The SI unit of energy is the Joule(J).Kilo Joules(kJ)and megaJoules(MJ) are also used. The Joule(J) is defined as the: (i) quantity of energy transferred when a force of one newton acts through a distance of one metre. 2 (ii) quantity of energy transferred when one coulomb of electric charge is passed through a potential difference of one volt. All thermochemical reactions should be carried out at standard conditions of: (i) 298K /25oC temperature (ii)101300Pa/101300N/m2 /760mmHg/1 atmosphere pressure. 2.Exothermic and endothermic processes/reactions Some reactions / processes take place with evolution/production of energy. They are said to be exothermic while others take place with absorption of energy. They are said to be endothermic. Practically exothermic reactions / processes cause a rise in temperature (by a rise in thermometer reading/mercury or alcohol level rise) Practically endothermic reactions / processes cause a fall in temperature (by a fall in thermometer reading/mercury or alcohol level decrease) To demonstrate/illustrate exothermic and endothermic processes/reactions a) Dissolving Potassium nitrate(V)/ammonium chloride crystals Procedure: Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Put about 1.0g of Potassium nitrate(V) crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using ammonium chloride in place of Potassium nitrate (V) crystals. Sample results Temperture (oC) Using Potassium nitrate(V) crystals Using Ammonium chloride crystals T2(Final temperature) 21.0 23.0 T1 (Initial temperature) 25.0 26.0 Change in temperature(T2 –T1) 4.0 3.0 Note: (i)Initial(T1) temperature of dissolution of both potassium nitrate(V) crystals and ammonium chloride crystals is higher than the final temperature(T2) (ii) Change in temperature(T2 –T1) is not a mathematical “-4.0” or “-3.0”.


Determine and record its temperature T1.Put about 1.0g of Potassium nitrate(V) crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using ammonium chloride in place of Potassium nitrate (V) crystals. Sample results Temperture (oC) Using Potassium nitrate(V) crystals Using Ammonium chloride crystals T2(Final temperature) 21.0 23.0 T1 (Initial temperature) 25.0 26.0 Change in temperature(T2 –T1) 4.0 3.0 Note: (i)Initial(T1) temperature of dissolution of both potassium nitrate(V) crystals and ammonium chloride crystals is higher than the final temperature(T2) (ii) Change in temperature(T2 –T1) is not a mathematical “-4.0” or “-3.0”. (iii)Dissolution of both potassium nitrate(V) and ammonium chloride crystals is an endothermic process because initial(T1) temperature is higher than the final temperature(T2) thus causes a fall/drop in temperature. b) Dissolving concentrated sulphuric(VI) acid/sodium hydroxide crystals
3 Procedure: Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Carefully put about 1.0g/four pellets of sodium hydroxide crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using 2cm3 of concentrated sulphuric(VI) acid in place of sodium hydroxide crystals. CAUTION: (i)Sodium hydroxide crystals are caustic and cause painful blisters on contact with skin. (ii) Concentrated sulphuric (VI) acid is corrosive and cause painful wounds on contact with skin. Sample results Temperture (oC) Using Sodium hydroxide pellets Using Concentrated sulphuric(VI) acid T2(Final temperature) 30.0 32.0 T1 (Initial temperature) 24.0 25.0 Change in temperature(T2 –T1) 6.0 7.0 Note: (i)Initial (T1) temperature of dissolution of both concentrated sulphuric (VI) acid and sodium hydroxide pellets is lower than the final temperature (T2).


CAUTION: (i)Sodium hydroxide crystals are caustic and cause painful blisters on contact with skin. (ii) Concentrated sulphuric (VI) acid is corrosive and cause painful wounds on contact with skin. Sample results Temperture (oC) Using Sodium hydroxide pellets Using Concentrated sulphuric(VI) acid T2(Final temperature) 30.0 32.0 T1 (Initial temperature) 24.0 25.0 Change in temperature(T2 –T1) 6.0 7.0 Note: (i)Initial (T1) temperature of dissolution of both concentrated sulphuric (VI) acid and sodium hydroxide pellets is lower than the final temperature (T2). (ii)Dissolution of both Sodium hydroxide pellets and concentrated sulphuric (VI) acid is an exothermic process because final (T2) temperature is higher than the initial temperature (T1) thus causes a rise in temperature. The above reactions show heat loss to and heat gain from the surrounding as illustrated by a rise and fall in temperature/thermometer readings. Dissolving both potassium nitrate(V) and ammonium chloride crystals causes heat gain from the surrounding that causes fall in thermometer reading. Dissolving both Sodium hydroxide pellets and concentrated sulphuric (VI) acid causes heat loss to the surrounding that causes rise in thermometer reading. At the same temperature and pressure ,heat absorbed and released is called enthalpy/ heat content denoted H. Energy change is measured from the heat content/enthalpy of the final and initial products. It is denoted ∆H(delta H).i.e.


At the same temperature and pressure ,heat absorbed and released is called enthalpy/ heat content denoted H. Energy change is measured from the heat content/enthalpy of the final and initial products. It is denoted ∆H(delta H).i.e. Enthalpy/energy/ change in heat content ∆H = Hfinal – Hinitial For chemical reactions: ∆H = Hproducts – Hreactants For exothermic reactions, the heat contents of the reactants is more than/higher than the heat contents of products, therefore the ∆H is negative (-∆H)
4 For endothermic reactions, the heat contents of the reactants is less than/lower than the heat contents of products, therefore the ∆H is negative (+∆H) Graphically, in a sketch energy level diagram: (i)For endothermic reactions the heat content of the reactants should be relatively/slightly lower than the heat content of the products (ii)For exothermic reactions the heat content of the reactants should be relatively/slightly higher than the heat content of the products Sketch energy level diagrams for endothermic dissolution Energy (kJ) H2 KNO3(aq) +∆H = H2 – H1 H1 KNO3(s) Reaction path/coordinate/progress Energy (kJ) H2 NH4Cl (aq) +∆H = H2 – H1 H1 NH4Cl (s) Reaction path/coordinate/progress Sketch energy level diagrams for exothermic dissolution
5 H2 NaOH (s) Energy(kJ) -∆H = H2 – H1 H1 NaOH (aq) Reaction path/coordinate/progress H2 H2SO4 (l) Energy (kJ) -∆H = H2 – H1 H1 H2SO4 (aq) Reaction path/coordinate/progress 3.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 is the physical change of a liquid to solid. Melting/freezing/fusion/solidification are therefore two opposite but same reversible physical processes.


Melting/fusion is the physical change of a solid to liquid. Freezing is the physical change of a liquid to solid. Melting/freezing/fusion/solidification are therefore two opposite but same reversible physical processes. i.e A (s) ========A(l)
6 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/vaporization/evaporation and condensation/liquidification are therefore two opposite but same reversible physical processes. i.e B (l) ========B(g) Practically (i) 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/freezing/fusion is an endothermic (+∆H)process that require/absorb energy from the surrounding. (ii)Freezing/fusion/solidification involves cooling a 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 (-∆H)process that require particles holding the liquid together to lose energy to the surrounding. (iii)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 (+∆H) process that require/absorb energy from the surrounding. (iv)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(+∆H) 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.


(iv)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(+∆H) 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) ∆H = +6.0kJ mole-1 (endothermic process) H2O(l) -> H2O(s) ∆H = -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. e.g. H2O(l) -> H2O(g) ∆H = +44.0kJ mole-1 (endothermic process) H2O(g) -> H2O(l) ∆H = -44.0kJ mole-1 (exothermic process)
7 The following experiments illustrate/demonstrate practical determination of melting and boiling a) 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 minutes. 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 Questions 1.Plot a graph of temperature against time(y-axis) Sketch graph of temperature against time boiling point 96 oC Temperature(0C) 25oC time(seconds) 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. The sample results above are from Kiriari Girls High School-Embu County on the slopes of Mt Kenya in Kenya. Water here boils at 96oC.


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 Questions 1.Plot a graph of temperature against time(y-axis) Sketch graph of temperature against time boiling point 96 oC Temperature(0C) 25oC time(seconds) 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. The sample results above are from Kiriari Girls High School-Embu County on the slopes of Mt Kenya in Kenya. Water here boils at 96oC. 8 3.Calculate the molar heat of vaporization of water.(H= 1.0,O= 16.O) Working: 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 Molar mass of H2O =>5.964kJ = 0.3313 kJ mole -1 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 30seconds for four minutes.


Insert a thermometer and remove the boiling tube from the flame. Stir continuously. Determine and record the temperature after every 30seconds for four minutes. Sample results 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 Questions 1.Plot a graph of temperature against time(y-axis) Sketch graph of temperature against time 93 oC Temperature(0C) melting point 69oC time(seconds) 2.From the graph show and determine the melting point of the candle wax
9 4.Energy changes in chemical processes Thermochemical reactions measured at standard conditions of 298K(25oC) and 101300Pa/101300Nm2/ 1 atmospheres/760mmHg/76cmHg produce standard enthalpies denoted ∆Hᶿ. Thermochemical reactions are named from the type of reaction producing the energy change. Below are some thermochemical reactions: (a) Standard enthalpy/heat of reaction ∆Hᶿr (b) Standard enthalpy/heat of combustion ∆Hᶿc (c) Standard enthalpy/heat of displacement ∆Hᶿd (d) Standard enthalpy/heat of neutralization ∆Hᶿn (e) Standard enthalpy/heat of solution/dissolution ∆Hᶿs (f) Standard enthalpy/heat of formation ∆Hᶿf (a)Standard enthalpy/heat of reaction ∆Hᶿr The molar standard enthalpy/heat of reaction may be defined as the energy/heat change when one mole of products is formed at standard conditions A chemical reaction involves the reactants forming products. For the reaction to take place the bonds holding the reactants must be broken so that new bonds of the products are formed. i.e. A-B + C-D -> A-C + B-D Old Bonds broken A-B and C-D on reactants New Bonds formed A-C and B-D on products The energy required to break one mole of a (covalent) bond is called bond dissociation energy.


For the reaction to take place the bonds holding the reactants must be broken so that new bonds of the products are formed. i.e. A-B + C-D -> A-C + B-D Old Bonds broken A-B and C-D on reactants New Bonds formed A-C and B-D on products The energy required to break one mole of a (covalent) bond is called bond dissociation energy. The SI unit of bond dissociation energy is kJmole-1 The higher the bond dissociation energy the stronger the (covalent)bond Bond dissociation energies of some (covalent)bonds Bond Bond dissociation energy (kJmole-1) Bond dissociation energy (kJmole-1) H-H 431 I-I 151 C-C 436 C-H 413 C=C 612 O-H 463 C = C 836 C-O 358 N = N 945 H-Cl 428 N-H 391 H-Br 366 F-F 158 C-Cl 346 Cl-Cl 239 C-Br 276 Br-Br 193 C-I 338 H-I 299 O=O 497 Si-Si 226 C-F 494
10 The molar enthalpy of reaction can be calculated from the bond dissociation energy by: (i)adding the total bond dissociation energy of the reactants(endothermic process/+∆H) and total bond dissociation energy of the products(exothermic process/-∆H). (ii)subtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the products(exothermic process/-∆H less/minus endothermic process/+∆H). Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic.


The SI unit of bond dissociation energy is kJmole-1 The higher the bond dissociation energy the stronger the (covalent)bond Bond dissociation energies of some (covalent)bonds Bond Bond dissociation energy (kJmole-1) Bond dissociation energy (kJmole-1) H-H 431 I-I 151 C-C 436 C-H 413 C=C 612 O-H 463 C = C 836 C-O 358 N = N 945 H-Cl 428 N-H 391 H-Br 366 F-F 158 C-Cl 346 Cl-Cl 239 C-Br 276 Br-Br 193 C-I 338 H-I 299 O=O 497 Si-Si 226 C-F 494
10 The molar enthalpy of reaction can be calculated from the bond dissociation energy by: (i)adding the total bond dissociation energy of the reactants(endothermic process/+∆H) and total bond dissociation energy of the products(exothermic process/-∆H). (ii)subtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the products(exothermic process/-∆H less/minus endothermic process/+∆H). Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic.


(ii)subtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the products(exothermic process/-∆H less/minus endothermic process/+∆H). Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)
11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic.


Practice examples/Calculating ∆Hr 1.Calculate ∆Hr from the following reaction: a) H2(g) + Cl2(g) -> 2HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (H-H + Cl-Cl) => (+431 + (+ 239)) = + 670kJ New bonds broken (exothermic process/-∆H ) = (2(H-Cl ) => (- 428 x 2)) = -856kJ ∆Hr =( + 670kJ + -856kJ) = 186 kJ = -93kJ mole-1 2 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)
11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic. The thermochemical reaction is thus: CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g) ∆H = +136 kJ Note that: (i)a reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products.


The thermochemical reaction is thus: ½ H2(g) + ½ Cl2(g) -> HCl(g) ∆Hr = -93kJ b) CH4(g) + Cl2(g) -> CH3Cl + HCl(g) Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl) => ((4 x +413) + (+ 239)) = + 1891kJ New bonds broken (exothermic process/-∆H ) = (3(C-H + H-Cl + C-Cl) => (( 3 x - 413) + 428 + 346) = -2013 kJ ∆Hr =( + 1891kJ + -2013 kJ) = -122 kJ mole-1 The above reaction has negative -∆H enthalpy change and is therefore practically exothermic. The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)
11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic. The thermochemical reaction is thus: CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g) ∆H = +136 kJ Note that: (i)a reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products. (ii)a reaction is endothermic if the bond dissociation energy of reactants is less than bond dissociation energy of products.


The thermochemical reaction is thus: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) ∆H = -122 kJ c) CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g)
11 Working Old bonds broken (endothermic process/+∆H ) = (4(C-H) + Cl-Cl + C=C) => ((4 x +413) + (+ 239) +(612)) = + 2503kJ New bonds broken (exothermic process/-∆H ) = (4(C-H + C-C + 2(C-Cl) ) => (( 3 x - 413) + -436 +2 x 346 = -2367 kJ ∆Hr =( + 2503kJ + -2367 kJ) = +136 kJ mole-1 The above reaction has negative +∆H enthalpy change and is therefore practically endothermic. The thermochemical reaction is thus: CH2CH2(g) + Cl2(g) -> CH3Cl CH3Cl (g) ∆H = +136 kJ Note that: (i)a reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products. (ii)a reaction is endothermic if the bond dissociation energy of reactants is less than bond dissociation energy of products. 12 (b)Standard enthalpy/heat of combustion ∆Hᶿc The molar standard enthalpy/heat of combustion(∆Hᶿc) is defined as the energy/heat change when one mole of a substance is burnt in oxygen/excess air at standard conditions. Burning is the reaction of a substance with oxygen/air. It is an exothermic process producing a lot of energy in form of heat. A substance that undergoes burning is called a fuel. A fuel is defined as the combustible substance which burns in air to give heat energy for domestic or industrial use.


It is an exothermic process producing a lot of energy in form of heat. A substance that undergoes burning is called a fuel. A fuel is defined as the combustible substance which burns in air to give heat energy for domestic or industrial use. A fuel may be solid (e.g coal, wood, charcoal) liquid (e.g petrol, paraffin, ethanol, kerosene) or gas (e.g liquefied petroleum gas/LPG, Water gasCO2/H2, biogas-methane, Natural gas-mixture of hydrocarbons) To determine the molar standard enthalpy/heat of combustion(∆Hᶿc) of ethanol Procedure Put 20cm3 of distilled water into a 50cm3 beaker. Clamp the beaker. Determine the temperature of the water T1.Weigh an empty burner(empty tin with wick). 13 Record its mass M1.Put some ethanol into the burner. Weigh again the burner with the ethanol and record its mass M2. Ignite the burner and place it below the clamped 50cm3 beaker. Heat the water in the beaker for about one minute. Put off the burner. Record the highest temperature rise of the water, T2. Weigh the burner again and record its mass M3 Sample results: Volume of water used 20cm3 Temperature of the water before heating T1 25.0oC Temperature of the water after heating T2 35.0oC Mass of empty burner M1 28.3g Mass of empty burner + ethanol before igniting M2 29.1g Mass of empty burner + ethanol after igniting M3 28.7g Sample calculations: 1.Calculate: (a) ∆T the change in temperature ∆T = T2 – T1 => (35.0oC – 25.0oC) = 10.0oC (b) the mass of ethanol used in burning mass of ethanol used = M2 – M1 => 29.1g – 28.7g = 0.4g (c) the number of moles of ethanol used in burning moles of ethanol = mass used => 0.4 = 0.0087 /8.7 x 10-3 moles molar mass of ethanol 46 2.


Put off the burner. Record the highest temperature rise of the water, T2. Weigh the burner again and record its mass M3 Sample results: Volume of water used 20cm3 Temperature of the water before heating T1 25.0oC Temperature of the water after heating T2 35.0oC Mass of empty burner M1 28.3g Mass of empty burner + ethanol before igniting M2 29.1g Mass of empty burner + ethanol after igniting M3 28.7g Sample calculations: 1.Calculate: (a) ∆T the change in temperature ∆T = T2 – T1 => (35.0oC – 25.0oC) = 10.0oC (b) the mass of ethanol used in burning mass of ethanol used = M2 – M1 => 29.1g – 28.7g = 0.4g (c) the number of moles of ethanol used in burning moles of ethanol = mass used => 0.4 = 0.0087 /8.7 x 10-3 moles molar mass of ethanol 46 2. Given that the specific heat capacity of water is 4.2 kJ-1kg-1K-1,determine the heat produced during the burning. Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 20 x 4.2 x 10 = 840 Joules = 0.84 kJ 1000 3.Calculate the molar heat of combustion of ethanol Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 0.84 kJ = 96.5517 kJmole-1 0.0087 /8.7 x 10-3 moles 4.List two sources of error in the above experiment. (i)Heat loss to the surrounding lowers the practical value of the molar heat of combustion of ethanol. 14 A draught shield tries to minimize the loss by protecting wind from wobbling the flame. (ii) Heat gain by reaction vessels/beaker lowers ∆T and hence ∆Hc 5.Calculate the heating value of the fuel.


(i)Heat loss to the surrounding lowers the practical value of the molar heat of combustion of ethanol. 14 A draught shield tries to minimize the loss by protecting wind from wobbling the flame. (ii) Heat gain by reaction vessels/beaker lowers ∆T and hence ∆Hc 5.Calculate the heating value of the fuel. Heating value = molar heat of combustion => 96.5517 kJmole-1 = 2.0989 kJg-1 Molar mass of fuel 46 g 6.Explain other factors used to determine the choice of fuel for domestic and industrial use. (i) availability and affordability-some fuels are more available cheaply in rural than in urban areas at a lower cost. (ii)cost of storage and transmission-a fuel should be easy to transport and store safely. e.g LPG is very convenient to store and use. Charcoal and wood are bulky. (iii)environmental effects –Most fuels after burning produce carbon(IV) oxide gas as a byproduct. Carbon(IV) oxide gas is green house gas that causes global warming. Some other fuel produce acidic gases like sulphur(IV) oxide ,and nitrogen(IV) oxide. These gases cause acid rain. Internal combustion engines exhaust produce lead vapour from leaded petrol and diesel. Lead is carcinogenic. (iv)ignition point-The temperature at which a fuel must be heated before it burns in air is the ignition point. Fuels like petrol have very low ignition point, making it highly flammable. Charcoal and wood have very high ignition point. 7.Explain the methods used to reduce pollution from common fuels. (i)Planting trees-Plants absorb excess carbon(IV)oxide for photosynthesis and release oxygen gas to the atmosphere. (ii)using catalytic converters in internal combustion engines that convert harmful/toxic/poisonous gases like carbon(II)oxide and nitrogen(IV)oxide to harmless non-poisonous carbon(IV)oxide, water and nitrogen gas by using platinum-rhodium catalyst along the engine exhaust pipes. Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0).


(i)Planting trees-Plants absorb excess carbon(IV)oxide for photosynthesis and release oxygen gas to the atmosphere. (ii)using catalytic converters in internal combustion engines that convert harmful/toxic/poisonous gases like carbon(II)oxide and nitrogen(IV)oxide to harmless non-poisonous carbon(IV)oxide, water and nitrogen gas by using platinum-rhodium catalyst along the engine exhaust pipes. Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0). Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000
15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,).


Further practice calculations 1.Calculate the heating value of methanol CH3OH given that 0.87g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oC.(C12.0,H=1.0 O=16.0). Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000
15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution.


Moles of methanol used = Mass of methanol used => 0.87 g = 0.02718 moles Molar mass of methanol 32 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 500 x 4.2 x 7 = 14700 Joules = 14.7 kJ 1000
15 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 14.7 kJ = 540.8389 kJmole-1 0.02718 moles Heating value = molar heat of combustion => 540.8389 kJmole-1 = 16.9012 kJg-1 Molar mass of fuel 32 g 2. 1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g.


1.0 g of carbon burn in excess air to raise the temperature of 400g of water by 18oC.Determine the molar heat of combustion and hence the heating value of carbon(C-12.0,). Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead.


Moles of carbon used = Mass of carbon used => 1.0 g = 0.0833 moles Molar mass of carbon 12 Heat produced ∆H = mass of water(m) x specific heat capacity (c)x ∆T => 400 x 4.2 x 18 = 30240 Joules = 30.24 kJ 1000 Molar heat of combustion ∆Hc = Heat produced ∆H Number of moles of fuel => 30.24 kJ = 363.0252 kJmole-1 0.0833 moles Heating value = molar heat of combustion => 363.0252 kJmole-1= 30.2521 kJg-1 Molar mass of fuel 12 g (c)Standard enthalpy/heat of displacement ∆Hᶿd The molar standard enthalpy/heat of displacement ∆Hᶿd is defined as the energy/heat change when one mole of a substance is displaced from its solution. A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead. (iv)Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq) Ionically: Cl2(g)+ 2Br- (aq) -> Br2(aq) + 2Cl- (aq)
16 Practically, a displacement reaction takes place when a known amount /volume of a solution is added excess of a more reactive metal.


A displacement reaction takes place when a more reactive element/with less electrode potential Eᶿ / negative Eᶿ /higher in the reactivity/electrochemical series remove/displace another with less reactive element/with higher electrode potential Eᶿ / positive Eᶿ /lower in the reactivity/electrochemical series from its solution.e.g. (i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq) Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq) (ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq) Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq) (iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s) This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead. (iv)Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq) Ionically: Cl2(g)+ 2Br- (aq) -> Br2(aq) + 2Cl- (aq)
16 Practically, a displacement reaction takes place when a known amount /volume of a solution is added excess of a more reactive metal. To determine the molar standard enthalpy/heat of displacement(∆Hᶿd) 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 .


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 Table 1 Experiment I II Final temperature of solution(T2) 30.0oC 31.0oC Final temperature of solution(T1) 25.0oC 24.0oC Change in temperature(∆T) 5.0 6.0 Questions 1.(a) Calculate: (i)average ∆T Average∆T = change in temperature in experiment I and II =>5.0 + 6.0 = 5.5oC 2 (ii)the number of moles of solution used Moles used = molarity x volume of solution = 0.2 x 20 = 0.004 moles 1000 1000 (iii)the enthalpy change ∆H for the reaction Heat produced ∆H = mass of solution(m) x specific heat capacity (c)x ∆T => 20 x 4.2 x 5.5 = 462 Joules = 0.462 kJ 1000 (iv)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 solution=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 ∆Hd = Heat produced ∆H
17 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.


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. Zn(s) + Cu2+(aq) Energy ∆H = -115.5 kJmole-1 (kJ) Cu(s) + Zn2+(aq) Reaction progress/path/coordinates 6. Iron is less reactive than Zinc. Explain the effect of using iron instead of Zinc on the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution. No effect. Cu2+ (aq) are displaced from their solution.The element used to displace it does not matter.The reaction however faster if a more reactive metal is used. 7.(a)If the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution is 209kJmole-1 calculate the temperature change if 50cm3 of 0.2M solution was displaced by excess magnesium. Moles used = molarity x volume of solution = 0.2 x 50 = 0.01 moles 1000 1000 Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>209kJmole-1x 0.01 moles = 2.09 kJ ∆T (change in temperature) = Heat produced ∆H Molar heat of displacement ∆Hd x Number of moles =>2.09 kJ = 9.9524Kelvin 0.01 moles (b)Draw an energy level diagram to show the above energy changes
18 Mg(s) + Cu2+(aq) Energy ∆H = -209 kJmole-1 (kJ) Cu(s) + Mg2+(aq) Reaction progress/path/coordinates 8. The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-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.


7.(a)If the standard molar heat of displacement ∆Hd of copper(II)sulphate (VI) solution is 209kJmole-1 calculate the temperature change if 50cm3 of 0.2M solution was displaced by excess magnesium. Moles used = molarity x volume of solution = 0.2 x 50 = 0.01 moles 1000 1000 Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>209kJmole-1x 0.01 moles = 2.09 kJ ∆T (change in temperature) = Heat produced ∆H Molar heat of displacement ∆Hd x Number of moles =>2.09 kJ = 9.9524Kelvin 0.01 moles (b)Draw an energy level diagram to show the above energy changes
18 Mg(s) + Cu2+(aq) Energy ∆H = -209 kJmole-1 (kJ) Cu(s) + Mg2+(aq) Reaction progress/path/coordinates 8. The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-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 ∆H Molar heat of displacement ∆Hd =>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 9. If the molar heat of displacement of Zinc(II)nitrate(V)by magnesium powder is 25.05kJmole-1 ,calculate the volume of solution which must be added 0.5 moles solution if there was a 3.0K rise in temperature.


The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-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 ∆H Molar heat of displacement ∆Hd =>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 9. If the molar heat of displacement of Zinc(II)nitrate(V)by magnesium powder is 25.05kJmole-1 ,calculate the volume of solution which must be added 0.5 moles solution if there was a 3.0K rise in temperature. Heat produced ∆H = Molar heat of displacement ∆Hd x Number of moles =>25.08kJmole-1x 0.5 moles = 1.254 kJ x 1000 =1254J Mass of solution (m) = Heat produced ∆H specific heat capacity (c)x ∆T => 1254J = 99.5238 g 4.2 x 3 Volume = mass x density = 99.5238 g x 1 = 99.5238cm3 Note: The solution assumes to be too dilute /infinite dilute such that the density and specific heat capacity is assumed to be that of water. Graphical determination of the molar enthalpy of displacement of copper Procedure:
19 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. 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.


Place all the (1.5g) Zinc powder provided. 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. Sample results 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 Sketch graph of temperature against time 36.5 Extrapolation Temperature point ∆T oC 130 Time(seconds) Questions 1. Show and determine the change in temperature ∆T From a well constructed graph ∆T= T2 –T1 at 150 second by extrapolation ∆T = 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 ∆H = mass of solution(m) x specific heat capacity (c)x ∆T => 20 x 4.2 x 11.5 = 966 Joules = 0.966 kJ 1000 Number of moles = Heat produced ∆H Molar heat of displacement ∆Hd =>.966 kJ = 0.007728moles 125 moles 7.728 x 10-3moles 2. What was the concentration of copper(II)sulphate(VI) in moles per litre. 20 Molarity = moles x 1000 => 7.728 x 10-3moles x 1000 = 0.3864M Volume used 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.


What was the concentration of copper(II)sulphate(VI) in moles per litre. 20 Molarity = moles x 1000 => 7.728 x 10-3moles x 1000 = 0.3864M Volume used 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 ∆T hence lowering the practical number of moles and molarity against the theoretical value 5.a) In an experiment to determine the molar heat of reaction when magnesium displaces copper ,0.15g of magnesium powder were added to 25.0cm3 of 2.0M copper (II) chloride solution. The temperature of copper (II) chloride solution was 25oC.While that of the mixture was 43oC. i)Other than increase in temperature, state and explain the observations which were made during the reaction.(3mks) ii)Calculate the heat change during the reaction (specific heat capacity of the solution = 4.2jg-1k-1and the density of the solution = 1g/cm3(2mks) iii)Determine the molar heat of displacement of copper by magnesium.(Mg=24.0). iv)Write the ionic equation for the reaction.(1mk) v)Sketch an energy level diagram for the reaction.(2mks) b)Use the reduction potentials given below to explain why a solution containing copper ions should not be stored in a container made of zinc. Zn2+(aq) + 2e -> Zn(s); Eø = -0.76v Cu2+(aq) + 2e -> Cu(s); Eø = +0.34v (2mks) (c)Standard enthalpy/heat of neutralization ∆Hᶿn The molar standard enthalpy/heat of neutralization ∆Hᶿn 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.


i)Other than increase in temperature, state and explain the observations which were made during the reaction.(3mks) ii)Calculate the heat change during the reaction (specific heat capacity of the solution = 4.2jg-1k-1and the density of the solution = 1g/cm3(2mks) iii)Determine the molar heat of displacement of copper by magnesium.(Mg=24.0). iv)Write the ionic equation for the reaction.(1mk) v)Sketch an energy level diagram for the reaction.(2mks) b)Use the reduction potentials given below to explain why a solution containing copper ions should not be stored in a container made of zinc. Zn2+(aq) + 2e -> Zn(s); Eø = -0.76v Cu2+(aq) + 2e -> Cu(s); Eø = +0.34v (2mks) (c)Standard enthalpy/heat of neutralization ∆Hᶿn The molar standard enthalpy/heat of neutralization ∆Hᶿn 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. 21 Strong acids/bases/alkalis are completely dissociated to many free ions(H+ /H3O+ and OH- ions). 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(-∆H) process.The enrgy produced during neutralization depend on the amount of free ions (H+ H3O+ and OH-)ions existing in the acid/base/alkali reactant: (i)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.


21 Strong acids/bases/alkalis are completely dissociated to many free ions(H+ /H3O+ and OH- ions). 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(-∆H) process.The enrgy produced during neutralization depend on the amount of free ions (H+ H3O+ and OH-)ions existing in the acid/base/alkali reactant: (i)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. (ii) (i)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 any wholly dissociated acid/base/alkali Practically ∆Hᶿn can be determined as in the examples below: To determine the molar enthalpy of neutralization ∆Hn of Hydrochloric acid Procedure Place 50cm3 of 2M hydrochloric acid into a calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper. Record its temperature T1.Using a clean measuring cylinder, measure another 50cm3 of 2M sodium hydroxide. Rinse the bulb of the thermometer in distilled water. Determine the temperature of the sodium hydroxide T2.Average T2 andT1 to get the initial temperature of the mixture T3. Carefully add all the alkali into the calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper containing the acid. Stir vigorously the mixture with the thermometer. Determine the highest temperature change to the nearest 0.5oC T4 as the final temperature of the mixture.


Carefully add all the alkali into the calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper containing the acid. Stir vigorously the mixture with the thermometer. Determine the highest temperature change to the nearest 0.5oC T4 as the final temperature of the mixture. Repeat the experiment to complete table 1. Table I . Sample results 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
22 (a)Calculate T6 the average temperature change T6 = 13.25 +13.75 = 13.5 oC 2 (b)Why should the apparatus be very clean? Impurities present in the apparatus reacts with acid /base lowering the overall temperature change and hence ∆Hᶿn. (c)Calculate the: (i)number of moles of the acid used number of moles = molarity x volume => 2 x 50 = 0.1moles 1000 1000 (ii)enthalpy change ∆H of neutralization. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T(T6) => (50 +50) x 4.2 x 13.5 = 5670Joules = 5.67kJ (iii) the molar heat of neutralization the acid. ∆Hn = Enthalpy change ∆H => 5.67kJ = 56.7kJ mole-1 Number of moles 0.1moles (c)Write the ionic equation for the reaction that takes place OH-(aq)+ H+(aq) -> H2O(l) (d)The theoretical enthalpy change is 57.4kJ. Explain the difference with the results above.


∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T(T6) => (50 +50) x 4.2 x 13.5 = 5670Joules = 5.67kJ (iii) the molar heat of neutralization the acid. ∆Hn = Enthalpy change ∆H => 5.67kJ = 56.7kJ mole-1 Number of moles 0.1moles (c)Write the ionic equation for the reaction that takes place OH-(aq)+ H+(aq) -> H2O(l) (d)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 ∆T/T6 and thus ∆Hn Heat/energy is absorbed by the reaction vessel/calorimeter/plastic cup lowers ∆T and hence ∆Hn (e)Compare the ∆Hn of the experiment above with similar experiment repeated with neutralization of a solution of: (i) 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. (ii) ammonia with ethanoic acid Temperature change( T5) 13.25 13.75
23 The results would be lower/∆Hn 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. (f)Draw an energy level diagram to illustrate the energy changes H2 H+ (aq)+OH (aq) Energy (kJ) ∆H = -56.7kJ H1 H2O (l) Reaction path/coordinate/progress 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.


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. (f)Draw an energy level diagram to illustrate the energy changes H2 H+ (aq)+OH (aq) Energy (kJ) ∆H = -56.7kJ H1 H2O (l) Reaction path/coordinate/progress 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? Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 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 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC.


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? Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 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 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn 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.


Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 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 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn 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.


∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn 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.


∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn 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 Table 1:Sample results.


(b)Plot a graph of volume of sodium hydroxide against temperature change. 28.7=T2 temperature(oC) ∆T From the graph show and determine : (i)the highest temperature change ∆T ∆T =T2-T1 => highest temperature-T2 (from extrapolating a correctly plotted graph) less lowest temperature at volume of base=0 :T1 =>∆T = 6.7 – 0.0 = 6.70C (ii)the volume of sodium hydroxide used for complete neutralization From a correctly plotted graph – 16.75cm3 (c)Calculate the number of moles of the alkali used Moles NaOH = molarity x volume =>2M x 16.75cm3 = 0.0335 moles 1000 1000 (d)Calculate ∆H for the reaction ∆H = mass of solution(acid+base) x c x ∆T =>(25.0 + 16.75) x 4.2 x 6.7 = 1174.845 J = 1.174845kJ 1000 (e)Calculate the molar enthalpy of neutralization of the alkali. ∆Hn = ∆Hn = 1.174845kJ = 35.0701kJ Number of moles 0.0335 Volume of sodium hydroxide(cm3) 22. oC T1
26 (d)Standard enthalpy/heat of solution ∆Hᶿs The standard enthalpy of solution ∆Hᶿsis defined as the energy change when one mole of a substance is dissolve in excess distilled water to form an infinite dilute solution. An infinite dilute solution is one which is too dilute to be diluted further. Dissolving a solid involves two processes: (i) breaking the crystal of the solid into free ions(cations and anion).This process is the opposite of the formation of the crystal itself. The energy required to form one mole of a crystal structure from its gaseous ions is called Lattice energy/heat/enthalpy of lattice (∆Hl). Lattice energy /heat/enthalpy of lattice (∆Hl) is an endothermic process (+∆Hl).


Dissolving a solid involves two processes: (i) breaking the crystal of the solid into free ions(cations and anion).This process is the opposite of the formation of the crystal itself. The energy required to form one mole of a crystal structure from its gaseous ions is called Lattice energy/heat/enthalpy of lattice (∆Hl). Lattice energy /heat/enthalpy of lattice (∆Hl) is an endothermic process (+∆Hl). The table below shows some ∆Hl in kJ for the process MX(s) -> M+ (g) + X- (g) Li Na K Ca Mg F +1022 +900 +800 +760 +631 Cl +846 +771 +690 +2237 +2493 Br +800 +733 +670 +2173 +2226 (ii)surrounding the free ions by polar water molecules. This process is called hydration. The energy produced when one mole of ions are completely hydrated is called hydration energy/ heat/enthalpy of hydration(∆Hh).Hydration energy /enthalpy of hydration(∆Hh) is an exothermic process(∆Hh). The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water.


The energy produced when one mole of ions are completely hydrated is called hydration energy/ heat/enthalpy of hydration(∆Hh).Hydration energy /enthalpy of hydration(∆Hh) is an exothermic process(∆Hh). The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature.


The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic .


(ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature.


(a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic .


Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of CaF2 (s) therefore decreases with increase in temperature.


Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of CaF2 (s) therefore decreases with increase in temperature. (d)Dissolving magnesium bromide crystal/s// MgBr2 (s) ->MgBr2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Br-(g) ∆Hl =+2226 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Br-(g) + aq -> 2Br-(aq) ∆Hh = (- 335x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 335 x 2 kJ)) + +2226 kJ = -364.0 kJmole-1 MgBr2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgBr2(s) therefore decreases with increase in temperature. 28 Practically the heat of solution can be determined from dissolving known amount /mass/volume of solute in known mass /volume of water/solvent. From the temperature of solvent before and after dissolving the change in temperature(∆T) during dissolution is determined.


Solubility of MgBr2(s) therefore decreases with increase in temperature. 28 Practically the heat of solution can be determined from dissolving known amount /mass/volume of solute in known mass /volume of water/solvent. From the temperature of solvent before and after dissolving the change in temperature(∆T) during dissolution is determined. To determine the ∆Hs ammonium nitrate Place 100cm3 of distilled water into a plastic beaker/calorimeter. Determine its temperature and record it at time =0 in table I below. Put all the 5.0g of ammonium nitrate (potassium nitrate/ammonium chloride can also be used)provided into the plastic beaker/calorimeter, stir using a thermometer and record the highest temperature change to the nearest 0.5oCafter every ½ minute to complete table I. Continue stirring the mixture throughout the experiment. Sample results: Table I Time (minutes) 0.0 ½ 1 1 ½ 2 2 ½ 3 3 ½ Temperature()oC 22.0 21.0 20.0 19.0 19.0 19.5 20.0 20.5 (a)Plot a graph of temperature against time(x-axis) 22.0=T1 temperature(oC) ∆T (b)From the graph show and determine the highest temperature change ∆T ∆T =T2-T1 => lowest temperature-T2 (from extrapolating a correctly plotted graph) less highest temperature at volume of base=0 :T1 Time (minutes) 18.7. oC T1
29 =>∆T =18.7 – 22.0 = 3.30C (c)Calculate the number of moles of ammonium nitrate(V) used Moles NH4NO3 = mass used => 5.0 = 0.0625 moles Molar mass 80 (d)Calculate ∆H for the reaction ∆H = mass of water x c x ∆T ->100 x 4.2 x 3.3 = +1386 J = +1.386kJ 1000 (e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V).


Continue stirring the mixture throughout the experiment. Sample results: Table I Time (minutes) 0.0 ½ 1 1 ½ 2 2 ½ 3 3 ½ Temperature()oC 22.0 21.0 20.0 19.0 19.0 19.5 20.0 20.5 (a)Plot a graph of temperature against time(x-axis) 22.0=T1 temperature(oC) ∆T (b)From the graph show and determine the highest temperature change ∆T ∆T =T2-T1 => lowest temperature-T2 (from extrapolating a correctly plotted graph) less highest temperature at volume of base=0 :T1 Time (minutes) 18.7. oC T1
29 =>∆T =18.7 – 22.0 = 3.30C (c)Calculate the number of moles of ammonium nitrate(V) used Moles NH4NO3 = mass used => 5.0 = 0.0625 moles Molar mass 80 (d)Calculate ∆H for the reaction ∆H = mass of water x c x ∆T ->100 x 4.2 x 3.3 = +1386 J = +1.386kJ 1000 (e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V). ∆Hs = ∆H = +1.386kJ = + 22.176kJ mole-1 Number of moles 0.0625 moles (f)What would happen if the distilled water was heated before the experiment was performed. The ammonium nitrate(V)would take less time to dissolves.


oC T1
29 =>∆T =18.7 – 22.0 = 3.30C (c)Calculate the number of moles of ammonium nitrate(V) used Moles NH4NO3 = mass used => 5.0 = 0.0625 moles Molar mass 80 (d)Calculate ∆H for the reaction ∆H = mass of water x c x ∆T ->100 x 4.2 x 3.3 = +1386 J = +1.386kJ 1000 (e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V). ∆Hs = ∆H = +1.386kJ = + 22.176kJ mole-1 Number of moles 0.0625 moles (f)What would happen if the distilled water was heated before the experiment was performed. The ammonium nitrate(V)would take less time to dissolves. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure.


∆Hs = ∆H = +1.386kJ = + 22.176kJ mole-1 Number of moles 0.0625 moles (f)What would happen if the distilled water was heated before the experiment was performed. The ammonium nitrate(V)would take less time to dissolves. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory.


The ammonium nitrate(V)would take less time to dissolves. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation.


Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation. Hess’ law of constant heat summation states that “the total enthalpy/heat/energy change of a reaction is the same regardless of the route taken from reactants to products at the same temperature and pressure”.


∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation. Hess’ law of constant heat summation states that “the total enthalpy/heat/energy change of a reaction is the same regardless of the route taken from reactants to products at the same temperature and pressure”. Hess’ law of constant heat summation is as a result of a series of experiments done by the German Scientist Henri Hess(1802-1850). He found that the total energy change from the reactants to products was the same irrespective of the intermediate products between. i.e. A(s) --∆H1-->C(s) = A(s) --∆H2-->B(s)--∆H3-->C(s) Applying Hess’ law of constant heat summation then:
31 A(s) ∆H2 B(s) ∆H1 ∆H3 C(s) The above is called an energy cycle diagram. It can be used to calculate any of the missing energy changes since: (i) ∆H1 =∆H2 + ∆H3 (ii) ∆H2 =∆H1 + -∆H3 (iii) ∆H3 = - ∆H1 + ∆H2 Examples of applying Hess’ law of constant heat summation 1.Calculate the molar enthalpy of formation of methane (CH4) given that ∆Hᶿc of carbon-graphite is -393.5kJmole-1,Hydrogen is -285.7 kJmole-1 and that of methane is 890 kJmole-1 Working Carbon-graphite ,hydrogen and oxygen can react to first form methane. Methane will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water.


Methane will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water. C(s)+ 2H2 (g)+2O2 (g) --∆H1--> CH4(g) +2O2(g) --∆H2--> CO2(g)+2H2O(l) C(s)+ 2H2 (g)+2O2 (g) --∆H3--> CO2(g)+2H2O(l) Energy cycle diagram C(s) + 2H2 (g) + 2O2(g) ∆H1=∆Hᶿc =-890.4kJ CH4(g)+2O2(g) ∆H3=∆Hᶿc =-393.5kJ ∆H3=∆Hᶿc =-285.7kJ x 2 ∆H2= ∆Hᶿf= x CO2(g) + 2H2O(l) Substituting: ∆H3 = ∆H1 + ∆H2 -393.5 + (-285.7 x 2) = -890.4kJ + x x = -74.5 kJ
32 Heat of formation ∆Hᶿf CH4 = -74.5 kJmole-1 2. Calculate the molar enthalpy of formation of ethyne (C2H2) given : ∆Hᶿc of carbon-graphite = -394kJmole-1,Hydrogen = -286 kJmole-1 , (C2H2) = -1300 kJmole-1 Working Carbon-graphite ,hydrogen and oxygen can react to first form ethyne. Ethyne will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water.


Ethyne will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water. 2C(s)+ H2 (g)+2 ½ O2 (g) --∆H1--> C2 H2 (g) +2 ½ O2(g) --∆H2--> CO2(g)+H2O(l) 2C(s)+ H2 (g)+ 2 ½ O2 (g) --∆H3--> 2CO2(g)+H2O(l) Energy cycle diagram 2C(s) + H2 (g) +2½O2(g) ∆H1=∆Hᶿf =x C2 H2+2½O2(g) ∆H3=∆Hᶿc =-394kJx 2 ∆H3=∆Hᶿc =-286kJ ∆H2= ∆Hᶿc= -1300kJ 2CO2(g) + H2O(l) Substituting: ∆H3 = ∆H1 + ∆H2 ( -394 x 2) + -286 = -1300kJ + x x = +244 kJ Heat of formation ∆Hᶿf CH4 = +244 kJmole-1 3. Calculate the molar enthalpy of formation of carbon(II)oxide (CO) given : ∆Hᶿc of carbon-graphite = -393.5kJmole-1, ∆Hᶿc of carbon(II)oxide (CO)= -283 kJmole-1 Working Carbon-graphite reacts with oxygen first to form carbon (II)oxide (CO). Carbon(II)oxide (CO) then burn in the excess oxygen to form carbon(IV)oxide. Carbon-graphite can burn in excess oxygen to form carbon (IV) oxide.


Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table 1. Table 1:Sample results
35 Experiment I II Highest /lowest temperature T2 27.0 29.0 Initial temperature T1 24.0 25.0 Change in temperature ∆T 3.0 4.0 Experiment II Weigh accurately 8.0g of anhydrous copper(II)sulphate(VI). Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II.


Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II. Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T
36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II.


Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II. Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T
36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4.


Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T
36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4. 5H2O (s)+ (aq) ∆H3= =-29.4kJ ∆H2= -12.6kJ CuSO4 .5H2O (aq) ∆H3 = ∆H1 +∆H2 =>-29.4kJ = -12.6kJ + x =>-29.4kJ - (+12.6kJ) = x x = 16.8kJ b)Practical example II Determination of enthalpy of solution of ammonium chloride Theoretical information.


∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4. 5H2O (s)+ (aq) ∆H3= =-29.4kJ ∆H2= -12.6kJ CuSO4 .5H2O (aq) ∆H3 = ∆H1 +∆H2 =>-29.4kJ = -12.6kJ + x =>-29.4kJ - (+12.6kJ) = x x = 16.8kJ b)Practical example II Determination of enthalpy of solution of ammonium chloride Theoretical information. Ammonium chloride dissolves in water to form ammonium chloride solution. Aqueous ammonia can react with excess dilute hydrochloric acid to form ammonium chloride solution. The heat change taking place can be calculated from the heat of reactions:
37 (i) NH3(aq) + HCl(aq) -> NH4Cl(s) (ii) NH4Cl(s) + (aq) -> NH4Cl(aq) (iii) NH3(aq) + HCl(aq) -> NH4Cl(aq) Experiment procedure I Measure 50cm3 of water into a 100cm3 beaker. Record its temperature T1 as initial temperature to the nearest 0.5oC in table I. Add exactly 5.0g of ammonium chloride crystals weighed carefully into the water. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table I.


Add exactly 5.0g of ammonium chloride crystals weighed carefully into the water. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table I. Sample results TableI Experiment I II final temperature(oC) 19.0 20.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 3.0 2.0 Experiment procedure II Measure 25cm3 of 2M aqueous ammonia into a 100cm3 beaker. Record its temperature T1 as initial temperature to the nearest 0.5oC in table II. Measure 25cm3 of 2M hydrochloric acid solution. Add the acid into the beaker containing aqueous ammonia. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II.


Add the acid into the beaker containing aqueous ammonia. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:
38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ.


Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:
38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ. Practice theoretical examples:
39 1.


Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:
38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ. Practice theoretical examples:
39 1. Using an energy level diagram calculate the ∆Hs of ammonium chloride crystals given that.


Using an energy level diagram calculate the ∆Hs of ammonium chloride crystals given that. ∆Hf of NH3 (aq) = -80.54kJ mole-1 ∆Hf of HCl (aq) = -164.46kJ mole-1 ∆Hf of NH4Cl (aq) = -261.7483kJ mole-1 ∆Hs of NH4Cl (aq) = -16.8517kJ mole-1 N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf=-261.7483kJ NH4Cl(s) + aq x=∆Hs (aq) (aq) - 80.54kJ=∆H1 -164.46kJ=∆H2 NH3 (aq) + HCl(aq) 16.8517kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 80.54kJ + -164.46kJ + -16.8517kJ = -261.7483kJ + ∆Hf =>∆Hf = -33.6kJmole-1. Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6
40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
2
20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed.


∆Hf of NH3 (aq) = -80.54kJ mole-1 ∆Hf of HCl (aq) = -164.46kJ mole-1 ∆Hf of NH4Cl (aq) = -261.7483kJ mole-1 ∆Hs of NH4Cl (aq) = -16.8517kJ mole-1 N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf=-261.7483kJ NH4Cl(s) + aq x=∆Hs (aq) (aq) - 80.54kJ=∆H1 -164.46kJ=∆H2 NH3 (aq) + HCl(aq) 16.8517kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 80.54kJ + -164.46kJ + -16.8517kJ = -261.7483kJ + ∆Hf =>∆Hf = -33.6kJmole-1. Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6
40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
2
20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up.


Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6
40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
2
20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up. Some reactions are too slow to be determined.