1. ENERGETICS /THERMOCHEMISTRY (AS)

2. 1.Often chemical changes are accompanied by changes in heat content / enthalpy of the materials reacting (H) 2. This change is shown by a change in temperature.

3. a. heat is lost to the surroundings Temperature of reaction mixture rises / increases H is negative Exothermic reaction

4. Enthalpy diagram / energy level diagram OR Reaction pathway/reaction coordinate/extent of reaction

6. b. heat is absorbed from the surroundings, temperature of reaction mixture decreases / falls H is positive Endothermic reaction

9. 3. a. The value of H depends on temp, pressure and concentrations of reactants. b. H are measured under standard conditions : Temperature = 298K ( 25 0 C ) Pressure = 1 atm / 1.01 x 10 5 Pa Concentrations = 1 mol dm -3

10. c. Any H measured under standard conditions is described as standard enthalpy change. Symbol : H 

11. STANDARD ENTHALPIES Definitions State symbols must be included in equations

12. STANDARD ENTHALPHY OF FORMATION ( H f  ) Enthalpy change when 1 mole of a compound is formed from its elements in their standard states, under standard conditions temperature 298 K and pressure of gases at 1 atm

13. Steps in writing equations: 1. write formula of compound formed 2. identify elements required to form the compound 3. balance equation to form 1 mole of the compound

14. Examples Mg (s) + ½O 2 (g) → 1MgO (s) -602 kJ mol -1 602 kJ evolved for every 1 mole of MgO formed ½ H 2 (g) + ½ Cl 2 (g) → 1HCl (g) - 92.3 kJ mol -1

15. Examples Notes : 1) H f  of elements is zero Eg : Cu(s)  Cu(s), H f  = 0 2) H f  are often theoretical only

16. Exercise : Write the equation for enthalpy of formation for : KMnO 4 (s) and H 2 O (l) K(s) + Mn(s) + 2O 2 (g) 1KMnO 4 (s) H 2 (g) + ½O 2 (g)  1H 2 O (l)

17. STANDARD ENTHALPY CHANGE OF COMBUSTION ( H c  ) Enthalpy evolved when 1 mole of the element or compound is completely burned in excess oxygen, under standard conditions

18. Steps in writing equations : 1. identify products formed from burning of compound in excess oxygen 2. balance equation for 1 mole of compound burnt

19. Examples 1 C (s) + O 2 (g) → CO 2 (g) H c  = - 394 H c  = - 394 kJ mol -1 394 kJ evolved for every 1 mole of carbon burnt 1 H 2 (g) + ½O 2 (g) → H 2 O (l)

20. Exercise : Write an equation for the enthalpy of combustion for C 3 H 6 and CH 3 OH

21. 1C 3 H 6 (g) + 9/2 O 2 (g)  3CO 2 (g) + 3H 2 O (l) 1CH 3 OH (l) + 3/2 O 2 (g)  CO 2 (g) + 2H 2 O (l)

22. STANDARD ENTHALPY OF ATOMISATION ( H at  ) Enthalpy required/absorbed 1 mole of gaseous atoms formed From the element in its standard state under standard conditions Examples : Fe (s)  1 Fe (g)

23. Example : ½Cl 2 (g)  1 Cl (g) + 122 kJ mol -1 molecules atoms Note : from Data Booklet Bond energy Cl-Cl = + 244 kJ mol -1 Cl-Cl  2 Cl (g) 2 moles atoms H at of Cl = ½ x bond energy Cl-Cl

24. STANDARD BOND DISSOCIATION ENTHALPY Also called bond energy Energy absorbed Separate the 2 atoms in a covalent bond in gaseous state under standard conditions, per mole of bond.

25. Examples 1) HCl (g) → H (g) + Cl (g), +431 kJmol -1 2) CH 4 contains 4 x C-H bond Total energy required to break all bonds in CH 4 = 1640 kJ ¼ CH 4 = ¼ ( 4 C-H bond ) = one C-H bond Average bond energy of one C-H bond = ¼ ( 1640 ) = +410 kJ ¼ CH 4 (g) → ¼ C (g) + H (g),+ 410 kJmol -1

26. STANDARD ENTHALPY OF NEUTRALIZATION Energy evolved Acid reacts with base to form 1 mole of water, under standard conditions Examples : NaOH + HCl → NaCl + 1 H 2 O KOH + HCl → KCl + 1 H 2 O Ionic equation(strong acid + strong base: H + (aq) + OH - (aq)  1 H 2 O (l),  H  neutralisation = -57 kJ mol -1

27. Generally, 1) Strong acid / strong alkali : H  = - 57 kJ mol -1 2) Weak acid / base : H  = -(<57) kJ mol -1 (eg. -54 kJmol -1 ) Reason : Certain amount of energy required to ionise the weak acid or base first

28. STANDARD ENTHALPY OF HYDRATION ( Hhyd ) Energy evolved 1 mole of separate gaseous ions dissolved in water under standard conditions Exothermic as attraction/bond forms between the ions and polar water molecules Called ion dipole attraction

29. Examples : 1 Na + (g) → Na + (aq) -406 kJmol -1 1 Cl - (g) → Cl - (aq) - 381 kJmol -1

30. H hyd for compounds = sum of H hyd of constituent ions Eg : H hyd MgCl 2 = H hyd Mg 2+ + 2 x H hyd Cl - = -1891 + 2(-381) = - 2653 kJ

31. H (hyd)  charge density of ions Charge density = charge/size Higher charge density, stronger ion dipole attraction, more exothermic H(hyd) Eg : H(hyd) Cl - > H(hyd) Br – -381 -351

32. STANDARD ENTHALPY OF SOLUTION ( H solution  ) Enthalpy change when 1 mole of a substance dissolved in a stated amount of solvent under standard conditions Example : 1 KOH (s)  K + (aq) + OH - (aq) or KOH (aq) - 57 kJ mol -1

33. STANDARD ENTHALPY CHANGE OF REACTION ( H r  ) Enthalpy change in a chemical reaction, for the number of moles of reactants as shown in a balanced chemical equation under standard conditions Example : 4 H 2 O + 3 Fe  Fe 3 O 4 + 4 H 2  H  r = x kJ when 4 moles H 2 O reacts with 3 moles Fe