PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Separation Methods Based on Distributions in Discrete Stages (9/20/13)1. Chemical Separations: The Big Picture Classification and comparison of methods2. Fundamentals of Distribution Separations3. Separation Methods Based on Distributions in Discrete Stages Such as solvent extraction and distillation4. Introduction to Distribution Separations in chromatographic methods. The plate theory, the rate theory; van Deemter's equation.1Counter-Current ExtractionSeparation of phasesExtraction 2Addition of fresh phases toBoth phase 1 and 2Extraction 3[A] = 0.01 M[B] = 1 MPhase 1Phase 2Phase 1Extraction 1V1=V2=10 mLDcA= 10, DcB=0.1Separation of phases1211111122112211221221 122Counter-Current ExtractionPhase 1Phase 2[A] = 0.01 M[B] = 1 MPhase 1Extraction 1Separation of phasesExtraction 2Addition of fresh phases toBoth phase 1 and 2111111221122V1=V2=10 mLDcA= 10, DcB=0.1fA1,1=0.091fB1,1=0.909fA2,1=0.909fB2,1=0.091fA2,2=0.826fB2,2=0.008fA1,2=0.008fB1,2=0.826fA1N,2=0.083fB1N,2=0.083Total A =0.909Total B =0.091Total A = 0.091Total B = 0.909fA2N,2=0.083fB2N,2=0.083321 12fA2,2=0.826fB2,2=0.008fA2N,3=0.151fB2N,3=0.015Separation of phases1122fA2,2=0.826fB2,2=0.008fA1,2=0.008fB1,2=0.826Extraction 31212fA2,2=0.826fB2,2=0.008fA1,2=0.008fB1,2=0.826fA1N,2=0.083fB2N,2=0.083fA2N,2=0.083fB2N,2=0.083Total A =0.166Total B =0.166fA2N,3=0.151fB2N,3=0.015fA1N,3=0.015fB1N,3=0.151fA1N,3=0.015fB1N,3=0.151Counter-Current Extraction[A] = 0.01 M[B] = 1 MV1=V2=10 mLDcA= 10, DcB=0.1fA1,2=0.008fB1,2=0.8264Counter-Current Extraction= 0.298=29.8%21 12fA2,2=0.826fB2,2=0.008fA2N,3=0.151fB2N,3=0.015fA1N,3=0.015fB1N,3=0.151fA1,2=0.826fB1,2=0.008Results: recovery of A in phase 2 = 0.826+0.151=0.977=97.7%Final purity of A in phase 2 =(0.01M)*(0.01L)*(0.977)(0.01M)*(0.01L)*(0.977) + (1.00 M)*(0.01L)*(0.023)Purification yield = 0.2980.0099= 30Recovery of A in phase 2 = 0.909 = 90.9%Final purity of A in phase 2= 0.091 (9.1%)Purification yield of A= 9.299.2%5.4%5.45One-stepTwo-stepCounter-current97.7%29.8%305F. Craig Apparatus and Craig Countercurrent distributionLyman C. Craig, Ph.D.Albert Lasker Award(1) Counter-current extraction are useful in that they improve both the recovery and purification yield of A. However, the technique is time-consuming and tedious to perform.(2) To overcome these difficulties L. C. Craig developed a device in 1994 to automate this method. Known as the Craig Apparatus, this device uses a series of “separatory funnels” to perform a counter-current extraction. The patern formed by the movement of a solute through the system is known as a counter-current distribution. 6Extraction 1Transfer 1aTransfer 1bExtraction 2Transfer 2Extraction 37http://www.chem.uoa.gr/Applets/AppletCraig/Appl_Craig2.html8(4) The result of this process is that solutes partition between the phases in each tube, but eventually all travel to the right and off of the apparatus, where they are collected. (5) Since this system involves both rate and phase separation processes (i.e., distribution of solutions between two phases affecting their rate of travel through the system), The Craig countercurrent distribution is often as a simple model to describe chromatography. In fact, anther term often used for countercurrent distribution is countercurrent chromatography (CCC). The Essence of Chromatography: p889 ~ 893 9H. Theory of Countercurrent distribution:(1) As in simple extraction, the distribution of A in any tube can be calculated based on it concentration distribution ratio, wherefphase1=(1 + Dc V2/V1)1fphase2 = 1- fphase1,1(fraction of A not removed from phase 1)(fraction of A extracted into phase 2)(2) In describing the Craig distribution, the terms fphase1 and fphase2 are often replaced with the terms q and p, whereq = fphase1=(1 + Dc V2/V1)1p = fphase2 = 1- fphase1,1= 1 - q(3) The ratio of q/p (i.e., the ration of the fraction (or moles) of A in the stationary phase to the faction (or moles) of A in the mobile phase at equilibrium) is known as the capacity factor k. k’ = p/q= mole Amobile phase/moles Astationary phase10(4) The equation for k = q/p may also be rewritten in terms of p and q, where p = k’/(1 + k’)q = 1/(1+k’)(5) k and concentration distribution ratio (Dc) are related by the expression k’ = Dc V2/V1In other works, k’ is another way to describe the distribution of A between two phase. Dc and k only differ in that k is based on the moles of A present rather than its concentration. For this reason, k’ is sometimes referred to as the mass distribution ratio.(6) The use of k’ to describe the distribution of a solute is particularly valuable in situations where the exact volumes of the mobile and stationary phases are not known. One common example of this is on chromatography (k=1/k’).(7) The value of k’, or p and q, can also be used to describe the distribution of a solute A in the Craig apparatus.11qDevelopment of solute distribution in Craig ApparatusExtraction 1p0 1234567Distribution of solutionTotal amountof solute1Transfer 1pqq(q+p)Total amountof solute in each stagepExtraction 2p2q2qpqp12Extraction 3p3q32q2p2qp2q2pqp2q3Transfer 3p32q2p2qp2qp2q2pTransfer 2p2q2qpqpDistribution of solution0 1234567(q+p)2Total amountof solute in each stageq22qpp2(q+p)3Total amountof solute in each stageq33q2p3q2pp313Development of solute distribution in Craig ApparatusTransfer 3Transfer 2Transfer 1pqDistribution of solution(q+p)p2q2qpqp(q+p)2p3q32q2p2qp2qp2q2p(q+p)30 123456714(8) The distribution of A in this system after r transfers is given by the binomial expression of the equation (q + p)r =1Where: (q+p)1 = q + p(q+p)2 = q2 + 2 qp + p2(q+p)3 = q3 + 3 q2p + 3qp2 + p3, etc(9) After given number of transfers (r), the relative amount of A in any tube n is Pr,n =r!n! (r-n)!pn qr-nWhere: Pr,n = Fraction of A in tube n after transfer r.Transfer 30 1234567(q+p)3q33q2p3q2pp3Mobile phaseStationary phaseGood news: We can get the distribution of solute among Craig tubes (chromatographic column)Bad news: give no distribution shape and position.http://www.chem.uoa.gr/Applets/AppletCraig/Appl_Craig2.html15http://www.chem.uoa.gr/Applets/AppletCraig/Appl_Craig2.html(10) The binomial can be expended as Gaussian distribution when n larger than 20 (rpq>3).Pr,n =2πrqp*1Exp [-(n-rp)2/2rpq)]Where: Pr,n =