THERMOCHEMISTRY

1 Exothermic reaction Is a chemical reaction that gives out heat to the surroundings. Chemical energy is converted to heat energy. The surroundings gain heat and as a result, the temperature of the surroundings increases. Others examples of exothermic reactions are shown below:(a) Respiration(b) Rusting of iron(c) Reaction between Group 1 elements with waterExample: 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)(d) Neutralisation reaction between acid and alkaliExample: HCl(aq) + NaCl(aq) NaCl(aq) + H2O(l)(e) Reaction of a dilute acid with metal carbonateExample: 2HCl(aq) + CaCO3(aq) CaCl(aq) + H2O(l) + CO2(g)(g) Displacement reaction of metalsExample: Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s) Some physical processes also release heat as shown below:(a) Dissolving of alkali in waterExample: NaOH(s) Na+(aq) + OH-(aq)(b) Dilution of concentrated acidsExample: H2SO4(l) 2H+(aq) + SO42-(aq)(c) Condensation and freezing process

2Endothermic reaction Is a chemical reaction that absorbs heat from the surroundings. Heat energy is converted to chemical energy and store it in the products. The surroundings lose heat energy and as a result the temperature of the surrounding decreases. Others examples of endothermic reactions are shown below:(a) Photosynthesis(b) Reaction between sodium hydrogen carbonate and dilute acidExample: NaHCO3(s) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g)(c) Decomposition of carbonate saltsExample: PbCO3(s) PbO(s) + CO2(g)(d) Decomposition of nitrate saltsExample: 2Zn(NO3)(s) 2ZnO(s) + 4NO2(g) + O2(g)(e) Decomposition of hydrated salt, such as hydrated copper(II) sulphateExample: CuSO4.5H2O(s) CuSO4(s) + 5H2O(l) Some physicals processes also absorb heat as shown below:(a) Dissolving of ammonium salts such as ammonium chloride, ammonium nitrate and ammonium sulphate in waterExample: NH4Cl(s) NH+(aq) + Cl-(aq)(b) Melting and boiling process3Heat of Reaction and Energy Level Diagrams-The total energy stored in a chemical substance is known as energy content.-It is impossible to measure the energy content for a particular substance but the changes in energy content that occurs when the reactants are converted to the products can be determined and is given the symbol, H.The changes in energy content = The amount of heat energy given out/absorbed during chemical reaction

H heat of reaction (the changes in energy)(delta) means change of and H is the heat content or enthalpy of the sytem

-The amount of heat change in a specific reaction, however, depends on how much of each reactant is present.- The more reactants are used, the more heat energy is given out/absorbed in the reaction. Thus, it is necessary to clearly states the amounts of reactants involved for the definition of the heat of reaction.The heat of reaction (H): is defined as the heat change which occurs when the numbers of moles of reactants indicated by the equation react together.

For a general reaction, A + B C + D H productsH reactants

H = (Energy content of products) (Energy content of reactants)H = H products H reactantsThe unit of H is kilojoules, kJ

When an exothermic reaction occurs, heat energy is released to the surroundings. Thus, the total energy content of the products is less than that of the reactants.

H products < H reactantsH = H products H reactants = negative value H of an exothermic reaction is given a negative sign.For example, when 1 mole of sodium hydroxide solution reacts with 1 mole of hydrochloric acid, the reaction releases 57.3 kJ heat energy. Thus, H = -57.3 kJ

A chemical equation together with H is called the thermochemical equation.For example, NaOH(aq) + HCl(l) NaCl(aq) + H2O(l) H = -57.3 kJ The heat energy changed in an exothermic reaction can be shown on an energy level diagrams. EnergyReactantsH = negativeProducts

Figure 1: The energy level diagram for an exothermic reaction

For example, the energy level for the reaction between sodium hydroxide solution and hydrochloric acid is shown in Figure 2.

EnergyNaOH(aq) + HCl(aq)H = -57.3 kJNaCl(aq) + H2O(l)

Figure 2: Energy level diagram for the reaction between sodium hydroxide solution and hydrochloric acid

When an endothermic reaction occurs, heat energy is absorbed from the surroundings. Thus, the total energy content of the products is higher than the reactants.

H products > H reactantsH = H products H reactants = positive value

H of an exothermic reaction is given a positive sign.For example, when 1 mole of sodium hydrogen carbonate solution reacts with 1 mole of hydrochloric acid, the reaction absorbs 11.8 kJ heat energy. Thus, H = +11.8 kJ The thermochemical equation for this reaction is:NaHCO3(aq) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g) H = +11.8 kJ The heat energy changes in an endothermic reaction can be shown on an energy level diagram. EnergyReactantsH = positiveProducts

Figure 3: The energy level diagram for an endothermic reaction

For example, the energy level for the reaction between sodium hydrogen carbonate solution and hydrochloric acid is shown in Figure 4. EnergyNaHCO3(aq) + HCl(aq)H = +11.8 kJNaCl(aq) + H2O(l) + CO2(g)

Figure 4: Energy level diagram for the reaction between sodium hydrogen carbonate solution and hydrochloric acid4Types of heat of reactions(a) Heat of precipitation Defined as the heat change when 1 mole of reactant reacts or one mole of product is formed. Therefore, the unit of H for these reactions is kJmol-1.

(b) Heat of displacement(c) Heat of neutralization(d) Heat of combustion

Heat of precipitation Precipitation reaction occurs when we add two solutions together to form a precipitate. For example, when lead(II) nitrate solution and potassium sulphate solution are added together, a white precipitate, lead(II) sulphate is formed. The heat of precipitate is the heat change when 1 mole of a precipitate is formed from their ions in aqueous solution. For example, 50.4 kJ of heat energy is given out when 1 mole of lead(II) sulphate, PbSO4 is formed. Thus, the heat of precipitation of lead(II) sulphate, PbSO4 is -50.4 kJ mol-1. Thermochemical equation:Pb2+(aq) + SO42-(aq) PbSO4(s)H = -50.4 kJ mol-1 The energy level diagram is shown below: EnergyPb2+(aq) + SO42-(aq)H = -50.4 kJ mol-1PbSO4(s)

Figure 5: Energy level diagram for the precipitation of lead(II) sulphateSolve numerical problems related to heat of precipitation In a chemical reaction involving solutions, the heat change of the solution can be determined by 2 method:(a) Using the formula mc if the value of m (mass or volume of solution) and the values of (change in temperature) are known.Thus, the heat change of a solution = mc (b) Using the heat of reaction, H if the value of H and the number of moles (n) of the reactant are given. Thus, the heat change in a reaction = n x H kJ

For example, Pb2+(aq) + SO42-(aq) PbSO4(s) H = -42 kJ mol-1When 2 moles of PbSO4 formed, the heat given out = 2 x 42 kJ = 84 kJWhen 0.5 moles of PbSO4 formed, the heat given out = 0.5 x 42 kJ = 21 kJ Since the heat change calculated by the 2 methods should be equal by theory, thenmc = n x H x 1000 Joule Heat of displacement Displacement reaction occurs when a more electropositive metal displace a less electropositive metal from its salt solution. For example, when zinc powder is added into lead(II) nitrate solution, lead is displaced and zinc nitrate solution is formed. The heat of displacement is the heat change when one mole of a metal is displaced from its salt solution by a more electropositive metal. For example, 112 kJ of heat energy is given out when one mole of lead, Pb is displaced from lead(II) nitrate, Pb(NO3) solution by zinc. Thus, the heat of displacement of lead is -112 kJ mol-1. Thermochemical equation:Zn(s) + Pb2+(aq) Pb(s) + Zn2+(aq) H = -112 kJ mol-1 The energy level diagram is shown below: EnergyZn(s) + Pb2+(aq)H = -112 kJ mol-1Pb(s) + Zn2+(aq)

Figure 6: Energy level diagram for the displacement of lead by zincSolve numerical problems related to heat of displacementExample 1:Excess zinc powder is added to 50 cm3 of 0.2 mol dm-3 copper(II) sulphate, CuSO4 solution in a polystyrene cup. The results of the experiment are shown below:Initial temperature of copper(II) sulphate,CuSO4 = 27.0 CHighest temperature of the mixture = 37.0 C(a) Write the ionic equation for the reaction.(b) Calculate the mass of zinc which reacts with 50 cm3 of 0.2 mol dm-3 copper(II) sulphate, CuSO4 solution.(c) Calculate the heat of displacement of copper by zinc[molar mass of Zn = 65 g mol-1; Specific heat capacity of solution = 4.2 J g-1C-1; density of solution = 1 g cm-3]Example 2:The thermochemical equation for the reaction between magnesium and iron(II) sulphate, FeSO4 solution is as follow: Mg(s) + Fe2+ (aq) Mg2+(aq) + Fe(s) H = -202 kJ mol-1In a experiment, excess magnesium powder is added to 100 cm3 of iron(II) sulphate, FeSO4 solution at 29 C in a polystyrene cup. It is found that 1.12 g of iron is displaced. Calculate (a) The highest temperature reached(b) The concentration of the iron(II) solution.[Specific heat capacity of solution = 4.2 J g-1C-1; density of solution = 1 g cm-3; molar mass of Fe = 56 g mol-1] Heat of neutralization Neutralization is a reaction between and acid and base or alkali to produce salt and water only. The chemical equations below represent some examples of neutralization reactions(a) HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)(b) HNO3 + KOH(aq) KNO3(aq) + H2O(l)(c) H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) In a neutralization between an acid and an alkali, the actual reaction that occurred is between hydrogen ions, H+ from the acid and hydroxide ions, OH- from the alkali to produce molecules of water, H2O. Ionic equation of neutralization: H+(aq) + OH-(aq) H2O(l) The heat of neutralization is the heat change when 1 mole of water is formed from the reaction between an acid and an alkali. Neutralization is an exothermic reaction. The heat of neutralization between a strong acid and a strong alkali is -57.3 kJ mol-1. All the neutralization reactions between a strong acid and a strong alkali can be represented by the thermochemical equation below:H+(aq) + OH-(aq) H2O(l)H = 57.3 kJ mol-1 The energy level diagram is shown below: EnergyH+(aq) + OH-(aq)H = -57.3 kJ mol-1H2O(l)

Figure 7: Energy level diagram for the neutralization reactionThe heat of neutralization between a diprotic acid and a strong alkali1) HCl and HNO3 are monoprotic acids. 1 mole of HCl or HNO3 produces 1 mole of hydrogen ions, H+ when ionized in water.HCl(aq) H+(aq) + Cl-(aq)HNO3(aq) H+(aq) + NO3-(aq)

2) Therefore, the heat of neutralization between a strong monoprotic acid and a strong alkali is always equal to -57.3 kJ mol-1. For example, HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H = -57.3 kJ mol-1

3) H2SO4 is a diprotic acid. 1 mole of sulphuric acid produces 2 moles of hydrogen ions, H+ when ionized in water.

H2SO4(aq) 2H+ (aq) + SO42-(aq)

The concentration of hydrogen ions, H+ of sulphuric acid is always double the concentration of H+ of the HCl and HNO3 of the same concentration.4) When 1 mole of H2SO4 is neutralized by NaOH solution, 114.6 kJ heat energy is given out because there are 2 moles of water formed.H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)H = -114.6 kJ mol-1 or 2H+(aq) + 2OH-(aq) 2H2O(l) H = -114.6 kJ mol-1

5) However, the heat given out by 1 mole of water formed is still 57.3 kJ as shown below:2 moles of H2O formed give out 114.6 kJ1 mole of H2O formed gives out Therefore, the thermochemical equation can be written as:H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)H = -57.3 kJ mol-1

The heat of neutralization involve weak acids and weak alkalis1) The heat of neutralization between a weak acid and a strong alkali is less than -57.3 kJ mol-1. 2) For example, the heat of neutralization between ethanoic acid, CH3COOH and sodium hydroxide, NaOH solution is -55 kJ mol-1. CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)H = -55 kJ mol-1 3) This is because most of the weak acids exist as molecules when they dissolve in water. They only ionize partially in water to produce low concentration of H+.4) Some of the heat given out during neutralization reaction is used to ionize the acid molecules completely to produce H+. As the result, the value of H is always less than -57.3 kJ mol-1.5) The heat of neutralization between a weak acid and a weak alkali is even much lower because more energy is required to ionize both the weak acid and weak alkali molecules.6) For example, the heat of neutralization between ethanoic acid and ammonia solution is -51.5 kJ mol-1.

Solve problems involving to heat of neutralizationExample 1:40 cm3 of 0.5 mol dm-3 sodium hydroxide, NaOH solution is added to 25 cm3 of 1.0 mol dm-3 nitric acid, HNO3 in a plastic cup. The results are shown below:Initial temperature of NaOH solution = 28.0 CInitial temperature of HNO3 = 28.0 CHighest temperature of the mixture = 32.0 C(a) Which solution is in excess?(b) Calculate the heat of neutralization between NaOH solution and HNO3?[Specific heat capacity of solution = 4.2 J g-1C-1; density of solution = 1 g cm-3]Example 2:When 50 cm3 of 1.0 mol dm-3 sodium hydroxide, NaOH is added to 50 cm3 of 1.0 mol dm-3 nitric acid, HNO3 the temperature rise of the mixture is 7 C. Predict the temperature rise in each of the following cases.(a) 100 cm3 of 1.0 mol dm-3 sodium hydroxide, NaOH is added to 100 cm3 of 1.0 mol dm-3 nitric acid, HNO3(b) 50 cm3 of 2.0 mol dm-3 sodium hydroxide, NaOH is added to 50 cm3 of 2.0 mol dm-3 nitric acid, HNO3

Heat of combustion Heat change when 1 mole of a substance is completely burnt in O2 under standard condition.At standard condition, the temperature is 25 C and atmospheric pressure is 1 atm

The combustion of a substance is an exothermic reaction which is heat is given out to the surroundings. For example, 394 kJ energy is given out when 1 mole of carbon burnt completely in excess of O2. Thus, the heat of combustion of carbon is -394 kJ mol-1. The thermochemical equation of the combustion of carbon is shown below:C(s) + O2(g) CO2(g) H = -394 kJ mol-1

Alcohol is a good fuel. The complete combustion of alcohol produces CO2, H2O and releases a lot of heat. For example, the heat of combustion of ethanol is shown by the thermochemical equation and the energy level diagram below:C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) H = -1376 kJ mol-1

EnergyC2H5OH(l) + 3O2(g)H = -1376 kJ mol-12CO2(g) + 3H2O(l)

Figure 8: Energy level diagram for the heat of combustion of ethanol Relationship between the heat of combustion of alcohol and the number of carbon atoms per alcohol molecule / relative molecular mass of alcohol The heat of combustion of alcohol increases from 1 member to the following member as shown in Table 1.

AlcoholMolecular formulaHeat of combustion(kJ mol-1)

MethanolCH3OH-728

EthanolC2H5OH-1376

PropanolC3H7OH-2026

ButanolC4H9OH-2678

PentanolC5H11OH-3332

During the combustion of alcohol, CO2 and H2O are formed. When the number of carbon atoms and hydrogen atoms per molecule of alcohol increases heat of combustion also increases because when more carbon atoms and hydrogen atoms are burnt, more CO2 and H2O are formed. The formation of chemical bonds in CO2 and H2O gives out heat energy. More CO2 and H2O formed will cause more chemical bonds to be formed, and more heat to be given out.

Solve problems involving heat of combustionExample 1:The thermochemical equation for the complete combustion of butanol is shown below:C4H9OH(l) + 6O2(g) 4CO2(g) + 5H2O(l)H = -2678 kJ mol-1Assume that there is no heat lost to the surroundings, calculate the mass of butanol needed to burn completely in excess oxygen in order to raise the temperature of 500 cm3 of water by 44 C.[Specific heat capacity of solution = 4.2 J g-1C-1; density of solution = 1 g cm-3; relative atomic mass: H,1: C,12: O,16]Example 2:The heat of combustion of methanol CH3OH is -728 kJ mol-1. What is its fuel value?[Relative atomic mass: H,1: C,12: O,16]