Experiment 6: Fractional Distillation Reading Assignment –Experiment 6 (pp. 58 -64) –Operation 29.

Slides:



Advertisements
Similar presentations
Gas Chromatography.
Advertisements

APPLICATIONS Applications of Raoult’s law
Chapter 12 Solutions.
FRACTIONAL DISTILLATION ORG I LAB Dr. W. J. KELLY.
Phase Equilibrium II - Two Component System. How many components and phases in this system? 2 components and 1 liquid phase Method to separate ethanol.
Chemistry.
DISTILLATION.
1 Lecture 7 Ch 6: Distillation & Boiling Points This Week In Lab: Ch 5: Extraction, Procedure 2 Ch 4 Final Report Due Next Week in Lab: Ch 6 PreLab due.
Colligative Properties Vapour pressure Boiling point Freezing point Osmotic pressure.
Colligative Properties are those properties of a liquid that may be altered by the presence of a solute. Examples vapor pressure melting point boiling.
Molality and Mole Fraction b In Chapter 5 we introduced two important concentration units. 1. % by mass of solute 2. Molarity.
SOLUTIONS SUROVIEC SPRING 2014 Chapter 12. I. Types of Solution Most chemical reaction take place between ions/molecules dissolved in water or a solvent.
Section 13.4 Colligative Properties Bill Vining SUNY Oneonta.
Phase diagram of Carbon Allotropes: elemental substances that occur in more than one crystalline form.
Chapter 6 PHASE EQUILIBRIA
Figure Figure 2 Chem 334 Expt. #3
Separating Azeotropic Mixtures CHEN 4460 – Process Synthesis, Simulation and Optimization Dr. Mario Richard Eden Department of Chemical Engineering Auburn.
Gas-Solution Processes Gas Solubility Raoult’s Law Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology.
Distillation Underlying Principles of Distillation
General Chemistry Principles & Modern Applications 9 th Edition Petrucci/Harwood/Herring/Madura Chapter 13 Solutions and their Physical Properties Dr.
Please Pick Up a Copy of Raoult’s Law and Vapor Pressure problem set Distillation problem set.
PHASE EQUILIBRIUM (GROUP SECOND) 1) Muhammad Shafique 2) Noor Yussuf 3) Noor Nadia Syahira 4) Norfaizah 5)Nur Alya 6) Nur Azahariah 7) Wan Muhammad Hakimi.
Bubble Point and dew point Calculations
Designing a Separations Process Without VLE Data by Thomas Schafer - Koch Modular Process Systems, LLC This presentation utilizes as it’s example a problem.
Chemical Thermodynamics 2013/ th Lecture: Mixtures of Volatile Liquids Valentim M B Nunes, UD de Engenharia.
Phase Equilibria Melting-Freezing Evaporation-Condensation Sublimation-Condensation Phase transition.
Chemistry Thermodynamics Lecture 10 : Phase Diagrams and Solubility Lecture 11 : Solubility (cont.) Lecture 12: Kinetic Coefficients for the Relaxation.
Separation Science Differences in IMFs can be used to separate chemical substances for further analysis. Differences in IMFs can be used to separate chemical.
Vapor Pressure of Solutions Chapter 13 Part 3. Vapor Pressure The pressure of the vapor present. Vapor is the liquid molecule in gas form over the liquid.
Chapter 12 Solutions. From Chapter 1: Classification of matter Matter Homogeneous (visibly indistinguishable) Heterogeneous (visibly distinguishable)
Vapor-Liquid Equilibrium (VLE) at Low Pressures
Experiment 2 DISTILLATION AND GAS CHROMATOGRAPHY OF ALKANES.
Pure Substances Elements
VAPOR PRESSURE The term "vapor" is applied to the gas of any compound that would normally be found as a liquid at room temperature and pressure For example,
Distillation ... A Separation Method.
Ethylene glycol, HOCH 2 CH 2 OH, boils at 198 o C and melts at -13 o C. What happens to the melting point of water if you add antifreeze? Melting point.
Colligative Properties. _______________ – physical properties of solutions that are affected only by the number of particles NOT the identity of the solute.
Effect of Pressure Copyright © 2011 Pearson Canada Inc. Slide 1 of 46 General Chemistry: Chapter 13 William Henry found that the solubility of a gas increases.
Colligative Properties of solutions A colligative property is that which depends on the relative number of solute and solvent molecules.
THERMODYNAMICS OF SEPARATION OPERATIONS
The Simplest Phase Equilibrium Examples and Some Simple Estimating Rules Chapter 3.
Distillation... A Separation Method. Background Concepts - Definitions Vapor Pressure – Gas pressure created by the molecules of a liquid which have acquired.
Molality and Mole Fraction Modified from: Chem%20102%20week%202.ppt Molality is a concentration unit based.
R AOULT ’ S L AW The partial vapour pressure of a component in a mixture is equal to the vapour pressure of the pure component at that temperature multiplied.
June 16, 2009 – Class 43 and 44 Overview
DISTILLATION PROCESS.
Vapor Pressure. Pressure of the vapor present when equilibrium is achieved between the rate of vaporization and the rate of condensation. At the boiling.
Multicomponent systems
Fractional Distillation Boiling point of mixtures Separation of mixtures by distillation.
Liquid-Vapor Phase Equilibrium Distillations will be performed in hood.
THERMODYNAMICS OF SEPARATION OPERATIONS
APPLICATIONS Applications of Raoult’s law
Introduction to phase equilibrium
1. Write down the vapor-liquid equilibrium (VLE) equations for a binary system assuming that the vapor phase is ideal and the liquid phase follows Raoult’s.
Noorulnajwa Diyana Yaacob PPK Bioproses Universiti Malaysia Perlis MULTIPHASE SYSTEM.
1 Classification and purification of Organic Compounds.
 During the dissolving process the solvent and solute go from a state of order, separate and particles arranged regularly, to a state of disorder. 
Chapter 11 Properties of Solutions. From Chapter 1: Classification of matter Matter Homogeneous (visibly indistinguishable) Heterogeneous (visibly distinguishable)
Which of the following concentration measures will change in value as the temperature of a solution changes? a) Mass percent b) Mole fraction c) Molality.
FRACTIONAL DISTILLATION L.E. College, Morbi Chemical Engineering Prepared by: Mehul Patel ( ) Sub: Organic Chemistry and Unit Process ( )
Chapters 8-9 Phase Diagrams
Ideal solutions for 3rd semester By T.Sweta Department of chemistry
Colligative Properties
By Moizul Hasan Assistant Professor
Boiling Points - Distillations
Distillation.
Lecture Notes Week 1 ChE 1008 Spring Term (03-2).
MASS TRANSFER II DISTILLATION.
Colligative Properties
Chapter 13 Solutions.
Presentation transcript:

Experiment 6: Fractional Distillation Reading Assignment –Experiment 6 (pp ) –Operation 29

Key Point! When conducting a distillation, the vapor should be richer in the lower boiling component than what you started with.

Simple Distillation: Apparatus Put in boiling stone!

Temperature Behavior During Distillation A.Single pure component B.Two components of similar boiling points C.Two components with widely different boiling points

Phase Diagram: Two Component Mixture of Liquids

Questions based upon the previous slide: a)What is the bp of pure A? b)What is the bp of pure B? c)What is the bp of a solution with the composition of 30 % B, assuming a simple distilllation apparatus? d) What is the composition of the vapor assuming a simple distillation apparatus? e) What is the composition of the distillate collected assuming a simple distillation apparatus? f) What does the “tie-line,” x-y represent? Hint: the upper curve is the vapor curve and the lower curve is the liquid curve. “Composition of the vapor and liquid that are in equilibriuim with each other at 130 o C.”

Vapor-Liquid Composition Curve (Benzene vs. Toluene) Vapor liquid

Questions based upon the previous slide: a)What is the bp of pure toluene? b)What is the bp of pure benzene? c)What is the bp of a solution with the composition of 50 % benzene, assuming a simple distilllation apparatus? d) What is the composition of the distillate assuming a simple distillation apparatus? e)How many theoretical plates would be necessary for a fractional distillation starting with a 50 % benzene solution?

When will simple distillation do a reasonable job of separating a mixture? 1)When the difference in boiling points is over 100 o 2)When the there is a fairly small amount of impurity, say less than 10 %. 3) When one of the components will not distil because of a lack of volatility (i.e. sugar dissolved in water).

Raoult’s Law

Fractional Distillation: Apparatus Put in boiling stone

Temperature vs. Volume: Fractional Distillation

Fractional Distillation Phase Diagram

How many theoretical plates are need to separate a mixture starting at L? Looks like about 5 plates are needed to separate the mixture on the previous slide! Count the “tie-lines” (horizontal lines) to come up with the 5 plates (labelled with arrows on the next slide)!

Fractional Distillation Phase Diagram. The arrows indicate a theoretical plate!

Theoretical Plates Required to Separate Mixtures based on BP Boiling Point Difference Theoretical Plates

Azeotrope Some mixtures of liquids, because of attractions or repulsions between the molecules, do not behave ideally These mixtures do not obey Raoult’s Law An azeotrope is a mixture with a fixed composition that cannot be altered by either simple or fractional distillation An azeotrope behaves as if it were a pure compound, and it distills from beginning to end at a constant temperature.

Types of Azeotropes There are two types of non-ideal behavior: –Minimum-boiling-point Boiling point of the mixture is lower than the boiling point of either pure component –Maximum-boiling-point Boiling point of the mixture is higher than the boiling point of either pure component

Maximum Boiling-Point Azeotrope

Observations with maximum boiling azeotrope On the right side of the diagram: Compound B will distill (lowest bp). Once B has been removed, the azeotrope will distill (highest bp). On the left side of the diagram: Compound A will distill (lowest bp) Once A has been removed, the azeotrope will distill. (highest bp) The azeotrope acts like a pure “compound”

Minimum Boiling-Point Azeotrope

Observations with minimum boiling azeotrope On the right side of the diagram: The azeotrope is the lower boiling “compound,” and it will be removed first. Pure ethanol will distill once the azeotrope has distilled. On the left side of the diagram: the azeotrope is the lower boiling “compound,” and it will distill first. Once the azeotrope has been removed, then pure water will distill. The azeotrope acts like a pure “compound”

The Gas Chromatograph

Gas Chromatography: Separation of a Mixture

Gas Chromatogram Lowest b.p. Highest b.p. Retention time

Triangulation of a Peak

Sample Percentage Composition Calculation

Gas Chromatography: Results In a modern gas chromatography instrument, the results are displayed and analyzed using a computerized data station. It is no longer necessary to calculate peak areas by triangulation; this determination is made electronically. Our analysis will be conducted on a modern data station.

Compounds in mixture: boiling points. Cyclohexane80 o C Toluene110 o C Mixture separates by distillation according to the boiling point. Compounds with the lower bp come off first! The same is true on the gas chromatographic column; the lower boiling compound comes off first!

How to identify the components in your unknown mixture Use the retention time information from your gas chromatograms to provide a positive identification of each of the components in the mixture. Don’t rely on the distillation plot to determine the composition of your mixture!

First Fraction: Cyclohexane/Toluene Chromatogram Solvents cyclohexane toluene

Data: Cyclohexane/Toluene First Fraction solvents cyclohexane toluene ?

Calculation of percentages from the data for fraction 2 area counts/response factor = adjusted area Cyclohexane area = 42795/1.133 = Toluene area = 18129/1.381 = Total area Note: this calculated area is different than that shown on the data sheet! Use this calculated area! Percent cyclohexane = 32104/45231 x 100 = 71.0% Percent toluene = 13127/45231 x 100 = 29.0 % Round off numbers so that the total equals 100%

Second Fraction: Cyclohexane/Toluene Chromatogram solvents cyclohexane toluene

Data: Cyclohexane/Toluene Second Fraction solvents cyclohexane toluene ?

Calculation of percentages from the data for fraction 4 area counts/ response factor = adjusted area Cyclohexane area = 57546/1.133 = Toluene area = /1.381 = Total area Note: this calculated area is different than that shown on the data sheet! Percent cyclohexane = 43170/ x 100 = 23.7 % Percent toluene =138981/ x 100 = 76.3 % Round off numbers so percentage = 100%