Equilibrium In Liquid Phase

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Physical Principles II CEBU CITY Chemical Engineering Department Name: Juphil Lamanilao Date: October 15, 2009 Course & Yr: BSChE-4 I.TITLE: Chemical Equilibrium in a Liquid Phase II.OBJECTIVE: To determine the equilibrium constant. III.APPARATUS: 50 ml burette, 8 test tubes with cork, 50 ml 6N HCl, 500 ml of std 1N NaOH, 15ml pure CH 3 COOC 2 H 5 , 15ml CH 3 COOH, 15ml absolute C 2 H 5 OH IV.SKETCH: Chemical Equilibrium in a Liquid Phase 1

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Transcript of Equilibrium In Liquid Phase

Page 1: Equilibrium In  Liquid  Phase

Physical Principles II

CEBU CITYChemical Engineering Department

Name: Juphil Lamanilao Date: October 15, 2009

Course & Yr: BSChE-4

I.TITLE: Chemical Equilibrium in a Liquid Phase

II.OBJECTIVE:

To determine the equilibrium constant.

III.APPARATUS:

50 ml burette, 8 test tubes with cork, 50 ml 6N HCl, 500 ml of std 1N NaOH, 15ml pure CH3COOC2H5, 15ml CH3COOH, 15ml absolute C2H5OH

IV.SKETCH:

Chemical Equilibrium in a Liquid Phase 1

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Physical Principles II

V.TABULATED DATA & RESULTS:

Test tube

Components of

Solution

Wt. of empty

tube (g)

Wt. of tube with

solution (g)

Wt. of Solution

Volume of NaOH used in titration

Kc

1 5ml 6M HCl + 5ml H2O 19.12 29.54 10.42 18.6 0.0118

25ml 6M HCl + 5ml H2O 17.73 29.17 11.44 18.1 0.0149

35ml 6M HCl + 5ml EtOAc

17.75 27.70 9.95 54 0.0334

45ml 6M HCl + 5ml EtOAc

17.41 27.40 9.99 54.4 0.0329

55ml 6M HCl + 3ml glacial + 2ml EtOH

17.62 27.70 10.08 65.7 0.0013

65ml 6M HCl + 3ml glacial + 2ml EtOH

17.62 27.79 10.11 65.6 0.0012

75ml 6M HCl + 3ml EtOH+ 2ml glacial

17.64 27.46 9.82 46.4 0.0016

85ml 6M HCl + 3ml EtOH+ 2ml glacial

19.23 29.90 10.61 41.6 0.0012

VI.SAMPLE COMPUTATIONS:

@ 25оC Kc= 5.38 ÷ (6 x 50.33) = 0.0178

HCl= 1.789 g/ml

Kc= CNaOH/ (CHCl x CH2O) = 7.77x10-3/M

CH2O= (10.42-5.89)g÷(18g/molx0.005L)= 50.33M

CNaOH= (1M)(100ml)÷ 18.6 ml = 5.38MChemical Equilibrium in a Liquid Phase 2

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Physical Principles II

VII.DATA ANALYSIS:

The experimental values gathered for the equilibrium constant approaches roughly to 0.0164 for HCl and H2O. Furthermore, 0.0379 was obtained for HCl and C2H5COOC2H5 mixture. For the HCl-CH3COOH- C2H5OH mixture, the equilibrium constant approaches 0.0013. From the data obtained, results showed that the concentration of NaOH decreases as the number of compounds in the mixture escalates. Consequently, the equilibrium constant will differ in all the components in the reaction mixture; it varies in both products and reactants. As observed, the equilibrium constant Kc is the same no matter how much is the starting amount.

VIII.APPLICATION TO ChE:

By way of attaining chemical equilibrium, constituents and properties of intricate mixtures can be obtained. Various applications include Liquid-liquid extraction, Ion exchange, Chromatography & Solubility product and in constructing bomb detonators which are commonly related to the field of chemical engineering.

IX.CONCLUSION:

In a chemical process, chemical equilibrium is the state in which the chemical activities or concentrations of the reactants and products have no net change over time. They proceed to a certain point and then apparently stop, often leaving considerable amounts of unaffected reactants. Under any given set of conditions of temperature, pressure, and concentration, the point at which any reaction seems to stop is always the same; there exists at this point among the concentrations of the various reactants and products of any reaction a reaction which is definitely fixed. When a reaction reaches this stage in its course, it is said to be in equilibrium. In this experiment, Kc is expressed by esterification reaction. Comparably with hydrolysis, esterification process is very slow at room temperature. Whatever the proportions of acid and alcohol that are mixed together, Kc always has the same value.

Chemical Equilibrium in a Liquid Phase 3

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Physical Principles II

Chemical Equilibrium in a Liquid Phase 4