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Surface and Interfacial Phenomena

Surface and Interfacial Phenomena. Surface is a boundary between solid or liquid phase and a vacuum. Interface is a boundary between two phases. Surface free energy is the work required to increase the area of the surface by 1cm2.

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Surface and Interfacial Phenomena

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  1. Surface and Interfacial Phenomena Surface is a boundary between solid or liquid phase and a vacuum. Interface is a boundary between two phases. Surface free energy is the work required to increase the area of the surface by 1cm2.

  2. Surface Tension is the force in dynes acting along the surface of a liquid at right angle to any line 1 cm in length. γ= W/ ΔA whereγis the surface tension , W is the work in erg required to generate one square cm of surface A. The units of γ iseither erg/cm2 or Dyne / cm As an erg is equal to a dyne.cm Surface tension is equal numerically and dimensionally to surface energy.

  3. The force of tension existing at the interface between two immiscible liquids is known as interfacial tension . There are two forces that affect the behavior of liquid on a solid: 1- Force of cohesion between the molecules of a liquid. 2- Force of adhesion between the liquid and solid. When the adhesion forces are stronger than the cohesion forces ,the liquid spread over the solid. e.g.water on glass.and the angle known as contact angle is smaller than 90.And vice versa.

  4. Factors affecting Surface Tension 1- Nature of the liquid. 2- Temperature : Surface tension decrease with increase of temp.,i.e.with an increase of kinetic energy of the molecules. γ = γ0(1- T/Tc)1.2 3- Solute The quantitative relationship between solute distribution and surface tension is expressed By Gibbs Adsorption Isotherm. Г = -c/RT. dγ/dc

  5. d Measurement of surface Tension 1- Thecapillary Rise Method: r2hdg= 2r γcosѲ Γ= rhdg/2 or ½ rhdg 2- The Drop weight Method: mg= 2r γ For relative measurement ,it easier to determine the No of drops formed by a given volume instead of determining the weight of drops. γ1/ γ2 = n2d1/n1d2

  6. 3-Ring Method: mg= 2∏ γ (r1+ r2) Spreading coefficient (s). S= Wa-Wc S= γs- (γL+ γLS) Example : if the surface tension of water is 72.8 dyne/cm , the surface tension of benzene is 28.9 and interfacial tension between them is 35.What is the initial spreading coefficient ? S = 72.8- (28.9+ 35) = 8.9 dyne/ cm.

  7. Adsorption Is the accumulation of a substance over the surface of another substance. e.g. Adsorption of alkaloid on charcoal. Absorption The penetration of a substance through another substance. e.g.water and sponge.

  8. The Characteristics of Physical Adsorption and Chemisorption Physical Adsorption Chemisorption 1- Involve sharing of electrons between adsorbent and adsorbed molecules . 2-non specific . 3- Irreversible. 4- Surface reaction only proceeds above certain temp. 5- Restricted to monolayer. 6-Rate increase rapidly by increase temp. 1- weak physical forces. 2- Non specific. 3- Reversible. 4- Process is exothermic.Amount of adsorption decreases with rise of temp. 5- More than one layer. 6- usually rapid at all temp.

  9. Adsorption Isotherm The relationship between the amount of gas (adsorbate ) physically adsorbed on a solid and equilibrium pressure or concentration at constant temperature. A- freundlich Adsorption Isotherm. Y= x/m= kc1/n Log x/m= log k + 1/n log c

  10. B- Langmuir Adsorption Isotherm. This equation is based on the theory that the molecules of the gas are adsorbed on active sites of the solid to form a layer one molecule thick. The fraction of the centers occupied by gas at pressure P is Ѳ . The fraction of the sites not occupied is 1- Ѳ Rate of adsorption r1 = K1(1- Ѳ) P Rate of evaporation r2 = K2Ѳ At equilibrium r1=r2 K1(1- Ѳ) P= K2Ѳ

  11. p/y = 1/b ym + p/ym Factor Affecting Adsorption From Solution : 1- Solute concentration Adsorption is more complete from dilute solution than from concentrated one. 2- Temperature:- the amount of adsorption will decrease as the temp. increase. 3- Surface area of adsorbent:- as surface area increased the amount of adsorption will increase. 4- PH of the solution:- depend on whether ionized or unionized species is already adsorbed.

  12. 5- Effect of Solvents:- adsorption of the solute will be at maximum if the solvent is inert ( has no affinity for adsorbent or for solute). 6- Removal of adsorbed impurities:- will increase the amount of adsorption e.g. activated charcoal .Heating it at 50-100oc. 7- Adsorbent - solute interaction and Solvent competition:- adsorption of a solute from dilute solution involve breaking solute solvent bond and adsorbent solvent bond and formation of adsorbent –solute bond. Selective adsorption : charcoal will adsorb magenta dye from solution however,ifsaponin is added to the system the dyeis released since saponin is preferentially adsorbed.

  13. Application of adsorption in pharmacy and allied field:- 1- Decolorizing agents. 2- Desiccant and drying agents . 3- Adsorption chromatography. 4- Medical adsorption. 5- Ion exchange. Adsorbent materials: Heavy kaolin, attapulgite ,bentonite , charcoal, diatomite , purified talc, veegum, alumina, silica Magnesium trisilicate ( florisil).

  14. Surface Active Agents SAA Are solute that cause decrease in the surface tension of the solvent. All types of soluble SAA contain: a- Lipophilic group b- Hydrophilic group A suitable balance between two opposing groups of SAA is necessary to ensure that surface active properties are obtained.

  15. Critical Micelle Concentration:- Is the concentration of surfactant that form micelle and at this concentration , dramatic changes in the physicochemical properties of the solution take place e.g. surface tension , conductivity, osmotic pressure. Classification of Surfactants: They classified on the bases of : 1- Chemical structure 2-Uses to which they are put 3- physical propertis

  16. Chemical classification of surfactants: Anionic Surface Active Agents : a- Soaps b- Organic sulfonate c- Organic Sulfate Cationic Surface Active Agents: a- Quaternary Ammonium Salts b- Pyridinium compounds

  17. Non Ionic Surfactants: Are classified according to the type of the linkage between the hydrophilic group and the lipophilic group in the molecules into: A- Ester –linked surfactants. b- Ether - linked surfactants. c- Ester- ether linked surfactants.

  18. Ampholytic Surface Active Agents: 2- Hydrophilic – Lipophilic Classification (HLB) hydrophilelipophile balance HLB value is the relative efficiency of the hydrophilic portion of the surfactant molecule to its lipophilic portion. The HLB system act as a guide to quantify all non –ionic surfactants to handful of general application.

  19. Pharmaceutical Application of Surfactants: Wetting Agents Emulsifying Agents Solubilizing Agents Foaming and Antifoaming Detergents Antibacterial Action of SAA.

  20. Dissolution Describe the process by which the drug particles dissolve. The dissolution of a drug is described in a simplified manner by the Noyes –Whitney equation : dm/ dt= kA(cs- c) dm/dt is the dissolution rate. K is the dissolution rate constant . A is the surface area of dissolving solids . Cs is the solubility of the drug . C is the concentration of drug in the dissolution medium at time t .

  21. K which incorporating the drug diffusion coefficient And diffusion layer thickness. The constant( k) is termed the intrinsic dissolution rate and is a characteristic of each solid drug compound in a given solvent under fixed hydrodynamic conditions.

  22. Buffer solution and Buffer capacity What is a buffer ? What is the use of buffer in the field of pharmacy? pH= pka + log [A- ]/ [HA] Buffer capacity( β) :- The ability of a buffer to withstand the effect of acid and bases . Buffer capacity is equal to the amount of strong acid or strong base, expressed as moles of H+ or oH- ions, required to change the pH of one liter of the buffer by one pH unit.

  23. It is clear that buffer capacity increases as the concentrations of buffer components increases • The capacity is also affected by the ratio of the concentration of weak acid and its salt. • Maximum capacity(β max)is obtained when the ratio of acid to salt = 1 i.e. pH = pka of the acid . • What is the characteristic of a suitable buffer? pKa value of the acid should be close to pH required. Its components must be compatible with other ingredients in the system. Toxicity of buffer components must be taken into account if the solution is to be used for medical purposes.

  24. Colligative Properties It is any physicochemical property of a solution which depend only on the number of dissolved particles and not on the nature of those particles. These include : 1- Vapor pressure lowering. 2- boiling point elevation. 3- Freezing point depression. 4- Osmotic pressure.

  25. Vapor pressure • The vapor pressure of the solution is lower than of pure solvent. Why? • It can be explained by Raoult,s law • P= xsolv x po • Xsolv = mole fraction of solvent. • Po = vapor pressure of pure solvent. • Since Xsolv< 1 in any solution P < P0

  26. Boiling Point : When boiling occur? As mentioned before the solute lower the vapor pressure of the solvent . In order to reach the atmospheric pressure ( boiling point ) . It is necessary to increase temperature. Freezing Point: The presence of solute molecules will interfere with the formation of crystals of the solvent . This change in the freezing process results in a depression of the freezing point for the solution relative to the pure solvent.

  27. Osmotic Pressure: Osmotic pressure of the solution is the external pressure that must be applied to the solution to prevent it being diluted by the entry of solvent via a process that is known as Osmosis. This process refers to the spontaneous diffusion of solvent from a solution of low solute concentration ( or pure solvent) into a more concentrated one through a semi permeable membrane. Such a membrane separate the two solutions and is permeable only to the solvent molecules. This process occurs spontaneously at constant temperature and pressure.

  28. Rate and order of reaction: The rate of a chemical reaction or process is the velocity with which it occurs. Order of reaction refers to the way in which the concentration of drug or reactant influences the rate of a chemical reaction or process. Zero order reaction : If the amount of a drug A is decreasing at a constant time interval t , then the rate of disappearance of drug A is expressed as follow: dA/dt = -Ko

  29. Where Ko is the zero order rate constant and its unit is mass/ time ( e.g. mg/min). Integration of equation yield the following expression A = Kot+ Ao where Ao is the amount of drug at time t=zero. First order reaction : If the amount of the drug A is decreasing at a rate that is proportion to the amount of drug A remaining , then the rate of disappearance of drug is expressed as follow: dA/dt= - kA where k is first order rate constant and is expressed in unit time -1 .

  30. Complex Reactions include 1-opposite or reverse reaction 2- consecutive reaction 3- Side reaction Along with main reaction . Reversible reaction : A--------------B A--------------B+ C A+B------------- C+D this reaction is example of second order reaction

  31. e.g. is the reaction between acid and alcohol to form ester and water . CH3CooH+ C2H5OH--------------CH3CooC2H5 +H2O The rate of reaction is proportion to the concentration of two reacting substance A and B in the forward reaction and C and D in the reverse reaction . If a and b represent the initial concn. Of two reacting substances and if x denote the moles of A and B in each liter reacting in interval of time t The velocity of reaction is expressed by this equation : dx/ dt= K (a-x) (b-x) (1) when a not equal b If a=b dx/ dt= K(a-x)2(2)

  32. Integration of equation (1)will yield : K= 2.303/ t(a-b)log b(a-x)/a(b-x) K can be determined by plotting t against 2.303/a-b log b(a-x)/a(b-x) The slope of the line equal K The unit for second order rate constant K is concentration -1time-1. Integration of equation( 2) will yield: K = 1/t.x/a(a-x) and k can be determined by plotting t against x/ a(a-x) the slope of the line = k and half life (t1/2) for the second order decomposition that obey this equation t1/2 = 1/aK

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