Energy Changes in Chemical and Physical Processes Questions and Answers - Chemistry Form 4 Topical Revision

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Questions

  1. 6g of Potassium nitrate solid was added to 120cm3 of water in a plastic beaker.
    The mixture was stirred gently and the following results were obtained.
    Initial temperature = 21.5oC
    Final temperature = 17.0oC
    1. Calculate the enthalpy change for the reaction (Density =1g/cm3, C= 4.2 Jg-1K-1)
    2. Calculate the molar enthalpy change for the dissolution of potassium nitrate (K=39, N= 14, O =16)
    1. The heat of combustion of ethanol, C2H5OH is 1370 KJ/mole.
      1. What is meant by heat of combustion?
      2. Calculate the heating value of ethanol (H = 1.0, C = 12.0, O = 16.0)
  2. Use the information below to answer the questions that follow:-
    Ca(s) + ½O2(g) → CaO(s) ΔH =-635KJ/mol
    C(s) + O2(g) → CO2(g) ΔH= -394KJ/mol
    Ca(s) + C(s) + 3/2O2(g) → CaCO3 ΔH = -1207KJ/mol
    Calculate the enthalpy change for the reaction:
    Ca(s) + CO2(g) → CaCO3(s)
  3. 0.92g of ethanol were found to burn in excess air producing a temperature rise of 32.5ºC in 200cm3 of water. 
    C=12.0 H=1.0 O=16.0
    Density of water 1g/cm3
    Specific heat capacity of water 4.2 KJ kg-1K-1
    1. Write the equation for combustion of ethanol
    2. Determine the molar heat of combustion of ethanol
  4. Study the information in the following table and answer the questions that follow. The letters do not represent the actual chemical symbols of the elements.
    ELEMENT U V X Y Z
    NUMBER OF PROTONS 18  20  6 16  19  17
    NUMBER OF NEUTRONS  22  20  8 16  20  20

    Which of the above elements are:
    1. Likely to be radioactive?
    2. Able to form a compound with the highest ionic character?
  5. The diagram below shows energy levels for the reaction
    ½ H2(g) + ½F2(g) → HF(g)
    energy level q6
    1. Work out the activation energy for the reaction
    2. Calculate the heat of formation of HF
    3. Is the reaction endothermic or exothermic?
  6. Using the heats of combustion of the following substances, calculate the heat of formation of ethanol
    C(s) + O2 (g) →CO2 (g); ΔH = -393KJmol-1
    H2 (g) + ½ O2 (g) → H2O(l); ΔH = -286KJmol-1
    CH3CH2OH(l) + O2 (g) → 2CO2 (g) + 3H2(l) H = 1386KJmol-1
  7. Nitrogen and hydrogen react reversibly according to the equation:-
    N2(g) + 3H2(g) → 2NH3(g); ΔH = -92kjmol-1
    The energy level diagram for the above reaction is shown below:-
    Energy level q8
    1. How would the yield of ammonia be affected by:
      1. A decrease in temperature
      2. An increase in pressure
    2. How does a catalyst affect reversible reaction already in equilibrium?
    3. On the above diagram, sketch the energy level diagram that would be obtained when iron catalyst is added to the reaction
  8. Study the table below and answer the questions that follow
    Bond type          bond energy kJmol-1
    C - C                           346
    C = C                          610
    C - H                           413
    C - Br                          280
    Br - Br                        193
    1. Calculate the enthalpy change for the following reaction
      C2H4(g) + Br2(g) → C2H4Br2(g)
    2. Name the type of reaction that took place in (a) above
  9. Bond energies for some bonds are tabulated below:-
    BOND BOND ENERGY KJ/mol
    H – H 436
    C = C 610
    C- H 410
    C - C 345
    Use the bond energies to estimate the enthalpy for the reaction
    C2H4(g) + H2(g) → C2H6(g)
  10. The able shows the results obtained when 20.2g of potassium nitrate was added in 50cm3 of water.
    Time in (min) 0.0 0.3 1.0 1.3 2.0 2.3 3.0 3.3 4.0
    Temperature (oC )  25.0  25.0  25.0  25.0  17.0  17.0  20.0  20.0  20.0
    1. Draw the graph of temperature against time
    2. Using the graph, determine the temperature change
    3. Calculate the heat change
    4. Find the molar heat of solution of potassium nitrate
  11. When 1.6g of ammonium nitrate were dissolved in 100cm3 of water, the temperature dropped by 6ºC. Calculate its enthalpy change. (Density of water = 1g/cm3specific heat capacity is 4.2kJ kg-1K-1)
  12. Sodium hydrogen carbonate was strongly heated.
    1. Write an equation for the reaction
    2. The grid below shows part of the periodic table. Use it to answer the questions that follow. The letters are not the actual symbols.
      periodictableq13
      1. Write the equation for the reaction that occurs between elements L and D
      2. The oxide of G reacts with both hydrochloric acid and sodium hydroxide. What is the nature of the oxide of G?
      3. Explain why elements H has a higher boiling points than element D.
      4. State one use of element E
      5. Compare and explain the atomic radius of B and C
      6. 11.5g of L was completely burnt in oxygen .Calculate the volume of gas that was used. (L = 23, molar gas volume at room temperature is 24dm3)
  13. A student has been provided with sodium hydroxide solution of 2M and hydrobromic acid of 4M. He was asked to investigate the equation for the reaction between these two substances and hence determine the molar enthalpy of neutralization. He carried out the reaction and obtained the following results:-
    Vol. of 4M Hydrobromic acid added to 20cm3 of 2M NaOH Temperature of the mixture (oC)
    4.0 26.8
    6.0 30.0
    8.0 33.2
    10.0 36.0
    12.0 35.2
    14.0 34.4
    20.0 30.8
    1. Draw a graph of the temperature of the mixture (vertical axis against the volume of the acid added)
    2. Using the graph estimate the temperature of the mixture when 17cm3 of the acid was added
    3. Both solutions were at room temperature at the start of the experiment. Use your graph to estimate the room temperature 
    4. What is the significance of the highest temperature of the solution mixture?
    5. The temperature of the mixture increased during the first additions of the acid. Why did the temperature increase?
    6. Suggest a reason why the temperature decreased during the latter part of the experiment
    7. Use your graph to determine the volume of 4M Hydrobromic acid which just neutralize 20cm3 of 2M NaOH 
    8. How many moles of Hydrobromic acid are present in your answer in (g) above? 
    9. How many moles of NaOH are present in 20cm3 of 2M of NaOH solution? 
    10. Use your answers in (h) and (i) above to write an equation of the reaction taking place in the experiment. Explain clearly how you have used your answers (1½mks)
    11. Determine the molar enthalpy of neutralization of hydrobromic acid (1½mks)
  14.   
    1. The following results were obtained in an experiment to determine the enthalpy of solution of sodium hydroxide
      Mass of plastic beaker = 8.0g
      Mass of plastic beaker + distilled water = 108.15g
      Mass of plastic beaker + distilled water + sodium hydroxide = 114.35g
      The table below shows the temperature at fixed times after mixing
      Time/seconds 0 30  60  90  120  150  180  210
      Temperature ( oC)  15  21  29  28  27 26 26 25

      1. Plot a graph of temperature (y-axis) against time (x-axis)
      2. From your graph, determine the maximum temperature attained
      3. Determine the temperature change of the reaction
      4. Calculate the number of moles of sodium hydroxide used in the experiment
        (Na = 11, H = 1, O = 16)
      5. Use your results to determine the molar enthalpy solution of sodium hydroxide. (Density of solution is 1g cm-3 , specific heat capacity of solution = 4.18 KJ-1K-1)
    2. Examine the energy level diagram below and use it to answer the questions that follow
      energy level diagram q15b
      1. Which ΔH values will have negative sign?
      2. What physical change is being represented where enthalpy change ΔH4 is involved? ( ½mk)
      3. In terms of ΔH1, ΔH2, ΔH3 and ΔH4, give the overall enthalpy change for the reaction:-
        H2(g) + ½O2(g) → H2O(l)
      4. Is the reaction in (iii) above exothermic or endothermic?
  15.   
    1. Distinguish between molar latent heat of fusion and molar latent heat of vaporization
    2. Study the graph below and answer the questions which follow:
      graph q16
      1. Explain the changes occurring between points
        BC ………………………………………… CD ………………………
      2. In an experiment to determine molar enthalpy of neutralization of hydrochloric acid using potassium hydroxide, the data below was obtained. The concentration of potassium hydroxide used was 0.5M
        Volume of 0.5M KOH (cm3 0 5 10  15  20  25  30  35
        Total volume of acid + Base  20  25  30  35  40  45  50  55
        Temperature (oC) 24  26  27  28  29  29  28  27
        1. Plot a graph of temperature (y-axis) against volume of potassium hydroxide used
        2. From your graph:
          1. Determine the temperature change
          2. Find the volume of potassium hydroxide which completely neutralized 20cm3 of the acid
        3. Calculate the heat change for the reaction (C = 4.2Jg-1K-1 density of solution = 1g/dm3)
        4. Calculate the molar enthalpy of neutralization of hydrochloric acid with potassium hydroxide
  16. A typical electrolysis cell uses a current of 40,000 amperes. Calculate the mass (in Kg of aluminium produced in one hour). (Al = 27) (Faraday = 96500 Coloumbs)
  17. Biogas is a mixture of mainly Carbon (IV) Oxide and methane.
    1. Give a reason why biogas can be used as a fuel
    2. Other than fractional distillation, describe a method that can be used to determine the percentage of methane in biogas
  18. Consider the following equilibrium reaction.
    H2(g) + Cl2(g) ⇌ 2HCl(g) ΔH= -74.4KJ
    State and explain the effect of formation of hydrogen chloride if pressure was increased in the equation above
  19. Turning of fossil fuels has adverse environmental effects:-
    1. Name two pollutants from the burning of petroleum products
    2. Give one precaution taken to minimise the pollution by fossil fuels
  20.   
    1. Define molar heat of neutralization
    2. The rise in temperature when 50cm3 of sodium hydroxide is reacted with two acids is given in the table below:-
      Acid 50cm3 of HCl  50cm of Oxalic acid
      Temp rise (oC)  7 4

      Explain the difference in the temperature.
  21. Calculate the latent heat of vaporization of water
    H2O(l) → H2O(g)
    Given the following thermo chemical equations:-
    H2(g) + ½O2(g) → H2O(g) ΔHθ= -242KJ/Mol
    H2(g) + ½O2(g) → H2O(l) ΔHθ= -286KJ/Mol
    1. Define the term fuel
    2. State four reasons why wood fuel is chosen for domestic cooking
  22. The setup bellow was used to investigate the changes that take place when sodium hydroxide pellets dissolve in water.
    dissolving sodium hydroxide in water
    1. Why is a plastic beaker used instead of a metallic beaker?
    2. State and explain the observations made in the above reaction
    1. What is a fuel? (1mark)
    2. Other than the cost, state two other factors to consider when choosing a fuel.
  23. The equation below represents changes in the physical state of ions metal:
    Fe(s) → Fe(l) ΔH= +15.4kjmol-1
    Fe(l) → Fe(g) ΔH=+354kjmol-1
    1. Calculate the amount of heat energy required to change 10kg of solid iron to gaseous iron (Fe = 56)
    2. Iodine can react with chlorine as shown below:-
      I2(g) + Cl(g) → 2lC(s) ΔH= -68kJ
      Determine the molar enthalpy change for this reaction
    3. Draw an energy level diagram for the reaction in (b) above
  24. Study the diagram below and answer the questions that follow:
    energy level q27
    1. What do ∆H1 and ∆H2 represent?
      ∆H1 …………………………………………………………………….
      ∆H2 ……………………………………………………………………..
    2. Write an expression to show the relationship between ∆H1, ∆H2 and ∆H3

Answers

  1.  
    1. ΔH = 120 x 4.2 x 4.5 ( ½mk)
                     1000
      = + 2.268KJ (½mk)
    2. RFM of KNO3 = 39 + 14 + 48 = 101
      6g → 2.268KJ
      101g → ?
      101 × 2.268 (½mk)
                         6
      = + 38.178KJ mol-1 (½mk
  2.  
    1. Heat evolved when one mole of a substance is completely burnt in oxygen
    2. RFM of C2H5OH = 46
      Molar mass = 46g
      Heating value = 1370 KJ
                               46g
      = 29.78KJ/g (with units)
  3. Ca(s) + C(s) + 3/2O2 (g)
  4.  
    1. C2H6O(l) + 3O(g) → 2CO2(g) + 3H2O
    2. ΔH = MCΔT 
      200 × 4.2 × 32.5 = -27.3Kj
      1000
      0.92g C2H6O = - 27.3Kj
      46g  C2H6O    = ?
      46g × 27.3Kj = -1365Kj
      0.92
      ΔHC C2H6O = -1365Kj mol
  5.  
    1. U,V,Y,Z All the 4 or nay 3 exclusively correct penalize ½ mk if wrong answer
    2. YZ is/are included any 2 correct ½ mk
  6.  
    1. 611 − 389 = +222 KJ
    2. H = +222 – (611 – 100)
      = -289KJ
    3. Exothermic reaction
  7.  
    energy flow diagram ans7
     âˆ†Hf + ∆H3 = ∆H1 + ∆H2
    ∴ ∆Hf = ∆H1 + ∆H2 - ∆H3 ½
    = -393 x 2 + -286 x 3 +1386 1
    = -786 – 858 + 1386
    = -1644 + 1386
    1
    ∆Hf = -258 KJmol-1 ½
  8.  
    1.  
      1. the yield of NH3 would be lowered √ ½ any supply of heat makes NH3 to decompose to N2 and H2
      2. the yield of NH3 would be increased
    2. a catalyst accelerate the rates of both forward and reverse reactions equally√ ½ . Equilibrium position is not affected by a catalyst√ ½
    3.  
      energy level diagram q8c
  9.  
    1. Breaking of ‘C = C’ =   +610 KJ
      Breaking of ‘Br – Br’ = +193
                                          803√
      Formation of 2C – Br = -560
      Formation of C - C       +243 Kj
                                         -346
                                       - 103KJ√ 2 marks
    2. Addition reaction/ halogenation √

  10. Bond breaking                       Bond formation
    4 C-H – 4 x 410 = 1640         6C – H    6 x 410
    C = C – 1 x 610 = 610                      = 2460
    H – H – 1 x 436 =  436             C – C –   345
                             2686                           2805
    ΔH = 2686 – 2805
    = -119 Kj/Mol
  11.  
    1. Graph
      labeling -*TZM*
      plotting – *TZM*
      scale – *TZM*
      line – *TZM*
      total 5mks
    2. Shown on the graph -*TZM*
    3. Heat change = MCT
      = 50 x 4.2 x 10.2
        100
      = 2.142kJ
    4. RFM of KNO3 = 39 + 14 + 48
      = 101
      H = 2.142 x 101 = -10.71Kjmol-1
                        20.2
  12. MCT = 100 × 4.2 × 6 = 2.52 Kj
              1000
    Moles of NH4NO3 = 1.6 = 0.02 moles
                                 80
    If 0.02 mol = 2.52 Kj
    1 mol = ?
     1 × 2.52 = +126KJ/ mol
      0.02
  13.  
    1. 2 NaHCO3 (g) → Na2CO3(g) + H2O(l) + CO2(g)
    2.  
      1. 2L(g) + D2(g)→ 2LD(g)
      2. Amphoteric oxide
      3. Element H has a giant atomic structure with strong covalent bonds throughout its structure while D has simple molecular structure with weak Vander wall forces (2 m)
      4.  - Used in advertising signs (Advertisements)
        - Used in florescent tubes (Any two correct use)
      5. C has a smaller atomic radius than B because it has stronger nuclear charge// more number of protons which attract the outer energy level electrons more firmly (2 mks)
      6. 4L(s) + O2(g) → 2 L2O(g)
        Moles of L = 11.5/23 = 0.5 moles
        Moles of O2 = 0.5/0.4 = 0.125 moles
        Volume of O2 = 0.125 mol X 24 = 3 dm3
        4L (s) + O2(g) → 2L2O(s)
        If 4 x 23g = 24dm3
        11.5g of L = ?
        11.5 x 24 = 3dm3
            4 x 23
  14.  
    1. Drawn on the graph
      A = ½ mk
      S = ½ mk
      P = ½ mk
      C = ½ mk
    2. 32.5oC + 1 Read from the student’s correctly plotted graph.
    3. 20oC + 0.5 Line is extrapolated downwards from the student’s correct graph.
    4. It is end point/ complete neutralization.
    5. The reaction is exothermic hence as reaction proceeded more heat was produced.
    6. Reaction was complete hence solution lost heat through radiation to the surrounding.
    7. 10.2 cm3 + 0.1. Read from the student’s correct graph.
    8. Moles = M x V
                  1000
      = 10.2 x 4 √ ½ = 0.0408 moles √ ½
            1000
    9. Moles = M x V
                  1000
      = 2 x 20 √ ½ = 0.04 moles √ ½
          1000
    10. HBr : NaOH
      0.0408 : 0.04
      0.0408 : 0.04
      0.04       0.04
      1 : 1
      HBr(aq) + NaOH(aq) → NaBr(aq) + H2O(l)
    11. ∆H = MC∆t
      = -30.2g x 4.2J x 16.3
                   g0c
      = -2067.49J √ ½
      Ans. in (h) = -2067.49 J.
      ∴ 1 Mole = 1 x 2067.49J √ ½ e.g. 1 x 2067.49
                       Ans in “h”                    0.0408
      = -Ans. e.g 50673.82 J mol-1
      Or 50.67382KJ mol-1 √ ½
  15.  
    1.  
      1. Max. temperature attained : 290c
      2. Temperature change of the reaction = (29-15)0c
        = 140c
      3. Mass of NaOH used = (114.35 – 108 .15)g
        = 6.2g
        R.F.M of NaOH = 40g
        Moles of NaOH used = 6.2/40 moles
        = 0.155moles
      4. Heat released = Mass × Specific Heat Capacity × Temperature change 

        Mass of water used = (108.15 – 8)g
        = 100.15g
        ∴ Heat released = 100.15/1000 × 4.18 × 14 kj
        =100.15kj
        0.155 moles NaOH → 5.861 kj
        1 mole NaOH → ?
        1 x 5.861 kjmol-1
          0.155
        = -37.8 kjmol-1
    2.  
      1. ΔH3 and ΔH4
      2. Condensation
      3. ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4
      4. Exothermic.
  16.  
    1. – Latent heat of fusion is the heat change that occurs when one mole of a solid substance changes into liquid at constant temperature.
      - Latent heat of vapourization is the heat change that occurs when one mole of liquid substance changes into gas at constant temperature.
    2.  
      1. – BC – The liquid loses heat as it cools hence decrease in kinetic energy of the particles
        - CD - The liquid changes to solid as temperature remains constant at freezing point.
      2.  
        1. Scale – *TZM*
          Plot – *TZM*
          Line
        2.  
          1. Should be shown on the graph – if not shown penalize ( ½ mk)
      3. Heat change = m x c x ΔT
        Where m = (vol. of acid (20cm3) + volume of bas in (b) above) x 1g/cm3
        ΔT-as read form the graph
      4. moles of acid
        Moles of base = 0.5 x volume in (b) above
                              1000
        Mole ratio acid: Base = 1:1
        Moles of acid → heat change in (iii)above
        1mole           →  ?
        Molar heat change = 1 x heat in (iii)
                                      Moles of acid
  17. Q = 40000 x 60 x 60 = 144000000c
    Mass of Al = 144000000 x 27
                            3 x 96500
    = 13.43kg


  18.  
    1.  
      1. Contains methane which is a fuel or contains methane which can burn
      2. Pass a known volume of biogas through Sodium hydroxide (Potassium hydroxide) solution to absorb Carbon (IV) Oxide. Measure the volume of remaining gas
        % = Volume of methane x 100
                Volume of Biogas
  19. No effect – Reaction is not accompanied by volume changes/ similar volumes of reactants and products
  20.  
    1. – carbon IV Oxide;
      - Sulphur IV Oxide;
      - Lead;
    2. Availed low sulphur diesel/ availed unleaded petrol
  21.  
    1. Heat change that occurs when one mole of hydrogen combines with one mole of hydroxide ions. //Heat evolved when one mole of water s formed during reaction of H+ and OH- ions
    2. HCl produces a higher temperature rise than oxalic acid;
      HCl is a stronger acid than oxalic acid;
  22.  
    energy flow diagram ans22
    ΔH2 = - ΔH1 + ΔH3
    = ΔH3 – ΔH1
    = -242 - -286
    = -242 + 286
    = +44KJ/mol (No units of sign = ½mk)
  23.  
    1. Chemical substance that burns to produce useful amount of heat.
    2. - Its cheap
      - Its readily available (½mk)
      - It burns slowly (½mk)
      - Does not produce poisonous gas. (½mk)
  24.  
    1. Metallic beaker would make most of the heat be lost to the environment
    2. - Thermometer reading increased
      - The reaction is exothermic
  25.  
    1. A substance that produce heat energy when burnt
    2. - Availability
      - ease of transport
  26.  
    1. 1 mole Fe (56) required → 15.4 + 354 = 396.5Kj
      10,000 (10 kg) → ?

      10,000g × 369.5 Kj
                 56g
      = 6596.285Kj
    2.  - 68Kj/2 = - 34 Kj ✓½
  27.  
    1. ∆H1 – Lattice energy ✓1
      ∆H2 – Hydrogen energy ✓1
    2. ∆H3 = ∆H2 + ∆H1 ✓1

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