In thermodynamics many definition has been made from partial derivatives at constant conditions for instance:

Let $U$ be $U:f(S,V,m_𝑘)$ then:

$$T = \left( \frac{\partial U}{\partial S} \right)_{V, m_{k}}; p = \left( \frac{\partial U}{\partial V} \right)_{S, m_{k}}; \mu_k = \left( \frac{\partial U}{\partial m_k} \right)_{V, S}$$

In a particular case, heat of reaction and van 't Hoff relation

Being $H:H(T,p,\xi)$

$$\mathrm dH= \left( \frac{\partial H}{\partial T} \right)_{p, \xi}\,\mathrm dT + \left( \frac{\partial H}{\partial p} \right)_{T, \xi}\,\mathrm dp + \left( \frac{\partial H}{\partial \xi} \right)_{T, p}\,\mathrm d\xi$$

Where $C_p$ is the specific heat at constant pressure, $h_T$ is the heat compressibility and $r_{T,p}$ is the heat of reaction at constant temperature and pressure:

$$C_p = \left( \frac{\partial H}{\partial T} \right)_{p, \xi} ; h_T = \left( \frac{\partial H}{\partial p} \right)_{T, \xi}; r_{T,p} = \left( \frac{\partial H}{\partial \xi} \right)_{T, p}$$

In terms of variation of affinity with temperature, the heat of reaction is given by (where $K$ is the equilibrium constant of a chemical reaction):

$$r_{T,p} = -RT^{2}\frac{\partial }{\partial T} \ln K(T,p)$$

So the question is: How can I define an equation where the left side needs constant temperature, but the right side implies the derivative of temperature?


2 Answers 2


The T that is held constant on the LHS of your equation has a different meaning from the T that is allowed to vary on the RHS.

The van 't Hoff equation answers the following:

Suppose I have a reaction that takes place at const T, p at T = T1. Now I change to a new temperature, T2, and allow that reaction to again take place at const T, p, except now T = T2. How does the equilibrium constant change?

Now that you understand this, you can see there is no inconsistency between "constant T" and "changing T", since they refer to two different things.

I.e., the fact that the reaction is constrained to take place at constant T in no way prevents me from testing the reaction at a different, but still constant, T.

The analogy I'd give is this. Suppose we want to determine the effect of altitude on 100 m sprint times. The 100 m is always run on a level surface (no change in altitude during the race). Yet this does not in any way prevent us from comparing 100 m run times at different altitudes (e.g, sea level vs. 10,000 feet).

So your question is like asking: "How can we determine the effect of changing altitude on 100 m run times, given that the 100 m run is always done at constant altitude?" Well, we can, of course.

  • $\begingroup$ Good point I was thinking about the same think, but I was getting trouble, for instance the temperature definition implies that V and chemical composition ($m_k$) must be constant hence Does it implie that chemical reaction with a change in volume has no temperature to be measurable? $\endgroup$
    – Ilya_Curie
    Commented May 28, 2020 at 5:22
  • $\begingroup$ @JuanSebastianGarciaPerez Not sure I understood your question -- there may be a language barrier -- but it seems you are saying that there will be no measurable temperature change because V and chem. comp. are constant. That's not the case. The reason there is no T change is because we've designed the experimental conditions that way: the reaction is taking place in a constant-T environment. $\endgroup$
    – theorist
    Commented May 28, 2020 at 5:31
  • $\begingroup$ Furthermore, it's not the case that V and chem. composition are constant. The van 't Hoff equation applies for reactions at constant T and p. This means that, typically, as the reaction takes place, V will not be constant (unless the molar volumes of the reactants and products are identical). And since a reaction is taking place, the composition will not be constant either. $\endgroup$
    – theorist
    Commented May 28, 2020 at 5:32
  • $\begingroup$ What I am trying to say is I understand your answer about Van't Hoff equation but now What happen with temperature definition? Does temperature only exist when V is constant? $\endgroup$
    – Ilya_Curie
    Commented May 28, 2020 at 6:31
  • 1
    $\begingroup$ But the zeroth law of thermodynamics is not actually a law, it implies the concept of equilibrium, temperature should be a transversal concept, exist in both equilibrium and non-equilibrium processes, how are we able to explain the heat flux by a temperature gradient? (which is a non-equilibrium process). $\endgroup$
    – Ilya_Curie
    Commented May 28, 2020 at 9:03

There are a few different ways to derive the van't Hoff equation. The question here however is to justify the mathematical form of the equation. A good (if not entirely satisfying) way to do so is to be (a little) more mathematically rigorous. You then find that the form of the equation follows from the fact that the Gibbs' free energy G is a thermodynamic potential with continuous second derivatives and so observes Schwartz's theorem, which in the context of thermodynamics is used to derive Maxwell's relations. Thermodynamic properties like the Gibbs free energy, pressure and temperature that obey Schwartz's theorem are called state functions. Algebraic combinations of these are also state functions.

In the case of the free energy divided by the temperature, the total differential can be written as

$$d(G/T) = \left(\frac{\partial (G/T)}{\partial P} \right)_{T,\xi}dP + \left(\frac{\partial (G/T)}{\partial T} \right)_{\xi,P}dT + \left(\frac{\partial (G/T)}{\partial \xi}\right)_{P,T}d\xi \tag{1}$$

Since G/T is a state function it follows that the order of differentiation of the first derivatives of the potential does not matter and that the following relation between second derivatives holds:

$$\left(\frac{\partial ^2 (G/T)}{\partial T \partial \xi} \right)_{P} = \left(\frac{\partial ^2 (G/T)}{\partial \xi \partial T}\right)_{P} \tag{2}$$

If you evaluate the first partial derivatives of (G/T) in this equation and perform appropriate manipulations you can arrive at the van't Hoff equation. The question then about how the odd relation arises in the van't Hoff equation, in which a derivative is taken wrt a variable that was previously held fixed, can be asked already here, because Equation 2 could be written as follows:

$$\left(\frac{\partial }{\partial T }\left(\frac{\partial (G/T)}{\partial \xi} \right)_{T,P}\right)_{P} = \left(\frac{\partial }{\partial \xi }\left(\frac{\partial (G/T)}{ \partial T}\right)_{\xi,P}\right)_{P} \tag{2'}$$

You see on the left-hand-side that you first hold T constant and later take the derivative wrt T.

  • $\begingroup$ How do you relate the heat of reaction $𝑟_{𝑇,𝑝}$ with an enthalpy accord to Maxwell's relations? $\endgroup$
    – Ilya_Curie
    Commented Jun 5, 2020 at 1:54
  • $\begingroup$ You can rewrite the mixed derivative on the rhs in equation 2' in terms of $r_{T,p}$ $\endgroup$
    – Buck Thorn
    Commented Jun 5, 2020 at 13:35
  • $\begingroup$ I edited my answer. As written earlier it was more difficult to perform the manipulations that lead to the van't Hoff equation. $\endgroup$
    – Buck Thorn
    Commented Jun 5, 2020 at 20:22

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