In the van't Hoff equation, why do we say that the change in enthalpy at standard pressure is constant? My book derives the equation for a mixture of ideal gases reacting with each other. The proof itself is good, until the authors declare that the equation says that the graph of $\ln K$ with $\frac{1}{T}$, where $K$ is the equilibrium constant and $T$ is the temperature of the mixture, will be a straight line. This would only be true if the change in enthalpy is constant (since the slope of the curve is proportional to this change). But isn't the enthalpy of an ideal gas proportional to temperature? So why is the change in enthalpy constant in $T$, then?

I've attached an example of this effect, as presented by the book. See the image below.

Van't Hoff equation graph

  • $\begingroup$ It is a simplifying assumption. Could you provide more details about the gases involved, the pressure, and the temperature window? $\endgroup$
    – Buck Thorn
    May 6 '21 at 11:53
  • $\begingroup$ @BuckThorn I don't have any particular gases or temperature windows in mind (simply because that wasn't mentioned in the book), but they do provide an example of this effect. Do you think it's a good idea to include that in my question? $\endgroup$ May 6 '21 at 11:55

Good question. Enthalpy of reaction dependence on temperature is given by Kirchoff's equation (you can check it) and it is true that enthalpy of reaction changes with temperature. For most reactions if temperature change isn't too big, enthalpy of reaction won't change much so we can regard it as constant in that temperature span on which we are plotting lnK vs 1/T for sake of simplicity.

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    $\begingroup$ Ah, okay, so it's just an approximation. But shouldn't the span of the temperature in the example I gave (see the edit) be big enough for the approximation to go wrong? $\endgroup$ May 6 '21 at 12:01
  • $\begingroup$ Well, it is pretty big temperature span. However, for this reaction difference in cp between products and reactants is probably small since you have only dissociation of hydrogen to atoms. Since products and reactants are similiar, I would say their cp isn't much different because of which enthalpy of reaction doesn't change much with temperature. How much enthalpy of reaction changes with temperature is determined by stochiometric difference in cp between products and reactants (Kirchoff's equation). $\endgroup$ May 6 '21 at 12:52
  • $\begingroup$ Thank you for that. Yeah, that makes sense. Also, thank you for the suggestion to look up Kirchhoff's equation. $\endgroup$ May 6 '21 at 13:40
  • $\begingroup$ The $\Delta C_p\Delta T$ is quite small compared the $\Delta H$ even though the temperature range is large. Typically $C_p$ is in tens of J/mol/K where as $\Delta H $ is in hundreds of kilo J/mol. $\endgroup$
    – porphyrin
    May 6 '21 at 17:10

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