chemistry research question?

Question

how does enthalpy and entropy typically influence the solubility of solids and liquids in solution?

Answer 1

it has to do with the Gibbs free energy equation



delta G=delta H -T delta S



for something to happen in nature (aka being spontaneous) delta G needs to be negative



there are 3 ways for this to happen



either enthalpy (delta H) is negative and entropy (delta S) is positive

this always happens



enthalpy (delta H) is negative and entropy (delta S) is negative and the temperature is low



enthalpy (delta H) is positive and entropy (delta S) is positive and the temperature is high



NEVER WORKS



enthalpy (delta H) is positive and entropy (delta S) is negative

(heating water and having it freeze)



with those rules out lets answer the question



dissolving is an increase in entropy s-->aq



SO

enthalpy (delta H) is positive and entropy (delta S) is positive and the temperature is high (explains why at higher temps some solutes are more soluble)



either enthalpy (delta H) is negative and entropy (delta S) is positive

this always happens (explans why solutes dissolve exothermically)



Kent

http://www.kentchemistry.com

Answer 2

Hi Sammy: (I answered this question for someone else yesterday!):



Let me illustrate the answer using sodium chloride in water. Let us initially consider the enthalpy and entropy changes separately.

The enthalpy change for salt dissolving in water can be divided into two steps. The first step is to split the ions in the solid state into gas phase ions:

NaCl(s) -> Na(g)+ + Cl(g)- -U

U is of course the lattice energy; because energy has to be put into the system the the -U term is +ve (the -ve sign for U is usually ignored).



We then plunge the ions into water:

Na(g)+ + nH2O -> Na(OH2)n+ +ΔHsolv(Na+)

Cl(g)- + nH2O -> Cl(H2O)n- +ΔHsolv(Cl-)

Δ = Capital Greek Delta = change



The problem is that for M+X- solids such as NaCl the lattice energy is approximately equal to the solvation energies (ion-dipole interactions of polar water molecules binding to the +ve and -ve ions) and it is hard to predict DHsoln. Ionic solids with large lattice energies (e.g., those with small multicharged ions such as M2+X2- as in MgO) the lattice energy dominates and the solid is usually insoluble. For some cases the solvation energies are small (e.g., "soft" ions such as Pb2+ and S2-; water is "hard") and do not offset the lattice energies and hence these compounds are extremely insoluble. On the other hand, solvation energies can be large and overwhelm -U (all nitrates are soluble).



The same is true for the ΔS term (change in entropy or change in degree of randomness of the system) for the dissolution process. There is a large increase in entropy when the ordered crystal lattice is broken up, but (and what is often not realized) there is loss of randomness when the water molecules solvate the ions. The ΔS term can be dominant. I illustrate this in lectures by dissolving NH4NO3 in water: the temp of the water drops significantly upon dissolution of the salt, that is, it is an endothermic process that is entropy driven.



Whether the compound will dissolve or not is given by the free energy change ΔG that must be -ve for the reaction to proceed:

ΔG = ΔH - TΔS

with, as I have said, the ΔH and ΔS terms being the difference of two large terms that are each difficult to predict



You can use similar arguments for two liquids being mixed together (like dissolves like!). cheers, dr p

Answer 3

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