the effect of a catalyst on reaction rate

Upload: roygunedwin

Post on 02-Jun-2018

222 views

Category:

Documents


0 download

TRANSCRIPT

  • 8/10/2019 The Effect of a Catalyst on Reaction Rate

    1/7

    Enthalpy Changes of Neutralisation

    By: Beatrice Valencia

    Class: IGCSE Year 1

    Chandra Kusuma School 2013

  • 8/10/2019 The Effect of a Catalyst on Reaction Rate

    2/7

    Introduction:

    When an alkali neutralizes an acid, a salt and water are formed.

    Aqueous hydrogen ions (H+(aq)) from the acid react with the hydroxideions(OH-(aq)) from the alkali, forming water. The identity of the salt will of

    course depend on the nature of the acid and alkali used. For hydrochloric

    acid and sodium hydroxide:

    H+(aq)+Cl-(aq) + Na+(aq)+OH-(aq) Na+(aq)+Cl-(aq)+H2O(I)

    Hydrochloric acid Sodium Hydroxide Sodium Chloride

    Notice in this equation the Na+ and Cl- ions are unchanged. The only

    chemical reaction occurring is between H+ and OH- ions:H+(aq) + OH-(aq) H2O(I)

    The combination of H+ and OH- ions in this way release energy.

    Purpose: to measure the enthalpy changes when the process of giving of

    heat(exothermic)

    No. Materials

    1 Hydrochloric acid

    2 Sodium Hydroxide3 Nitric acid

    4 Sulphuric acid

    5 Measuring cylinder

    6 Dropper

    7 Beakers

    8 Thermometer

  • 8/10/2019 The Effect of a Catalyst on Reaction Rate

    3/7

    Procedure/ Method :

    1) Hydrochloric acid with sodium hydroxide

    Picture Observation

    Measure 25cm3 of 2.0

    mol dm-3 hydrochloric

    acid into a polystyrene

    beaker.

    Record its temperature.

    Put 25cm3 of 2.0 dm-3

    sodium hydroxide

    solution in a measuring

    cylinder and take its

    temperature.

    Now pour this into the

    acid, stir and take the

    final temperatue.

  • 8/10/2019 The Effect of a Catalyst on Reaction Rate

    4/7

    Data:

    Temp. HCL: 29o

    Temp.NaOH: 33o

    Temp.Final: 41o

    Question #1 :A. What was the temperature rise? (If the acid and alkali were not at the same

    temperature initially, use the mean of their initial temperatures as the starting

    temperature.)

    29+33/2=31o

    B. How much energy was delivered to the 50cm3 of final solution during thereaction? (Assume the specific heating capacity of the polystyrene cup is

    negligible and that the specific heating capacity of the solution is 4.2JK-1

    g-1

    .)Q=mc.t

    =50g. 4.2J/kg=2.1Kj

    = 2.1 kJ/0.05mol=42kJ/mol

    C. How many moles of water were formed by the reaction you have just carried out?25cm3=0.025 dm3

    Mole HCL=2mol/dm3 x 0.025 dm3 =0.05mol

    Mole of NaOH,NaCl,H2O=0.05

    D. Work out the enthalpy change of neutralisation for this reaction in kilojoules permole of water formed.

    NaOH + HCL NaCL + H2O

    2) Nitric acid and sodium hydroxide

    Picture Observations

    Measure 25cm3of 2.0 mol

    dm-3nitric acid into a

    polystyrene beaker.

  • 8/10/2019 The Effect of a Catalyst on Reaction Rate

    5/7

    Record its temperature.

    Put 25cm3of 2.0 dm-3

    sodium hydroxide

    solution in a measuring

    cylinder and take its

    temperature.

    Now pour this into the

    acid, stir and take the final

    temperatue.

  • 8/10/2019 The Effect of a Catalyst on Reaction Rate

    6/7

    Question:1. Write an ionic equation to represent the reaction that has occurred.

    H+(aq) + NO3 -(aq) Na+(aq) + OH-(aq)2. Work out the enthalpy change of neutralisation for this reaction.3. Compare the values you have obtained for the enthalpy changes of neutralisation

    of the three reactions and explain the way the values compare.

    Q=mc.t

    =50g. 4.2J. 10,5=2205kJ

    = 2205 kJ/0.05mol=44.1kJ/mol

    4. How do the values for the first two reactions in this table compare with your ownexperimental results for the same reactions? Account for any errors in your

    results.

    The enthalpy change in experiment 2 is greater than experiment 1

    5. Do the values for the first three reactions in the table agree with the explanationyou have given in question 4?

    %error= [Theory-Lab/Theory]/theory x 100%

    = [-57.9 42]/-57.9 x 100% HCL with NaOH

    = 172.5%

    %error= [Theory-Lab/Theory]/theory x 100%

    = [-57.6 44.1]/-57.6 x 100% HNO3 with NaOH

    = 176.5%

    6. Suggest why the enthalpy change of neutralisation for the reaction involvingethanoic acid is slightly lower than the values for the first three reactions, and

    why the enthalpy change of neutralisation for the reaction involving hydrogen

    sulphide is substantially lower. (The dissociation constants, Ka, for hydrochloric

    acid, hydrobromic acid and nitric acid are very large; that of ethanoic acid is 1.7 x

    10-5 mol dm-3 and that of hydrogen sulphide is

    8.9 x 10 -8mol dm-3.)

    Because the dissociation constant is larger so its make the enthalpy of HCL is

    larger than the ethanoic acid

    3) Sulphuric acid and sodium hydroxide

    Picture Observations

    Measure 25cm3of 2.0mol dm-3sulphuric acid

    into a polystyrene

    beaker.

  • 8/10/2019 The Effect of a Catalyst on Reaction Rate

    7/7

    Record its temperature.

    Put 25cm3of 2.0 dm-3

    sodium hydroxide

    solution in a measuringcylinder and take its

    temperature.

    Now pour this into the

    acid, stir and take the

    final temperatue.

    Question:

    1. Compare the temperature rises for the two experiments and explain theirrelative values.