Chapter 06: Chemical Energetics

Energy Profile Diagrams Bond Breaking and Making in Reactions: All chemical reactions involve breaking of bonds in reactant molécules. Followed by the formation of new bonds in product molecules. Activation Energy Definition: The minimum amount of energy that reactant molecules must possess for effective collisions to overcome the energy barrier and start a chemical reaction is called activation energy. Symbol: Ea Without enough activation energy, collisions between molecules will not lead to a reaction Energy profile diagram ,Ea HR- reactants Enthalpy H ple 5Jd ree products Progress of Reaction Exothermic reaction Fig: Energy profile diagram in terms of AH and Ea This diagram shows how energy changes during a chemical reaction. • Reactant energy level • Activation energy Peak or transition state • Product energy level • Enthalpy change (AH) Exothermic Reactions • Reactant have more energy than products • AH is negative • Heat is released • Low activation energy give more stable products Endothermic Reactions • Products have more energy than reactants. • AH is positive • Heat is absorbed • Ea is high less stable products are obtained. products Hp Enthalpy H E: - -L - - HR- reactants Progress of Reaction Endothermic reaction Quick Check 6.1 Draw the energy profile diagrams for the following reactions: (1) CH 4(g) + 202(g) →CO2(B) +2H2O(e) (11) → CaO(s) + CO 2(g) CaCO 3(s) Ans. CH4(g) + 202(g) - →→CO 2(g) + 2H, Oie) Combustion of methane • (exothermic reaction) AH = -890.3 kJ mol-1 This is an exothermic reaction. • . Products have lower energy than reactants. • Energy is released to the surroundings. Energy profile diagram: Ea HR Enthalpy free Hp.- Progress of Reaction

Standard Enthalpy changes (AH°) Factors affecting of enthalpy of substance The enthalpy of a substance depends upon the following factors: • Physical state (solid, liquid, gas) • Pressure • Temperature To make accurate comparisons between different reactions, we must define standard conditions. Conditions of standard enthalpy changes: AH = -890.3 kJ mol-1 AH = +572 kJ mol-1 CaCO3(s) → CaO(s) + CO2(g) carbonate Formation of calcium (endothermic reaction) AH = +572 kJ mol • This is an endothermic reaction. • Products have higher energy than reactants. • Energy is absorbed from surroundings. Energy profile diagram: Enthalpy" Hp H CaCO 3(8 Progress of Reaction (a) Temperature 25°C or 298K (b) Pressure latm or 101kPa When enthalpy is measured under these conditions, it is called standard enthalpy change, denoted as: AH° Did you Know? S.Q. What is standard state of an element? Ans. The standard state of an element is its most stable form at 298 K and I atm pressure. For example, the standard state of C is graphite not diamond. By definition, the standard enthalpy change of formation of any element in its standard state is zero. (i) Enthalpy change of.reaction (AH,) Definition: The standard enthalpy of a reaction is the enthalpy change involved when stoichiometric amounts of reactants in their standard states react together completely to form products under standard conditions. Example: The reaction between hydrogen and oxygen gases to form 1 mole of water. H2(g) + 02(g) → H2O) AH; =-286kJmoll 2 (ii) Enthalpy change of combustion (AH.) Definition: The standard enthalpy change of combustion of a substance is the enthalpy change involved when one mole of the substance is completely burnt in excess of oxygen, under standard conditions. It is always exothermic. Example: Standard enthalpy of combustion AH, of ethanol is - 1368kJmoll Example: CHOH (e) + 302(g) 2CO2(g) + 3H2O(e) Enthalpy change of combustion is useful in calculating calorie content of foods and fuels. (ili) Enthalpy change of formation (AH,) Definition: Standard enthalpy change of formation of a compound is the enthalpy change involved when one mole of the compound is formed from its elements under standard conditions. It can be exothermic or endothermic. Example: AH; for methane is given below. C(s) + 2H2(g) CH 4(8) (iv) Enthalpy change of atomization (AH*, ) Definition: The standard enthalpy change of atomization of an element is the enthalpy change involved when one mole of gaseous atoms are formed from the element, under standard conditions. Example: Standard enthalpy of atomization of His given below: H2(g) 2H (g) 1 - H 2(g) → H (8) 2 AH: =-1368kJmoll AH; =-74.8kJ mol-1 EH-H= +436kJmoll AH.. = +218kJmol-1 (v) Enthalpy change of neutralization (AH,) Definition: The standard enthalpy change of neutralization is the enthalpy change involved When one mole of water is formed by the reaction of an acid with an alkali under standard conditions. It is always exothermic. Explanation: The enthalpy of neutralization of NaOH by HCl is -57 1kJmol:!. When these solutions are mixed together during the process of neutralization, the only change that occurs is the formation of water molecules leaving Na and Cl as free ions in solution. Example: The enthalpy of neutralization is merely the heat of formation of one mole of liquid water from its ionic components and the actual neutralization reaction is: For all strong acid-base reactions AH, is always near - 57.1 kJmol-1. When acid or base is weak, the value of enthalpy of neutralization will be less than 57.1 kJ/mol. (vi) Electron Affinity (AHca) Definition: The first electron affinity is the enthalpy change involved when 1 mole of electron is added to 1 mole of gaseous atoms to form 1 mole of gaseous uni-negative ions under standard conditions. Example: Electron attinity of chlorine atom → Cl • SH =-348.8kJmol-1 Cl (g) + e Since energy is released, so first electron affinity carries negative sign. Summary Table of Enthalpy changes Exo/ Enthalpy Endotherm Definition change type ic Stoichiometric amount of reactants react Reaction AH, Exoor Endo #2175026→1,010 to form products Combustion 1 mole burned Always Exo AH® in excess O2 1 mole compound Formation AH; Exo or Endo formed from elements 1 mole of Atomization Always gaseous atoms Endo AH:, formed from element Example C,H, OH (e) + 302 (g)2CO 2(g) + 3H, 0(() C(s) + 212(g) →CH 4(g) 1 1 mole of H20 Neutralization formed from Always Exo AH: acid and base 1 mole of Electron Affinity electrons added Usually Exo to gaseous AHCa atoms Quick Check 6.2 (a) Write equations, including state symbols, that represent the enthalpy change of atomization of: (i) Oxygen (ii) Barium (iii) Bromine Ans. Equations representing enthalpy change of atomization Definition: The enthalpy change of atomization is the enthalpy change when 1 mole of gaseous atoms is formed form the element is its standard state. 1 (i) Oxygen (O2) 2(g) → 0(8) 2 (ii) Barium (Ba) Ba (s) ™ → Ba (g) (iii) Bromine (Br) (b) Name the enthalpy changes that occur in each of the following reactions: (1) (graphite) + 02(g) → CO2 (g) tree 1lm. → NH 4Cl (s) (1) HCl (g) + NH 3(g) - (ii) Н2(g) + ½02(g) Ans. (i) C(graphite) + O2(g)→CO2(g) Type: Standard enthalpy of combustion. One mole of carbon burnt completely in oxygen. (i) HC (g)+NH3(g) →NH4C1(s) Type: Enthalpy of neutralization / lattice formation. Formation of an ionic solid from gaseous molecules. 1 (ili) H2(g) + - 2(g) → HI.,010) 2 Type: Standard enthalpy of formation. One mole of water is formed from its elements in standard states. BOND ENERGY (BOND DISSOCIATION ENERGY AND ENTHALPY CHANGES)

Definition: Bond energy (or Bond dissociation energy) is the average amount of energy required to break (dissociate) one mole of a particular bond in a substance. Symbol: E Units: kJmol →H, °(e) H(aq) + OH (aq) → Cl (g) Cl (g) +e • Breaking a bond requires energy (endothermic) • Forming a bond releases energy (exothermic) • Exact bond energy refers to a bond in a specific molecule (C-C in ethane) Average bond energy is the mean value taken over different compounds containing the same bond type. If bond energy of particular substance is determined, it is exact bond energy. Bond energies vary even for the same type of bond depending on neighboring atoms and the molecular structure. Examples of exact bond energies: C-C bonds usually have B.E values of approximately 350- 380 kJ/ mol. In ethane: [H,C-CH,] Ec-c=376kJmol-1 In propane: [H,C-CH,-CH] Ec-c= 356kJmoll In butane: [H,C-CH, -CH, -CH] Ecc= 352kJmoll In these examples Ec-c for ethane, propane and butane are exact bond energies. When average of all Ec-c energies is taken, it is average C-C bond energy from different compounds is 348 kJmoll Table 6.1: Average bond energies of some important bonds (kJmol) 391 436 413 292 615 812 348 413 615 161 891 418 292 615 391 607 728 222 351 463 259 477 339 1. 270 563 441 CI 200 328 432 366 276 BI 240 299 Enthalpy change of reaction (AH, ) and chemical bonds A chemical bond represents a form of energy known as chemical energy, which is interconvertible to all forms of energy. A chemical reaction is a process in which: • Old bonds are broken in the reactants • New bonds are formed in the products Bond Breaking Requires energy → Endothermic → AH = +ve Bond formation Release energy → Exothermic → AH = - ve Source of AH • The energy required to break bonds in reactants. • The energy released when new bonds form in products. AH; = Total energy of bonds broken - Total energy of bonds formed C1 0 F Br NE 463 891 728 351 607 945 222 498 139 213 251 327 347 255 159 185 218 209 203 243 192 180 201 151 Difference between these deciding whether a reaction is exothermic or endothermic. Condition Energy released > absorbed Exothermic Energy absorbed > released Endothermic Example: Formation of water Reaction →2H,0(g) 2H2(g) +02(g) Break: H-H and O=O bonds (energy absorbed) Form: H-O bonds (energy released) Since the energy released in forming H-O bonds is more than the energy absorbed in breaking H-H and O=0 bonds, the reaction is exothermic. AH, is negative. Sample example 6.1 With the help of the following bond energy data; calculate the enthalpy change of the following reaction: E#= 436kJmol; Eoo = 495kJmol\ andEs-o= 463kJmoll 12(g) + 1/202(g) H,0(g) Solution: 436 + ½(495) (bond forming energy) O (bond breaking energy) = [Ен-н + Eo-o]-[2Е#-o] tree. = [436+247.5]-[2(463)] AH: = -242.5KJmol-1 MEASUREMENT OF ENTHALPY CHANGE OF A REACTION

Measurement of enthalpy change of a reaction Calorimetry Definition: Calorimetry is the measurement of heat evolved or absorbed during a physical or chemical process. The device used for this purpose is called a calorimeter. Heat capacity: The amount of heat required to raise the temperature of a substance of mass 1kg though 1K (or 1°C) is known as specific heat capacity. Example: Heat capacity of H2O is = 4.18 Jg 'K' (or Jg| °C-') rounded of 4.2 Jg k" which is rounded off to 4.2 Jg K Type of calorimeter • Many types exist, but here we focus on the glass calorimeter, and phenomenon is called calorimetry. Basic components of a simple calorimeter • Insulated vessel: Prevents heat exchange with surroundings. • Stirrer: Ensures uniform temperature distribution. • Thermometer: Measures temperature change (AT). Result (AH is negative) (AH is positive) 2 x (463) GLASS CALORIMETER Glass Calorimeter Construction: A glass calorimeter is a simple, insulated vessel used to measure enthalpy changes (AH) for reactions in aqueous solutions. • Beaker (reaction vessel) • Thermometer • Stirrer • Loose-fitting lid • Insulating material (cotton wool) Limitations of glass calorimeter • Reactions involving gases (gases may escape). • Reactions producing very high temperatures (can damage glass or escape heat). • Cotton wool insulation minimizes heat exchange with the surrounding air. Working • The reaction occurs inside the beaker. • The temperature change (AT) is measured. • Pressure is constant, the change in heat energy is equal to enthalpy change (AH). Equations for heat calculations: Heat energy absorbed or evolved. AH is calculated as follows: (i) 9= mwater X Cwater X AT = mass of water (or solution)in grams| Mo water = 4.18J/g°C AT = Tf.- Ti (ii) Convert to kilojoules (kJ: By dividing with 1000. (iii) Calculate AH for the reaction using relation. (enthalpy change per mole of reactant) ДН = - 1 - (kJ/ mol) n * 9= mcAT mcAT AH = - - (kJ/ mol) n The solutions we are using here are so dilute that almost all of their mass consists of water therefore, we can simply use specific heat capacity of water. Such a calorimeter could be used to measure the heat of neutralization (AH"). Sample Problem 6.2 Neutralization of 100cm? of 0.5 moldm NaOH at 25°C with 100 cm of 0.5 moldm Hcl at 25°C raised the temperature of the reaction neutralization. Specific heat of water = 4.2Jg K Solution Density of H2O is around 1gcm3 ., so total volume of solution which is 200 cm? in temperature, AT = 28.5 - 25.0 = 3.5°C = 3.5K •: Amount of total heat evolved, q = m X c x AT = 200 × 4.2 × 3.5 = 2940J = 2.94kJ Thermometer Stirrer Hydrochloric acid + sodium hydroxide Beakers Cotton wool Fig: Glass calorimeter to measure enthalpy change of reactions mixture to 28.5°C. Find the enthalpy of = 200g Rise Calculation of number of moles of H2O formed (n) Using mole = concentration (mol/dm?) 0.5×100 - = 0.05 mol 1(HCe)"(NaOH) = 1000 Using equation, Number of moles of water formed, n(N20) = 0.05 mol Heat evolved in the formation of 0.05 mole of water, q = -2.94 kJ -2.94 kJ AH; = 9 n 0.05 mol So, Enthalpy of neutralization, AH; =-58.8kJ mal Quick Check 6.3 Calculate AH" of the reaction of 50 cm of 1.5 mol dm HNO with 50 cm? of 1.5 mol/dm of NaOH. The change in temperature is 4°C? Ans. To calculate the standard enthalpy change of neutralization (AH,) from the data given in we'll use the formula: 9 = mcAT • q = heat evolved (in joules) • c = specific heat capacity of water = 4.18 J/g°C • AT = change in temperature = 4°C Step 1: Calculate heat evolved (g) Since 50 cm' of HNO and 50 cm? of NaOH are mixed: Total volume = 50 + 50 = 100cm 3, Total mass = 100g (assuming density = 1g/cm') g= mcAT = 100x4.18x4=1672J = 1.672kJ Step 2: Calculate moles of NaOH 1.5mol / dm' × 50cm? Moles of NaOH = Since HNO3 iș also 50 cm of the same molarity (or neutralizing amount), reaction is complete. Step 3: Calculate AH, (in kJ/mol) -9 moles of water formed Answer: AH, = -22.3kJ/ mol = 0.075mol 1000 -1.672 - =-22.29kJ/ mol 0.075 ENTHALPY CHANGEAND CALORIE CONTENT OF FOOD

Calorie content of food The calorie content of food is a measure of the energy 'released' when the food is completely consumed in the body. This energy is typically expressed in units of kilocalories (k cal) or joules (J). Food as a source of energy When food is digested, chemical energy stored in food (also called calorie content) in it is released as heat energy. This is similar to combustion; the enthalpy of combustion (AH?) of food is equal to caloric content and is measured using a bomb calorimeter. Caloric content is usually expressed as: Kilocalories (kcal), Joules (J) or kilojoules (kJ 1kcal = 4.184 kJ. Relation between enthalpy change and calorie content The enthalpy of combustion of food (AH.) is the calorie content of that food when it is translated or converted into kilocalories per gram. AH° (kJ/ g) Calorie content- 4.184 Example: Combustion of glucose. Energy provided from glucose. tre → 6СО 2(g) + 6H,0(1) C6H1206(s) + 602(g) The calorie content of glucose can be calculated as follows: Step by Step: Find calorie content AH per gram of glucose is measured • AH° = -2803 kJ/mol • Molar mass of glucose = 180 g/mol 180.0g of glucose burns to produce energy - 2803 kJ -2803kJ/ mol 1. Convert AH per mole of per gram: 180g/ mol 2. Convert kJ/g to kcal/g using: 1 kcal = 4.184 kJ Using the relation AH (kJ/g) = - 4.184 x calorie content AH° (kJ / g) Calorie content= 4.184 - 15.57kJ/ g 3. Calorie content: Calorie content = Final result Calorie content of glucose = 3.72 kcal/g Remember the calorie content we take from food must be balanced by working exercising and doing positive activities. Otherwise, our bodies will have imbalanced growth and maintenance. - - 15.57kJ/g •=-3.72kcal/ g 4.184 HESS'S LAW OF HEAT SUMMATION

Hess's law of constant heat summation Statement: "The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place as long as the initial and final conditions are the same." Enthalpies cannot be measured experimentally Germain Henri Hess applied the law of conservation of energy to enthalpy changes. There are many reactions, for which AH cannot be measured directly by calorimetric method Example: Tetrachloromethane CCl4 cannot be prepared directly by combining carbon and chlorine. Hess's law helps us to calculate the enthalpy changes for such reaction or processes. It states: First law of thermodynamics (law of conservation of energy): "Energy can neither be created nor be destroyed, but can be converted from on form to another." Energy Cycle: Hess's law can be illustrated by drawing enthalpy cycles, often called energy cycles or Hess cycles. Let's say a reaction goes from A → B: • Direct route 1: A → B • Indirect routes 2: A → intermediates → B • The products formed in these routes (M, N and X, Y, Z) are called reaction intermediates. By Hess's Law For the indirect route 1, we can write For the indirect route 2, we can write AH; = ΔH3 + ΔH4 + AHs Hess's Law, which states: "The total enthalpy change for a reaction is the same, no matter how it is carried out, provided the initial and final conditions are the same." (i) Calculating enthalpy of formation (AH;) using enthalpy of combustion (AH®) Sample problem 6.3 → 2(g) C (graphite) + 02(g) - 02(g) → CO2(g) AH, = -283kJmol (O(g) + 2 C (graphite) + - 02(g) → CO (8) 2 Solution: Applying Hess' law, we can write =ΔH1 + 4H2 ДН1 = ДН - ΔH2 = - 393.5 - (- 283) ΔH =- 110.5kJmoll H2 H Indirect route H Direct route H: Indirect route 2, Figure 6.4: Hess's cycle AH = -393.5kJmol Route 1 AH; AH, =? C + 02 CO, ДН, AH, C+20 Route 2 (ii) Calculating enthalpy change of reaction (AH,) using enthalpies of formation (AH;). Sample Problem 6.4 Calculate the enthalpy change of reaction of Using Hess cycle with the help of following combustion data AH; of C,H = +52.2K.Jmol AH; of Hcl = -92.3KJmol-1 = -109[+52.2 + (92.3)] AH =-68.9 KJmol Quick Check 6.4 Calculate the standard enthalpy change for the formation of methane (AH;): 6С (s) + 3Н 2(g)* The standard enthalpies of combustion of C(s), H2(g) are-394 kJ mol 1, -286 kJ mol: and enthalpy of formation of CoH64)-3267 kJ mol respectively. Given Ans. We are to calculate the standard enthalpy of formation (AH;) of benzene (CoHe)) Given standard enthalpies of combustion AH, of C(s) = -394 kJ/mol tree lim. AH, of H2(g) =-286 kJ/mol AH. of (CHe)) = -3267 kJ/mol Step by Step using Hess's Law Combustion equations Combustion of 6C(s) 6C (s) + 602(g) 6CO2(g) H =6x(-394) = -2365kJ Combustion of 3H2(g) 3Н 2(g) + 02(g) 3Н,0(e) AH = 3x (-286) = -858kJ c) Combustion of CoHole): 02(g) →6C02(g) +3H2O(ı) 2. Apply Hess's Law To find: 6C (s) +312(g) → C,H6(e) AH; = [AH, (reactants)] - [AH. (products)] ДН; = [-2364 + (-858)] - (-3267) AH, = -3222 + 3267 = +45kJ/mol Answer: AH; (CH.) = +45kJ / mol • C,H,CI(g) C,H, (g) + Hcl (g) -109 +52.2 1-92.3 -CL(g) → C, 16(e) AH = -3267 kJ

Hess's law of constant heat summation Statement; "The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place as long as the initial and final conditions are the same." (i) Calculating enthalpy change of formation of a substance (AH,) using Enthalpy of Combustion and Enthalpies of formation of other substances. Sample problem 6.5 Propane (CHs(g)) burns in oxygen according to the equation: → 3C02 (g) + 4H, 0(g) CH 8(g) + 502(g) When 14:64g off propane is burned in an excess of oxygen in a calorimeter at 25°C and 1 atm pressure, 678.6 kJ of heat is evolved. Calculate the standard enthalpy of formation of propane. The standard enthalpies of formation of CO2(g) and H20(g) are -393.51 kJmol ' and - 241.82kJmol ' respectively. Solution: The first step is to find the enthalpy change when one mole of propane is burned. Molar mass of propane, C3H8 = 3(12.011) + 8(1.0079) = 36.033 + 8.0632 = 44.096 No. of moles of propane = 14.64/44.096 = 0.3320 mol So, 0.3320 mol propane evolves heat = 678.6kJ 1.0 mol propane evolves heat = 678.6/0.3320 kJmol = -2044kJmoll According to Hess' law AH, = [3×AH; (CO,) +4×AH; (HO)]-[I×ДН:(CH) + 5×ДН° (02)] -2044kJ= [3x (-393.51) + 4x (-241.82)] - [1x H, (CHg) + 5x (0)] Taking the heat of combustion of 02(g) to be zero and solving this equation for AH, (CHs) gives. AH; (CHs) =-104kJmoll Quick Check 6.5 Draw enthalpy cycle of sample problem 6.5 according to Hess'law to validate the above calculation. •Ans. Elements (3C (graphite) + 4H 2(g)) AH*(Cills) (CH8)(g) ДН, = ЗДН; (СО2) + ДН; (H2O) (g) 3CO 2(g) + 4H2O 02(g) (excess) (i) Route 1 (Formation): Elements C3H8(g) AH; (СзН8) = ? (i) Route 2 (combustion): Combustion of eléments to form 3CO2 + 4CO2 directly. Д1 = 3(-393.51) + 4(-241.82) = - 2147.81 kJ/mol. (iii) Hess's Law validation: АН; (СзН8) + ДНсоть = ДНІ ДН; (CзН8) + (-2044) =-2147.81 AH; (CH8) = -2047.81+2044=-103.81 kJ/mol (ii) Calculate the enthalpy change of reactions using bond energies. Hess's Law statement A special case of Hess's law allows us to calculate the enthalpy change of a reaction (AHr) using average bond energies (B.E.), particularly for gas-phase reactions. Step 1: Breaking bonds in reactants • All bonds in the reactant molecules are broken. • This process requires energy input. • The enthalpy change for this step is positive, since bond breaking is endothermic. • Use the average bond enthalpies from Table 6.1 to calculate the energy required. Step 2: Forming Bonds in Products • Atoms recombine to form the product molecules. • The enthalpy change for this step can again be estimated from the bond enthalpies of table 6.1 • These values now be taken with minus sign because the bonds are being formed instead of being broken down. In general, the heat of reaction for any gaseous chemical reaction can be calculated from average bond energies by use of the following version of Hess's law. This formula gives an estimated enthalpy change using average bond energies. Bonds Energies of Products AH, = ≥ Bond Energies of Reactants Reactants E(R) > Atoms Fig: Hess's cycle; showing the relationship between bond energies and Hr. = LER-ZEp Sample problem 6.6 In the case of formation of HCl(g) from H2(g) and Cl2(g), use B.E. data from Table 6.1 to estimate AH for the reaction: 12(g) + Cl 2(g) → 2HCl (g) And finally calculate the heat of formation of HCl, we replace this reaction by a hypothetical two-step process. The bonds in all the reactant molecules are first broken, and then the atoms are combined to make the products. First Step: →→ 2HCl (g) 12(g) + Cl 2(g) Ен-н = 436kJ mol-' kJ 1 mol x 242- 1 mol × 436.- ZER = mOl moT) = 436 kJ + 242 kJ = 678 kJ Second step: → Products E(p) AH? → 2HCl H2(g) + Cl 2(g) + • (g) 2H1 (g) 2HCl (g) H 2(g) + Cl 2(g) EH-a= 431kJmol Ep = 2mol× 431kJmol-1= 862kJ The standard enthalpy change in the reaction is obtained by the following formula: AH; = LER-LEp AH; =678-862 AH=- =-184KJ(for 2 mol of Hcl) Enthalpy of formation of HCl, = -184/2kJ= -92kJmol Quick Check 6.6 (a) The reaction for the Haber process is: 1 2(g) + 312(g) 2N3(8) The relevant bond energies are: ENoN 945kJ mol ,EN-H = 391 kJ mol-1 Calculate the enthalpy change of the above reaction. Ans. Haber Process Step 1: Bonds broken (Reactants) •1x N= N=945kJ/ mol • H-H=3x436=1308kJ H-H=3×436=1308kJ/ mol . Total energy to break bonds = 945 + 1308 = 2253 kJ/mol Step 2: Bonds formed (Products) In 2NH3, each NH3 has 3 N-H bonds → 2 X 3 = 6 N-H bonds 6 × N-H = 6 × 391 = 2346 kJ/mol Step 3: Enthalpy change AH = Bonds broken - Bonds formed AH = 2253 - 2346 = -93 kJ/mol Answer (a): AH = -93kJmol (b) Calculate the enthalpy change for the following reaction CH2OH (e) + 302(g) 2CO2(g) +3H, O(e). The bond energies of various bonds (in kJ mol 1) are given below: Ecc=347,Ec.=+410,Eco=+336,Ec.o=+496,Ec-o=+805,Eo = +465 Ans. Combustion of Ethanol: C,H, OH ) + 302(g) → 2C0 2(g) + 3H, 0(1) Step 1: Bonds broken (Reactants) In CzHsOH • 5 C-H bonds = 5 ×410 = 2050 • 1 C-C bond = 347 1 C-0 bond = 336 • 10-H bond = 465 In 3 O2: • 3 × 0 = 0 = 3 X 496 = 1488 Total = 2050 + 347 + 465 + 1488 = 4686kJ/mol Step 2: Bonds formed (Products) In 2 CO2: • 2 0-H bonds per H20 × 3 = 6 × 465 = 2790 Total = 3220 + 2790 = 6010 kJ/mol Step 3: Enthalpy change AH = Bonds broken - Bonds formed AH = 4686 - 6010 = -1324kJ/mol Answer (b): AH = - 1324kJ/mol ENERGETICS OF SOLUTION

Energetics of solution Dissolution: The process of dissolving a solute in a solvent is called dissolution. The formation of a solution takes place in three main steps. Three steps in dissolution process tree (i) Expanding the solvent Overcoming intermolecular forces in the solvent to make space for the solute. (ii) Expanding the solute Breaking the solute into individual ions or molecules (ill) Interaction between solute and solvent Solute and solvent molecules interact to form a homogenous solution Heat involved in dissolution During dissolution, heat may be absorbed or released. The total heat change is known as the standard enthalpy change of solution (AHs) Definition of AH° - sol Standard Enthalpy of Solution "The standard enthalpy of solution is the amount of heat absorbed or evolved when one mole of a substance is dissolved in a solvent to give an infinitely dilute solution" Symbol: AHol Unit: kJmol Nature: It may be exothermic (-ve) or endothermic (+ve) -sol =-25kJmol →2 Na (ag) + CO3(ag) AH. Na, CO3(s) + aq- → NH 4(aq) +Clag) AH.=+16.2kJmol NH, Cls) + ag Factors affecting dissolution of ionic Table 6.2 Heats of solution of some compounds (i) Lattice Energy • Energy required to break the ionic lattice. • Higher lattice energy means more energy required to dissolve. (ii) Hydration Energy • Energy released when ions are surrounded by water molecules. • Higher hydration energy favors dissolution HYDRATION

Hydration Definition: Hydration is the process when ionic compounds dissolve in water, they dissociate into individual ions. These ions are then surrounded by water molecules. This process is called hydration. "The process in which water molecules surround and interact with the solute ions is called 15J- hydration" Nature of Hydration forces tree 11 Ion-dipole forces are formed between: • Ions from the solute • Polar water molecules These forces cause ions to become hydrated in aqueous solution. Enthalpy of hydration Definition: The enthalpy change involved when one mole of a solute is dissolved in excess water to make an infinitely dilute solution under standard conditions is called enthalpy of hydration. Symbol: AHhyd This energy is always released (exothermic) because new ion-dipole bonds are formed. Factors Affecting the magnitude of Hydration Energy The heat of hydration depends on following factors: (i) Charge on the ion • Higher charge and stronger attraction to water molecules gives greater hydration energy. (il) Size of the ion • Smaller ions and higher charge density gives stronger hydration energy. Charge density is defined as charge per unit area. • Greater charge density gives greater hydration energy. important ionic solids (kJ mol") of Heat Substance solution +4.98 NaCl +17.8 Kcl +19.9 KBr +21.4 KI +16.2 NH4CI +26.0 NH4NO3 Figure 6.6: Dissolution of an ionic compound through the process of hydration. Order of Hydration Energies: Enthalpies of hydration of following ions are in the order: Na < Cat Mg? > The smaller and more highly charged the ion, the greater is its hydration energy. Hydration of Anions Smaller more negatively charged anions (F) have higher hydration energy than larger ones (I).

Lattice energy Definition: The lattice energy of an ionic crystal is the enthalpy change involved when one mole of the ionic compound is formed gaseous ions under standard conditions. Naig) + Clg) → NaCls) Factors affecting lattice energy (i) Charge on the lon • Greater the ionic charge, greater is the lattice energy. • This is because higher charged ions attract each other more strongly. • Examples: AHar (MgO) = - 3923 kJmol cEC AHlatt (LiF) = -1049 kJmol1 Mg? and O2 have higher charges compared to Lit and F- (ii) Size of the lon • Smaller ions means closer packing and stronger attraction gives higher lattice energy. • Lattice energy decreases with increasing size of either cation or anion. Example: Alkali Metal Halides When we go down the group of alkali metals, lattice energy decreases.

Solubility trends of group hydroxides and sulfates AHol Depends on both lattice energy and hydration enthalpy. AHsol = Hydration enthalpy - Lattice energy (i) Lattice energy (AHlatt) (ii) Hydration Enthalpy (ARhyd) Al3+ < (Mg?+) Sr2+ > AH® lar =-787kJmol-1 1100 Li -1000- Na -900 - 1- 10ш -800 Rb Cs -700 - 600 - Figure 6.7: Lattice energy of alkali metal halides Cation Lattice Energy Lit Highest Nat K+ Rb+ Cst Lowest 2 (i) Group 2 Hydroxides: Increasing solubility of hydroxides of group 2 • Both lattice energy and hydration enthalpy decrease down in the group 2. • But lattice energy decreases faster, making AHol more exothermic. (ii) Group 2 Sulfates: Decreasing solubility of sulfates of group 2. Now look at the solubility of group 2 metal sulfates. • SO? is large, so lattice energy doesn't decrease much. • Hydration energy decreases significantly down the group. • AHo becomes more endothermic, and solubility decreases down the group. Calculating enthalpy change of solution AHol • We can calculate the enthalpy change of solution or the enthalpy change of hydration by constructing an enthalpy cycle and using Hess's enthalpy change law Figure 6.8: of solution • From this enthalpy cycle: Hool AH® Patt t AH:oL = AHhyd This energy cycle used to calculate enthalpy change of solution. Fig: Energy cycle of formation of an aqueous solution of Sample problem 6.7 Determine the enthalpy change of solution AHiyd Of sodium fluoride (NaF) using the following data: Lattice energy of sodium fluoride (NaF) = -902kJmol-1 Heat of hydration of sodium ions (Nat) =-406kJmoll Heat of hydration of fluoride ions (F) = -506kJmoll Solution: Step 1: Draw the enthalpy cycle NaF(s) AHim - 202 kJ mol Na@+F®) taq ДН° sol Step 2: Rearrange the équation and substitute the values to find AHol • AH at + AHsol = AHhyd AH® -Sol = (-406) + (-506) - (-902) AHnya[NaF] = +10 kJ mol-1 Solubility Compound Lowest Mg(OH)2 Ca(OH)2 Sr (OH)2 Ba(OH)2 Highest Solubility Compound Highest MgSO4 CaSO4 SrSO4 Lowest BaSO4 lattice energy ionic solid gaseous ions HO latt enthalpy change of hydration of cation and anion, Mind ions in aqueous solution an ionic solid using Hess's law Hiya [Nat] = -406 kJ mol-i AHhya [F] = -506 kJ mol BORN-HABER CYCLE

Theoretical calculation of AH att • It is impossible to determine the lattice energy of a compound by a single direct experiment. • We can calculate the value for AH at using several experimental data and an energy cycle called the Born-Haber cycle • Born-Haber cycle is an application of Hess's law. It is used to measure lattice energy of binary ionic compounds M X Example: Consider the calculation of lattice energy of sodium Chloride using Hess's law and Born-Haber cycle. AH;: Standard enthalpy of formation of NaCl can be measured conveniently in a calorimeter. AH;: Total energy involved in changing sodium and Na(s) + ½C12(g) chlorine from their standard physical states to gaseous ions. Applying Hess's law on the above energy cycle. AH; = AH, + AH att The above energy triangle has been extended to show Figure 6.9 Energy cycle of sodium chloride the various stages involved in finding AH,. The complete energy cycle is called the Born-Haber cycle and it is presented in Figure 6.10. Nat '(g) AHa, of C/(g) = + 121 kJ mol A Hea of Cl=-349 kJ mol-1 Na(g) + e +1/2 Cl2() AH, of Nag) = + 496 kJ mol 1 Na (g) + 1/2 C12(g) AHar of Nag) = + 108 kJ mol ' Na(s) + 1/2 C12(g) AH, of Nat Cl (g) =- 411 kJ mol ' Nat Cle Fig: Born-Haber's cycle of sodium chłoride Calculation of AH This energy can be roken into several steps using Hess's Law. Enthalpy Terms Involved AH° Naig) + CLe AH° latt Na Clg) (g) Nat (8) + CI (8) (Lattice Energy of Nacl) - (REACTANTS) (FINAL PRODUCT) Process Sublimation of Na (Na(s) → Na(g)) Ionization of Na (Na → Na* + e) Dissociation of Cl2 (½ Cl2 → CI) Electron gain by CI (Cl + e' → CI) Using Hess's Law: Calculation of AH From above cycle we have, AH; = AHat(Na) + 4Hil(Na) + AHat(C12) + AHeal(C12) AH; = 376 kJmol-! AH patt = 4Н°f- AН* AH att = - 411 - (+376) AH att =-787kJmoll Conclusion: • The Born - Haber cycle allows indirect calculation of lattice energy using other known enthalpy changes. • It is a stepwise application of Hess's Law for ionic compounds. Sample problem 6.8 Calculate the heat of formation of şodium fluoride which crystallizes in the sodium chloride lattice. The heat of atomization of Na(s) is 108kJ/mol, half the bond energy of F2(g) is 79kJ/mol, the ionization energy of sodium atoms is 494 kJ/mol, the electron affinity of fluorine atoms is -328 kJ/mol, and the lattice energy is -939 kJ/mol. Solution: Values are given for all the quantitates in the Born-Haber cycle, so we can apply Hess' law: AH; = AH at lat + AH° =(-939 + 108 + 494 + 79-328)kJmoti AH; =-584kJmol Quick Check 6.7 (a) Draw Born-Haber cycle for sample Problem 6.8. Ans. Na(g) + € + F(g) AH, of F2(g) =+79 kJ mol-l A Hea of F=-328 kJ mol-1 Na g) + e +1/2 F2(g) AH; of Na() = + 496 kJ mol- Na (g) + 1/2 F 2(g) AHa, of Na(g) = + 108 kJ mol 1 Nas) + 1/2 F2(g) = -584 kJ mol-! AH, of Nat F (s) Nat Fl6) Symbol AHat (Na) AHil (Na) AHeal(Ce,) Nat (8) + F(8) A Hart Of Nat F (= -939 kJ/mol (Lattice Energy of NaF) (REACTANTS) - (FINAL PRODUCT) (b) Calculate the heat of formation of lithium fluoride. The heat of atomization of Li(s) is 161 kJ/mol, half the bond energy of Fz(g) is 79 kJ/mol, the ionization energy of lithium atoms is 520 kJ/mol, the electron affinity of fluorine atoms is -328 kJ/mol, and the lattice energy is 1107 kJ/mol. Ans. Born-Haber Cycle for LiF(s): (atomization of Li) AH = +161 kJ/mol Li(s) + ½ F2(g) (bond dissociation of F2) AH = +79 kJ/mol Li(g) + ½ F 2(g) (ionization energy of Li) AH = +520 kJ/mol Li(g) + F(g) Lit (g)t F(g) (electron affinity of F) AH = -328 kJ/mol (lattice energy) AH = -1107 kJ/mol Li (g) +F (g) LiF (s) Calculation of enthalpy of formation (AH°+) Step-by-step values: Enthalpy Change (kJ/mol) Step +161 Atomization of Li +79 ½ Bond energy of F2 +520 Ionization energy of Li -328 Electron affinity of F -1107 Lattice energy of LiF Apply Hess's Law: AH°f= Sum of all steps above AH°f= 161 + 79 + 520 - 328 - 1107 . AH°f= -675kJ/mol Answer (b): The standard enthalpy of formation of LiFs) is -675kJ/mol.. NTROPY

Entropy Definition: Entropy is a measure of the number of ways energy can be distributed within a system at a specific temperature. Explanation Systems naturally evolve towards states of higher entropy. Entropy can also be thought of as a measure or the randomness or dısorder of a system. The higher the randomness or disorder the greater the entropy of the system. Entropy, Diffusion and Number of Ways of arrangement Diffusion and Entropy • Diffusion is driven by random motion of particles and high probability of new arrangements. • As particles spread out, the number of possible microstates increases, so entropy increases. partition gas jar A gas jar B Figure 6.11: Diffusion and number of possible arrangements There are 8 different ways for these molecules to arrange themselves in two jars by diffusion from Jar A to jar B. that is This is calculated as under Number of molecules in A Number of Jars =23=8 Number of possible arrangements • If there were 5 molecules initially in jar A, the possible ways of arrangement will be 2%., i.e. 32 • If there were 100 molecules, the probability will be 2100 • A general formula x can be written for the calculation of probability, where x is the number of places and y is the number f particles to be arranged. • Diffusion happens because there is a large number of ways of arranging the molecules. • The concept of the 'number of ways' of arrangement either particles or the energy within these particles helps predict whether it can happen or not. General formula Number of arrangements = x Where x = number of places, y = number of particles. Factors Affecting Entropy (i) Number of particles: More particles, more microstates and higher entropy (ii) Hardness of substance: Harder substances have lower entropy (due to restricted vibrations). Example: Diamond < Graphite (iii)Physical state: Solids < Liquids < Gases (in terms of entropy). Example: Water states near transition points. Comparison of entropy values Entropies of different substances can be compared on the basis of following factors. (i) Number of particles: Small number of particles means low entropy and vice versa. Example: CaCO; has higher standard entropy (92.9 JK-' mol) than CaO (39.7 Jk' mol). This is because the number of possible arrangements is lower when the number of particles is smaller. (ii) Physical properties: The entropy of substances having șimilar chemical nature is dictated by their hardness. Harder substances have lower entropy than softer ones. Diamond has lower entropy than graphite because it is much harder. Stronger forces result in limited vibrations in harder substances which decrease the probability of disorder. (iii) Physical states: A substance has lower entropy in solid state than in liquid and gaseous states. The entropy of ice near its melting point is 48.0 JK 'mol-1 'mol-1; whereas water vapor just above the boiling point is 188.7 JK 'mol-1 Quick Check 6.8 Explain the difference in the entropy of each of the following pairs of substances in terms of their states and structures. Ans. (i) Brz(e) S°= =151.6JK-1, mol and 2(s) S°= 116.8JK"mol" = 3 = 2 , for water, it is 69.9 JK Reason: Br is a liquid: particles are more mobile than in a solid (12), so entropy is higher. I2 is a solid, with tightly packed particles and low randomness → lower entropy. Conclusion: Br, (l) has higher entropy due to the more disordered liquid state. (i) 2(g) S° = 130.6JK mol and CHa(g) 5°= 1862JK mol-1 Reason: • Both are gases, but CH4 is a larger and more complex molecule with more vibrational and rational motions → higher entropy. • H2 is a small diatomic molecule → fewer possible motions → Lower entropy. Conclusion: CH4(g) has higher entropy due to greater molecular complexity. (iii) Hg(,) S°= 'mol-1 mol-1 and Na)S° = 51.2JK' mol) Reason: • Hg is a liquid, so its atoms are freer to move → more disorder • Na is a solid, so its atoms are arranged in a rigid lattice → less disorder. Conclusion: Hg(e) has higher entropy due to its liquid state. (iv) SO 2(g) S° = 248.1JK mol and S03(0) S° = 95.6JK mol-1 Reason: • SO2 is a gas → high particle randomness and motion → high entropy. • SO2 is a liquid → less freedom of motion than gas → lower entropy. Conclusion: SO2(g) has significantly higher entropy due to its gaseous state. ENTROPY CHANGES IN REACTIONS

Entropy changes in reactions In a chemical reaction, comparing the entropy of reactants and products helps determine whether the entropy change (AS°) is increased or decreased General rules to predict entropy change: The following rules must be followed to calculate change in entropy of a reaction. (i) Solid → Liquid or Gas • When a solid is converted to a liquid or a gas in the product, the entropy change is positive. • Entropy increases (AS° is positive). Example AS° > 0 H2O(s) → H2O(i) (i) Increase in Number of Moles (especially gases) • If there is a change in the number of gaseous molecules in a reaction, due to high values of entropy are associated with gases. • More the gas molecules, greater is the number of ways of arranging them and the higher the entropy. • More moles in products than reactants shows increase in entropy. • Especially important for gaseous molecules, due to their high entropy. (iii) Gas formation from solid or liquid • Entropy significantly increases. • There is an increase in entropy of the above system because the gas is being produced (high entropy) but the reactant, calcium carbonate, is a solid (low entropy). Example 1 CaCO3(5) ™ → CaO (s) + CO2(g) • Solid to solid or gas gives entropy increases (4S°>0) • Spontaneous reaction Example 2 2N,0s(g) 4N02(g) + 02(g) • An increase of entropy of the system because there are a greater number of moles of gas molecules in the products (5 molecules) than in the reactants (2 molecules). • Reactants: 2 gas molecules • Products: 5 gas molecules • AS° is positive, more disorder • Reaction is spontaneous Sign of Entropy Change (AS°) and spontaneity AS® Sign Interpretation Positive (+) Increased disorder: more ways to arrange Negative (-) Increased order: fewer arrangements possible No, may not be spontaneous Entropy alone cannot always predict spontaneity. We need Gibbs Free Energy (AG) for final decision. Calculating the entropy change of the system (reaction) In order to calculate the entropy, change of the system we use the relationship: AS° system =S° Let us calculate the entropy change of the system for the reaction: 2Ca (s) +02(g) The standard entropy values are: SO [a(s) = 41.40JK ' mol-1 S° 103(8)) = 205.0JK ' mol-1 S° [Cao(s) =39.70JK 'mol-1 As AS° = S° - SO system reactants products (Ca(s)) + 5° [Cao(s)) - {2x5° = 2x - {2x41.40+205.03 = 79.40-287.8 AS° system = -208.4JK-'mol-' Spontaneous Yes, usually spontaneous - SO products reactants →2CaO(s) Interpretation • AS° system = -208.4 JK-1 mol-1 → Entropy decreases • Even though entropy decreases, the reaction is spontaneous. • This means we must also consider entropy of the surroundings or Gibbs Free Energy (AG) to fully judge spontaneity (feasibility).

Total entropy change, AS Definition: AStotal is sum of standard entropy change of system and standard entropy change of surrounding. AS° for system can be found out by subtracting the standard entropies of reactants from the standard entropies of products (similar to the way you obtained AH°). ASsystem = 2 S"products) - 2S"reactants) Respective number of moles of reactants and products in a balanced chemical equation are multiplied with entropy of reactants and products. AS® for surrounding is calculated suing following equation: AS surrounding = T Where, The value of AH, is in KJ/mol so must be multiplied with 1000. This is because entropy changes are measured in J/Kmol. The negative sign is part of equation and not the sign of AH, total is calculated using following equation: AStotal = 453y surrounding system + AS° -AH? ASO SO (products) (reactants) + system T It is useful to be able to predict about spontaneity of reaction using the sign of AStotal AStotal is tve → entropy increases → reaction is spontaneous AStotal is tve → entropy decreases → reaction is nonspontaneous Sample problem 6.9 2Ca (s) + 02(g) → 2CaO(s) The standard entropy values are: S02(g) = 205.0JK mol-1 As * (reac tan ts) System = 2S° (products) = 2x39.70-1 (2x41.40) + 205:03 = 79.40-287.8 SO =-208.4JK ' mol-1 system Sample problem 6.10 Calculate the entropy change of the surroundings ASg 2Ca (s) + 02(g) → 2Ca (s) Step 1 convert the enthalpy change into J mol-1 by multiplying by 1000. - 1270.2 × 1000 = - 1270200J mol Step 2 Apply the relationship reaction AS® surrounding T -1270200 AS surrounding 298 ASO = + 4262.4 J K' mol-1 surrounding Sample problem 6.11 Calculate the total entropy change of the. AStotal for the reaction and check its spontaneity. 2Ca (s) + 02(g) →2CaO(s) ASa = - 208:4 J K-' mol-1 system ASO = + 4262.4 J K-' mol-1 surrounding As, AStotal = - 208.4 + 4262.4 . AStotal = + 4054.0 J K-' mol-1 The total entropy change is positive and the reaction is feasible. THE FREE ENERGY CHANGE

The Free energy change, AG In determining whether a chemical reaction is likely to be spontaneous, we use the new quantity Gibbs free energy change, 4G. the Gibbs free energy change is given by the relationship. AG = - TAStotal Gibb's Equation Multiplying the ASotal expression by - T, we can also get the expression without having to consider the entropy changes of the surroundings. This expression is called Gibb's equation. AH + -T× AStotal = ASsygtem -TASiotal = TASsystem + AH; AG® = AH; - TAS° surrounding for the reaction at 290k: AHreaction =- 1270.2 kJ mol Extend your knowledge S.Q. What is unit of temperature into entropy? Ans. The unit of temperature times entropy (TAS) is Joule. AS® represents AS system. Conclusion: The following conclusion can be drawn from it. (i) If AG < 0 = (-ve), the given process may occur spontaneously. (ii) It AG > 0 (tve), the indicated process cannot occur spontaneously; instead, the reverse of it may occur. (iii) If AG = 0 neither the indicated process nor reverse of it can occur spontaneously. The system is in a state of equilibrium. Calculating 4G° for a reaction Sample Problem 6.12 For the reaction: →Ca + (ag) CaSO4(s) 9(ag) + 504 Calculate AH°, AS° and 4G® at 25°C using the following data; and discuss its spontaneity. Heat of Formation: AH;[CaS04(s)1=-1432.7kJ,AH;[Ca(ag)1=-543.0kJ,AH;[S04 (aq)]=-907.5kJ Standard entropy.: S°[CaSAs]= 106J/ K,S°[Ca?g)]=-55.2J | K,S°[SO* (ag)] = +17.2J / K Solution (reactants) = [-543.0-907.5]-[-1432.7] =-17.8kJ = [(-55.2+17.2) = (106.7)]J/ K=1447J/ K S" (reactants) (products) AS° = Es° AS° = -0.1447kJ K AG®=AH°-TAS®=(-17.8kJ)-(298K) × (-0.1447kJ/K)=+25.3kJ Result: We conclude that this process is nonspontaneous at standard conditions at 25°C Note that you substitute AS° in units of kJ/K in the formula of AG®. Sample Problem 6.13 What is the standard free-energy, 4G° for the following reaction (Haber's process) at 25°C? N2(g) + 312(g) 2NH 3(8) Also discuss its spontaneity Standard entropy: S° [N2(g)] = 191.5J / K, S°[H2(g)] = 130.6J / K, 5°[NH3(8)] = +193J / K Solution AH= 2H, (products) - A (reactants) = [2x (-45.9)]-[0 + 0] =-91.8kJ AS° = L nS" (products) - ¿ mS® (reactants) = [2x (193)]-[191.5+3x130.6]J/K K 4S° =- 0.197kJ/ K AG° = AH° -TAS° = (-91.88kJ) - (298K)x (-0.197kJ/ K) =-33.1kJ Result Since 4G° is a negative value thus it is concluded that Haber's process is spontaneous at standard conditions' i.e. at 25°C.

Spontaneity and temperature change The free-energy for a reaction at 25°C, is calculated. For the measurement of 4G° at any temperature, following equation is used. AG° = AH° - T AS° We get the value of AG at any temperature by substituting values of AH° and AS® at 25°C. But in this method, we assume that AH° and AS® are essentially constant w.r.t. temperature (this is only approximately true). Remember that the superscript degree sign (°) refers to substances in standard states, which are substances at 1 atm and at the specified temperature. Although until new this was 25°C (298 K), we now consider other temperatures. Note that in general, AG depends strongly on temperature. In table 6:3, all of the four possible choices of signs for AH° and AS®, give different temperature behaviors for 4G° are being discussed. In all this discussion temperature is always positive as it is taken on Kelvin scale. Table 6.3 Effect of temperature on the spontaneity of reactions AH" PlE GAG" Description © hand - + Spontaneous at all T + + Non- spontaneous at all T + or - Spontaneous at low T; non-spontaneous at high 1 + + + or - Non-spontaneous at low T; spontaneous at high T The term low temperature and high temperature are relative. For a particular reaction, high temperature could mean room temperature. Sample Problem 6.14 Calculate 4G° at 1000 °C for the following reaction and discuss its spontaneity. CaCO 3(s) → CaO(s) + CO2 Solution: For this reaction, AH° = + 178.0 kJ and AS° = + 160.4 J/K = 0.1604 kJ/K Substituting T = 1000 + 273 = 1273 K AG = AH - T AS = (+178.0 kJ) - (1273 K) x (+0.1604 k J/K) = + 178.0 kJ - 204.2 kJ AG = - 26.2 kJ We conclude that this reaction is spontaneous at 1000 °C and 1 atm, since 4G has negative sign. In other words, at 1000 °C, limestone (CaCO3) will decompose in an open container. Quick Check 6.9 (a) What do you expect about the entropy value of the following reactions, whether it would be positive or negative? (i) N2(g) + 3H 2(g) 2N3(8) (i) 2H1 (g) H2(g) + 12(g) Ans. (i) N2(g) +3H2(g) > 2NH 3(8) Predicting the sign of entropy change (AS.) • Reactants: 1 + 3 = 4 moles of gas • Products: 2 moles of gas Entropy decreases (system becomes more ordered). AS is negative (-) (i) 2H1 (g) →H2(g) + 12(g) • Reactants: 2 moles of gas • Products: 1 + 1 = 2 moles of gas Number of gas molecules remains same, but product molecules are simpler. AS is slightly positive or nearly neutral, since more types of particles are formed. AS is slightly positive (+) (b) Which of the following changes are likely to be spontaneous? (i) The smell from an open bottle of aqueous ammonia diffusing throughout a room. (i) Water turning to ice at - 10°C (iii) Ethanol vaporizing at 20°C (iv) Water mixing completely with salt (v) Limestone (CaCO3) decomposing at room temperature Ans: (1) The smell from an open bottle of aqueous ammonia diffusing throughout a room Particles spread out, increasing entropy. This process is spontaneous. (il) Water turning to ice at - 10°C Freezing at temperature below 0°C is natural. It is spontaneous (even though entropy decreases, AH is strongly negative and T is low). (iii) Ethanol vaporizing at 20°C Vaporization increases entropy, and ethanol has a low boiling point. It is spontaneous at room temperature. (iv) Water mixing completely with salt. Dissolving salt increases disorder. It is spontaneous. (v) Limestone (CaCO3) decomposing at room temperature CaCO3(s) → CaO(s) +CO2(g) Requires heat (endothermic) and doesn't occur at room temperature. So, it is non- spontaneous at room temperature.