Liquid-Liquid Equilibrium and Phase Diagram PDF [PDF]

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Ayu Dahliyanti, M.Eng. Physical Chemistry



Chemical Engineering Universitas Pertamina



BINARY LIQUIDLIQUID & LIQUID-SOLID EQUILIBRIA



OUTLINE



OBJECTIVES



• Binary Liquid – Liquid phase diagram for partially miscible liquid • Lever rule



• Binary Liquid – Solid phase diagram



REFERENCES



• •



• To interpret the phase change using binary liquid-liquid and solid-liquid phase diagram. • To determine the composition and amount of each phase based on binary liquid-liquid and solid-liquid phase diagram



Atkins & de Paula. Atkin’s Physical Chemistry, 10th Ed– chapter 5C & 5D Levine, Physical Chemistry, 6th Ed – Chapter 12



LIQUID-LIQUID EQUILIBRIUM Phase Diagram Tc • When two partially miscible liquid (A and B) are mixed together, the will form two separate phases until it exceeds critical solution temperature (Tc)



One phase region



One aa



B-rich (A-rich) Two phase region



• First phase will be A-rich • Second phase will be B-rich • Above Tc, two liquid will form one phase (solution). Temperature – composition diagram for A-B liquid-liquid equilibrium at constant P



LIQUID-LIQUID EQUILIBRIUM Phase Diagram



Systems with lower Tc (a) and both upper and lower Tc (b)



LIQUID-LIQUID EQUILIBRIUM



Nitrobenzene (C6H5NO2) – Hexane (C6H14) system at 1 atm



Lever Rule



Where: nα = amount of phase α (in mole or mass) nβ = amount of phase β (in mole or mass) l α = length of line l α l β = length of line l β Example: When xN=0.41 at 290 K; exist two phases with compositions xN



xH



phase α



0.35



0.65



phase β



0.83



0.17



Self test. Repeat the calculation for 50 g of hexane and 100 g of nitrobenzene at 280 K!



Nitrobenzene (C6H5NO2) – Hexane (C6H14) system at 1 atm



LIQUID-LIQUID EQUILIBRIUM Lever Rule Self test. Repeat the calculation for 50 g of hexane and 100 g of nitrobenzene at 280 K! Solution: Calculate xN=



100Τ 123.11 100Τ 50 123.11+ Τ86.18



= 0.58



When xN=0.58 at 280 K; exist two phases with compositions phase α (hexane rich) phase β (nitrobenzene rich)



xN



xH



0.1



0.9



0.92



0.08



𝑛𝛼 0.92 − 0.58 = = 0.71 𝑛𝛽 0.58 − 0.1 Since the total mol in the mixture is 1.4 mol: 𝑛𝛼 = 0.71𝑛𝛽 → 𝑛𝛼 = 0.71(1.4 − 𝑛𝛼 ) 𝑛𝛼 =0.58 mol; 𝑛𝛽 = 0.82 mol



LIQUID-LIQUID EQUILIBRIUM Lever Rule Lever rule is also applicable for other phase diagram (eg. Vapor liquid equilibrium phase diagram) Self test. At 80oC there are 5 mol methanol and 5 mol water in equilibrium. How many phases are present in the system? Find the amount (in mole) and composition of each phase. Xmethanol = 0.5 ; two phase L-V mixture xM



xW



phase α (liquid)



0.23



0.77



phase β (vapor)



0.6



0.4



𝑛𝑙𝑖𝑞𝑢𝑖𝑑 0.6 − 0.5 = = 0.37 𝑛𝑣𝑎𝑝𝑜𝑟 0.5 − 0.23 Since the total mol in the mixture is 10 mol: 𝑛𝛼 = 0.37𝑛𝛽 → 𝑛𝛼 = 0.37(10 − 𝑛𝛼 ) 𝑛𝛼 = mol; 𝑛𝛽 = mol