Ionic Strength, Activity Corrections Ionic Strength: measure of electrostatic field Typically 1‐10 mmole/L 1 2 ICz= ∑ ii 2 i Activity: Effective concentration or conc. in ideal solution Actually ratio of conc. to conc. at standard state Dimensionless []Ci {}Aii= γ []Cstd. State Because [Cstd.State] = 1, {A} = γ·[C] Activity Coefficient Name Equation Conc. Range Debye‐Hückel I < 5 mM log(γ ) =−AzI2 Extended Debye‐Hückel I I < 0.1 M log ()γ =−Az2 1+ BaI Güntelberg I I < 0.1 M log ()γ =−Az2 1+ I Davies I < 0.5 M 2 I log()γ =−AzI − 0.2 1+ I In all equations A = 0.5 for water. Standard States Reference state conditions Std.Conc. Temp. Pressure Convention for real Conc. Solid Conc. in pure 25oC 1 atm 1.0 solid (X = 1.0) Liquid Conc. in pure 25oC 1 atm 1.0 liquid (X = 1.0) o Gas Conc. in pure 25 C 1 atm Pi gas (Pi = 1 atm) Solute 1.0 M 25oC 1 atm [i] (molar) 00 µµii=+RTAln{ i } =+ µ i RT ln( γ i) + RTC ln ([ i]) Free Energy Relationships kJ ∂GkJ ∆=G mol dξ mole 0 ∆=∆+GGRTQrxn rxn ln ( ) νi A cd ∏{ i, products} {}{}CD Q == Reaction Quotient= i νi ABab ∏{}Ai,reactants {}{} i 00 0 ∆=GGrxn∑∑νµ i i = ν i ∆ f, i ii Equilibrium Expressions At equilibrium ΔG = 0 0 ∆=∆+GGRTQln( eqm ) = 0 0 ∆=−GRTQln ()eqm ∆=−GRTK0 ln () −∆G0 or K = exp RT Combinations of Reactions aA+→ bB cC Consider: aA+ bB+→ dD eE cC+→ dD eE Recall that G is extrinsic property and is additive GG= ∆⋅ξ Add ΔG GGGGnet =+1 22 =∆+∆1ξξ12 G 00 Gnet =∆ G11ξξ + 1 RTln() Q 1 +∆ G 22ξξ + 2 RT ln () Q 2 00 ξξ12 GGnet =∆11ξξ +∆GRTQQ 22 + ln () 1 2 At equlibrium: Gnet ∆==Gnet 0 ξnet 1 RT 00 ξξ12 ∴∆+∆GG11ξξ 22 =−ln () QQ 1,eqm 2, eqm ξξnet net 00 ξξ12 0 ∆+∆GG11ξξ2 2 =−RTln () K1 K2 =∆Gnet ξξ12 Multiply component K values KKnet = 12K Temperature corrections ∆G0 ∂ 00 RECALL: GHTS=− T 11000∂∆GH 1 −∆ =−222 ∆GGTS + =−() ∆ + ∆ = ∂∂TT TTT T dG TT ∴ =−S 22∆−∆GH00 ddT= 2 dT ∫∫TT TT22 0 00 dG∆ 0 ∆∆GG21 0 11 −=∆−H Also =−∆S TT TT dT 21 12 Similarly, for equlibrium constants (K), use the Van’t Hoff equation: K ∆H 0 11 ln 2 =− KRTT112 Balancing reaction equations Step 1. Write known reactants on the left and known products on the right −−2 −− CH343 COO+⇔ SO HCO + HS Step 2. Adjust stoichiometric coefficients to balance all elements except H and O −−2 −− CH343 COO+⇔ SO2 HCO + HS Step 3. Add water (H2O) to balance oxygen (O) 6 O’s on left = 6 O’s on right, no change needed Step 4. Add H+ to balance hydrogen (H) 3 H’s on left = 3 H’s on right, no change needed Step 5. Add electrons (e‐) to balance charges Charge of ‐3 on left = charge of ‐3 on right, no change needed.