Thermochemistry / energy changes - KCSE Chemistry Paper 3 - Practicals

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Energy is the capacity to do work which is measured in Joules(J) or(kJ) .

Chemical/physical changes take place with absorption (Endothermic) or evolution/ production (Exothermic) of heat.

Practically:

  1. endothermic changes show absorption of heat by a fall / drop in temperature and has a +ΔH
  2. exothermic changes show evolution/ production of heat by a rise in temperature and has a -ΔH
  3. temperature is measured using a thermometer.
  4. a school thermometer is either coloured (alcohol) or colourless(mercury)
  5. For accuracy, candidates in the same practical session should use the same type of thermometer.
  6. fall / drop (+ΔH) in temperature is movement of thermometer level downward.
  7. rise (-ΔH) in temperature is movement of thermometer level upwards.

Physical changes mainly involve melting/ freezing/ fusion and boiling / vapourization.

Chemical changes mainly involve displacement, dissolving, neutralization

Glossary terms and their definations are highlighted all throughout the text.

Energy changes in physical processes

Melting/freezing/fusion/solidification and boiling/ vaporization/evaporation are the two physical processes.

Melting /freezing point of pure substances is fixed/ constant.

The boiling point of pure substance depend on external atmospheric pressure.

Melting/fusion is the physical change of a solid to liquid. Freezing/fusion is the physical change of a liquid to solid.

Melting/freezing/fusion/solidification are therefore two opposite but same reversible physical processes.

             A (s)   <========> A(l)

Boiling/vaporization/evaporation is the physical change of a liquid to gas/vapour.

Condensation/liquidification is the physical change of gas/vapour to liquid.

Boiling and condensation are therefore two opposite but same reversible physical processes. i.e

               B (l)   <========> B(g)

Practically

  • Melting/liquidification/fusion involves heating a solid to weaken the strong bonds holding the solid particles together.
    Solids are made up of very strong bonds holding the particles very close to each other.
    (Kinetic Theory of matter)
    On heating these particles gain energy/heat from the surrounding heat source to form a liquid with weaker bonds holding the particles close together but with some degree of freedom.
    Melting/fusion is an endothermic (+ΔH)process that require/ absorb energy from the surrounding.
  • Freezing/fusion/solidification involves cooling a liquid to reform /rejoin the very strong bonds to hold the particles very close to each other as solid and thus lose their degree of freedom (Kinetic Theory of matter).
    Freezing /fusion / solidification is an exothermic (-ΔH)process that require particles holding the liquid together to lose energy to the surrounding.

  • Boiling/vaporization/evaporation involves heating a liquid to completely break/free the bonds holding the liquid particles together.
    Gaseous particles have high degree of freedom
    (Kinetic Theory of matter).
    Boiling /vaporization / evaporation is an endothermic (+ΔH) process that require/absorb energy from the surrounding.

  • Condensation/liquidification is reverse process of boiling /vaporization / evaporation.
    It involves gaseous particles losing energy to the surrounding to form a liquid.
    It is an exothermic(-ΔH) process.
    The quantity of energy required to change one mole of a solid to liquid or to form one mole of a solid from liquid at constant temperature is called molar enthalpy/latent heat of fusion. e.g.
      H2O(s)   ---> H2O(l)  ΔH = +6.0kJ mole-1              (endothermic process)

        H2O(l)  ---> H2O(s)  ΔH = -6.0kJ mole-1              (exothermic process)

    The quantity of energy required to change one mole of a liquid to gas/vapour or to form one mole of a liquid from gas/vapour at constant temperature is called molar enthalpy/latent heat of vapourization. 

    H2O(l)   -> H2O(g)  ΔH = +44.0kJ mole-1              (endothermic process) 
    H2O(g)   -> H2O(l)  ΔH = -44.0kJ mole-1               (exothermic process)

 

To determine the boiling point of water

Procedure:
Measure 20cm3 of tap water into a 50cm3 glass beaker.
Determine and record its temperature.
Heat the water on a strong Bunsen burner flame and record its temperature after every thirty seconds for four minute

Questions

1. Plot a graph of temperature against time(y-axis)

Sample results

Time (seconds)  30  60  90  120  150 180 210 240
Temperature (oC) 25.0 45.0 85.0 95.0 96.0 96.0 96.0 97.0 98.0

graph of temperature against time

2. From the graph show and determine the boiling point of water

Note:

Water boils at 100oC at sea level/one atmosphere pressure/ 101300Pa but boils at below 100oC at higher altitudes.

Results above are from Kiriari Girls High School-Embu County on the slopes of Mt Kenya. Water here boils at 96oC.

3. Calculate the molar heat of vaporization of water. (H= 1.0,O= 16.O)

Mass of water = density x volume =>(20  x  1) /1000 = 0.02kg

Quantity of heat produced =  mass of water x specific heat capacity of water x temperature change

  =>0.02kg  x  4.2  x ( 96  –  25 ) = -5.964kJ 

Heat of vaporization of one mole H2O

   =   Quantity of heat    => -5.964kJ = -0.3313 kJ mole -1
         
Molar mass of H2O       18 

 

To determine the melting point of candle wax

Procedure

Weigh exactly 5.0 g of candle wax into a boiling tube.

Heat it on a strongly Bunsen burner flame until it completely melts.

Insert a thermometer and remove the boiling tube from the flame. Stir continuously.

Determine and record the temperature after every 30 seconds for four minutes.

Time(seconds) 0 30 60 90 120 150 180 210 240
Temperature (oC) 93.0  85.0 78.0  70.0  69.0  69.0  69.0  67.0  65.0

1. Plot a graph of temperature against time(y-axis)

melting candle wax

Energy changes in chemical processes

  1. Standard enthalpy/heat of displacement ΔHod
  2. Standard enthalpy/heat of neutralization ΔHon
  3. Standard enthalpy/heat of solution/dissolution ΔHos

The molar standard enthalpy/heat of displacement may be defined as the energy/heat change when one mole of substance is displaced /removed from its solution at standard conditions.

Standard enthalpy/heat of displacement ΔHod

Some displacement reactions

  1. Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq)
    Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq)

  2. Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq)
    Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq)

  3. Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s)
    This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead.

  4. Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq)
    Ionically: Cl2(g)+ 2Br- (aq) -> Br2(aq) + 2Cl- (aq)

To determine the molar standard enthalpy/heat of displacement (ΔHod) of copper

Procedure

Place 20cm3 of 0.2M copper(II)sulphate(VI)solution into a 50cm3 plastic beaker/calorimeter.

Determine and record the temperature of the solution T1.

Put all the Zinc powder provided into the plastic beaker. Stir the mixture using the thermometer.

Determine and record the highest temperature change to the nearest 0.5oC- T2 .

Repeat the experiment to complete table 1 below

Experiment

 
I II 
 

Final temperature of solution(T2)

30.0oC

 

31.0oC

 

Initial temperature of solution(T1)

 

25.0oC

 

24.0oC

 

Change in temperature(ΔT)

 

5.0

 

6.0

 

     Questions

  1. Calculate:
    1. average ΔT
      Average ΔT = change in temperature in experiment I and II
                  =>5.0 + 6.0    =   5.5oC
                            2
    2. the number of moles of solution used

      Moles used = molarity x volume of solution   =   0.2 x 20 
                                          1000                            1000
            =  0.004 moles

    3. the enthalpy change ΔH for the reaction
      Heat produced ΔH = mass of solution(m) x specific heat capacity (c)x ΔT
        =>  20 x 4.2 x 5.5 =  462 Joules =  -0.462 kJ
                                         1000                  
    4. State two assumptions made in the above calculations.
      Density of solution = density of water = 1gcm-3
      Specific heat capacity of solution=Specific heat capacity of water   = 4.2 kJ-1kg-1K
      This is because the solution is assumed to be infinite dilute.
  2. Calculate the enthalpy change for one mole of displacement of Cu2+ (aq) ions

    Molar heat of displacement ΔHd = Heat produced ΔH
                                               Number of moles of fuel

    =>         0.462 kJ         = -115.5 kJmole-1
                    0.004

  3. Write an ionic equation for the reaction taking place.

         Zn(s) + Cu2+(aq) ->   Cu(s) + Zn2+(aq)

  4.  State the observation made during the reaction.

    Blue colour of copper(II)sulphate(VI) fades/becomes less blue/colourless.
    Brown solid deposits are formed at the bottom of reaction vessel/ beaker.

  5.  Illustrate the above reaction using an energy level diagram.

  6. The enthalpy of displacement ΔHd of copper(II)sulphate (VI) solution is 126kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings.

    Number of moles = Heat produced ΔH   
                       Molar heat of displacement ΔHd

      =>     2.204 kJ        =   0.0206moles
                126 moles

    Molarity of the solution = moles x 1000 
                                  Volume of solution used

      =     0.0206moles x 1000  = 0.5167 M
                             40

    Number of moles = Heat produced ΔH 
                       Molar heat of displacement ΔHd

      =>     2.204 kJ            =   0.0206moles
              126 moles

    Molarity of the solution = moles x 1000          =>   0.0206 moles x 1000            
                                        Volume of solution used                  40

           = 0.5167 M

     

Graphical determination molar enthalpy of displacement of copper

Procedure:

Place 20cm3 of 0.2M copper(II)sulphate (VI) solution into a calorimeter/50cm3 of plastic beaker wrapped in cotton wool/tissue paper. Record its temperature at time T= 0. Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds. Place all the (1.5g) Zinc powder provided after 1 ½ minutes. Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds for five minutes. Determine the highest temperature change to the nearest 0.5oC.

Time oC

 
0.0  30.0  60.0 90.0 120.0 150.0 180.0 210.0 240.0 270.0
 

Temperature

 25.0  25.0  25.0  25.0 25.0   xxx  36.0  35.5  35.0  34.5

molar enthalpy of displacement copper graph

Questions

  1. Show and determine the change in temperature ΔT
    From a well constructed graph ΔT= T2 –T1 at 150 second by extrapolation
        ΔT = 36.5 – 25.0    =   11.5oC
  2. Calculate the number of moles of copper(II) sulphate(VI)used given the molar heat of displacement of Cu2+ (aq)ions is 125kJmole-1 
    Heat produced ΔH = mass of solution(m) x specific heat capacity (c)x ΔT
          =>  20 x 4.2 x 11.5 =  966 Joules =  -0.966 kJ
                                            1000

    Number of moles  =              Heat produced ΔH   
                                      Molar heat of displacement ΔH
                        =>  -0.966 kJ              =   -0.007728moles = -7.728 x 10-3moles
                                125 moles   

  3. What was the concentration of copper(II)sulphate(VI) in moles per litre.

    Molarity = moles x 1000

                    Volume used

            =>7.728 x 10-3moles x 1000 = 0.3864M

                        20
  4. The actual concentration of copper(II) Sulphate (VI) solution was 0.4M. Explain the differences between the two

    Practical value is lower than theoretical

    Heat/energy loss to the surrounding and that absorbed by the reaction vessel decreases ΔT hence lowering the practical number of moles and molarity against the theoretical value

 Standard enthalpy/heat of neutralization ΔHon

The molar standard enthalpy/heat of neutralization ΔHon is defined as the energy/heat change when one mole of a H+ (H3O+)ions react completely with one mole of OH- ions to form one mole of H2O/water.

Neutralization is thus a reaction of an acid /H+ (H3O+)ions with a base/alkali/ OH- ions to form salt and water only.

Strong acids/bases/alkalis are completely/fully/wholly dissociated to many free ions(H+ /H3O+ and OH-  ions).

for strong acid/base/alkali neutralization, no energy is used to dissociate /ionize since molecule is wholly/fully dissociated/ionized into free H+ H3O+ and OH- ions.

The overall energy evolved is comparatively higher / more than weak acid-base/ alkali neutralizations.

For strong acid-base/alkali neutralization, the enthalpy of neutralization is constant at about 57.3kJmole-1 irrespective of the acid-base used. This is because ionically:

     OH-(aq)+  H+(aq)  ->   H2O(l)

 for all wholly/fully /completely dissociated acid/base/alkali

Weak acids/bases/alkalis are partially dissociated to few free ions(H+ (H3O+ and OH-  ions) and exist more as molecules.

Neutralization is an  exothermic(-ΔH) process.

The energy produced during neutralization depend on the amount of  free ions (H+ H3O+ and OH-)ions existing in the acid/base/alkali reactant:

for weak acid-base/alkali neutralization,some of the energy is used to dissociate /ionize the molecule into free H+ H3O+ and OH- ions therefore the overall energy evolved is comparatively lower/lesser/smaller than strong acid / base/ alkali neutralizations. 

Practically ΔHᶿn can be determined as in the examples below:

To determine the molar enthalpy of neutralization ΔHn of Hydrochloric acid

Procedure

  1. Place 50cm3 of 2M hydrochloric acid into a calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper.
  2. Record its temperature T1.
  3. Using a clean measuring cylinder, measure another 50cm3 of 2M sodium hydroxide.
  4. Rinse the bulb of the thermometer in distilled water.
  5. Determine the temperature of the sodium hydroxide T2.
  6. Average T2 and T1 to get the initial temperature of the mixture T3.
  7. Carefully add all the alkali into the calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper containing the acid.
  8. Stir vigorously the mixture with the thermometer.
  9. Determine the highest temperature change to the nearest 0.5oC T4 as the final temperature of the mixture.
  10. Repeat the experiment to complete table 1.

Enthalpy change ΔH of neutralization.

ΔH = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ΔT(T6)   => (50 +50) x 4.2 x 13.5 = 5670Joules = 5.67kJ

The molar heat of neutralization the acid.

     ΔHn = Enthalpy change ΔH => 5.67kJ = 56.7kJ mole-1  
                  Number of moles       0.1 moles

Write ionic equation for the reaction that takes place

  OH-(aq)+ H+(aq) ->   H2O(l)

The theoretical enthalpy change is 57.4kJ. Explain the difference with the results above.

The theoretical value is higher
Heat/energy loss to the surrounding/environment lowers ΔT/T6 and thus ΔH
Heat/energy is absorbed by the reaction vessel/ calorimeter /plastic cup lowers ΔT and hence ΔHn

Experiment

 
II 

Temperature of acid T1 (oC)

22.5

 

22.5

 

Temperature of base T2 (oC)

22.0

23.0

Final temperature of solution T4(oC)

35.5

36.0

Initial temperature of solution T3(oC)

22.25

22.75

 
 

Temperature change( T5)

13.25

 13.75

 

  1. Calculate T6 the average temperature change    
    T6   =   13.25 +13.75      =      13.5 oC
                      2
  2. Why should the apparatus be very clean?
    Impurities present in the apparatus reacts with acid /base lowering the overall temperature change and hence ΔHᶿn
  3. Calculate the:
    1. number of moles of the acid used
      number of moles = molarity x volume  =>  2 x 50    =   0.1moles
                                             1000                 1000
  4. Compare the ΔHn of the experiment above with similar experiment repeated with neutralization of a solution of:
    1. potassium hydroxide with nitric(V) acid

      The results would be the same/similar.
      Both are neutralization reactions of strong acids and bases/alkalis that are fully /wholly dissociated into many free H+ / H3O+ and OH- ions.
    2. ammonia with ethanoic acid
      The results would be lower/ΔHn would be less.
      Both are neutralization reactions of weak acids and bases/alkalis that are partially /partly dissociated into few free H+ / H3O+ and OH- ions. Some energy is used to ionize the molecule.
  5. Draw an energy level diagram to illustrate the energy changes
    enthalpy change of neutralization

Theoretical examples

  1. The molar enthalpy of neutralization was experimentary shown to be 51.5kJ per mole of 0.5M hydrochloric acid and 0.5M sodium hydroxide. If the volume of sodium hydroxide was 20cm3, what was the volume of hydrochloric acid used if the reaction produced a 5.0oC rise in temperature?

    Moles of sodium hydroxide = molarity x volume
                                                    1000

    => 0.5 M x 20cm3 = 0.01 moles
               1000

    Enthalpy change ΔH =     ΔHn           =>       51.5   = 0.515kJ

    Moles sodium hydroxide   0.01 moles

    Mass of base + acid =       Enthalpy change ΔH in Joules
                                              Specific heat capacity x ΔT

     =>   0.515kJ   x 1000     =   24.5238g
                     4.2 x 5

    Mass/volume of HCl = Total volume – volume of NaOH
                             =>24.5238   - 20.0 =   4.5238 cm

    Graphically ΔHn can be determined as in the example below:

    Procedure

    Place 8 test tubes in a test tube rack.
    Put 5cm3 of 2M sodium hydroxide solution into each test tube. Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker.
    Record its initial temperature at volume of base =0.
    Put one portion of the base into the beaker containing the acid.
    Stir carefully with the thermometer and record the highest temperature change to the nearest 0.5oC.
    Repeat the procedure above with other portions of the base to complete table 1 below

    Volume of acid(cm3)

    25.0  25.0  25.0  25.0  25.0  25.0 25.0 25.0 25.0
     

    Volume of alkali(cm3)

    0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0

    Final temperature(oC)

    22.0 24.0 26.0 28.0 28.0 27.0 26.0 25.0 24.0

    Initial temperature(oC)

    22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0 22.0

    Change in temperature

    0.0 2.0 4.0 6.0 6.0 5.0 4.0 3.0 2.0


    Complete the table to determine the change in temperature.

    Plot a graph of volume of sodium hydroxide against temperature change.

    From the graph show and determine :

     (i)the highest temperature change  ΔT

         ΔT =T2-T1 : highest temperature-T2 (from extrapolating a correctly plotted graph)
                              less lowest temperature at volume of base=0-T1

        => 28.7 – 22.0 = 6.70 oC

    enthalpy of neutralization

    The volume of sodium hydroxide used for complete neutralization

    From correctly plotted graph = 16.75 cm3

    Calculate the number of moles of the alkali used

    Moles NaOH = molarity x volume ()Vn=
                                    1000

       =>   2  x  16.75 = 0.0335 moles
                  1000

    Calculate ΔH for the reaction.

    ΔH = mass of solution mixture   x c x ΔT
              => (25.0 + 16.75) x 4.2 x 6.7

        =  1174.845 J    =   1.174845 kJ
                1000

    (iii) Calculate the molar enthalpy of the alkali:

      ΔHn = Heat change    =>  1.174845 kJ
             number of moles      0.0335 moles

               = 35.0699kJ mole-1  



Standard enthalpy/heat of solution/dissolution ΔHos

The standard enthalpy of solution ΔHos is defined as the energy change when one mole of a substance is dissolved in excess distilled water to form an infinite dilute solution.
An infinite dilute solution is one which is too dilute to be diluted further.
Practically the heat of solution is determined by dissolving a known mass /volume of a solute in known mass/volume of water/solvent and determining the temperature change.

To determine the heat of dissolution of ammonium nitrate(V)

Place 100cm3 of  distilled water into a plastic cup/beaker/calorimeter
Put all the 5.0g of ammonium nitrate(v)/potassium nitrate(V)/ ammonium chloride into the water.
Stir the mixture using the thermometer and record the temperature change after every ½ minute to complete table1.
Continue stirring throughout the experiment.

Sample results table 1

Time(minutes)

 

0

 
½

1

1½ 

 

2

 

2½ 

 

Temperature (oC)

22.0  21.0 20.0 19.0 19.0 19.5 20.0 20.5

Plot a graph of temperature (y-axis )against temperature

heat of dissolution of ammonium nitrate

 

  1. From the graph show and determine :
    1. the highest temperature change  ΔT
      ΔT =T2-T1 : highest temperature-T2 (from extrapolating a correctly plotted graph) less lowest temperature at volume of base=0-T1
          => 18.7 – 22.0 = 3.3oC        ( not -3.3oC)
  2. Calculate the total energy change ΔH during the reaction
     ΔH = mass of water x c x ΔT
    =>ΔH=100 x4.2 x 3.3oC  =  + 1386 J    =   + 1.386 kJ
                                                    1000
  3. Calculate the number of moles of ammonium nitrate(v) used
    Moles   =      mass        =>     5.0      =      0.0625 moles
                    molar mass             80
  4. What is the molar heat of dissolution of ammonium nitrate(V)

    ΔH = Heat change   =   + 1.386 kJ = + 22.176 kJmole-1
           Number of mole    0.0625 moles

  5. What would happen if the distilled water is heated before experiment was performed .

    The ammonium nitrate(V) would take less time to dissolve.
    Increase in temperature reduces lattice energy causing endothermic dissollution to be faster.

  6. Illustrate the above process on an energy level diagram

    Graphically ΔHs can be represented in an energy level diagram
    Endothermic process

    energy level diagram

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