2012 15 1 and 15 2

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2012 15 1 and 15 2

  1. 1. IB Chemistry Power Points Topic 15 Energeticswww.pedagogics.ca LECTURE Standard Enthalpies Born-Haber Cycle
  2. 2. Much taken fromENTHALPY CHANGES Great thanks toJONATHAN HOPTON & KNOCKHARDY PUBLISHING www.knockhardy.org.uk/sci.htm
  3. 3. Enthalpy of reaction from enthalpies of formationConsider a reaction as reactants  elements  productsStep 1 Energy is required as reactants ELEMENTS are broken into their elements. NEGATIVE SUM OFStep 2 Energy is released as products THE ENTHALPIES OF are formed from their elements. FORMATION OF THE SUM OF THE REACTANTS ENTHALPIES OF FORMATION OF H = Step 2 - Step 1 THE PRODUCTS REACTANTSIn Step 1 the route involves going in theOPPOSITE DIRECTION to the defined Henthalpy change, it’s value is subtracted (oralternatively the sign of the enthalpies isreversed) PRODUCTS ΔH = Σ ΔHf products - Σ ΔHf reactants
  4. 4. Enthalpy of reaction from enthalpies of formationSample calculationCalculate the standard enthalpy change for the following reaction, given that thestandard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286, -1+33 and -173 kJ mol respectively; 2H2O(l) + 4NO2(g) + O2(g) ———> 4HNO3(l)
  5. 5. Enthalpy of reaction from enthalpies of formation 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, -1 +33 and -173 kJ mol respectively; 2H2O(l) + 4NO2(g) + O2(g) ———> 4HNO3(l)By applying Hess’s Law ... The Standard Enthalpy of Reaction will be... PRODUCTS REACTANTSΔH° = 4 x (-173) MINUS 2 x (-286) + 4 x (+33) + 0 ANSWER = - 252 kJthe value for the enthalpy of formation for oxygen is ZERO as it is already in elemental form
  6. 6. Enthalpy of reaction from enthalpies of combustion
  7. 7. Enthalpy of reaction from enthalpies of combustion
  8. 8. Enthalpy of reaction from enthalpies of combustion
  9. 9. Enthalpy of reaction from enthalpies of combustion
  10. 10. The Born–Haber cycle
  11. 11. Background Information (from Wikipedia)“The Born–Haber cycle is an approach to analyzingreaction energies. It was named after and developed by thetwo German scientists Max Born and Fritz Haber.The Born–Haber cycle involves the formation of an ioniccompound from the reaction of a metal (often a Group I or GroupII element) with a non-metal. Born–Haber cycles are used primarilyas a means of calculating lattice energies (or more precisely [1]enthalpies ) which cannot otherwise be measured directly.The lattice enthalpy is the enthalpy change involved in formationof the ionic compound from gaseous ions. Some chemists define itas the energy to break the ionic compound into gaseous ions. Theformer definition is invariably exothermic and the latter isendothermic.”
  12. 12. Background Information (from Wikipedia) continued“A Born–Haber cycle calculates the lattice enthalpy by comparingthe standard enthalpy change of formation of the ionic compound(from the elements) to the enthalpy required to make gaseous ionsfrom the elements. This is an application of Hesss Law.This latter calculation is complex. To make gaseous ions fromelements it is necessary to atomise the elements (turn each intogaseous atoms) and then to ionise the atoms. If the element isnormally a molecule then we have to consider its bond dissociationenthalpy (see also bond energy). The energy required to removeone or more electrons to make a cation is a sum ofsuccessive ionization energies; for example the energy needed to 2+form Mg is the first plus the second ionization energies of Mg.The energy released when one electron is added to an atom tomake it an anion is called the electron affinity.”
  13. 13. A schematic of a Born-Haber cycleThe sum of the enthalpies on any two sides must equalthe enthalpy change of the remaining side
  14. 14. There are many differentstandard enthalpy valuesdepending on what is beingmeasured
  15. 15. Here is a cycle for NaCl. Make sure you understandthe changes occurring and which enthalpy values areused for each reaction.
  16. 16. 774 kJ/mol
  17. 17. Factors affecting lattice enthalpyObservation: A decrease in the size of any ion increasesthe lattice enthalpy. (more positive/endothermic).Explanation: This is because small ions can be closetogether and the smaller distance of separation thelarger the attractive force between the ions.
  18. 18. Factors affecting lattice enthalpyObservation: An increase in charge also increases latticeenthalpy. (more positive/endothermic).Explanation: This is because the force of attractionbetween ions increases as the charge on the ionincreases.
  19. 19. Differences between experimental and theoretical valuesin lattice energy.In general, the larger the difference between empiricaland theoretical value, the more covalent character thebond has.You should be able to relate this to electronegativity

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