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Ideal solution
The document discusses the thermodynamics of ideal solutions, focusing on the definitions, mathematical formulations, and laws governing the behavior of solutions. Key concepts include mole fractions, Raoult’s law, and the change in free energy, enthalpy, and entropy during mixing processes. The implications of these principles for various types of solutions, including solid in liquid, liquid in liquid, and gas in liquid, are also examined.
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Ideal solution
- 1. Thermodynamics of idealsolution Dr.P.GOVINDARAJ Associate Professor & Head , Department of Chemistry SAIVA BHANU KSHATRIYA COLLEGE ARUPPUKOTTAI - 626101 Virudhunagar District, Tamil Nadu, India
- 2. Thermodynamics of idealsolution Solution Solvent + Solute = Solution (large amount) (small amount) Mole fraction Let n1 moles of solvent and n2 moles of solute present in the solution Total number of moles = n1 + n2 Mole fraction of solute = n1 n1+n2 = 𝑥1 Mole fraction of solvent = n2 n1+n2 = 𝑥2 Total mole fraction of the solution = n1 n1+n2 + n2 n1+n2 = n1+n2 n1+n2 = 1
- 3. Types of solution 1.Solid in liquid solution Ex: NaCl solution (solute) (solvent) 2. Liquid in liquid solution Ex: dil HCl (solute) (solvent) 3. Gas in liquid solution Ex: Soda drinks (solute) (solvent) Ideal solution Let A liquid + B liquid → Solution • The above solution is ideal solution if the solution obey Raoult’s law at all conditions Thermodynamics of ideal solution
- 4. • Mathematical statementof Raoult’s law is 𝑃𝐴 = 𝑥 𝐴 𝑃𝐴 𝑜 𝑃𝐵 = 𝑥 𝐵 𝑃𝐵 𝑜 Where 𝑃𝐴 ----- Partial pressure of A in solution 𝑃𝐴 𝑜 ----- Vapour pressure of pure A 𝑥 𝐴 ----- Mole fraction of A in solution 𝑃𝐵 ----- Partial pressure of B in solution 𝑃𝐵 𝑜 ----- Vapour pressure of Pure B 𝑥 𝐵 ----- Mole fraction of B in solution Thermodynamics of ideal solution
- 5. It is thepressure of the vapour which is equilibrium with its liquid state Vapour pressure Thermodynamics of ideal solution Vapour Liquid Liquid ⇌ Vapour
- 6. Change of freeenergy of mixing for an ideal solution • Consider a formation of solution nAA + nB B Solution (∆G) mixing = (Free energy of solution) – (Sum of free energies of the pure A and B) = G solution – (GA + GB ) --------(1) • Since G = (n1μ1 + n2μ2), equation (1) becomes (∆G) mixing = (nA μA + nB μB) – (nA μ 𝐴 𝑜 + nB μ 𝐵 𝑜 ) (∆G) mixing = nA (μA – μ 𝐴 𝑜 ) + nB (μB – μ 𝐵 𝑜 ) ---------(2) Thermodynamics of ideal solution • The change in free energy for the formation of solution by mixing nA moles of A and nB moles of B is
- 7. where μA and μBare the Chemical potential of A and B in solution μ 𝐴 𝑜 and μ 𝐵 𝑜 are the Chemical potential of pure A and pure B • Substitute μ = μo + RT ln a in equation (2) (∆G) mixing = nA (μ 𝐴 𝑜 + RT ln aA – μ 𝐴 𝑜 ) + nB (μ 𝐵 𝑜 + RTln aB – μ 𝐵 𝑜 ) (∆G) mixing = nA RT ln aA + nB RT ln aB ---------(3) • For an ideal solution, the activity of each component (aA & aB )is equal to its mole fraction Thermodynamics of ideal solution i.e., aA = 𝑥A & aB = 𝑥 B
- 8. • So, theequation (3) becomes (∆G) mixing = nA RT ln 𝑥A + nB RT ln 𝑥B ---------(4) • If the solution contains more than two components, equation (4) becomes (∆G) mixing = nA RT ln 𝑥A + nB RT ln 𝑥B + nC RT ln 𝑥C + ….. (∆G) mixing = RT ni ln 𝑥i Enthalpy change of mixing for an ideal solution • The change in free energy for the formation of solution by mixing nA moles of A and nB moles of B is (∆G) mixing = nA RT ln 𝑥A + nB RT ln 𝑥B where 𝑥A and 𝑥A are the mole fraction of A and B in solution (∆G) mixing = T(nA R ln 𝑥A + nB R ln 𝑥B ) (∆G) mixing 𝑇 = (nA R ln 𝑥A + nB R ln 𝑥B ) Thermodynamics of ideal solution
- 9. • At constantpressure (∆G) mixing 𝑇 p = (nA Rln xA + nB Rln xB ) -------(1) • Differentiate equation (1) w.r.t Temperature 𝜕{∆Gmixing /𝑇} 𝜕𝑇 p = 0 ------(2) 𝑇 𝜕(∆G) mixing 𝜕𝑇 p− (∆G) mixing 𝑇2 = 0 ------(3) • We know that the Gibbs- Helmholtz equation is ∆G - ∆H = 𝑇 𝜕(∆G) 𝜕𝑇 p ------(4) Thermodynamics of ideal solution
- 10. • Substitute (5)in (3) we get ∆Gmixing− ∆ Hmixing− (∆G) mixing 𝑇2 = 0 − ∆ Hmixing 𝑇2 = 0 ∆ Hmixing = 0 • i.e., if two pure liquids are mixed together, the entropy change (∆H mixing ) is zero Thermodynamics of ideal solution ∆Gmixing - ∆Hmixing = 𝑇 𝜕(∆G)mixing 𝜕𝑇 p ------(5) • For the formation of solution by mixing of two liquids, the equation (4) becomes
- 11. • Since ∆Hmixing = 0 for ideal solution, equation (2) becomes ∆G mixing = - T∆S mixing ------(3) • Substitute ∆G mixing = nA RT ln 𝑥 A + nB RT ln 𝑥 B in equation (3) we get nA RT ln 𝑥 A + nB RT ln 𝑥 B = - T∆S mixing - (nA R ln 𝑥 A + nB R ln 𝑥 B ) = ∆S mixing • For more than two components ∆S mixing = - (nA R ln 𝑥 A + nB R ln 𝑥 B + nC R ln 𝑥 C +… ) ∆S mixing = - R niln𝑥i Thermodynamics of ideal solution Entropy change of mixing for an ideal solution • As per the second law of thermodynamics the relation between ∆G , ∆H , ∆S is ∆G = ∆H - T∆S ------(1) • For mixing of two liquids for making ideal solution, equation (1) becomes ∆G mixing = ∆H mixing - T∆S mixing ------(2)
- 12.