going from reactants at T2 to products at T2. And your eqn is essentially correct, except it makes the common simplifying assuption that the diff. in Cp between reactants and products is T-independent. – theorist Dec 20 '20 at 19:17
• If you're looking at the change at two different temperatures, then the initial states and final states are different? – Zhe Dec 20 '20 at 19:42
• @Alchimista You'd need to look at a derivation of the eqn. to understand more fully, but Kirchoff's eqn does do that, i.e., it is obtained by integrating delta_Cp from T1 to T2 (the equations you see here are obtained by assuming delta_Cp is T-independent, so the delta_Cp can be taken out of the integral). – theorist Dec 22 '20 at 8:57

From Hess' law, the heat of reaction at temperatures $$T_1$$ and $$T_2$$ are related by $$\Delta H_{T_1}=C_{p,\mathrm r}(T_2-T_1)+\Delta H_{T_2}+C_{p,\mathrm p}(T_1-T_2),$$where $$C_{p,\mathrm r}$$ is the heat capacity of a stoichiometric molar mixture of reactants and $$C_{p,\mathrm p}$$ is the heat capacity of a stoichiometric molar mixture of products.