Thermochemistry / energy changes - KCSE Chemistry Paper 3 - Practicals

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)	0	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

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

Note:

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

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

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)

Energy changes in chemical processes

1. Standard enthalpy/heat of displacement ΔH^o_d
2. Standard enthalpy/heat of neutralization ΔH^o_n
3. Standard enthalpy/heat of solution/dissolution ΔH^o_s

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 ΔH^o_d

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 (ΔH^o_d) 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 0.5^oC- 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    =   5.5^oC
                            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 ΔH_d = 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 ΔH_d 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 ΔH_d

     =>     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 0.5^oC.

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

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    =   11.5^oC
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_d 
                       =>  -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 ΔH^o_n

The molar standard enthalpy/heat of neutralization ΔH^on 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 ΔH_n 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 0.5^oC 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.

     ΔH_n = 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_n 
Heat/energy is absorbed by the reaction vessel/ calorimeter /plastic cup lowers ΔT and hence ΔH_n

Experiment	I	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 ΔHᶿ_n. 
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 ΔH_n 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/ΔH_n 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
   
   
   

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 5.0^oC rise in temperature?
   
   Moles of sodium hydroxide = molarity x volume
                                                   1000
   
   => 0.5 M x 20cm³ = 0.01 moles
              1000
   
   Enthalpy change ΔH =     ΔH_n           =>       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 cm³ 
   
   

   Graphically ΔH_n 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 0.5^oC.
   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

   
   
   

   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:

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

              = 35.0699kJ mole^-1  

   
   
   

Standard enthalpy/heat of solution/dissolution ΔH^o_s

The standard enthalpy of solution ΔH^o_s 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	3	3½
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

 

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 = 3.3^oC        ( not -3.3^oC)
2. Calculate the total energy change ΔH during the reaction
    ΔH = mass of water x c x ΔT
   =>ΔH=100 x4.2 x 3.3^oC  =  + 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

   

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