Ppt 13 R1.2 Energy cycles in reactions.pptx

the ppt is on energy cycles in reactions and deal with Hess's law of constant heat summation where the law of conservation of mass is obeyed. The enthalpy change can be calculated no matter whatever path the reaction has taken place as long as reactants and products are same the enthalpy change can be determined. Thus if enthalpy change of reaction is to be calculated than we can use enthalpy of combustion and formation to calculate it.

1. Ppt 13 R 1.2 Energy cycles in reaction. Ppt 13 Reactivity 1.2 Energy cycles in reactions SL &HL Reactivity 1.2.1 Bond-breaking absorbs and bond-forming releases energy SL &HL Reactivity 1.2.2 Hess’s law states that the enthalpy change for a reaction is independent of the pathway between the initial and final states. How does application of the law of conservation of energy help us to predict energy changes during reactions?

2. Reinforce : Enthalpy diagram and Reaction profile diagram

3. Average Bond enthalpy Definition Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms. Values Endothermic - Energy must be put in to break any chemical bond Examples C l 2 (g) ———> 2C l (g) O-H(g) ———> O(g) + H(g) Notes • strength of bonds also depends on environment; MEAN values quoted • making bonds is exothermic as it is the opposite of breaking a bond • for diatomic gases, bond enthalpy = 2 x enthalpy of atomisation • smaller bond enthalpy = weaker bond = easier to break

4. Average bond enthalpy Bond dissociation- endothermic Bond making - exothermic

5. Enthalpy change of the reaction using bond enthalpy table

6. Practise questions 1) Calculate the enthalpy change for the reaction N 2 + 3H 2  2NH 3 2) Calculate the enthalpy change for the burning of 11 grams of propane ( Use average bond enthalpies table) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) Answers 1) - 90 kJ/mol 2) Enthalpy of reaction = -2046 kJmol -1 Now 11 grams = 0.25 mole of propane (11 g/44 g mol -1 ) (0.25 mol )(-2046 kJ mol -1 ) = - 511.5 kJ

7. Practice questions

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9. Hess’s law

10. Hess’ Law: Example 1 N 2 (g) +2O 2 (g)  + 2 NO 2 (g) The required equation is really the sum of the two given equations Solution: N 2 (g) + O 2 (g)  2 NO (g) D H 1 = +181 kJ 2 NO (g) + O 2 (g)  2 NO 2 (g) D H 2 = -113 kJ ------------------------------------------------------------ N 2 (g) +2O 2 (g)+ 2 NO (g)  2 NO (g) + 2 NO 2 (g) N 2 (g) +2O 2 (g)  + 2 NO 2 (g) D H = D H 1 + D H 2 = +181 kJ +(-113) = + 68 kJ

11. Hess Law Example 2 From the following reactions and enthalpy changes: 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) D H = -196 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g) D H = -790 kJ Find the enthalpy change for the following reaction: S (s) + O 2 (g)  SO 2 (g) 2 SO 3 (g)  2 SO 2 (g) + O 2 (g) D H = +196 kJ 2 S (s) +3 O 2 (g)  2 SO 3 (g) D H = -790 kJ -------------------------------------------------------------------------------------------------------------- 2 SO 3 (g) +2 S(s) + 2 3 O 2 (g)  2 SO 3 (g) +2 SO 2 (g) + O 2 (g) D H = -594 kJ 2 S(s) + 2 O 2 (g)  2 SO 2 (g) D H = -594 kJ S(s) + O 2 (g)  SO 2 (g) D H = -297 kJ

12. Calculate ∆Hf of methane using given data

13. Calculate ∆H and C-H bond enthalpy

14. Practise questions

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16. Practise

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18. HL AHL Reactivity 1.2.3 Standard enthalpy changes of combustion,ΔHc ⦵, and formation,ΔHf ⦵, data are used in thermodynamic calculations. AHL Reactivity 1.2.4 An application of Hess’s law uses enthalpy of formation data or enthalpy of combustion data to calculate the enthalpy change of a reaction. AHL Reactivity 1.2.5 A Born–Haber cycle is an application of Hess’s law, used to show energy changes in the formation of an ionic compound.

19. Standard enthalpy change of reaction

20. Enthalpy of formation and enthalpy of combustion

21. Practise question

22. Sample calculation Calculate the standard enthalpy change for the following reaction, given that the standard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286, +33 and -173 kJ mol -1 respectively; the value for oxygen is ZERO as it is an element 2H 2 O(l) + 4NO 2 (g) + O 2 (g) ———> 4HNO 3 (l) By applying Hess’s Law ... The Standard Enthalpy of Reaction DH ° r will be... PRODUCTS REACTANTS [ 4 x D f H of HNO 3 ] minus [ (2 x D f H of H 2 O) + (4 x D f H of NO 2 ) + (1 x D f H of O 2 ) ] DH ° r = 4 x (-173) - 2 x (-286) + 4 x (+33) + 0 ANSWER = - 252 kJ Enthalpy of reaction from enthalpies of formation DH =  D f H of products –  D f H of reactants

23. Enthalpy of reaction from enthalpies of combustion Sample calculation Calculate the standard enthalpy of formation of methane; the standard enthalpies of combustion of carbon, hydrogen and methane are -394, -286 and -890 kJ mol -1 . C(graphite) + 2H 2 (g) ———> CH 4 (g) By applying Hess’s Law ... The Standard Enthalpy of Reaction DH ° r will be... REACTANTS PRODUCTS [ (1 x D c H of C) + (2 x D c H of H 2 ) ] minus [ 1 x D c H of CH 4 ] DH ° r = 1 x (-394) + 2 x (-286) - 1 x (-890) ANSWER = - 76 kJ mol -1 DH =  D c H of reactants –  D c H of products

24. Sample Problem 1 Calcium carbonate reacts with hydrochloric acid according to the following equation: CaCO 3 (s) + 2HCl (aq)  CaCl 2 (aq) + H 2 O (l) + CO 2 (g) Calculate the enthalpy change for this reaction D H o reaction = S D H o products – SD H o reactants D H o CaCO 3 -1207 D H o HCl ( aq ) -167 D H o CaCl 2 -796 D H o H 2 O (l) -286 D H o CO 2 (g) -394 Solution D H o products =(-796)+(-286)+(-394) = -1476 kJ D H o reactants =(-1207)+(2)(-167) = -1541 kJ D H o reaction = -1476-(-1541) = +65 kJ

25. Sample Problem 2 Calculate the enthalpy change for the burning of 11 grams of propane C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) D H o reaction = S D H o products – SD H o reactants D H o C 3 H 8 -104 D H o O 2 (g) D H o H 2 O (g) -242 D H o CO 2 (g) -394 Solution D H o products =(3)(-394)+(4)(-242) = -2150 kJ D H o reactants =(-104)+(5)(0) = -104 kJ D H o reaction = -2150-(-104) = -2046 kJmol -1 Now 11 grams = 0.25 mole of propane (11 g/44 g mol -1 ) (0.25 mol )(-2046 kJ mol -1 ) = - 511.5 kJ

26. Practise

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28. Practise questions

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30. Born haber cycle for ionic compounds,

31. Enthalpy of atomisation: Enthalpy change when one mole of gaseous atoms is formed from its elements under standard conditions. Note: if direct atomisation values are not given especially in case of non metals than we may need to use half the value of bond energies. Ex: Na (s) Na (g) ½ Cl 2(g ) Cl (g ) ΔH˚ atm = 108 KJ/mol ΔH˚ atm = 121 KJ/mol Electron affinity: Enthalpy change when one mole of electrons is added to one mole of gaseous atoms to form one mole of negative ions under standard conditions. Note: First electron affinity is exothermic. Second and third electron affinity are endothermic because when electrons are added to negative ion repulsion increases therefore requires energy to add electron. Ex: Cl (g ) + e - Cl - (g) ΔH˚ ea1 = -342 KJ/mol

32. ENTHALPY CHANGES Lattice energy: Enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions. Ex: Na + (g) + Cl - (g) NaCl (s ) Energy is always released in this process because of the strong electrostatic force between oppositely charged ions. It is an exothermic process . Product that is the ionic compound has less energy than the reactants, that is its ions in their Gaseous state. Lesser the energy of the compound, more stable is the compound More the energy released i.e. more exothermic the process, stable is the compound. ΔH˚ latt = - 787 KJ/mol Lattice Dissociation energy: Enthalpy change when one mole of an ionic compound is dissociated into its gaseous ions under standard conditions. Ex: NaCl (s) Na + (g) + Cl - (g) ΔH˚ = + 787 KJ/mol Enthalpy of formation: Enthalpy change when one mole of a compound is formed from its elements under standard state and under standard conditions. Ex: Na (s) + ½ Cl 2(g) NaCl (s) ΔH˚ f = - 412 KJ/mol Ionisation energy: It is the energy needed to remove one electron from each atom in one mole of atoms of the element in the gaseous state to form one mole of gaseous ions under standard conditions. Ex: Na (g) Na + (g) + e - ΔH˚ i1 = + 494 KJ/mol

33. Born - haber cycle Born – Haber cycle: It is the enthalpy cycle used to calculate the lattice energy of any given ionic compound. Ionic compound Ions in gaseous state Elements in their standard state Lattice energy Enthalpy of formation ΔH˚ 1 ΔH˚ latt = ΔH˚ f - ΔH˚ 1 ΔH˚ 1 = ΔH˚ atm + ΔH˚ i + ΔH˚ ea

34. Born haber cycle

35. solution

36. Example2: Calculate the lattice energy of CaO. ΔH˚ f [CaO] = - 635 KJ/mol ΔH˚ at [Ca] = +178 KJ/mol ΔH˚ atm [O] = +248 KJ/mol ΔH i1 [Ca] = +590 KJ/mol ΔH i2 [Ca] = +1150 KJ/mol ΔH ea1 [O] = -141 KJ/mol ΔH ea2 [O] = +844 KJ/mol Ca (s) + ½ O 2(g) CaO (s) ΔH˚ f Ca (g) + ½ O 2(g) ΔH˚ at [Ca] Ca (s) + O (g) ΔH˚ at [O] Ca 1+ (s) + O (g) ΔH˚ i1 [Ca] Ca 2+ (g) + O (g) ΔH˚ i2 [Ca] Ca 2+ (g) + O - (g) ΔH˚ ea1 [O] Ca 2+ (g) + O 2- (g) ΔH˚ ea2 [O] ΔH˚ latt ΔH˚ latt = ΔH˚ f - ΔH˚ 1 = [-635] - [ ΔH˚ at [Ca] + ΔH˚ atm [O] + ΔH i1 [Ca] + ΔH i2 [Ca] + ΔH ea1 [O] + ΔH ea2 [O] ] = [ - 635] - [ 178 + 248 + 590 + 1150 + ( - 141 ) + 844 ] = [ - 635] - [ 2869 ] = - 3504 KJ/ mol

37. Try this

38. Factors affecting Lattice energy

39. Practise questions

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41. Enthalpy of solution (Not on IB) but can be used to understand Lattice enthalpy

42. E nthalpy of hydration ( Not on IB syllabus) Calculate Enthalpy of solution for Srcl2

43. Practise questions

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45. Complete end of topic questions