Dissolution can be either endothermic or exothermic. At higher temperature dissolution is much faster. However, supplying heat also leads to an increase in the internal energy of the system. Ultimately, this affects the enthalphy of product and reactant. I want to know whether an increase in temperature affects the enthalphy of dissolution?


1 Answer 1


I want to know whether an increase in temperature affects the enthalphy of dissolution?

It certainly can. You can treat the temperature dependence of dissolution with a formalism analogous to that applied to chemical reactions. The key parameter is the difference in heat capacity of the solid and dissolved solute, since the enthalpy of solubilization can be written as

$$\Delta H(T) = \Delta H(T_\mathrm{ref}) + \int_{T_\mathrm{ref}}^{T} \Delta C_p\,\mathrm{d}T$$

where $\Delta C_p = C_{p,\mathrm{solute}} - C_{p,\mathrm{s}}$. The $\Delta C_p$ term accounts for the different temperature sensitivities of the enthalpy (which for condensed phases approximates the energy) of solid and solute. If the heat capacity of the solute is greater than that of the solid, the enthalpy of dissolution will increase (become more positive) with an increase in $T.$

Note that if we associate a solubility constant $K_\mathrm{sp}$ with the dissolution process

$$\ce{A(solid) <=> A(solution)}$$

then we can write

$$\frac{\mathrm{d}\log K_\mathrm{sp}}{\mathrm{d}T}=\frac{\Delta H^\circ}{RT^2}$$

as suggested in a comment. Here both $K_\mathrm{sp}$ and $\Delta H^\circ$ are functions of $T.$

  • $\begingroup$ This is a great answer, but I think it's worth mentioning that from a practical standpoint, the change in enthalpy is typically very small unless $\Delta T$ is quite large (greater than 100 K is a good ballpark), so you can usually ignore it. This is why, for example, you can treat the standard enthalpy change as a constant in the van't Hoff equation even though it's technically only accurate for 298 K. Similarly, you can do the correction described here under the assumption that $\Delta C_P$ is constant over T or you can be even more precise and include the variation of $\Delta C_P$ with T. $\endgroup$
    – Andrew
    Mar 7, 2019 at 18:38
  • $\begingroup$ @Andrew Thanks for the comment. I thought of what you mention, but decided to provide a general answer. $\endgroup$
    – Buck Thorn
    Mar 7, 2019 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.