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Standard enthalpy of formation is a value that represents the enthalpy variation in the formation of compounds with the purpose of using this values to know the enthalpy of a reaction where these compounds participate.

Those values are referred to the formation of said compounts at 273 K and 1 atm (known as standard conditions).

My question is:

Does the value change, for distinct pressures or temperatures? If yes, how?

Why am I asking this?

Let's suppose I have a reaction that goes like this:

$A + B \rightarrow C$

At 1 atm and 273 K (standard conditions).

I can say that in this reaction $\Delta H = Hf^0(C) - Hf^0(A) - Hf^0(B)$

Where $Hf^0$ means standard enthalpy of formation.

Because the reaction happens at 1 atm and 273 K.

But, what if the reaction happens at 3 atm and 350 K?

I asked my University teacher and he said: "You don't need that because we will always be working at 1 atm and 273 K".

What tool do I have to know the variation of enthalpy in this case?

A small extra confusion is the fact that in some examples I see standard conditions being 1 atm and 273 K, and some other times it's at 298 K.

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  • $\begingroup$ Not my downvote, but generally on the site you are expected to show what you have tried to solve the problem. This not only avoids answers repeating information that is readily available elsewhere (or is already here on the site), but it also helps us better understand what exactly you want clarified. In regards to your question, this previous answer may get you started on the relationship of enthalpy and pressure chemistry.stackexchange.com/a/73226/41556 $\endgroup$ – Tyberius Apr 27 at 15:04
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    $\begingroup$ @Tyberius Thanks, I can understand the point. Maybe I did not make the question clear enough, but I am surely not the kind of person who just asks without researching first. Before I asked, I really did my research, but have not been able to find an answer. I updated the question's title and content to try to make it more accurate. Your linked answer is good information and I thank you for it, but I still cannot undertsand how I would face the problem I mention in the question. Always appreciate the effort people do on StackExchange, probably the most useful websites in the net. $\endgroup$ – Alvaro Franz Apr 28 at 10:50
  • $\begingroup$ What do you mean by "tool"? $\endgroup$ – Buck Thorn Apr 29 at 13:21
  • $\begingroup$ @BuckThorn Thanks for your attention. By "tool" I mean a set of true affirmations accompained by some math, to be able to get the result I am looking for. I updated my question to make clearer what it is that I want to calculate. (I saw your edit to your answer, I am trying to dive into it and learn from it). $\endgroup$ – Alvaro Franz Apr 30 at 10:03
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    $\begingroup$ Yes, standard states can vary (see e.g. these wikipedia pages. The important thing is to know what standard state is associated with the data which sometimes requires a little care. $\endgroup$ – Buck Thorn Apr 30 at 15:51
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For a pure substance you can write generally that the enthalpy depends on pressure as follows:

$$\left(\frac{\partial H}{\partial p}\right)_T = V - T \left(\frac{\partial V}{\partial T}\right)_p \tag{1}$$

If you evaluate this for an ideal gas you find that $\left(\frac{\partial H}{\partial p}\right)_T = 0$. If the substance is present in a condensed phase with a density (or molar volume) that is weakly dependent on the temperature, then roughly

$$\left(\frac{\partial H}{\partial p}\right)_T = V$$

If you speak of standard enthalpies then by definition there is no pressure dependence, as they are defined at a particular pressure.

If your reaction involves gases which can be assumed ideal, then you can use the standard enthalpy for all reagents and products, since according to my answer the enthalpy of an ideal gas is independent of pressure. On the other hand if it involves condensed phases you need to evaluate the molar volume and perhaps also the thermal expansion coefficient $\alpha$, since Eq. (1) can be rewritten as

$$\left(\frac{\partial H}{\partial p}\right)_T = V (1- T\alpha)$$

For the enthalpy of reaction you might write

$$\left(\frac{\partial \Delta_r H}{\partial p}\right)_T = \sum_i \nu_i V_{m,i} (1- T\alpha_i)$$

where $\nu_i$ are stoichiometric coefficients. In the summation in the rhs account should be taken for sign changes depending on whether species are reactants or products.

If you want to evaluate the enthalpy of reaction at one pressure given the value is known at another pressure, and the molar volumes are known, you can integrate the previous equation, giving (at constant T)

$$\Delta_r H(p) = \Delta_r H(p^\circ) + \int_{p^\circ}^{p} \sum \nu_i V_{m ,i}(1- T\alpha_i) dp$$

where $p^\circ$ is the reference pressure at which the enthalpy is already known. This last expression may be simplified in various ways depending on the state of the species and the pressure integration range. If the molar volumes and expansion coefficients are approximately constant in the pressure range then

$$\Delta_r H(p) = \Delta_r H(p^\circ) + (p -p^\circ) \sum \nu_i V_{m,i} (1- T\alpha_i)$$

If the species have negligible thermal expansion coefficients then

$$\Delta_r H(p) = \Delta_r H(p^\circ) + (p -p^\circ)\sum \nu_i V_{m,i}$$

and so on. For instance, for a reaction converting a solid between two forms $\text{react}$ and $\text{prod}$ having constant molar volume and negligible thermal expansion coefficients in the pressure range $p^\circ$ to $p$,

$$\ce{\text{react}<=>\text{prod}}$$

$$\Delta_r H(p) = \Delta_r H(p^\circ) + (p -p^\circ)(V_m(\text{prod})-V_m(\text{react}))$$

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  • $\begingroup$ I had to re-formulate my question because it was closed as off-topic. Now it has been reopened but (even though your answer is very helpful) it does not answer the question. That is why I unmarked the "accepted" checkbox. Thank you for your help. $\endgroup$ – Alvaro Franz Apr 28 at 15:39
  • $\begingroup$ @AlvaroFranz Please see my edit. $\endgroup$ – Buck Thorn Apr 28 at 16:09
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    $\begingroup$ Thank you for your great help. $\endgroup$ – Alvaro Franz May 1 at 8:43

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