Chem 300 - Ch 26/#3 Today’s To Do List

• Metallurgy and K• Ellingham Diagrams• Ionic Strength & Solubility• Life & K

Metallurgy: Extraction of Metals from Oxides

Reduction with C(s) or CO(g)• MO(s) + C(s) M(s) + CO(g)

• MO(s) + ½ C(s) M(s) + ½ CO2(g)

• MO(s) + CO(g) M(s) + CO2(g)

K > 1 Consider some related reactions

4 Oxidation Reactions

1. M(s) + ½ O2(g) MO(s)

2. ½ C(s) + ½ O2(g) ½ CO2(g)

3. C(s) + ½ O2(g) CO(g)

4. CO(g) + ½ O2(g) CO2(g)

T-Dependence of rGo:

• d( rGo) = - ( rSo)dT

Ellingham Diagram

Returning to Reduction

• MO(s) + C(s) M(s) + CO(g)rGo = rGo(3) - rGo(1)

• MO(s) + ½ C(s) M(s) + ½ CO2(g) rGo = rGo(2) - rGo(1)

• MO(s) + CO(g) M(s) + CO2(g) rGo = rGo(4) - rGo(1)

• When rGo < 0, the line for rxn (1) lies below the line for rxn (2) – (4).

Ellingham Diagram

Ionic Strength & Solubility

M+X - (s) = +Mz+ + -Xz-

Solubility Product constant:• Ksp = a+

+ a- - = m+

+ m- -

( + + -)

Through the solubility is affected by other non-chemically-active ions.

ln = -(1.177 (Z+ )(Z-) I1/2)/(1 + I1/2)

ln = -1.173(Z+ )(Z-) I1/2 (if I <<1)

Example: AgCl in 1.5 mmol/kg Ba(NO3)2

Ka = 1.60 x 10-10 = [Ag+][Cl-] = S2 2

I = 3(1.50 x 10-3) + S = 0.00450 + S Solubility = S = (Ka

1/2)/ What is S of AgCl in plain water? Ideal solution: = 1.00

S = (1.60x10-10)1/2 = 1.26x10-5

What is effect of Ba(NO3)2 on S of AgCl?

With other ions added

1st approx: • I = 0.00450 + 1.26x10-5 = 0.0045126

ln = -(1.177 I1/2)/(1 + I1/2) = -0.07409 = 0.9286

S = (Ka1/2)/ = (1.26x10-5)/0.9286 =

1.36x10-5 mol/kg This value is ~8% higher

• (measured S = 1.372x10-5)

A BioPhysChem App.

Ca3(PO4)2(s) = 3Ca2+ + 2PO43-

Ka = [Ca2+]3[PO43-]2

5 = S5 5

Bone/teeth deposition of phosphate/carbonate controlled by Ka relationships

S is influenced by presence of other ions

Next Time

• Start Chapter 27 – Kinetic Theory of Gases• Average Translational KE & T• The Gaussian Distribution & Molecular Speeds