An ionic compound has a solubility of $1\ \mathrm M$ in water at $25\ ^\circ \mathrm C$ and its solubility increases as the temperature is raised. What are the signs of $\Delta H^\circ$ and $\Delta S^\circ$ for the dissolving process?

Since the the solubility increases as the temperature is raised ($\Delta G^\circ$ becomes more negative), I know that $\Delta S^\circ$ is positive. However, I am unsure of how to determine the sign of $\Delta H^\circ$. The answer is $\Delta H^\circ > 0$.

  • 1
    $\begingroup$ Hint: What happens to the bonds in the substance as it dissolves... $\endgroup$
    – Buck Thorn
    Commented Jan 20, 2019 at 20:06
  • $\begingroup$ I know the bonds break (delta H is positive), but I thought that some substances, like NaOH, dissolve exothermically (so delta H is negative)? $\endgroup$ Commented Jan 20, 2019 at 23:38
  • $\begingroup$ I am not sure the proposer of the exercise want all this but your doubt is legitimate. See here: chemistry.stackexchange.com/questions/51862/… $\endgroup$
    – Alchimista
    Commented Jan 21, 2019 at 10:09
  • $\begingroup$ See answer about Van't Hoff equation here. $\endgroup$
    – Karsten
    Commented Jan 21, 2019 at 15:02
  • 1
    $\begingroup$ The statement about $\Delta G^\circ$ is not quite true. K becomes larger, and $\frac{\Delta G^\circ}{T}$ becomes more negative as the temperature increases. $\endgroup$
    – Karsten
    Commented Jan 21, 2019 at 15:17

2 Answers 2


I would solve this question rather simply.

Let's consider the general equation $\ce{ X-Y_{(s)} \rightarrow X^+_{(aq)} + Y^-_{(aq)}}$ with a $\Delta H = C$. Intuitively, if the reaction is endothermic ($\Delta H > 0$), an increase in temperature should help the dissolution (more heat to be absorbed by the salt while dissolving). You can rationalize this further using the Le Châtelier's principle:

$\ce{ X-Y_{(s)} + C \rightarrow X^+_{(aq)} + Y^-_{(aq)}}$

If you add more heat ($C$), the dissolution will be more favored.

More rigorously, you can use Van't Hoff equation (assuming a dissolution equilibrium): $\frac{d(ln(K))}{dT} = \frac{\Delta H}{RT^2}$. If $\Delta H > 0$, then $ln(K)$ increases with $T$, i.e. dissolution is more favored.


@Raphaël did the hard part. ΔH is larger than zero.

[OP] Since the the solubility increases as the temperature is raised (ΔG∘ becomes more negative), I know that ΔS∘ is positive

This argument linking solubility and Gibbs energy is incorrect. Le Chatelier or the Van't Hoff equation connects the temperature-dependence to the reaction enthalpy, not to the entropy. So how can we make a statement about the entropy? You might think the entropy of dissolution must be positive because the solid ionic compound is more ordered than the individual ions in solution. This is a good guess, but sometimes the solvent loses degrees of freedom, so it is not a sure bet.

Instead, let's consider the Gibbs energy of reaction. It is zero, because the reaction is at equilibrium. What about the standard Gibbs energy of reaction? Because all species are approximately at standard state, it is also zero (or near zero). With a positive change in enthalpy, the change in entropy has to be positive as well to get a Gibbs energy of zero.


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