ABSTRACT EALUATION OF MINIMUM INHIBITORY CONCENTRATION FOR SOME OF THE AEROBIC BACTERIA AND ITS ANTI-MICROBIAL ACTIVITY. DEVI ALLAM,MALLESWATRA RAO.P, Ch.NEERAJA, S.ASHUTOSH KUMAR A.K.R.G college of Pharmacy,Nallajerla, W.G, 534112, A.P A.K.R.G college of Pharmacy,Nallajerla, W.G, 534112, A.P A.K.R.G college of Pharmacy,Nallajerla, W.G, 534112, A.P A.K.R.G college of Pharmacy,Nallajerla, W.G, 534112, A.P Author for Correspondence : email@example.com A soil microbe infected seed of palm tree was collected streptomyceses like organism AIM: To determine reaction rate constant of ethyl acetate in sodium hydroxide solution at room temperature when the initial molar concentration of both the reactants are same. REQUIREMENTS: 0.05N Ethyl acetate, 0.05N NaOH, 0.05N HCl, beaker, conical flask, burette. PRINCIPLE: In a second order reaction , the rate of the reaction depends on the concentration of both the reactants. The alkaline hydrolysis of ethyl acetate (saponification) follows second order kinetics at 250C. CH3COOC2H5 + NaOH CH3COONa + C2H5OH If the initial concentration of ethyl acetate and sodium hydroxide are equal i.e., ‘a’ and the amount hydrolysed in time ‘t’ is ‘x’. then the amount un-hydrolysed is a-x, since the reaction is equimolar, then the rate is; dx/dt = k (a-x)2. Integration of the above equation will give ; K = 1/at X [x/a(a-x)] ‘K’ value will give the specific reaction rate constant and its unit is Lit mole-1 time-1. The half-life is given as t1/2=1/ak where a is the initial concentration and K is the specific rate constant. Half-life is inversely proportional to the initial concentration. PROCEDURE: Preparation of solutions: 1. 0.05N HCl: Add 4.35ml of conc. HCl in 1000ml of distilled water taken in volumetric flask and shake well. 2 Sodium hydroxide (0.05 N) solution IP:- Molecular weight of Sodium hydroxide=40 40g of NaOH in 1000ml of water give 1N solution. 2g of NaOH in 1000ml of water give 0.05N solution. Transfer 2g of NaOH into a 1000ml volumetric flask. Add water to make 1000ml. Allow it to stand overnight and decant the clear liquid into a bottle. 2. 0.05N ethyl acetate:- Molecular weight of ethyl acetate = 88.11 88.11g in 1000ml = 1N 8.811g in 1000ml = 0.1N 0.44055g in 100ml = 0.05N 0.9g = 1 ml (1 X 0.44055)/0.9 =0.4895 ml Determination of rate constant: Mix 50ml of 0.05N ethyl acetate and 0.05N NaOH solution. Withdraw 10ml of the mixture immediately in to a conical flask containing 10ml of ice cold water. Add few drops of phenolphthalein as indicator and titrate against 0.05N HCl. The titer value is ‘a’. Similarly withdraw 10ml of the mixture at 5,10,15,20,25,30min intervals and titrate the samples as done earlier . the titer value is ‘á-x’. Substitute these in the second order rate reaction and calculate the specific rate constant. Plot a graph of x/a(a-x) Vs. time ‘t’. the slope gives the value ’K’. Calculate t1/2 using the equation. OBSERVATIONS: Volume Amount (a-x)/a S.NO Time(min) Of HCl decomposed x/a(a-x) K=(1/t) X [x/a(a-x)] (ml) X=a-(a-x) Average K = millimeter mole-1 min-1. t1/2 = min from graph: X1 = X2 = Y1= Y2 = Slope = Y2-Y1 X2-X1 REPORT: Reaction rate constant from the graph: K= millimeter mole-1 min-1. t1/2 = min Reaction rate constant using the equation: K= millimeter mole-1 min-1. t1/2 = min.
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