4
$\begingroup$

We know that dissolution of sugar in water is a spontaneous process.

So, change in Gibbs free energy $(ΔG)$ must be negative for the overall process:

$$ΔG = ΔH - TΔS < 0$$

Hence either the enthalpy or entropy must drive the reaction.

Now, after searching for the values, most websites suggest the process is endothermic (i.e. $ΔH > 0).$

But Jan's answer has me convinced that the overall entropy of a system should decrease (i.e. increase in entropy of sugar molecules is outweighed by decrease in entropy of water molecules).

So, what's the real reason for reaction to be spontaneous if neither entropy and enthalpy are driving the reaction?

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ This doesn't answer your question, but serves as an important point of information: You can have a reaction where the inherent enthalpy and entropy are both unfavorable, yet it will still proceed spontaneously if you are away from equilibrium, because of the entropy of mixing. See: chemistry.stackexchange.com/questions/129589/… $\endgroup$
    – theorist
    Feb 2, 2021 at 7:25
  • $\begingroup$ How can you say that neither energy nor enthalpy drive the reaction? $\endgroup$
    – ultralegend5385
    Feb 2, 2021 at 8:40
  • $\begingroup$ @ultralegend5385 For a reaction to be driven in a direction, ΔG must be negative in that direction. For ΔG to be negative, either ΔH must be -ve or ΔS must be +ve. (Neither of which is happening in this situation) $\endgroup$
    – InfiniteCool23
    Feb 2, 2021 at 8:52
  • 3
    $\begingroup$ I think that no matter what, since the process is spontaneous, $\Delta G<0$ . As you said, $\Delta H>0$ is true for certain. So, that implies $\Delta S$ must be positive. I think that @Jan's reasoning has some flaws. Read the comments and other answers in the link that you mentioned, they do point out that it is wrong. $\endgroup$
    – V.G
    Feb 2, 2021 at 9:30
  • 2
    $\begingroup$ What about the entropy change of mixing? $\endgroup$
    – Chet Miller
    Feb 2, 2021 at 12:32

2 Answers 2

Reset to default
0
$\begingroup$

Since solubility of sugar in water increases with temperature, the change in enthalpy of the sugar-water system is positive. Since the sugar-water system is absorbing heat(at constant pressure, change in enthalpy is the heat absorbed) and since entropy change of an irreversible process is greater than that in an reversible process, the entropy of the sugar-water system will increase. So at constant temperature and pressure the enthalpy change is positive, entropy change is positive and the the change in Gibbs free energy is negative(before equilibrium is attained) . The process is spontaneous. By the time the system reaches equilibrium the enthalpy change will have been balanced by the change in entropy and so the change in Gibbs free energy is 0.

Improve this answer
$\endgroup$
0
$\begingroup$

The entropy change for ideal mixing is $\Delta S=-R(x_A\ln{x_A}+x_B\ln{x_B})$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Here, 'x' refers to mole fraction, right? $\endgroup$
    – InfiniteCool23
    Feb 3, 2021 at 7:00
  • $\begingroup$ Yes, that is correct. $\endgroup$
    – Chet Miller
    Feb 3, 2021 at 17:12

Your Answer

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

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