
361Lec25 HENRY’S LAW CONSTANT Tue 16oct18 is the PARTITION COEFFICENT for a gas between gas phase and a liquid B(gas) B(liquid) B(liquid) B(gas) partial pressure of gas pB pB K( forward) = = or concentration dissolved gas [B]eq XB eq OR concentration dissolved gas [B] XB K(backward) = = or partial pressure of gas pB eq pB eq so, HENRY’S LAW CONSTANT= an equilibrium constant ALL of the above = “HENRY’S LAW” 1 Vapor Pressure of dissolved gas k = K = B eq Solubility of dissolved gas Why are they so large??? They are the vapor pressure of the “pure dissolved gas” i.e. Xgas = 1 Hypothetical standard state of pure aqueous gas with a ∆H0 = to that of the gas surrounded by the solvent. The numbers are straight-line EXTRAPOLATIONS of values at very low mole fraction up to X = 1 for the dissolved gas. 2 Looks exactly like Rauolt’s Law: X PB= XBK 3 Table 6.1 Looks exactly like Rauolt’s Law: X X PB= XBK K = PB when XB=1 (the VERY hypthetical “100% gas in liquid” –but surrounded by liquid”) More reasonable is using the usual 1M standard state 4 361Lec25 Tue 16oct18 Partition Coefficient(s) Most common: B(solvent 1 B(solvent 2) Keq= [B(2)]/[B(1)] e.g., using “sep funnel” Which of the following are partition coefficients? Name (Keq) Partition Coeff. Henry’s Law constant B(solvent) gas(vacuum) PB/[B] or PB/XB Evaporation/Condensing A(liquid) gas(vacuum) PA/XA = Vapor Press Dissolving/Crystallizing crystal any solvent e.g., [B(aq)]/1 (solubility) Melting/freezing A(s,Xs,A) A(liquid,Xliq,A ) X liq,A/Xs,A Answer: All of the above! For example, Henry’s Law constant is actually the PARTITION COEF. of a gas between vacuum and liquid 5 At equilibrium the activity of a solute or solvent will be the same in ALL phases if the same standard state is used in each phase Phase 1 Phase 2 Phase 3 A A A A B B B B If A in equil between 1 and 2 and between 2 and 3 THEN also at equil between 1 and 3, even if not touching 6 Gibbs phase rule: F= C – P + 2 F= number of “free” intensive variables =how many variables (e.g.,T, p, conci) may be changed without changing the number of phases Phase 1 Phase 2 Phase 3 C = number of “components” A A A A B B B B P = number of phases C(P-1) equilibrum constants (constraints) why P-1 ? Because K13 =K12 x K23 P(C-1) + 2 variables why C-1? what are the 2? Because the mole fractions must add to 1 F = number of extra variables = variables – constraints F = P(C-1) – C(P - 1) + 2 F = PC –P – PC + C + 2 = C – P + 2 7 Gibbs phase rule: F= C – P + 2 Example 10 independent things dissolved in water P=1 C=10 F = 10-1 + 2 = 11 = 9 of the 10 things , temperature and pressure. If you lower the temperature until one of the things X precipitates, number of phases is suddenly 2 F becomes 10 because you can no longer control the concentration of X. Why? because of the equilibrium X(s)Xaq) If you try to add more X(aq) it would just precipitate and adding more X(s) just makes more X(s), which changes nothing 8 Shaking will create an emulsion: like mayonnaise. 9 Take a different route 10 At equilibrium the activity of a solute or solvent will be the same in ALL phases if the same standard state is used in each phase 11 12.
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