Chapter 13 The Solution Process - Kimika · Thermodynamics – the study of heat ... The solution...
Transcript of Chapter 13 The Solution Process - Kimika · Thermodynamics – the study of heat ... The solution...
Homework
13.115 13.122 13.130 13.133 13.139 13.143
DUE MONDAY DECEMBER 15, 2008
House-keeping
Make-up Class (WHEN?!) Monday 4:30 – 6:00 PM Possibly C-114
Long test: December 17, 2008 Wednesday 4:30 – 6:00 PM or 6:00 – 7:30 PM
a thermodynamic approach
A few key terms
Thermodynamics – the study of heat (“energy in transit”) and its interconversion.
We talk of movement of energy in a system
The solution process
Intermolecular Forces have to be broken Solute-solute Solvent-solvent
Intermolecular Forces have to be created Solute-solvent
Just like phase changes, breaking and creating forces require an input and output of energy
The solution process
∆Hsol'n = ∆Hsolvent + ∆Hsolute + ∆Hmix
Breaking IMFs between solvent molecules: Endothermic, ΔHsolvent > 0
Breaking IMFs between solute molecules: Endothermic, ΔHsolute > 0
Forming IMFs between solute-solvent molecules: Exothermic, ΔHmix < 0
Energy Changes
Aqueous solutions salts
ΔHlattice
It is the energy needed to completely liberate ions from an ionic solid
It is endothermic!
ΔHlattice = ΔHsolute
Aqueous solutions salts
ΔHhydration
Solvation (Hydration) is the process of surrounding a solute particle with solvent (water)
It is the combined process of disturbing your solvent particles and mixing solute with solvent
It is exothermic!
ΔHhydration = ΔHmix+ ΔHsolvent
Aqueous solutions salts
ΔHhydration = ΔHmix+ ΔHsolvent
ΔHlattice = ΔHsolute
∆Hsol'n = ∆Hsolvent + ∆Hsolute + ∆Hmix
∆Hsol'n = ∆Hlattice + ∆Hhydration
Aqueous solutions salts
The solution process
Statement 1: The solution process is accompanied by heat changes (enthalpy). Exothermic processes are favored.
However!
Entropy
Entropy (S) is directly related to the number of ways that a system can distribute its energy.
Entropy is directly related to the freedom of motion of particles and the number of ways they can be arranged
Sgas > Sliquid > Ssolid
Entropy of Solution
There are more ways you can distribute the energy in a solution than individual solute
Ssol’n > (Ssolute+Ssolvent) ∆Ssol’n> 0
The solution process
Statement 1: The solution process is accompanied by heat changes (enthalpy)
Statement 2: The solution process is also accompanied by the greatest tendency to distribute energy in as many ways possible (entropy)
∆Hsol’n ∆Ssol’n
So why does like dissolve like?
Case 1: NaOH in water
So why does like dissolve like?
Case 2: octane in hexane
Enthalpy of solution is almost zero, the solution process is entropy driven
So why does like dissolve like?
Case 3: NaCl in hexane
A solution will not form because the entropy increase is not enough to compensate for the huge enthalpy change
So why does like dissolve like?
Case 4: NH4NO3 in water
The increase in entropy during crystal break down and the favorable mixing of ions in polar water solvent compensates for increase in enthalpy
Thermodynamics of solution process
How is ∆G related to enthalpy (∆H) and entropy (∆S)?
For solutions to occur, ∆Gsol’n<0
∆Hsol’n ∆Ssol’n Temp
Exercise
Sketch a qualitative enthalpy diagram for the process of dissolving KCl (s) in H2O (endothermic)? NaI(s) in H2O (exothermic)?