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The Determination of an Equilibrium Constant.

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Introduction

The Determination of an Equilibrium Constant Introduction: PARTI: Theory Many chemical reactions are irreversible reactions, for example, if magnesium is burnt in air a brilliant white flame is observed as the white powdery solid magnesium oxide is formed: 2Mg(s) + O2(g) 2MgO(s) We also say that the reaction goes to completion (as Fig. 1 shows), and for practical purpose it is. Some apparently irreversible reactions are reversible to such a small extent that we can ignore it. Concentration Reaction stops when concentration reaches zero Time Fig. 1 A reversible reaction is one that can take place in both directions and so is incomplete (thus this symbol is used). Take the reaction between hydrogen and iodine for example: H2(g) + I2(g) 2HI(g) 0.5 moles of hydrogen gas and 0.5 moles of iodine gas react in sealed glass bulbs and the temperature is kept constant at 445 �C. One may assume that there would be 0.1 mole of hydrogen iodide present when the reaction has completed but this is not the case. After 84 minutes, there are only 0.78 moles of hydrogen iodide present. No matter how much time elapsed after 84 minutes, the amounts of all three chemicals were identical: the reaction has reached its equilibrium position (Fig. 2). Equilibrium reached: no further changes Concentration Concentration of product Concentration of reactant Time Fig. 2 However, it does not mean that the reaction has stopped. In fact both the forward and the reverse reactions still continue, which is why we use the term dynamic equilibrium: When a reaction is in dynamic equilibrium, the forward and reverse reactions are occurring at the same rate (Fig. 3). Forward rate Equilibrium reached Rate of reaction Back rate Time Fig. 3 Equilibrium is maintained only in a closed system, where there is no exchange with the surroundings. There is therefore no loss of reactant or product materials to affect their equilibrium concentrations. ...read more.

Middle

+ 0.4675 = 1.8565 mole 4 [CH3COOC2H5] = 1.2515 / 0.25 mole dm-3 [H2O] = 1.8565 / 0.25 mole dm-3 [CH3COOH] = 0.845 / 0.25 mole dm-3 [C2H5OH] = 0.8205 / 0.25 mole dm-3 ?Kc1 = [CH3COOC2H5] * [H2O] / [CH3COOH] * [C2H5OH] = (1.2515/0.25) * (1.8565/0.25) / (0.845/0.25) * (0.8205/0.25) =1.2515 * 1.8565 / 0.845 * 0.8205 = 3.35 Mixture 2: 1 * Mass of CH3COOH = 1.05 * 75 = 78.75 g Molecular mass of CH3COOH = 12 + 3 +12 + 32 + 1 = 60 g/mol Number of moles of CH3COOH = 78.75 / 60 = 1.3125 mole * Mass of C2H5OH = 0.79 * 75 = 59.25 g Molecular mass of C2H5OH = 24 + 5 + 16 + 1 = 46 g/mol Number of moles of C2H5OH = 59.25 / 46 = 1.288 mole * Mass of CH3COOC2H5 = 0 g Number of moles of CH3COOC2H5 = 0 mole * Mass of H2O = 1 * 100= 100 g Molecular mass of H2O = 18 g/mol Number of moles of H2O = 100 / 18 = 5.556 mole 2 Volume of NaOH used in titration is 17.05 cm3 Number of moles of OH- used = 0.2 / 1000 * 17.05 = 3.4 * 10-3 mole ?1 mole of H+ react with I mole of OH- ?There are 3.4 * 10-3 * 250 = 0.85 mole H+ in the 250 cm3 mixture 3 Number of moles of HCl in the mixture = 1 /1000 * 25 = 0.025 mole Number of moles of CH3COOH at equilibrium = 0.85 - 0.025 = 0.825 mole Number of moles of CH3COOH reacted = 1.3125 - 0.825 = 0.4875 mole According to the equation CH3COOH + C2H5OH CH3COOC2H5 + H2O The amount of C2H5OH reacted equals the amount of CH3COOH reacted ?Number of moles of C2H5OH at equilibrium = 1.288 - 0.4875 = 0.8005 mole The amount of CH3COOC2H5 (H2O) ...read more.

Conclusion

It has led to errors. However the experiment can be improved if the following actions can be taken into account: * Find out the actual density of the hydrochloric acid. It can be done easily by dividing the mass of a fixed amount of acid by its volume; * A more closed container is recommended to keep the mixtures, e.g. a vacuumed cylinder. The exchange of materials between the mixture and the surroundings can therefore be minimised; * A PH metre can be used during titration so the equivalence point can be detected much more accurately than simply being judged by eyes, which means the accuracy of the titration can be greatly improved; * Ensure that the mixtures were not subjected to any change of temperature and/or pressure which may have an effect on the position of the equilibrium. This can be achieved by using more sophisticated measuring equipments; * Repeat the titration several times and find out the average reading. Therefore mistakes made by one single titration can be minimised. Finally, we do not know whether seven days have been long enough for the mixtures to reach equilibrium. If they were not long enough then no matter how accurate we were we still cannot find the actual value of Kc. Therefore it is essential to: Divide each mixture into several samples for separate analyses at different times, say we can divide mixture 1 into four samples and label them one to four. Sample one can be analysed after six days; sample two after eight days; sample three after ten days and sample four after twelve days. Then different Kc1 values can be compared. Whenever Kc1 becomes constant we know that equilibrium has been achieved. Repeat this procedure with mixtures 2, 3 and 4 so an actual actual Kc2, Kc3 and Kc4 values can be obtained, from which we can calculate the actual Kc value for the equilibrium. ...read more.

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