Ideal solution

Thermodynamics of ideal solution - change in free energy - change in enthalpy - change in entropy of mixing for an ideal solution

1. Thermodynamics of ideal solution Dr.P.GOVINDARAJ   Associate Professor & Head , Department of Chemistry SAIVA BHANU KSHATRIYA COLLEGE ARUPPUKOTTAI - 626101 Virudhunagar District, Tamil Nadu, India

2. Thermodynamics of ideal solution Solution Solvent + Solute = Solution ( large amount) ( small amount) Mole fraction Let n 1 moles of solvent and n 2 moles of solute present in the solution Total number of moles = n 1 + n 2 Mole fraction of solute = = 1   Mole fraction of solvent = = 2   Total mole fraction of the solution = + = = 1  

3. Types of solution Solid in liquid solution Ex: NaCl solution (solute) ( solvent) Liquid in liquid solution Ex: dil HCl (solute) ( solvent) 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 statement of Raoult’s law is =   =   Where ----- Partial pressure of A in solution ----- Vapour pressure of pure A ----- Mole fraction of A in solution ----- Partial pressure of B in solution ----- Vapour pressure of Pure B ----- Mole fraction of B in solution   Thermodynamics of ideal solution

5. It is the pressure of the vapour which is equilibrium with its liquid state Vapour pressure Thermodynamics of ideal solution Vapour Liquid Liquid ⇌ Vapour

6. Change of free energy of mixing for an ideal solution Consider a formation of solution n A A + n B B Solution (∆G) mixing = (Free energy of solution) – (Sum of free energies of the pure A and B) = G solution – (G A + G B ) --------(1) Since G = (n 1 μ 1 + n 2 μ 2 ), e quation (1) becomes (∆G) mixing = ( n A μ A + n B μ B ) – ( n A + n B )   (∆G) mixing = n A ( μ A – ) + n B ( μ B – ) ---------(2)   Thermodynamics of ideal solution The chang e in free energy for the formation of solution by mixing n A moles of A and n B moles of B is

7. where μ A and μ B are the Chemical potential of A and B in solution and are the Chemical potential of pure A and pure B   Substitute μ = μ o + RT ln a in equation (2) (∆G) mixing = n A ( + RT ln a A – ) + n B ( + RTln a B – )   (∆G) mixing = n A RT ln a A + n B RT ln a B ---------( 3) For an ideal solution, the activity of each component ( a A & a B ) is equal to its mole fraction Thermodynamics of ideal solution i.e., a A = A & a B = B  

8. So, the equation (3) becomes (∆G) mixing = n A RT ln A + n B RT ln B ---------(4)   If the solution contains more than two components, equation (4) becomes (∆G) mixing = n A RT ln A + n B RT ln B + n C RT ln C + …..   (∆G) mixing = RT   Enthalpy change of mixing for an ideal solution The change in free energy for the formation of solution by mixing n A moles of A and n B moles of B is (∆G) mixing = n A RT ln A + n B RT ln B where A and A are the mole fraction of A and B in solution ( ∆G) mixing = T( n A R ln A + n B R ln B ) = ( n A R ln A + n B R ln B )   Thermodynamics of ideal solution

9. At constant pressure p = ( n A Rln x A + n B Rln x B ) -------(1)   Differentiate equation (1) w.r.t Temperature p = 0 ------(2)   = 0 ------( 3 )   We know that the Gibbs- Helmholtz equation is ∆ G - ∆ H = ------(4)   Thermodynamics of ideal solution

10. Substitute (5) in (3) we get = 0   = 0   = 0   i.e., if two pure liquids are mixed together, the entropy change ( ∆H mixing ) is zero Thermodynamics of ideal solution ∆ G mixing - ∆H mixing = ------( 5 )   For the formation of solution by mixing of two liquids, the equation (4) becomes

11. Since ∆H mixing = 0 for ideal solution, equation (2) becomes ∆G mixing = - T∆S mixing ------(3) Substitute ∆G mixing = n A RT ln A + n B RT ln B in equation (3) we get   n A RT ln A + n B RT ln B = - T∆S mixing   - ( n A R ln A + n B R ln B ) = ∆ S mixing   For mor e than two components ∆ S mixing = - ( n A R ln A + n B R ln B + n C R ln C +… )   ∆ S mixing = - R   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)

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