JK-TK/Newdata huggingface dataset:
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.