2024 Chemical Energetics Lecture Notes (Teacher) PDF

Chemical Energetics 2024 Semester 1 CATHOLIC JUNIOR COLLEGE H2 CHEMISTRY (Syllabus 9729) CHEMICAL ENER...

Chemical Energetics 2024 Semester 1 CATHOLIC JUNIOR COLLEGE H2 CHEMISTRY (Syllabus 9729) CHEMICAL ENERGETICS Learning Outcomes Students should be able to: (a) explain that most chemical reactions are accompanied by energy changes, principally in the form of heat, usually associated with the breaking and forming of chemical bonds; the reaction can be exothermic. (H negative) or endothermic (H positive) (b) construct and interpret an energy profile diagram, in terms of the enthalpy change of the reaction and of the activation energy (See also Section 8) (c) explain and use the terms: (i) enthalpy change of reaction and standard conditions, with particular reference to: formation; combustion; hydration; solution; neutralisation; atomization (ii) bond energy (H positive, i.e. bond breaking) (see also Section 2) (iii) lattice energy (H negative, i.e. gaseous ions to solid lattice) (d) calculate enthalpy changes from appropriate experimental results, including the use of the relationship: heat change = mcT (e) explain, in qualitative terms, the effect of ionic charge and of ionic radius on the numerical magnitude of a lattice energy (f) apply Hess’ Law to construct simple energy cycles, e.g. Born–Haber cycle, and carry out calculations involving such cycles and relevant energy terms (including ionisation energy and electron affinity), with particular reference to: (i) determining enthalpy changes that cannot be found by direct experiment, e.g. an enthalpy change of formation from enthalpy changes of combustion (ii) the formation of a simple ionic solid and of its aqueous solution (iii) average bond energies (g) explain and use the term entropy (h) discuss the effects on the entropy of a chemical system by the following: change in temperature (i) (ii) change in phase (iii) change in the number of particles (especially for gaseous systems) (iv) mixing of particles [quantitative treatment is not required] (i) predict whether the entropy change for a given process or reaction is positive or negative (j) state and use the equation involving standard Gibbs free energy change of reaction, ΔG o : ΔG o = ΔH o – TΔS o [The calculation of standard entropy change, ΔS o , for a reaction using standard entropies, S o, is not required] (k) state whether a reaction or process will be spontaneous by using the sign of ΔG o (l) understand the limitations in the use of ΔG o to predict the spontaneity of a reaction (m) predict the effect of temperature change on the spontaneity of a reaction, given standard enthalpy and entropy changes P a g e | 5-1 Chemical Energetics 2024 Semester 1 Contents 1. INTRODUCTION................................................................................................. 3 1.1 Enthalpy and Enthalpy Change..................................................................... 3 1.1 Standard Conditions...................................................................................... 4 2. ENTHALPY CHANGE OF REACTION, H.......................................................... 5 2.1 Exothermic and Endothermic Reactions........................................................ 5 2.2 Thermochemical Equations........................................................................... 6 3. CALCULATION OF ENTHALPY CHANGES, H.............................................. 8 3.1 Calorimetry (Experimental Determination of Enthalpy Changes using the Calorimeter)............................................................................................................ 8 (A) Standard Enthalpy Change of Combustion, ΔHco........................................... 12 (B) Standard Enthalpy Change of Neutralisation, Hno........................................ 15 3.2 Bond Energy (only for covalent bonds)........................................................ 19 3.3 Use of Hess’ Law......................................................................................... 21 (C) Standard Enthalpy Change of Formation, Hfo.............................................. 21 3.3.1 Hess’ Law............................................................................................. 22 3.3.2 Other Energy Changes......................................................................... 27 (D) Standard Enthalpy Change of Hydration, Hhydo............................................ 27 (E) Standard Enthalpy Change of Solution, Hsolo................................................ 29 (F) Lattice Energy (Revision: Chemical Bonding)................................................. 32 (G) Standard Enthalpy Change of Atomisation Hato........................................... 34 (H) Ionisation Energy (Revision: Atomic Structure).............................................. 35 (I) Electron Affinity................................................................................................ 35 3.3.3 Born–Haber Cycle................................................................................ 36 3.3.4 Energy Level Diagram.......................................................................... 38 Summary of methods to find enthalpy change......................................................... 41 4. Idea of Spontaneous Processes....................................................................... 42 4.1 Entropy........................................................................................................ 42 4.2 Gibbs Free Energy...................................................................................... 48 Answers to Thinking Questions............................................................................ 54 Chemical Energetics summary:............................................................................ 55 P a g e | 5-2 Chemical Energetics 2024 Semester 1 1. INTRODUCTION Learning Outcome: a) explain that most chemical reactions are accompanied by energy changes, principally in the form of heat Processed food that is sold in supermarkets normally carries a nutrition label that lists the energy value of the food in kilojoules (kJ) or calories. What does this “energy value” refer to? This is the chemical energy made available to the human body when it digests the food. How do we know how much energy is available in food? We will need to study the ENERGY CHANGES (principally in the Nutrition label on a pack of rice (UK) form of the flow of thermal energy or HEAT) that accompany chemical reactions and processes such as human digestion. Heat can be released or absorbed during a chemical reaction. In Chemical Energetics, this heat change is termed the ENTHALPY CHANGE. 1.1 Enthalpy and Enthalpy Change Note: The lower the Enthalpy, H, represents the total heat content of a substance or system heat content of that is held at constant pressure. a substance, more stable the What is “heat content”? substance is. Heat content includes the potential and kinetic energy of particles. Heat content cannot be measured directly. What is a “system”? A “system” is a collection of things that are under study. For example, this can be a sample of solid (or liquid or gas), or the reactants and products in a reaction mixture. The “system” is surrounded by space that we call the “surroundings”, and the two are separated by a “boundary”. Illustration of system, SURROUNDINGS boundary and surroundings: SURROUNDINGS system = reaction mixture in SYSTEM beaker; boundary = dotted BOUNDARY SYSTEM line representing edges of the BOUNDARY reaction mixture; surroundings = beaker and air. P a g e | 5-3 Chemical Energetics 2024 Semester 1 For a system which undergoes a chemical reaction, as reactants change to products, there is an enthalpy change, ∆H, which is the heat absorbed or released in the chemical reaction, and is given by: ΔH = H – H products reactants Learning Outcome c) explain and use the term: i) standard conditions 1.1 Standard Conditions Since enthalpy varies under different temperatures, pressures and physical states of the reactants and products, an enthalpy change is usually stated under standard conditions; i.e. standard enthalpy change, ∆Ho. The superscript o denotes “at standard conditions”. Standard conditions: (i) Temperature: 298 K (25 oC) (ii) Pressure: 1 bar (105 Pa) (iii) Substance in its most stable physical form e.g. NaCl(s); H2(g); Br2(l); H2O(l) (iv) Concentration (when dealing with solutions): 1 mol dm–3 Note: Standard conditions is NOT standard temperature and pressure (s.t.p.), 273 K, 1 bar! S.t.p and r.t.p details can be found in the Data Booklet, Section 1 (Important values, constants and standards). Look for ‘molar volume of gas’. Standard Room Standard Temperature and Temperature and Conditions Pressure Pressure Pressure 1 bar 1 bar 1 atm Temperature 25 oC 0 oC 20 oC Substance in its Additional 1 mole of gas 1 mole of gas most stable 3 information occupies 22.7 dm. occupies 24.0 dm3. physical form Topic in H2 Chemical Chemistry this is The Mole Concept and Stoichiometry Energetics used in P a g e | 5-4 Chemical Energetics 2024 Semester 1 2. ENTHALPY CHANGE OF REACTION, H The enthalpy change of a reaction, H, (units: kJ mol–1) is defined as the heat change when molar quantities of reactants as stated in the chemical equation react together. Units of H, kJ mol–1 The unit for energy is the joule, J. As values of H are usually rather large, enthalpy change is expressed in units of kilojoules (kJ), where 1 kJ = 1000 J. Because this change in energy depends on the number of particles taking part in the reaction, it is convenient to consider the energy changes associated with one mole of the reaction equation. [This is similar to how the “energy value” of a food product is given per 100 g of food (or sometimes, per serving] Learning Outcome: Hence the unit of H is in kJ mol–1. (a) Able to identify exothermic (ΔH, negative) or endothermic (ΔH, positive) energy 2.1 Exothermic and Endothermic Reactions changes in chemical Chemical reactions can be classified as either exothermic or endothermic. reactions. Exothermic Endothermic Exothermic (Greek: exo means “out”) reaction Endothermic (Greek: “endo” means “in”) is one which releases heat to the surroundings. reaction is one which absorbs heat from the surroundings. System: System: Heat is Heat is evolved/ absorbed released (to Surroundings (from Surroundings surroundings), gains heat, surroundings), loses heat, enthalpy temperature enthalpy temperature decreases. increases increases. decreases Since heat is released to the surroundings, the Since heat is absorbed from the surroundings, surrounding temperature increases. the surrounding temperature decreases. Hence, enthalpy of the system decreases. Hence, enthalpy of the system increases. H is negative (< 0) H is positive (> 0) e.g., C(s) + O2(g) → CO2(g) Ho = –393.4 kJ mol–1 e.g., H2(g) + I2(s) → 2HI(g) Ho = +51.9 kJ mol–1 P a g e | 5-5 Chemical Energetics 2024 Semester 1 Learning Outcome: b) construct and interpret an energy profile diagram, in terms of the enthalpy change of the reaction and of the activation energy Energy profile diagrams: EXOTHERMIC REACTION ENDOTHEMIC REACTION enthalpy enthalpy Ea Ea H1 reactants products H2 ΔH H2 products ΔH H1 reactants Progress of reaction \ Progress of reaction H1 = enthalpy of reactants H2 = enthalpy of products H = enthalpy change of reaction = Hproducts – Hreactants = H2 – H1 Ea = activation energy (minimum amount of energy reactants must possess for reaction to occur) A higher enthalpy means a less stable system. A lower enthalpy means a more stable system. Therefore, exothermic processes are energetically more favourable than endothermic ones. 2.2 Thermochemical Equations A thermochemical equation is used to represent the reaction or process being studied in Chemical Energetics. It is essentially an equation with stoichiometric ratio and its corresponding H indicated. Note: Physical state symbols MUST be given in a thermochemical equation. The magnitude of H is directly proportional to the amount of reactant consumed in the process. If we multiply both sides of a thermochemical equation by a factor n, then H must change by the same factor. e.g., CH4(g) + 2O2(g) → CO2 (g) + 2H2O(l) Ho = –890.4 kJ mol–1 2CH4(g) + 4O2(g) → 2CO2(g) + 4H2O(l) Ho = 2(–890.4) kJ mol–1 P a g e | 5-6 Chemical Energetics 2024 Semester 1 Worked Example 1 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) Ho = –1260 kJ mol–1 (a) Which has the greater enthalpy, the products or the reactants? Reactants (because reaction is exothermic) (b) What is the value of Ho for: 3 1 3 (i) NH3(g) + 4O2(g) → + N2(g) + H2O(l) 2 2 1260 Ho = − = – 315 kJ mol–1 4 (ii) 2N2(g) + 6H2O(l) → 4NH3(g) + 3O2(g) Ho = +1260 kJ mol–1 Lecture Practice 1 2ZnO(s) + 2SO2(g) → 2ZnS(s) + 3O2(g) Hθ = +890 kJ mol–1 (a) Which has the greater enthalpy, the products or the reactants? Products (because reaction is endothermic) (b) What is the value of Ho for 3 ZnS(s) + O2(g) → ZnO(s) + SO2(g) [–445 kJ mol–1] 2 890 Hθ = − = –445 kJ mol–1 2 (c) Write down the thermochemical equation for this reaction for which Ho = +2670 kJ mol–1. 6ZnO(s) + 6SO2(g) → 6ZnS(s) + 9O2(g) P a g e | 5-7 Chemical Energetics 2024 Semester 1 3. CALCULATION OF ENTHALPY CHANGES, H There are three main methods of calculating the enthalpy change of a reaction: Calculation of Enthalpy Changes, H 3.1 Calorimetry 3.2 Bond Energies 3.3 Use of Hess’ Law (experimental method) Values obtained The drawing of energy cycles, Used when ΔT and from Data Booklet. Born–Haber cycles or energy mass (o volume) are Only used for level diagrams is necessary to given in the question. simple covalent calculate the enthalpy change reactants and that cannot be directly obtained products. by experiment. Learning Outcome: d) calculate enthalpy changes from appropriate experimental results, including the use of the relationship: heat change = mcΔT 3.1 Calorimetry (Experimental Determination of Enthalpy Changes using the Calorimeter) The calorimeter is an apparatus used to determine the enthalpy change of a reaction. A simple calorimeter is shown below: thermometer Insulated plastic bottle Reactant B (usually in slight excess to ensure complete reaction of A) Reactant A (amount in moles accurately measured) Thinking Question 1: What is the purpose of insulating the plastic bottle? To ensure negligible heat loss to or gain from the surroundings (external environment). Title: Calorimetry https://youtu.be/EAgbknIDKNo Duration: watch the video till 2:55 min P a g e | 5-8 Chemical Energetics 2024 Semester 1 Calculations using experimental results: Heat change in the reaction = mass of solution × specific heat capacity of solution × temperature change = mcT where: m = mass of the solution (units: g, sometimes given in cm3) Note: –1 –1 –3 –1 Sometimes, you c = specific heat capacity of solution (units: J g K , or J cm K ) may encounter C Specific heat capacity is the amount of heat required to raise the (capital letter). C denotes “heat temperature of 1 g (or 1 cm3) of a substance by 1 K (or 1 oC). capacity” and has units J K–1. e.g., 4.18 J g–1 K–1 for water, 2.1 J g–1 K–1 for petroleum. C = mc T = temperature change (units can be either K or oC) Since we define enthalpy change of reaction (Hro) for every one mole of the thermochemical equation, if we have one mole of limiting reagent in the equation, then we can calculate enthalpy change of reaction (Hro) as follows: Hro = enthalpy change of reaction 𝑚𝑐∆𝑇 =± × coefficient of the limiting reagent no of moles of limiting reagent HOW TO… Calculate H from Calorimetric Data (1) Determine from question, values of m, c, T and n. Note: T in oC = T in K, ‘m' refers to mass of total aqueous solution (assume solution has a density equal to that of water = 1 g cm–3), ignore mass of solid for reactions where a solid is added to water or solution. (2) Calculate heat evolved or absorbed (mcT). Units will be in J → convert to kJ (Note: 1000 J = 1 kJ) (3) Calculate the no. of moles (amount) of limiting reagent per equation giving rise to Ho. (4) Calculate Hro.  If reaction is exothermic, include ‘–’ sign; If endothermic, include ‘+’ sign.  If the question states x% efficiency, it means that (100–x)% heat is lost/gained by the external environment. Hence, divide (mcT) by the efficiency (i.e. x%) to find H. P a g e | 5-9 Chemical Energetics 2024 Semester 1 Worked Example 2 Thermometer – measures In an experiment, 2.00 g of powdered zinc is added temperature of surroundings to 100 cm3 of 0.2 mol dm–3 copper(II) nitrate solution in an insulated container. The maximum temperature rise recorded is 5.3 °C. Calculate the enthalpy change for the reaction: Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) Surroundings (specific heat capacity of solution = 4.18 J g–1 K–1; – solution in Ar of Zn = 65.4) beaker which is surrounding Assumptions: System – reaction the reacting Solution is dilute between Zn and Cu2+ particles Since density of water = 1 g cm–3, mass of 100 cm3 =100 g Heat evolved in the reaction is only absorbed by the water and not anything else Volume does not change (1) Determine from question, values of m, c, T and n. Note: T in oC = T in K, hence T = 5.3 °C (same as 5.3 K) m = 100 cm3 (same as 100 g); c = 4.18 J g–1 K–1, n to be calculated in step (3) Recall: ignore mass of solid for reactions where a solid is added to water or solution. (2) Calculate heat evolved (mcT). Units will be in J → convert to kJ Heat evolved in the reaction = mcT = 100 × 4.18 × 5.3 = 2215.4 J = 2.22 kJ (Note: 1000 J = 1 kJ) [NO SIGN] (3) Calculate the no. of moles (amount) of limiting reagent giving rise to H. 2.00 Amount of Zn used = = 0.0306 mol 65.4 100 Amount of Cu2+(aq) used = 0.2 × = 0.0200 mol (limiting) 1000 Hence, n = 0.0200 mol (4) Calculate Ho  If reaction is exothermic, include ‘–’ sign; If endothermic, include ‘+’ sign. 𝑚𝑐∆𝑇  Hro =± × coefficient of the limiting reagent no of moles of limiting reagent 2.22 = − 0.0200 × 1 = –111 kJ mol–1 (3 sf, sign and units) P a g e | 5-10 Chemical Energetics 2024 Semester 1 Lecture Practice 2 In an experiment, 100 cm3 of 0.5 mol dm–3 AgNO3(aq) was placed in a plastic bottle. 100 cm3 of 0.5 mol dm–3 KCl(aq) was then added and the mixture stirred. Both solutions were initially at the same temperature. After mixing, the temperature of the mixture was found to have risen by 7.5 oC. Calculate the enthalpy change of the reaction. [–125 kJ mol–1] (specific heat capacity of solution = 4.18 J g–1 K–1) AgNO3 + KCl → AgCl + KNO3 Balanced equation for the reaction: ________________________________ (1) Determine from question, values of m, c, T and n. Note: T in oC = T in K, hence T = ___________ 7.5 K 100 + 100 = 200 cm = 200 g ; c = 4.18 J g–1 K–1, n to be calculated in step (3) 3 m = _________________________ (2) Calculate heat evolved (mcT). Units will be in J → convert to kJ Heat evolved in the reaction = mcT = (100 + 100) × 4.18 × 7.5 = 6270 J = 6.27 kJ (NO SIGN) (3) Calculate the no. of moles of limiting reagent giving rise to H. Amount of AgNO3 used = Amount of KCl used 100 = 0.5 × 1000 = 0.0500 mol (4) Calculate Hro  If reaction is exothermic, include ‘–’ sign; If endothermic, include ‘+’ sign. 6.27  Hro = − 0.05 × 1 = –125 kJ mol–1 (3 sf, sign and units) Related video: Title: Energetics Worked Example 3 and 4 learning to use mc delta T https://youtu.be/2LMWnyqw–vs Duration: Example 3(a) till 6:05min. Slight difference: The c value used was 4.20 J g–1 K–1 P a g e | 5-11 Chemical Energetics 2024 Semester 1 (A) Standard Enthalpy Change of Combustion, ΔHco Learning Outcome c) explain and use the term: i) enthalpy change of combustion The standard enthalpy change of combustion, Hco, is defined as the energy released when one mole of a substance is completely burnt in oxygen under standard conditions of 298 K and 1 bar. e.g. S(s) + O2(g) → SO2(g) Hco = –297 kJ mol–1 Always 1 mol exothermic! Hco is always exothermic (negative value), as heat is always evolved in the combustion. Use of Hco : it gives the energy values of fuels and foods. Experimental measurements of enthalpy change of combustion can be made in a bomb calorimeter. A bomb calorimeter is built to withstand high pressures arising due to combustion in a limited space. The heat released in the reaction heats up a volume of water in the calorimeter. If you are interested to find out more: Title: How a bomb calorimeter works https://youtu.be/VG9YG0VviHc Duration: 2:41 min The typical experimental set–up for schools is as shown below: P a g e | 5-12 Chemical Energetics 2024 Semester 1 Calculations using experimental results: Heat change in the reaction = mass of water × specific heat capacity of water × temperature change = mcT Note: Sometimes, you where: m = mass of the water heated up (units: g, sometimes given in cm3) may encounter C c = specific heat capacity of water (units: J g–1 K–1, or J cm–3 K–1) (capital letter). C denotes “heat T = temperature change (units can be either K or oC) capacity” and has units J K–1. C = mc Since the thermochemical equation for enthalpy change of combustion is defined per mole of substance burnt, mc∆T enthalpy change of combustion, ∆𝐻𝑐𝜃 = − no of moles of substance burnt Note: Enthalpy change of combustion should always be a negative value because the reaction is always exothermic. Experimental Limitation of Determination of Hco (Efficiency) In a combustion reaction, it is unlikely that all the heat released on burning a substance is used to raise the temperature of water. This is due to heat loss to the surroundings during the course of experiment (insulation to prevent heat loss is very difficult to carry out). Hence the value obtained for Hco may be inaccurate and we need to consider the efficiency. For example, if only 70% of the heat released from the combustion of a substance was transferred to the water, it is described as 70% efficient. 100 100 Heat released by combustion = heat absorbed by water × = mc∆T × 70 70 mc∆T 100 Enthalpy change of combustion, ∆𝐻𝑐𝜃 = − × no of moles of substance burnt 70 Some assumptions in the calculations: 1. Heat capacity of the calorimeter is negligible (taken to be zero). i.e. the cup that contains the reaction mixture does not absorb any heat. 2. No gain or loss of heat from the calorimeter to its surroundings. i.e. the calorimeter is well–insulated. 3. Density and specific heat capacity of the solution is the same as that of water (1.0 g cm −3 and 4.18 J g–1 K–1 respectively). i.e. the solution used must be dilute aqueous solutions. In reality, the heat capacity of the calorimeter can be determined experimentally. Details of such an experiment are in Practical Handbook 3, section 4.2.3 Calibration of Calorimeter. P a g e | 5-13 Chemical Energetics 2024 Semester 1 Worked Example 3 1.80 g of glucose was completely burnt in oxygen. The heat produced was found to raise the temperature of 5.0 dm3 of water by 1.34 oC. Calculate the enthalpy change of combustion of glucose. [Mr of glucose = 180; specific heat capacity of water = 4.18 J K–1 cm–3] Determine the values of m, c and T and calculate the heat evolved: Heat evolved in the reaction = mcT = (5000)(4.18)(1.34) = 28006 J = 28.0 kJ (NO SIGN) Calculate the amount of substance burnt: 1.80 Amount of glucose = 180 = 0.0100 mol Calculate H 28.0 Hco of glucose = − 0.0100 = –2.80  103 kJ mol–1 (3 sf, sign and units) Lecture Practice 3 Two campers are desperately short of butane fuel, yet they badly need a hot drink. They estimate that they have 0.05 mol of fuel in their ‘butane gas bottle’. What is the maximum volume of water (at 20oC) which they could heat to boiling point in order to make some hot coffee? [Hco(butane(g)) = –3000 kJ mol–1; specific heat capacity of water = 4.18 J K–1 cm–3] [449 cm3] mc∆T Hco of butane(g) = − = –3000 kJ mol–1 no. of moles of butane burnt m(4.18)(100–20) =− 0.05 = − 3000 ×103 (3 sf, sign and units) m = 449 g vol = 449 cm3 of water (since the density of water is 1 g cm–3) In reality, the amount of butane left was not enough to heat the calculated volume of water to boiling point. Assuming that the campers estimated the amount of fuel correctly, why do you think this is so? Heat was lost to the surroundings so not all the heat released was absorbed by the water. Did you know? Unlike cars or motorcycles, when airplanes refuel, they almost never fill up to “full tank”. Fuel is expensive, and the more excess fuel an airplane carries, the more fuel is used (wasted) to carry the excess weight. The flight engineer is responsible for determining the amount of fuel to carry on an airplane flight, which depends on the energy released from the combustion of the fuel, as well as the engine’s ability to convert the energy to mechanical energy (work). P a g e | 5-14 Chemical Energetics 2024 Semester 1 (B) Standard Enthalpy Change of Neutralisation, Hno Learning Outcome c) explain and use the term: i) enthalpy change of neutralisation The standard enthalpy change of neutralisation, Hno, is defined as the enthalpy change when one mole of water is formed in the neutralisation between an acid and an alkali, the reaction being carried out in aqueous solution under standard conditions of 298 K and 1 bar. e.g. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Hno = –57.1 kJ mol–1 1 mol Always exothermic 1 NaOH(aq) + __ __ ½ H2SO4 → __ ½ Na2SO4(aq) + H2O(l) Hno = –57.3 kJ mol–1 for strong 1 mol acids and alkalis! Hno is always exothermic (negative value) for strong acids and alkalis Enthalpy change of neutralisation can be measured by mixing solutions of acids and alkalis in a calorimeter and measuring the rise in temperature. Experimental Determination of Hno Because the thermochemical equation for enthalpy change of neutralisation is defined per mole of water formed, mc∆T enthalpy change of neutralisation, ∆𝐻𝑛𝜃 = − no of moles of water formed P a g e | 5-15 Chemical Energetics 2024 Semester 1 Thermometer – measures Worked Example 4 temperature of surroundings In an experiment, 100 cm3 of 0.500 mol dm–3 of HCl(aq) was placed in a plastic beaker. 100 cm3 of 0.500 mol dm–3 NaOH(aq) was then added and the mixture stirred. Both solutions were initially at the Surroundings – same temperature. After mixing, the temperature of the solution in mixture was found to have risen by 3.4 oC. beaker which is surrounding the reacting particles Calculate the enthalpy change of neutralisation. [Assume the specific heat capacities of all solutions are 4.18 J g–1 K–1 and that all solutions have a density of System – reaction 1.0 g cm–3.] between HCl and NaOH Write a balanced equation for the reaction: HCl + NaOH → NaCl + H2O (1) Determine from question, values of m, c, T and n. Note: T in oC = T in K, hence T = 3.4 K m = 100 + 100 = 200 cm3 = 200 g; c = 4.18 J g–1 K–1, n to be calculated in step (3) (2) Calculate heat evolved (mcT). Units will be in J → convert to kJ Heat evolved in the reaction = mcT = (100 + 100) × 4.18 × 3.4 = 2842.4 J = 2.84 kJ (NO SIGN) (3) Calculate the no. of moles of water giving rise to H. 100 Amount of HCl = 0.500 × = 0.0500 mol 1000 100 Amount of NaOH = 0.500 × = 0.0500 mol 1000 There is no limiting reagent as the amounts used are exact. HCl ≡ NaOH ≡ H2O, hence amount of H2O = 0.0500 mol (4) Calculate H 2.84 Hno = – = –56.8 kJ mol–1 (3 sf, sign and units.) 0.0500 P a g e | 5-16 Chemical Energetics 2024 Semester 1 Lecture Practice 4 NaOH(aq) + ½ H2SO4(aq) → ½ Na2SO4(aq) + H2O(l) The standard enthalpy change of the above reaction was determined experimentally by mixing known volumes of 1.0 mol dm–3 NaOH(aq) and 1.0 mol dm–3 H2SO4(aq). The following results were obtained: Volume of NaOH(aq) used = 40.0 cm3 Increase in temperature = 9.0 oC Volume of H2SO4(aq) used = 20.0 cm3 Use the data given to calculate the standard enthalpy change of neutralisation. [Assume the heat capacity of all solutions = 4.18 J K–1 cm–3] [–113 kJ mol–1] (1) Determine from question, values of m, c, T and n. Note: T in oC = T in K, hence T = ___________ 9.0 K 40.0 + 20.0 = 60.0 cm = 60.0 g ; c = 4.18 J K–1 cm–3, n to be calculated in step 3 m = __________________________ (3) (2) Calculate heat evolved (mcT). Units will be in J → convert to kJ Heat evolved in the reaction = mcT = (40.0 + 20.0) × 4.18 × 9.0 = 2257.2 J = 2.26 kJ (NO SIGN) (3) Calculate the no. of moles of water formed, giving rise to H. 40.0 Amount of NaOH= 1.0 × 1000 =0.0400 mol 20.0 Amount of H2SO4 = 1.0 × 1000 =0.0200 mol NaOH and H2SO4 react exactly. There is no limiting reagent. NaOH ≡ ½ H2SO4 ≡ H2O, hence amount of H2O = 0.0400 mol (4) Calculate Hno 2.26 Hno = − 0.0400 = –56.5 kJ mol–1 (3 sf, sign and units.) Extension question: Under the same experimental circumstances, what is the enthalpy change for this reaction: 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) 2NaOH ≡ H2SO4, and the amount of NaOH = 0.0400 mol Recall page 5–9: Hro = enthalpy change of reaction 𝑚𝑐∆𝑇 =± × coefficient of the limiting reagent no. of moles of limiting reagent 2.26 Hro = − 0.0400 ×2 = − 113 kJ mol–1(3 sf, sign and units.) P a g e | 5-17 Chemical Energetics 2024 Semester 1 The enthalpies of neutralisation for some strong acids and strong alkalis are given below: Strong Acid Strong Alkali Hno / kJ mol–1 HNO3 NaOH –57.3 HCl KOH –57.3 HNO3 KOH –57.3 Example: The reaction between HNO3 and NaOH is: HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l) HNO3 is a strong acid and NaOH is a strong base and they are fully dissociated in dilute solutions. HNO3 → H+ + NO3– NaOH → Na+ + OH– Hence, the reaction between any strong acid and any strong base involves simply the formation of water from H+ and OH– ions. H+(aq) + OH–(aq) → H2O(l) Hno = –57.3 kJ mol–1 For the neutralisation of strong acids with strong bases, the standard enthalpy change of neutralisation is almost constant (Hno = –57.3 kJ mol–1). The enthalpies of neutralisation involving weak acids / weak base are given below: Weak Strong Hno Strong Weak Hno Acid Alkali / kJ mol–1 Acid Alkali / kJ mol–1 CH3CO2H NaOH –55.0 HCl NH3(aq) –51.5 HCN NaOH –12.0 If a weak acid or a weak base is used, or if both acid and base are weak, then the standard enthalpy of neutralisation deviates from –57.3 kJ mol–1. Reason: Weak acids and bases only ionise partially in dilute aqueous solution. e.g. CH3CO2H(aq) ⇌ CH3CO2−(aq) + H+(aq) H > 0 Ionisation, which involves bond–breaking, is an endothermic process. During neutralisation, some heat released from the neutralisation reaction is absorbed to further ionise the weak acid/base completely.  Overall, less heat is released. Hence, for neutralisation reactions involving weak acid and/ or weak base, Hno will be less exothermic (less negative than –57.3 kJ mol−1) than that of strong acid−strong base neutralisation reactions. It is incorrect to say  Hno will be "lower/smaller". It should be "less exothermic". When H is less exothermic (less negative), less heat is released. P a g e | 5-18 Chemical Energetics 2024 Semester 1 The actual value for a particular neutralisation will depend on the degree of ionisation of the weak acid/base. 3.2 Bond Energy (only for covalent bonds) Learning Outcome: c) explain and use the term: ii) bond energy (ΔH positive, i.e. bond breaking) Bond energy (of dissociation) is the energy required to break one mole of a covalent bond between two atoms in the gaseous state. Always e.g. A ⎯ B(g) → A(g) + B(g) Ho > 0 (positive) in kJ mol–1 endothermic! Energy must be absorbed in order to break a bond between two atoms. Bond–breaking is an endothermic process. Energy is released when a bond forms between two atoms. Bond–forming (or bond– making) is an exothermic process. Since chemical reactions involve bond–breaking followed by bond forming, the enthalpy change of a reaction is, therefore, the energy difference between the bond–breaking and bond–forming processes. Uses of Bond Energies 1. Bond energies are useful in comparing the strength of covalent bonds. The greater the bond energy, the stronger the bond. Very large bond energies can make molecules inert. e.g. NN molecules are generally unreactive due to the high NN bond energy. 2. Bond energies can also be used to estimate the enthalpy changes of reactions involving SIMPLE COVALENT MOLECULES. These estimations are particularly useful when calorimetric or other experimental measurements cannot be made. The table below shows a list of some bond energies from the Data Booklet: Bond Energy Bond Energy Bond Bond / kJ per mole of bonds / kJ per mole of bonds C–H 410 O–O 150 C–C 350 O=O 496 C=C 610 N–N 160 CC 840 NN 944 F–F 158 H–F 562 Cl–Cl 244 H–Cl 431 Br–Br 193 H–Br 366 Bond energies in the Data Booklet used for calculation are merely average bond energy values and may not be a true reflection of actual bond energies in the molecules. Thus the enthalpy changes calculated using bond energies are only an approximation. P a g e | 5-19 Chemical Energetics 2024 Semester 1 HOW TO… Calculate H using Bond Energies NOTE: Bond energies can ONLY be USED in reactions involving ONLY SIMPLE COVALENT MOLECULES!! (1) Draw structures of the compounds involved. Recall: O2 is O=O; CO2 is O=C=O (2) List and count the no. of bonds to be BROKEN (bond–breaking is an endothermic process. Hence put a “+” sign) (3) List and count the no. of bonds to be FORMED (bond–forming is an exothermic process. Hence put a “–” sign) (4) Add up the enthalpy changes from total bonds broken and total bonds formed = Hro Worked Example 5 Hydrazine is often used as a rocket fuel as it can be stored conveniently as a liquid and it reacts very exothermically with oxygen forming purely gaseous products. With reference to the bond energies of the molecules, what is Hro for this reaction per mole of hydrazine? N2H4(g) + O2(g) → N2(g) + 2H2O(g) Hro = ? (1) Draw structures of the compounds involved: (2) List and count the no. of bonds to be BROKEN (3) List and count the no. of bonds FORMED bonds broken (endothermic) bonds formed (exothermic) 1 N−N (+160) 1 NN (–944) 4 N−H 4(+390) 4 O−H 4(–460) 1 O=O (+496) +2216 –2784 (4) Add up the enthalpy changes from total bonds broken and total bonds formed = H ΔHro = +2216 + (–2784) = –568 kJ mol–1 (3 sf, sign and units.) Note: Parts (2) to (4) in general is: ΔHro = sum of all bond energies in reactants – sum of all bond energies in products (Since B.E. are all positive values in Data Booklet) P a g e | 5-20 Chemical Energetics 2024 Semester 1 Lecture Practice 5 Calculate Ho for the hydrogenation of ethene. H H H H C C + H H H C C H H H H H (1) Structures of the compounds involved have been given. [–124 kJ mol–1] (2) List and count the no. of bonds to be BROKEN (3) List and count the no. of bonds FORMED bonds broken (endothermic) bonds formed (exothermic) 1 C=C (+610) 1 C−C (–350) 4 C−H 4(+410) 6 C−H 6(–410) 1 H−H (+436) +2686 –2810 (4) Add up the enthalpy changes from total bonds broken and total bonds formed = H ΔHro = +2686 + (–2810) = –124 kJ mol–1 (3 sf, sign and units.) Recall: Parts (2) to (4) in general is: ΔHro = sum of all bond energies in reactants – sum of all bond energies in products (Since B.E. are all positive values in Data Booklet) 3.3 Use of Hess’ Law (C) Standard Enthalpy Change of Formation, Hfo Learning Outcome c) explain and use the terms: i) enthalpy change of formation The standard enthalpy change of formation, Hfo, of a substance (usually a compound) is defined as the enthalpy change when one mole of the substance is formed from its elements under standard conditions of 298 K and 1 bar. (Elements must be in most stable physical form.) e.g. H2(g) + ½ O2(g) → H2O(l) Hfo = –268 kJ mol–1 1 mol If Hfo is exothermic, the substance has lower heat content than its elements and hence, is energetically more stable than its constituent elements. Hfo of any element in its normal physical state is zero. e.g. Cl2(g) → Cl2(g) Hfo = 0 Thinking Question 2: What enthalpy changes can be used to describe the following equation? C(s) + O2(g) → CO2(g) The equation represents the enthalpy change of combustion of C(s) and also the enthalpy change of formation of CO2(g). P a g e | 5-21 Chemical Energetics 2024 Semester 1 3.3.1 Hess’ Law Learning Outcome: f) apply Hess’ Law to construct simple energy cycles, and carry out calculations involving such cycles and relevant energy terms We have seen definitions of various enthalpy changes of reactions so far. Some can be easily calculated once we do the necessary measurements by conducting experiments: Hco and Hno. [Refer to Section 3.1(A) and 3.1(B)] 𝑚𝑐∆𝑇 𝑚𝑐∆𝑇 ∆𝐻𝑐𝜃 = − ∆𝐻𝑛𝜃 = − no of moles of substance burnt no of moles of water formed Some of the enthalpy changes can be calculated by considering the bonds broken and formed through the use of bond energies [Refer to Section 3.2] H = sum of all bond energies in reactants – sum of all bond energies in products (Since B.E. are positive values in Data Booklet) But… Many of the other defined enthalpy changes are difficult to find by experimental methods because the reactions are almost impossible to carry out. Example: The enthalpy change of formation of carbon monoxide, CO. C(s, graphite) + ½ O2(g) → CO(g) The above reaction can never occur alone, as some CO2 will always be produced at the same time: C(s, graphite) + O2(g) → CO2(g) Hence the temperature change measured will never be accurate! Fortunately we can find such enthalpy changes indirectly, using Hess’ Law. Hess’ Law states that the enthalpy change of a chemical reaction depends only on the final and initial states of the system, and is independent of the reaction pathway taken. This means that the enthalpy change is independent of its path of reaction as long as the reactants and the products are the same (i.e., same p, T and physical state). The enthalpy change of a chemical reaction is the SAME whether the change is brought about in one stage or through intermediate stages. P a g e | 5-22  Chemical Energetics 2024 Semester 1 How is Hess’ Law applied? To apply Hess’ Law, you usually need to construct an energy cycle. Consider the energy cycle below: By Hess’s Law, the overall enthalpy change will be the same whether the direct route or indirect route is used as long as the initial and final states of the system are the same. By applying Hess’s Law, H1 = H2 + H3 Consider another energy cycle below: For the indirect route, A and B are first converted to E and F respectively, which then react to form C. Given that H for C → E + F is H6, hence H for the reverse reaction E + F → C is −H6. By applying Hess’s Law, H1 = H4 + H5 − H6 Hess’s Law may not be a representation of actual processes. It is simply an application of the Law of Conservation of Energy. Diagrams taken from: We can draw an energy cycle to find the enthalpy change of reaction based on Lewis, M.; AS & A (i) enthalpies of formation: (ii) enthalpies of combustion: Level Chemistry through diagrams; Oxford University Press, 2001 (i) indirect pathway = direct pathway (ii) indirect pathway = direct pathway Hfo (reactants) + Horeaction = Hfo(products) Horeaction + Hco (products) = Hco (reactants) rearranging, rearranging, Horeaction = Hfo (products) – Hfo (reactants) Horeaction = Hco (reactants) – Hco (products) P a g e | 5-23 Chemical Energetics 2024 Semester 1 HOW TO… Calculate H using Energy Cycles (1) Write the chemical equation (with state symbols) asked in the question. (2) Complete the energy cycle with the other information given from the question. Some tricks: If Hco values are given, arrows POINT AWAY FROM substance. If Hfo values are given, arrows POINT TOWARDS compounds. (refer to energy cycles shown earlier for illustration) (3) Fill the H values into the cycle. Note: If no. of moles > 1, you need to multiply the given H value by the correct no. of moles as in the enthalpy change definition. Note: Some values may not be given. So you might need to obtain them from Data Booklet, usually IE and Bond Energy. (4) Apply Hess’ Law to calculate Hro of the unknown reaction. A useful formula derived from Hess’ Law for working with enthalpy changes of formation: Hro = (Sum of Hfo of products) – (Sum of Hfo of reactants) Worked Example 6 What is the standard enthalpy change of combustion of propyne, C3H4 C3H4(g) + 4O2(g) → 3CO2(g) + 2H2O(l) given that the standard enthalpy changes of formation of propyne, carbon dioxide and water are –1258 kJ mol–1, –394 kJ mol–1 and –286 kJ mol–1 respectively? Energy cycle method: (1) Write the chemical equation (with state Hco = ? C3H4(g) + 4O2(g) 3CO2(g) + 2H2O(l) symbols) asked in the question. (2) Since Hfo values are given, arrows –1258 0 3(–394) 2(–286) POINT TOWARDS compounds from elements. (3) Fill the H values into 3 C(s) + 2 H2(g) + 4 O2(g) the cycle. (Hfo O2 = 0 kJ mol–1) Using Hess' Law, (4) Apply Hess’ Law to calculate Hro Hco + (–1258) + 0 = 3(–394) + 2(–286) of the unknown reaction. Hco = 3(–394) + 2(–286) – (–1258) = –496 kJ mol–1 (3 sf, sign and units.) Or, using formula above: Hco = [3(–394) + 2(–286)] – [(–1258) + 0] = –496 kJ mol–1 P a g e | 5-24 Chemical Energetics 2024 Semester 1 Lecture Practice 6 Calculate the enthalpy change for the reaction: CaF2(s) + H2SO4(l) → 2HF(g) + CaSO4(s) given that the enthalpy changes of formation of CaF2(s), H2SO4(l), HF(g) and CaSO4(s) are –1220 kJ mol–1, –814 kJ mol–1 , –271 kJ mol–1 and –1434 kJ mol–1 respectively? [+58.0 kJ mol–1] Energy cycle method: Hr = ? CaF2(s) + H2SO4(l) 2HF(g) + CaSO4(s) –1220 –814 2(–271) –1434 Ca(s) + F2(g) + H2(g) + S(s) + 2O2(g) Using Hess’ Law, Hro = – (–1220 – 814) + 2(–271) + (–1434) = +58.0 kJ mol–1 (3 sf, sign and units.) Or, using formula above: Hro = [2(–271) + (–1434)] – [(–1220) + (–814)] = +58.0 kJ mol–1 P a g e | 5-25 Chemical Energetics 2024 Semester 1 Learning Outcome f) apply Hess’ Law to construct simple energy cycles, particularly: (i) determining enthalpy changes that cannot be found by direct experiment, e.g. an enthalpy change of formation from enthalpy changes of combustion A useful formula derived from Hess’ Law for working with enthalpy changes of combustion: Hro = (Sum of Hco of reactants) – (Sum of Hco of products) Worked Example 7 The standard enthalpy change of combustion of graphite is –393.4 kJ mol–1 and that of CO is –282.9 kJ mol–1. Calculate the standard enthalpy change of formation of CO. Energy cycle method: (1) Write the chemical equation (with state Hfo = ? C(s) + ½O2(g) CO(g) symbols) asked in the question. + ½O2(g) + ½O2(g) (2) Since Hco values are given, arrows –393.4 –282.9 POINT AWAY from substance. (3) Fill the H values into the cycle. CO2(g) (4) Apply Hess’ Law to calculate ΔHfo Using Hess’ Law, ΔHfo + (–282.9) = –393.4 of the unknown reaction. ΔHfo = –110.5 kJ mol–1 (3 sf, sign and units.) Or, using formula above: ΔHfo = –393.4 – (–282.9) = –110.5 kJ mol–1 Lecture Practice 7 The standard enthalpies of combustion of methane, graphite and hydrogen are –890.2 kJ mol–1, –393.4 kJ mol–1 and –285.7 kJ mol–1 respectively. Calculate the standard enthalpy change of formation of methane, CH4. [–74.6 kJ mol–1] Energy cycle method: Hfo = ? C(s) + 2 H2(g) CH4(g) + O2(g) + O2(g) + 2O2(g) –393.4 2(–285.7) –890.2 CO2(g) + 2 H2O(l) Using Hess’ Law, Hfo + (–890.2) = –393.4 + 2(–285.7) Hfo = –74.6 kJ mol–1 (3 sf, sign and units.) Or, using formula above: ΔHfo = –393.4 + 2(–285.7) – (–890.2) = –74.6 kJ mol–1 P a g e | 5-26 Chemical Energetics 2024 Semester 1 3.3.2 Other Energy Changes (D) Standard Enthalpy Change of Hydration, Hhydo Learning Outcome c) explain and use the term: i) enthalpy change of hydration The standard enthalpy change of hydration, Hhydo, of an ion is defined as the enthalpy change when one mole of the gaseous ions is dissolved in such a large amount of water that addition of more water produces no further heat change under standard conditions of 298 K and 1 bar. (i.e. further dilution has no effect) Always exothermic! e.g. Na+(g) + aq → Na+(aq) Hhydo = –390 kJ mol–1 Cl– (g) + aq → Cl – (aq) Hhydo = –381 kJ mol–1 Note: Bond–forming takes place between gaseous ions and water molecules due to ion– dipole attractions. Since there is no bond–breaking, the enthalpy change is always exothermic. hydrated Na+(aq) ion–dipole attractions hydrated Cl–(aq) Hhydo is always exothermic (negative value), as heat is produced when ion–dipole attractions are formed between the ions and the dipoles on the water. The hydration energy of an ion is proportional to the charge density of the ion, i.e. ionic charge (q) and size (radius, r) of the ion. 𝜃 𝑞 hydration energy, ∆𝐻ℎ𝑦𝑑 ∝ 𝑟 The higher the charge and the smaller the size (radius) of the ion, the more exothermic the hydration energy will be. P a g e | 5-27 Chemical Energetics 2024 Semester 1 Worked Example 8 The hydration energies for various cations and anions are shown in the following table: Ion ∆Hhydo / kJ mol–1 Ion ∆Hhydo / kJ mol–1 Na+ –390 Cl– –381 – Mg 2+ –1891 Br –351 Al 3+ –4613 I– –307 (a) Why does the hydration energy get progressively less exothermic down the series Cl–, Br–, I–? Although they all have the same charge, the ionic size (radius) is increasing from Cl– to I–. Hence the ion–dipole attractions formed between the ions and water molecules become increasingly weaker, and thus the hydration energy becomes less exothermic. (b) What is the total hydration energy of MgCl2? Hhyd of MgCl2 = (–1891) + 2(–381) = –2653 kJmol–1 = –2650 kJmol–1 (3 sf, sign and units.) Lecture Practice 8 Refer to the table in Worked Example 8. (a) Why does the hydration energy get progressively more exothermic across the series Na+, Mg2+ and Al3+? The ions from Na+ to Al3+ have increasing charges and decreasing ionic size. Hence the ion–dipole attractions formed between the ions and water molecules become increasingly stronger, and thus the hydration energy becomes more exothermic. (b) What is the ionic compound that has the least exothermic hydration energy? [–697 kJ mol–1] NaI Hhyd of NaI = (–390) + (–307) = –697 kJ mol–1 (3 sf, sign and units.) P a g e | 5-28 Chemical Energetics 2024 Semester 1 (E) Standard Enthalpy Change of Solution, Hsolo Learning Outcome c) explain and use the term: i) enthalpy change of solution The standard enthalpy change of solution, Hsolo, is defined as the enthalpy change when one mole of a substance is dissolved in such a large volume of solvent that addition of more solvent produces no further heat change under standard conditions of 298 K and 1 bar. (i.e. further dilution has no additional effect.) e.g. NaCl(s) + aq → Na+(aq) + Cl–(aq) Hsolo = = + 3.9 kJ mol–1 NH3(g) + aq → NH3(aq) Hsolo = –35.2 kJ mol–1 Hsolo can be exothermic or endothermic (negative or positive values). Despite its common reference to ionic compounds, standard enthalpy change of solution is also defined for covalent compounds (such as ethanol or sugar), and gases (such as carbon dioxide or ammonia). (Refer to Chemical Bonding for more information on the process of dissolution.) Did you know? Instant cold packs can be used to treat fever, burns, and swelling due to bruises. They are useful in situations such as outdoor camping or trekking, where ice is not readily available. How do instant cold packs work? Many cold packs contain ammonium nitrate as the active ingredient. The enthalpy change of solution of ammonium nitrate is endothermic: NH4NO3(s) + aq → NH4+(aq) + NO3–(aq) Hsolo = +25.4 kJ mol–1 In instant cold packs, a sachet of ammonium nitrate solid is sealed inside a bag of water. Hitting the cold pack breaks the sachet, causing the ammonium nitrate to mix with and dissolve in the water. Since the dissolution of ammonium nitrate is endothermic, heat is absorbed from the surroundings and the pack becomes cold. P a g e | 5-29 Chemical Energetics 2024 Semester 1 Energy Changes Involving Aqueous Solutions of Ionic Compounds Learning Outcome: f) apply Hess’ Law and carry out calculations with particular reference to: (ii) the formation of a simple ionic solid and of its aqueous solution When sodium chloride dissolves in water, the following process occurs: Na+Cl–(s) + aq → Na+(aq) + Cl–(aq) Hsolo We can see how the enthalpy terms: lattice energy and Hhydo are related to Hsolo by constructing the following energy cycle: Solid ionic Hsolo Aqueous Recall: compound ions Lattice energy is the energy evolved when lattice energy one mole of Hhydo ionic solid is formed from Gaseous its separate ions gaseous ions. By Hess’ Law, Hsolo = – lattice energy + ΣHhydo Using NaCl to illustrate: Hsolo NaCl (s) Na+(aq) + Cl–(aq) lattice energy Hhydo Na+(g) + Cl–(g) Diagrammatically, we can see how the processes are related: Diagram modified from JGR Briggs, Longman ‘A’ Level Course in Chemistry, Longman 2005 P a g e | 5-30 Chemical Energetics 2024 Semester 1 In general, if Hhydo > lattice energy (in magnitude) then Hsol is negative, the salt likely dissolves readily in water. If Hhydo < lattice energy (in magnitude) then Hsolo is positive, the salt may be relatively insoluble. (This may not always be true and Gsolo (see 4.2) has to be considered) Worked Example 9 Draw an energy cycle relating the following enthalpy changes for magnesium chloride: Hsolo (MgCl2) = –132 kJ mol–1 Hhydo (Mg2+) = –1921 kJ mol–1 Hhydo (Cl–) = –364 kJ mol–1 Lattice energy of magnesium chloride Hence, calculate the lattice energy of magnesium chloride. Lattice energy Mg2+(g) + 2Cl–(g) MgCl2(s) –1921 + 2(–364) –132 Mg2+(aq) + 2Cl–(aq) By Hess’ Law, Lattice energy = –1921 + 2(–364) – (–132) = –2517 = –2520 kJ mol–1 (3sf, sign and units) Lecture Practice 9 Draw an energy cycle relating the following enthalpy changes for iron(III) chloride: Hsolo (FeCl3) = –928 kJ mol–1 Hhydo (Fe3+) Hhydo (Cl–) = –364 kJ mol–1 Lattice energy of iron(III) chloride = –3865 kJ mol–1 Hence, calculate the enthalpy change of hydration of Fe3+. [–3700 kJ mol–1] –928 FeCl3(s) Fe3+(aq) + 3Cl–(aq) Hhydo 3(–364) –3865 3+ – Fe (g) + 3Cl (g) By Hess’ Law, –3865 – 928 = Hhydo (Fe3+) + 3(–364) Hhydo (Fe3+) = –3701 kJ mol–1 = –3700 kJ mol–1 (3 sf, sign and units) P a g e | 5-31 Chemical Energetics 2024 Semester 1 (F) Lattice Energy (Revision: Chemical Bonding) Learning Outcome: c) explain and use the term: (iii) lattice energy (H negative, i.e., gaseous ions to solid lattice) e) explain, in qualitative terms, the effect of ionic charge and of ionic radius on the numerical magnitude of a lattice energy It is very important not to confuse bond energy with lattice energy. Bond energy can only be used for covalent bonds, whereas lattice energy can only be used for ionic compounds. What is lattice energy? The ions in an ionic crystal are bound together by electrostatic attractive forces between the ions and there is a considerable release of energy when such a crystal is formed. e.g., Na+(g) + Cl–(g) → NaCl (s) H = lattice energy The lattice energy of an ionic solid is the energy evolved when one mole of the ionic solid is formed from its separate gaseous ions. e.g., K+(g) + Cl–(g) → KCl(s) Ho = –701 kJ mol–1 Always All lattice energies are exothermic (negative values) because heat is evolved exothermic! when the ions come together to form ionic bonds. What is the physical significance of lattice energy? It provides a measure of the attractive force holding the ions in their positions in a crystal lattice, i.e., it provides a measure of the strength of the ionic bonds. The more exothermic the lattice energy, the stronger the ionic bond strength. Factors affecting the magnitude of lattice energy: q+ q– since lattice energy  r+ +r– (q+ = charge on cation, q− = charge on anion, r+ = cationic radius, r− = anionic radius) Recall: The higher the charges on the ions, the more exothermic the lattice energy. The smaller the radii of the ions, the more exothermic the lattice energy. The arrangement of the ions in the crystal (crystal structure) also affects the magnitude of the lattice energy, but this effect is small. P a g e | 5-32 Chemical Energetics 2024 Semester 1 Worked Example 10 Explain why the melting point of MgO is higher than that of CaO. Both MgO and CaO have giant ionic structures. Both Mg2+ and Ca2+ have the same ionic charge, however Mg2+ is smaller than Ca2+. Hence, the lattice energy of MgO is more exothermic / greater in numerical magnitude than CaO. More energy is required to break the stronger ionic bonds between Mg2+ and O2– as compared to the energy required for Ca2+ and O2–. Therefore MgO has a higher melting point than CaO. Lecture Practice 10 Which of the following compounds would have the largest lattice energy in magnitude? Na2O MgO NaCl MgCl2 Explain your choice. q+ q– Using lattice energy  , r+ +r– Mg2+ has a higher charge than Na+, and O2– has higher charge than Cl–. Mg2+ is smaller than Na+, and Cl– is smaller than O2–. (Charge is the more dominant factor affecting lattice energies, if ions concerned are from same period.) Ans: MgO would have the largest lattice energy in magnitude. Theoretical and Experimental Lattice Energy Theoretical calculation of lattice energies assumes that compounds are purely ionic. The table below shows some comparisons: Calculated L.E. / Difference as Observed L.E. kJ mol–1 a % of /kJ mol–1 Compound (theoretical, purely Difference observed (experimental) ionic) value NaCl –776 –786 10 1.3 NaBr –733 –744 11 1.5 NaI –684 –697 13 1.9 AgCl –768 –890 122 15.9 AgBr –759 –877 118 15.5 AgI –736 –867 131 17.8 Observations: Slight discrepancies between the theoretical and experimental lattice energies for NaCl, NaBr, NaI. This close agreement provides strong evidence in favour of the ionic model i.e. alkali metal halides form ionic crystals with little or no covalent character. Large discrepancies between the theoretical and experimental lattice energies for AgCl, AgBr, AgI. This difference is attributed to polarisation of the halide ions i.e. silver halides form ionic crystals with a considerable percentage of covalent character. P a g e | 5-33 Chemical Energetics 2024 Semester 1 (G) Standard Enthalpy Change of Atomisation Hato Learning Outcome c) explain and use the term: i) enthalpy change of atomisation The standard enthalpy change of atomisation, Hato, is defined as the enthalpy change when one mole of separate gaseous atoms is formed from its element under standard conditions of 298 K and 1 bar. e.g. Na(s) → Na(g) Hato = +108 kJ mol–1 1 mol ½ H2(g) ___ → 1 H(g) ___ Hato = +218 kJ mol–1 1 mol Always endothermic! I. Hato is always endothermic (positive value), because energy must be absorbed to pull the atoms far apart and to break all the bonds between them. II. The enthalpy change of atomisation is not the same as the enthalpy change of vaporisation. Note: Enthalpy change of atomisation involves the breaking of intramolecular bonds as well. Enthalpy change of vaporisation involves only the breaking of intermolecular bonds, i.e. a change of state. Br2(l) → Br2(g) Hence when an element is vaporised, the gas particles are usually not separate atoms. III. For all noble gases (e.g., Ne, Ar), the enthalpy change of atomisation is zero. This is because the elements are already in the form of separate gaseous atoms under standard conditions. IV. For simple gaseous diatomic molecules, Hato = ½ Bond Energy Example using chlorine gas: Hato is given by ½ Cl2(g) → Cl(g) Bond Energy is given Cl2(g) → 2Cl(g) Thinking Question 3: Would Hato of Br2 = ½ Bond Energy of Br2? No. Br2 is a liquid under standard conditions. Hence Hato of Br2 = Hvapo of Br2 + ½ Bond Energy of Br2 P a g e | 5-34 Chemical Energetics 2024 Semester 1 (H) Ionisation Energy (Revision: Atomic Structure) Ionisation energy (I.E.) is the energy required to remove one mole of electrons from one mole of gaseous atoms (or ions) to produce one mole of gaseous cations. Always endothermic! I.E. is always endothermic (positive value) since energy is absorbed in ionisation. The successive ionisation energies of an element increase with removal of each electron because the remaining electrons are attracted more strongly (due to less shielding effect) to the constant positive charge on the nucleus. e.g. X(g) → X+(g) + e– H1 = 1st I.E. of X(g) X (g) → X (g) + e + 2+ – H2 = 2nd I.E. of X(g) (>H1) The energy needed to ionise one mole of X(g) atoms to its divalent ions (i.e. X(g) → X2+(g) + 2e–) = H1 + H2 Thinking Question 4: What is the definition of the second ionisation energy? Energy required to remove one mole of electrons from one mole of gaseous X+ cations to form one mole of gaseous X2+ cations. (I) Electron Affinity The electron affinity of an element is the energy change when one mole of electrons is added to one mole of gaseous atoms to produce one mole of gaseous anions. e.g. Cl(g) + e– → Cl–(g) Ho = –364 kJ mol–1 (first electron affinity) It is a measure of the attraction of the atom for additional electrons. The first E.A. of an element is usually exothermic (negative value) as the incoming electron is attracted to the nuclear charge of the atom. Thinking Question 5: Would you expect the 2nd electron affinity of O to be exothermic or endothermic? Why? O–(g) + e– → O2–(g) For second EA, energy is always absorbed (endothermic) because the incoming second electron must overcome the repulsion from the (negatively charged) anion. P a g e | 5-35 Chemical Energetics 2024 Semester 1 3.3.3 Born–Haber Cycle Learning Outcome: f) apply Hess’ Law to construct simple energy cycles, e.g., Born–Haber cycle, and carry out calculations involving such cycles and relevant energy terms (including ionisation energy and electron affinity) The Born–Haber Cycle is an energy cycle commonly used to calculate the lattice energy of an ionic solid as lattice energy cannot be directly measured. The cycle gives the relationship between the standard enthalpy change of formation of the ionic solid and the various enthalpy changes leading to its formation. When the various enthalpy changes and lattice energy are put in a cycle, it may appear something like this: Gaseous ions IE, EA Hlatto (Lattice Energy) Gaseous atoms Hato, BE Hfo elements solid ionic compound Worked Example 11 Lattice energies are not measured directly. By using Hess' Law and the Born–Haber cycle, they can be obtained from experimental data. Construct a Born–Haber cycle which can be used to calculate the lattice energy of sodium chloride. Draw arrows representing the energy changes involved, and label them clearly. Use words and symbols to represent these energy terms. By Hess’ Law, – Na+ (g) + Cl (g) Hfo = H1o + H2o + H3o + H4o + H5o H3o H4o H5o where Na(g) + Cl(g) H1o = enthalpy change of atomisation of Na H2o = enthalpy change of atomisation of Cl H1o H2o H3o = first ionisation energy of Na Hfo Na (s) + ½ Cl2(g) NaCl (s) H4o = first electron affinity of Cl H5o = lattice energy of NaCl Title: Energetics Lecture 2013 How to draw a BH Cycle https://youtu.be/pqBGu2LEbbU Duration: 5:55 min (Lecturer: Mrs Angela Tie) P a g e | 5-36 Chemical Energetics 2024 Semester 1 HOW TO… Draw a Born–Haber Cycle 1. Anchor the Born–Haber cycle with two H equations: Hf o & Hlatt o of the ionic compound. e.g., Hf o of NaCl (s) & Hlatt o of NaCl : Hf o of NaCl (s): Na(s) + ½ Cl2(g) → NaCl (s) Hlatt o of NaCl (s): Na+(g) + Cl–(g) → NaCl (s) 2. Connect the two equations. 3. Complete the cycle by filling in the remaining H (moving from elements → ionic compound) 4. Fill in the H values from data given. Note: Some values may not be given. So you might need to obtain them from Data Booklet, usually IE and Bond Energy. 5. Apply Hess’ Law to calculate H of the unknown reaction Lecture Practice 11 By drawing a Born–Haber cycle, calculate the lattice energy of calcium fluoride given the following information: H / kJ mol–1 Enthalpy change of atomisation of calcium +184 Enthalpy change of atomisation of fluorine +79 First ionisation energy of calcium +590 Second ionisation energy of calcium +1150 Electron affinity of fluorine –328 Enthalpy change of formation of calcium fluoride –1220 [–2650 kJ mol–1] Born–Haber Cycle for formation of CaF2 Ca2+(g) + 2 F–(g) +1150 Ca+(g) 2(–328) H = ? +590 Ca(g) 2 F(g) +184 2(+79) –1220 Ca(s) + F2(g) CaF2(s) By Hess’ Law, –1220 = 184 + 2(79) + 590 + 1150 + 2(–328) + H H = –2646 kJ mol–1 = –2650 kJ mol–1 (3 sf, sign and units) P a g e | 5-37 Chemical Energetics 2024 Semester 1 3.3.4 Energy Level Diagram Alternatively, the reactions making up the Born–Haber cycle can also be represented using an energy level diagram. Unlike the Born–Haber Cycle, the energy level diagram must have: 1 y–axis representing energy 2 A datum line if elements are involved. A datum line is the arbitrary zero enthalpy content of any pure element. (I.e. Pure elements in their standard states are defined as having zero enthalpy content) 3  arrows for endothermic reaction (+ve) and  arrows for exothermic reaction (–ve). It will be useful to note that: Standard enthalpy change of formation of ionic compound:  Standard enthalpy change of atomization of metal:  Standard enthalpy change of atomization of non–metal:  Ionisation energy of metal:  Electron affinity of non–metal: usually 1st E.A.  and 2nd E.A.  Lattice energy:  Example: For an ionic compound, MX, the energy level diagram looks like this: energy M+ (g) + e– + X (g) 1st IE (M) 1st EA (X) M+ (g) + X– (g) M (g) + X (g) Hato (X) = ½ BE(X–X) M (g) + ½ X2 (g) Hlatto (MX) Hato (M) Each change in level represents M (s) + ½ X2 (g) only 1 chemical 0 change, i.e., no Hf o (MX) “parallel” changes are allowed. MX (s) Hess’ Law can be applied to energy level diagrams as well. For example, in the above diagram, Sum of H in clockwise direction = Sum of H in anticlockwise direction Hf o (MX) = Hato (M) + Hato (X) + 1st IE (M) + 1st EA (X) + Hlatto (MX) P a g e | 5-38  Chemical Energetics 2024 Semester 1 An energy level diagram for the formation of NaCl: Energy Na+(g) + Cl(g) Ho3 Na(g) + Cl (g) o Ho4 o Ho2o Na+(g) + Cl–(g) Na(g) + ½ Cl2(g) Ho1 0 Na(s) + ½ Cl2(g) o Ho5o Hof NaCl (s) Energy level diagrams can be drawn in place of all types of energy cycles. For example, enthalpy changes involving aqueous ionic compounds: Energy Na+(g) + Cl–(g) Hohyd Na+(aq) + Cl–(aq) Holatto Hosol NaCl (s) In the above diagram, there is no datum line as pure elements are not involved. Note here that Hosol for NaCl(s) is endothermic, thus aqueous NaCl has higher enthalpy than NaCl(s) and so the Hosol arrow points upwards. Hosol for many other ionic compounds such as KCl(s) is exothermic, however, so the aqueous solution has a lower enthalpy than the ionic solid, resulting in the energy level diagram below: Energy K+(g) + Cl–(g) Hohyd Holatt KCl (s) Hosol K+(aq) + Cl–(aq) P a g e | 5-39 Chemical Energetics 2024 Semester 1 Lecture Practice 12 Fill in the blanks in the energy level diagram with the names of the appropriate enthalpy changes, e.g. lattice energy, including coefficients if necessary. Energy Al3+(g) + 3F(g) + 3e– Al (g) + 3F(g) (c) (d) 3 (b) Al3+(g) + 3F–(g) Al (g) + 2 F2(g) 3 (a) Al (s) + 2 F2(g) lattice 0 energy Hof = ‒1510 kJ mol–1 AlF3 (s) (a) Hatmo (Al) 3 (b) 3 Hatmo (F2) = BE (F2) 2 (c) 1st IE + 2nd IE + 3rd IE of Al (d) 3 x 1st EA of F Calculate the lattice energy of aluminium fluoride using the following data any other data that can be obtained from the Data Booklet. enthalpy change of atomisation of aluminium = +326 kJ mol‒1 first electron affinity of fluorine = ‒328 kJ mol‒1 Sum of H in clockwise direction = Sum of H in anticlockwise direction 3 ‒1510 = +326 + 2(158) + 577 + 1820 +2740 + 3(‒328) + Hlatto (AlF3(s)) Hlatto (AlF3(s)) = ‒6226 kJ mol–1 = ‒6230 kJ mol–1 (3 sf, sign and units) [‒6230 kJ mol–1] P a g e | 5-40  Chemical Energetics 2024 Semester 1 SUMMARY OF METHODS TO FIND ENTHALPY CHANGE 3.1 Calorimetry (experimental method) Used when ΔT and mass are given in the question. Relevant formulae: 𝑚𝑐∆𝑇 𝑚𝑐∆𝑇 ∆𝐻𝑐𝜃 = −