A concentrated solution of copper sulphate, which is dark blue in colour, is mixed at room temperature with a dilute solution of copper sulphate, which is light blue. For this process:

  1. Entropy change is positive, but enthalpy change is negative.
  2. Entropy change and enthalpy changes are both positive.
  3. Entropy change is positive and enthalpy does not change.
  4. Entropy change is negative and enthalpy change is positive.

The final conc. of the mixture would be less than the initial concentration of the concentration $\ce{CuSO4}$ given.
Therefore I think the entropy change is positive. But, what about enthalpy? What factors should I look into to find answer to these types of questions?

  • $\begingroup$ Dilution in general is a process involving negative enthalpy change. Therefore, since the final concentration is lower than the initial, therefore the enthalpy change is negative, making the mixing favourable under any temperature. $\endgroup$ Aug 24, 2013 at 16:40
  • $\begingroup$ The enthalpy change could be positive or negative, just like the enthalpy change when dissolving a solid in a liquid. Of course, for the specific case here, you could look it up. $\endgroup$
    – Karsten
    Apr 25, 2019 at 19:37

2 Answers 2


This is difficult, because we do not know what is your current understanding and at what “level” of approximation you expect the question to be answered.

At a undergraduate level, or in the absence of any details (or thermochemical data or access to thermochemical tables), I would go with option (3).

  • Enthalpy: we are not given any indication at what is the enthalpy of mixing. In the simplest approximation, the enthalpy of mixing is zero, as in an ideal solution. It would correspond to a regime where the enthalpy of dissolution ($\Delta H_{\text{sol}}$) does not depend on the concentration of the solution. It's probably a good approximation, until you reach high concentrations.

  • Entropy: the entropy of mixture is positive, i.e. the process of mixing the two solutions is spontaneous. It can be explained by this reasoning.


You could argue that the copper and sulphate ions are (on average) better solvated in a more dilute solution. Furthermore, the ions with the same charge can "go out of each other's way" more easily in a dilute solution, so that the correlations between the ions are weaker. Both of these effects would lead to an energetically more favourable state, so that the enthalpy change should be negative (exothermic reaction).

  • $\begingroup$ To the extent that there is an effect, I don't think the sign of the enthalpy change can be argued so easily. It depends on the nature of the intermolecular interactions, and there may be other important effects than those you describe (pairing, for example). $\endgroup$
    – F'x
    Aug 24, 2013 at 17:57
  • $\begingroup$ @F'x Admittedly, my answer might be based too much on intuition. I certainly haven't considered all possible effects. Would you really say, that the enthalpy of dissolution doesn't depend on the concentration for the case at hand? Maybe, I should be more careful with my intuition then. $\endgroup$
    – Philipp
    Aug 24, 2013 at 18:07
  • $\begingroup$ On first approximation, it's constant. Then comes a first-order term in $c$, but its sign is not necessarily clear-cut. Just like enthalpy of mixing in real life can be either positive or negative, thermochemistry is full of weird corner cases :) $\endgroup$
    – F'x
    Aug 24, 2013 at 19:12

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