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Introduction

Date: 4/3/2011 Exp. No.: 14 Title: Determining an equilibrium constant Aim: The aim of this experiment is to calculate the equilibrium constant (kc) of the reaction: CH3CO2C2H5(l) + H2O(l) C2H5OH(l) + CH3CO2H(l) Ethyl ethanoate Ethanol Ethanoic acid Introduction: The above reaction is very slow, so it takes a lot of time to attain equilibrium. However, by using dilute hydrochloric acid as a catalyst, the reaction rate will become much fast and equilibrium can be attained in a few days. This experiment is separated into 2 parts. The first part of the experiment is to prepare mixture of different proportions of the reactants in sealed containers using dilute hydrochloric acid as a catalyst. These mixtures will be allowed to react for a week at room temperature before doing the second part to ensure that the reaction mixtures have attained equilibrium. The second part of the experiment, the mixtures will be titrated one by one using standard sodium hydroxide. Sodium hydroxide reacts with acid in the mixture including ethanoic acid(which is a product of the reaction) and also hydrochloric acid(catalyst of the reaction). Finally, using the amounts of ethanoic acid produced, we can calculate the equilibrium concentration of all four components and they can be used to determine the equilibrium constant of the reaction. Apparatus and chemicals: For part A: 5 boiling tubes with stoppers, electronic balance, air displacement pipette, about 30 cm3 of 2M HCl, two 10 cm3 measuring cylinder, about 15 cm3 of ethyl ethanoate, deionized water, plastic films(wrap) ...read more.

Middle

The amount of HCl present in each tube = � 1.0086 = 9.81�10-3 For tube 2 Total amount of acid at eqm (hydrochloric acid and ethanol acid) in tube 2 (Calculated using the titre) = = 3.74�10-2 mol Eqm amount of ethanoic acid in the mixture = 3.74�10-2 -9.81�10-3 = 2.76�10-2 mol From the equation of this reaction, we know that the mole ratio between ethanol and ethanoic acid = 1:1, so the eqm amount of ethanol produced = eqm amount of ethanoic acid produced = 2.76�10-2 mol Number of moles of ethyl ethanoate added to tube 2 = = = 5.04�10-2 mol Since from the equation, the mole ratio between ethyl ethanoate and ethanoic acid = 1:1, so the amount of ethyl ethanoate used = amount of ethanoic acid produced Eqm amount of ethyl ethanoate = 5.04�10-2 - 2.76�10-2 = 2.28�10-2 mol Mass of pure HCl in HCl(aq) = Amount of HCl � Molar mass of HCl = 9.81�10-3�36.5 = 0.358g Mass of water in HCl(aq) = 5.0 - mass of pure HCl = 5.0 - 0.358 = 4.64g Initial amount of water = = = 0.254 mol Since from the equation, the mole ratio between water and ethanoic acid = 1:1, so the amount of water used = amount of ethanoic acid produced = Initial amount of water - Eqm amount of ethanoic acid = 0.254 - 2.76�10-2 = 0.226 mol By equilibrium law, the equilibrium constant (Kc) ...read more.

Conclusion

The number values for equilibrium constants are tied to the nature of reactants and products in a reaction. The number values for "K" are gotten from experiments measuring equilibrium concentrations. The number value tells the equilibrium ratio of products to reactants. In an equilibrium mixture both reactants and products coexist. The value for K is large when products dominate the mixture. The value for K is small when the reactants dominate the mixture. The equilibrium constant may be expressed in the form where [C] represents the molar concentration of C at equilibrium. For a given reaction, the concentrations at equilibrium would have to be determined experimentally. The value of Kc is ONLY affected by temperature. For gaseous reactants it is more convenient to express the equilibrium condition in terms of the partial pressures of the reactants and products. For this case the equilibrium constant is defined by where P denotes the partial pressure, usually in atmospheres. The two forms of the equilibrium constant are related by where ?n is the sum of the coefficients of the gaseous products in the chemical equation minus the sum of the coefficients of the gaseous reactants. What is chemical equilibrium? Chemical equilibrium means a reaction has reached a state that the forward reaction rate and the backward reaction rate are the same and there is no net gain or loss of a reactant or product. Chemical equilibrium can be changed or shifted by changing the temperature, pressure, amount of reactants/products and volume of the system. Reference: http://www.800mainstreet.com/7/0007-007-Equi_exp_k.html http://en.wikipedia.org/wiki/Equilibrium_constant http://en.wikipedia.org/wiki/Chemical_equilibrium http://www.avogadro.co.uk/chemeqm/chemeqm.htm http://www.chemistrytutorials.org/index.php/content/chemical-equilibrium/factors-affecting-chemical-equilibrium http://en.wikipedia.org/wiki/Pipette http://chemwiki.ucdavis.edu/Physical_Chemistry/Chemical_Equilibrium/The_Equilibrium_Constant http://wn.com/chemical_equilibrium http://wiki.answers.com/Q/What_is_chemical_equilibrium ?? ?? ?? ?? ...read more.

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