• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Determine the equilibrium constant - Kc; of ethanoic acid reacting with ethanol producing an equilibrium to form ethyl ethanoate and water.

Extracts from this document...

Introduction

The Determination of an Equilibrium Constant Determine the equilibrium constant - Kc; of ethanoic acid reacting with ethanol producing an equilibrium to form ethyl ethanoate and water. CH3COOH(aq) + C2H5OH(aq) <=> CH3COOC2H5(aq) + H2O(l) Following the method as detailed, I conducted experiment 4 and these results were obtained: Titration Trial Volume of Sodium Hydroxide Neutralised (cm3) 1 7.65 2 7.75 3 7.80 4 7.70 5 7.75 � 7.75 To calculate Kc, the concentrations of each reactant must be calculated from the point of equilibrium at which the titration was taken. Therefore, using 7.75cm3 as the average titre, Moles = 0.2M x (7.75cm3 / 1000 cm3) = 1.55 x 10-3. However, this represents the total amount of acid in the system which included ethanoic acid but also the acid catalyst: hydrochloric acid. Therefore this must be taken away from the number of moles of acid to calculate number of moles of ethanoic acid. Since 25 cm3 of 1.0M HCl was included in the initial 250 cm3 mixture, Moles = 1 x (25 / 250) = 0.1 moles. But 1cm3 samples were taken at a time, therefore: Moles = 0.1 x (1 cm3 / 1000 cm3) = 1 x 10-4. Taking this away from the total amount of acid = 1.55 x 10-3 - 0.1 x 10-3 = 1.45 x 10-3 (moles of ethanoic acid). ...read more.

Middle

Conducting experiments 1, 2 and 3 as detailed on the enclosed sheet, provided such comparative values for Kc: Experiment Number Value of Kc 1 5.1 2 4.39 3 4.5 4 3.62 Calculating the percentage error for all of these results gives: -9.5% < Theoretical Value < 27.5% This clearly shows some varying results, all of which are distributed around the theoretical value of the equilibrium. However the deviations of these results when considering the size of the results involved is rather large. This may be due to a range of reasons concerning the reliability of the experiment. The concentration of the alkali used was 0.2M, this may be considered low but because accuracy is of utmost concern, a larger sample of more dilute alkali would yield a larger more accurate titre of the number of moles present in the solution. This is despite the fact that the titres taken were within 0.5cm3 of each other. Also if a 2cm3 sample were taken instead of simply 1cm3 then there would twice the number of moles of acid present and the titre would be sufficiently large to be considered accurate. The indicator chosen was phenolphthalein, which has a transition pH of 8 - 9.6. ...read more.

Conclusion

However this source of error is questionable because all experiments were conducted in the same laboratory in the same interval in time. Titration is a method which releases a precise quantity of fluid into a reactant containing vessel. This is done by viewing the bottom of the meniscus and controlling the flow of fluid out of the burette. This technique is flawed only by the measurement of the level of the meniscus and the fact that the device works on a large bore of tubing which allows for more variation with approximately the same result. However the method of using a volumetric pipette with a 1cm3 measuring cylinder provides a hugely accurate volume of solution to which is mixed a large quantity of water which ensures all solution has been absorbed in. To increase reliability of the overall method, a larger sample of the solution would have to be taken to ensure an accurate titre was achieved. An indicator which worked exactly within the pH range of the given experiment was chosen. The dissociation of the measured acid would have to be taken into account in order to provide an accurate concentration and there Kc value. A more accurate burette would have to be used which had smaller graduations and a smaller bore of tubing to reduce the size of the meniscus, and the experiment would have to be conducted under standard conditions. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Aqueous Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Aqueous Chemistry essays

  1. The aim of this experiment is to answer the following question: What is the ...

    work out the value of which I am investigating; the equilibrium constant. The equilibrium constant is worked out with this equation: Kc = Concentration products Concentration reactants In the case of this investigation the value of the equilibrium constant is found by this equation: Kc = [Ester] [Water] [Acid][Alcohol] In

  2. Analysing the ethanoic acid concentration in different types of vinegars.

    0.15/7 x 100 = 2.14% WWV distillate= 0.15/8.27 x 100 = 1.81% Total Maximum % error MV = 1.78 + 0.015 + 0.084 +0.27 +3 = 5.15% CV = 1.61 + 3.369 = 4.98% WWV = 1.39 + 3.369 = 4.76% MV distillate = 2.26 + 3.369 = 5.63% CV

  1. Determining the equilibrium constant for the hydrolysis of Ethyl Ethanoate Definition

    Hydrolysis is therefore the reaction of an organic compound with water. The hydrolysis of an ester forms a carboxylic acid and an alcohol. Few organic compounds react readily with water. When water is added to an ester, a carboxylic acid and alcohol are formed.

  2. Determination of the equilibrium constant for esterification of ethanoic acid and propan-1-ol by using ...

    Moreover, to make sure that the result is concise, we have to continue refluxing for an additional 1/2 hour and titrate another 1 cm3 sample. If the two titrations agree to within 0.2 cm3, it shows that the reaction has reached the equilibrium point.

  1. Obtain pure samples of Ethanol (CH3CH2OH) and Ethanoic Acid (CH3COOH) from fermented Yeast (Saccharomyces ...

    C6H12O6 (aq) 2CH3CH2OH (aq) + 2CO2 (g) (Glucose) (Ethanol) (Carbon dioxide) The reaction is catalysed by Zymase, an enzyme present in yeast. Yeast is a microorganism, which obtains its energy from what is available during the reaction. All alcoholic drinks are products of this reaction.

  2. The Determination of an Equilibrium Constant.

    * Pressure and equilibrium In a gas pressure depends on the number of molecules acting on a particular area. If the amount of a gas is increased (and so the pressure is increased), the equilibrium shifts to wherever there are fewer amounts of gases, in other words, less pressure; or

  1. Titration with a primary standard.

    Calculating % error: Calculating the percentage uncertainty (often called percentage error) .. To calculate the error in using different measuring instruments, * Look at the graduations on the measuring scale. * Decide what reading you can be certain about. * Usually the error is half the size of the graduation.

  2. Determining an equilibrium constant

    added Mass of water added/g 5.105 � 0.002 2.946 � 0.002 PART B: Tube number Final burette reading/ cm3 Initial burette reading/ cm3 Titre/ cm3 1a 0 � 0.05 9.61 � 0.05 9.61 � 0.05 1b 0 � 0.05 9.62 � 0.05 9.62 � 0.05 2 9.61 � 0.05 35.10

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work