# Is Kirchoffs law valid at all pressures?

$$\Large H_{T_\mathrm{f}, p}=H_{T_\mathrm{i},p}+\int_{T_\mathrm{f}}^{T_\mathrm{i}}c_p(T)dT$$ This can be derived by integrating: $$C_v =\left(\frac{\mathrm{d}U}{\mathrm{d}T}\right)_V$$ Applying this to both reactants and products in a reaction, and subtracting reactans from products, we can arrive to

$$\Delta H_{T_\mathrm{f},p}=\Delta H_{T_\mathrm{i},p}+\int_{T_{\mathrm{f}}}^{T_{\mathrm{i}}}\Delta c_p(T)\mathrm{d}T \tag{1}\label{a}$$ Where $$c_p(T)=\sum_i\nu_\mathrm{i}c_{p,\mathrm{i}}(T)$$ This is essentially Kirchoffs law how enthalpy varies with temperature. However, whenever I see the the equation $\ref{a}$, it always deals with standard enthalpy changes, i.e: $$\Delta H_{T_\mathrm{f}}^{⊖}=\Delta H_{T_\mathrm{i}}^{⊖}+\int_{T_{\mathrm{f}}}^{T_{\mathrm{i}}}\Delta c_{p^{⊖}}(T)\mathrm{d}T$$ Is this law not valid for all reactions? If so, what is the mistake in the above derivation?

• Enthalpy and heat capacity are functions of both temperature and pressure. But, at the low pressures of the standard state, it can be treated as exclusively a function of temperature. Why do you feel that it would not be valid for all reactions? – Chet Miller Apr 16 '18 at 20:59
• @ChesterMiller Personally i am not convinced, but everywhere it is referenced. E.g Atkins Physical chemistry 10th edition, our lecture notes etc. it is only stated for standard pressure. It states the top equation (how it varies for a single species) as true, but when it is applied to a reaction, it is only given for standard pressure. – Adroit Apr 17 '18 at 6:42

It would say that you are making three separate mistakes:

1. The standard state pressure is any arbitrary pressure, not necessarily 1 bar, so there is no implication from the equation specifying standard state that it is only true at one particular pressure.

2. The important specification of being in the standard state is not the value of arbitrary pressure chosen, but that the substances are not mixed. They are pure or in solvent as the single solute. Also, for gases and solutions the states are fictitious. In your derivation, you are not accounting for changes due to mixing and any changes due to being in real, rather than fictitious, states.

3. I don't see why you say to integrate Cv rather than Cp.

• Very helpful, thank you. The non mixing thig is something that was mentioned in the very lecture we learned about this topic. So, just to be clear: The standard state(as indicated by the symbol in the quoted equation) implies that no mixing is occuring? – Adroit Apr 17 '18 at 21:20
• Also, the Cv instead of Cp was simply a typo, i copied the maths language from a related thread but forgot to edit – Adroit Apr 17 '18 at 21:21
• @Adroit yes, it means each reactant or product existing separately. – DavePhD Apr 17 '18 at 22:22
• Please provide a reference for item 1. above. In every source I have seen, the standard state pressure for gases is taken as 1 bar. – Chet Miller Apr 18 '18 at 11:01
• @ChesterMiller The statement is already hyperlinked to the IUPAC definition which says "a well defined but arbitrarily chosen standard pressure". But if that is insufficient see "The standard state with respect to pressure or concentration for each state of aggregation is arbitrarily chosen as some value that can be conveniently measured" books.google.com/… – DavePhD Apr 18 '18 at 11:27

If the question is, "Is the heat of reaction a function of temperature and pressure?" the answer is Yes. If the question is, "Is the standard heat of reaction a function of temperature and pressure?," the answer is No; the standard heat of reaction applies only for reactants and products at 1 bar. If the question is, "Are the heat capacities of the pure reactants and products functions of temperature and pressure?", the answer is Yes. So are you asking, "How do I determine the heat of reaction at a specified temperature and an arbitrary pressure higher than 1 bar (without knowing details on how the heat capacities vary with pressure)?"