2. New Way Chemistry for Hong Kong A-Level Book 11 Chapter 6 Energetics

3. New Way Chemistry for Hong Kong A-Level Book 12 Standard Enthalpy Changes As enthalpy changes depend on temperature and pressure. In order to standardize the data recorded, it is necessary to define the standard conditions: 1. elements or compounds in their normal physical states; 2. a pressure of 1 atm (101325 Nm -2 ); and 3. a temperature of 25 0 C (298 K) 4. concentration of solution = 1 mol dm -3 6.3 Standard Enthalpy Changes (SB p.141) Enthalpy change under standard conditions denoted by symbol:  H ø

4. New Way Chemistry for Hong Kong A-Level Book 13 Standard Enthalpy Change of Neutralization 6.3 Standard Enthalpy Changes (SB p.142) The standard enthalpy change of neutralization (  H neut ) is the enthalpy change when one mole of water is formed from the neutralization of an acid by an alkali under standard conditions. ø H + (aq) + OH - (aq)  H 2 O(l)  H  neut = -57.3 kJ mol -1

5. New Way Chemistry for Hong Kong A-Level Book 14 Standard Enthalpy Changes of neutralization NH 3 (aq) + H 2 O(l) === NH 4 + (aq) + OH - (aq) 6.3 Standard Enthalpy Changes (SB p.145) -57.1 -57.3 -52.2 NaOH KOH NH 3 HCl HNO 3 HCl  H neu AlkaliAcid ø The value is smaller if weak acids/alkalis are used because some energy is used for the complete dissociation/ionization of the weak acids/alkalis.

6. New Way Chemistry for Hong Kong A-Level Book 15 Standard Enthalpy Change of Formation 6.3 Standard Enthalpy Changes (SB p.144) The standard enthalpy change of formation (  H f ) is the enthalpy change of the reaction when one mole of the compound in its standard state is formed from its constituent elements under standard conditions. ø Na(s) + ½Cl 2 (g)  NaCl(s)  H f = -411 kJ mol -1 1 mole ø

7. New Way Chemistry for Hong Kong A-Level Book 16 Standard Enthalpy Change of Combustion 6.3 Standard Enthalpy Changes (SB p.146) The standard enthalpy change of combustion (  H c ) of a substance is the enthalpy change when one mole of the substance is burnt completely in oxygen under standard conditions. ø e.g. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l)  H c = -2220 kJ mol -1 ø

8. New Way Chemistry for Hong Kong A-Level Book 17 Experimental Determination of Enthalpy Changes Calorimeter = a container used for measuring the temperature change of solution 6.4 Experimental Determination of Enthalpy Changes by Calorimetry (p.148)

9. New Way Chemistry for Hong Kong A-Level Book 18 Hess’s Law A + BC + D Route 1 H1H1 E H2H2 H3H3 Route 2  H 1 =  H 2 +  H 3 Hess’s Law states that the total enthalpy change accompanying a chemical reaction is independent of the route by which the chemical reaction takes place. Why? 6.5 Hess’s Law (p. 153) Conservation of energy

10. New Way Chemistry for Hong Kong A-Level Book 19 Importance of Hess’s Law the reactions cannot be performed in the laboratory the reaction rates are too slow the reactions may involve the formation of side products The enthalpy change of some chemical reactions cannot be determined directly from experiment because: But the enthalpy change of such reactions can be determined indirectly by applying Hess’s Law. 6.5 Hess’s Law (p. 155) Q. 5

11. New Way Chemistry for Hong Kong A-Level Book 110 Energetics of Formation of Ionic Compound Na(s) + ½Cl 2 (g)  NaCl(s)  H f = formation of ionic bond? 7.2 Energetics of Formation of Ionic Compounds (SB p. 189)  Hf Hf ø Na(g) Cl(g) Na + (g) Cl - (g)  H f  = -411 kJ mol -1

12. New Way Chemistry for Hong Kong A-Level Book 111 The enthalpy change when one mole of gaseous atoms is formed from its elements in the defined physical state under standard conditions. Questions:Why are the changes endothermic? What type of bond is broken in each case? 7.2 Some important enthalpy terms (SB p. 190) Na(s) Na(g)  H atom [Na(s)] = +109 kJ mol -1 ø 1/2 Cl 2 (g) Cl(g)  H atom [1/2Cl 2 (g)] = +121 kJ mol -1 ø Standard Enthalpy Change of Atomization (  H atom ) ø

13. New Way Chemistry for Hong Kong A-Level Book 112 The amount of energy required to remove one mole of electrons from one mole of atoms or ions in the gaseous state. Questions: Why are the changes endothermic? Energy is needed to overcome the attractive force between the positive nucleus and the negatively charged electrons. 7.2 Energetics of Formation of Ionic Compounds (SB p. 191) Ionization Enthalpy (  H I.E. ) Na(g) Na + (g) + e -  H I.E [Na(g)] = +494 kJ mol -1 ø Mg(g) Mg + (g) + e -  H I.E [Mg(g)] = +736 kJ mol -1 ø Mg + (g) Mg 2+ (g) + e -  H I.E [Mg + (g)] = +1 450 kJ mol -1 ø

14. New Way Chemistry for Hong Kong A-Level Book 113 The energy change when one mole of electrons is added to one mole of atoms or ions in the gaseous state. Questions: Why does the 1 st E.A. of O is negative while the second one is positive? 7.2 Energetics of Formation of Ionic Compounds (SB p. 191) Electron Affinity (  H E.A. ) ø First electron affinity O(g) + e - O - (g)  H E.A [O(g)] = - 142 kJ mol -1 Second electron affinity O - (g) + e - O 2- (g)  H E.A [O - (g)] = + 844 kJ mol -1 ø ø

15. New Way Chemistry for Hong Kong A-Level Book 114 The energy change when one mole of an ionic crystal is formed from its constituent ions in the gaseous state under standard conditions 7.2 Energetics of Formation of Ionic Compounds (SB p. 185) Lattice Enthalpy (  H L.E. ) ø Na+ (g) + Cl - (g) NaCl(s)  H lattice [Na + Cl - (s)] ø It is a measure of the strength of ionic bond. + – –+

16. New Way Chemistry for Hong Kong A-Level Book 115 L.E. can be determined indirectly by either: (1) calculations basing on the knowledge of electrostatics in Physics (assuming ions are point charges) – Theoretical Value or (2) calculations basing on Hess’s Law --- Experimental Value 7.2 Energetics of Formation of Ionic Compounds (SB p. 192) Na+ (g) + Cl - (g) NaCl(s)  H lattice [Na + Cl - (s)] ø Q. 8

17. New Way Chemistry for Hong Kong A-Level Book 116 Born-Haber Cycle for the formation of sodium chloride 7.2 Energetics of Formation of Ionic Compounds (SB p. 193)  H  atom [Na(s)]  H I.E. Do Q. 6(a) on p. 181 Do. Q. 49 on p. 325 Do Q. 9(c) i. on p. 316

18. New Way Chemistry for Hong Kong A-Level Book 117 Strength of ionic lattice 7.2 Strength of Ionic lattice (SB p. 196) 1.Ionic Size Greater the size  lower charge density  weaker attraction  lower (less negative) lattice enthalpy NaCl: -771 kJ mol -1 ;KCl: -701 kJ mol -1 2.Ionic Charge Higher charge  stronger attraction  higher lattice enthalpy CaO: -3513 kJ mol -1 ;CaCl 2 : -2237 kJ mol -1

19. New Way Chemistry for Hong Kong A-Level Book 118 Discrepancy between calculated and experimental value 7.2 Strength of Ionic lattice (not typically mentioned in book) Assumption in calculating lattice enthalpy:.Ions are hard sphere.Completely transfer of electrons.Charge density distributes evenly on the sphere compoundcalculatedexperimentaldifference NaCl -770-78010 KCl -702-7119 AgCl -833-90572 AgI -778-889111 Incomplete transfer of electrons  covalent characters

20. New Way Chemistry for Hong Kong A-Level Book 119 Bond Enthalpy (for covalent bond) is the energy associated with a chemical bond. When a chemical bond is broken, a certain amount of energy is absorbed. When a chemical bond is formed, a certain amount of energy is released. Bond Enthalpies – Covalent Compounds 8.3 Bond Enthalpies (SB p.220)

21. New Way Chemistry for Hong Kong A-Level Book 120 8.3 Bond Enthalpies (SB p.221) Bond Dissociation Enthalpies B.D.E of a certain bond is the amount of energy required to break one mole of that bond in a particular compound under standard conditions. Bond Enthalpies Average bond enthalpy (or simply bond enthalpy) is the average of the bond dissociation enthalpies required to break a particular chemical bond. Q. 12 b, Q. 13

22. New Way Chemistry for Hong Kong A-Level Book 121 Bond length (for covalent bond) 8.6 Relationship between Bond Enthalpies and Bond Lengths (SB p.228) Bond Lengths

23. New Way Chemistry for Hong Kong A-Level Book 122 Any conclusion for the relationship between bond length & bond enthalpy? Usually a longer bond length corresponds to a lower value of bond enthalpy (weaker bond) 8.6 Relationship between Bond Enthalpies and Bond Lengths (SB p.228) BondBond length (nm)Bond enthalpy (kJ mol -1 ) H-H Cl-Cl Br-Br I-I H-F H-Cl H-Br H-I 0.074 0.199 0.228 0.266 0.092 0.127 0.141 0.161 436 242 193 151 565 431 364 299 Bond enthalpies and bond lengths Distance between shared electrons pair and nuclei increases  attraction decreases  bond strength decreases

24. New Way Chemistry for Hong Kong A-Level Book 123 (often referred as ‘Atomic radius’ ???) The space occupied by an atom in a covalently bonded molecule in the direction of the covalent bond (generally taken as half of the bond length) 8.6 Relationship between Bond Enthalpies and Bond Lengths (SB p.228) Covalent Radius

25. New Way Chemistry for Hong Kong A-Level Book 124 8.6 Relationship between Bond Enthalpies and Bond Lengths (SB p.230) BondCalculated bond length (nm)Experimentally determined bond length (nm) C-O C-F C-Cl C-Br C-C H-Cl C-H N-Cl 0.150 0.149 0.176 0.191 0.154 0.136 0.114 0.173 0.143 0.138 0.177 0.193 0.154 0.128 0.109 0.174 Calculated and experimentally determined bond length Difference in electronegativities  polar bond  ionic character

26. New Way Chemistry for Hong Kong A-Level Book 125 Standard Enthalpy Change of Solution -- dissolving Note that enthalpy changes of solution associate with physical changes. 6.3 Standard Enthalpy Changes (SB p.146, NB p. 33)) The standard enthalpy change of solution (  H soln ) is the enthalpy change when one mole of a solute is completely dissolved in a sufficient large volume of solvent to form an infinitely dilute solution under standard conditions. ø NaCl(s) + water  Na + (aq)+Cl - (aq)  H soln =+3.9 kJ mol -1 ø LiCl(s) + water  Li + (aq) + Cl - (aq)  H soln =-37.2 kJ mol -1 ø

27. New Way Chemistry for Hong Kong A-Level Book 126 According to Hess’s law, The enthalpy change of solution is :  H soln =  H hyd –  H lattice

28. New Way Chemistry for Hong Kong A-Level Book 127 This change involves 2 processes: 1st : NaCl(s)  Na + (g) + Cl – (g)  H = +776 kJ mol –1 The enthalpy change involved in this process is the reverse of lattice enthalpy. The lattice enthalpy is –776 kJ mol –1. 2nd : hydration enthalpy Na + (g) + Cl – (g)  Na + (aq) + Cl – (aq)  H hyd = –772 kJ mol –1 NaCl(s)  Na+(aq) + Cl–(aq)  H soln = +776 – 772 = +4 kJ mol –1

29. New Way Chemistry for Hong Kong A-Level Book 128 Effect of charge and size of ions on  H hyd and  H lattice 6.3 Standard Enthalpy Changes (NB p. 34)

30. New Way Chemistry for Hong Kong A-Level Book 129 Generalizations (Not group trend): When there is a mismatch in ionic size, the salt is expected to be fairly soluble. Reasons: i.The magnitude of lattice enthalpy is not great because of the large ion, making (r + + r - ) relatively large. ii.The magnitude of the hydration enthalpy is still large due to the presence of the small counter ion. Fast Prediction of solubility (NB. P.38)

31. New Way Chemistry for Hong Kong A-Level Book 130 Relative Solubility of the alkalis metal halides Group trend of solubility (NB. P.36) LiI NaI KI RbI CsI less negative  weaker bond  more soluble less negative  smaller attraction to water  less soluble

32. New Way Chemistry for Hong Kong A-Level Book 131 For Group I iodide, cations are much smaller than anions The  H lattice is determined by the reciprocal of the sum of cationic and anionic radii (i.e. )  Large anionic radius makes the relatively small cationic radius insignificant w.r.t the sum of r + and r – r + + r - ~ r -  Down the group, increase in cationic size does not make a significant decrease in the magnitude of  H lattice Solubility of Group I metal iodides – group trend Group trend of solubility (NB. P.36)

33. New Way Chemistry for Hong Kong A-Level Book 132 The hydration enthalpy is determined by An increase in cationic size causes  H hyd to become less and less negative significantly   H soln becomes less and less exothermic  The solubility of Group I metal iodides decreases down the group Group trend of solubility (NB. P.36)

34. New Way Chemistry for Hong Kong A-Level Book 133 How about the solubility trend of Group I metal Fluorides? Group trend of solubility (NB. P.37) For small anion like F -, as the size of the cation increases, the magnitude of the lattice enthalpies decrease quickly (become less negative). On the other hand, the increase in cationic size does not cause a great decrease in the sum of the hydration enthalpies because of the great magnitude of hydration enthalpy of the small anion. As a result, the solubility increases with an increase in cationic radius.

35. New Way Chemistry for Hong Kong A-Level Book 134 Questions: 1.The solubility of CsF is relatively high. Explain. 2.The solubility of LiF is relatively low. Explain. 3.The solubility of CsI is relatively low. Explain. Group trend of solubility (NB. P.38)

36. New Way Chemistry for Hong Kong A-Level Book 135 Entropy 6.7 Entropy Change (p. 164) Scientists want to find out what governs a spontaneous reaction: an exothermic reaction is a spontaneous reaction while an endothermic reaction is not (exothermicity) any exception? Some spontaneous change is endothermic, e.g. melting of ice at room temperature.  besides enthalpy change, there is another factor that determine a chemical reaction.  Entropy Entropy is a measure of the randomness or the degree of disorder of a system. Q. 11

37. New Way Chemistry for Hong Kong A-Level Book 136 Entropy 6.7 Entropy Change (p. 164) In any spontaneous process, there is always an increase in the entropy (disorder) of the system and its surroundings --- The second law of thermodynamics. Melting of ice at room temperature  increase in entropy of water molecules Dissolving of sodium chloride in water  increase in entropy

38. New Way Chemistry for Hong Kong A-Level Book 137 6.7 Entropy Change (p. 164) Entropy Change  S  S = S final - S initial Entropy change is temperature dependent: High temperature  entropy increases Low temperature  entropy decreases Unit of entropy: kJ mol -1 K -1

39. New Way Chemistry for Hong Kong A-Level Book 138 Free Energy 6.7 Free Energy Change (p. 168) When predicting whether a reaction is spontaneous, we need to consider both enthalpy change and entropy change (at constant temperature). A new term is developed to include both enthalpy and entropy  free energy:  G =  H – T  Swhere  G = change in free energy A process is spontaneous when  G is negative (  H is –ve,  S is +ve) A process is not spontaneous when  G is positive.

40. New Way Chemistry for Hong Kong A-Level Book 139 The END