1 / 25

Topic 15 Energetics (HL)

Topic 15 Energetics (HL). 15.1 Standard enthalpy changes of reaction 15.2 Born-Haber cycle 15.3 Entropy 15.4 Spontaneity. 15.1 Standard Enthalpy Change of reaction . Standard state: 101kPa, 298K. Standard enthalpy of formation D H f q.

catrin
Download Presentation

Topic 15 Energetics (HL)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Topic 15 Energetics (HL) • 15.1 Standard enthalpy changes of reaction • 15.2 Born-Haber cycle • 15.3 Entropy • 15.4 Spontaneity

  2. 15.1 Standard Enthalpy Change of reaction Standard state: 101kPa, 298K

  3. Standard enthalpy of formation DHfq DHfq: The energy absorbed or evolved when 1 mol of the substance is formed from its elements in their standard states. The enthalpy of formation of any element is zero. ½ H2(g) + O2(g) H2O(l)DHfq= -285 kJ/mol DH = SDHf(products) - SDHf(reactants)

  4. Decomposition of ammonium nitrate NH4NO3(s) N2O(g) + 2 H2O(l) • NH4NO3(s) DHfq= -366 kJ/mol • N2O(g) DHfq= +82 kJ/mol • H2O(l)DHfq= -285 kJ/mol DH = [DHf(N2O(g)) + DHf(H2O(l))] – [DHf(NH4NO3(g)] = =[82 + 2(-285)] - [-366] = -122 kJ/mol

  5. Standard enthalpy of combustion,DHcq • When a substance is fully combusted in oxygen CH4+ 2O2 CO2 + 2H2O DHcq= ?

  6. 15.2 Born-Haber cycle • To determine the lattice enthalpy and the degree of ionic character of a salt • To find an unknown value (Hess’ law)

  7. Lattice enthalpy, DHlattice • Relates to the endothermic process MX(s) M+(g) + X-(g) in which the gaseous ions of a crystal are separated to an infinitive distance from each other. • NaCl(s) Na+(g) + Cl-(g) DHlattice= 769 kJ/molEndothermic reactions

  8. Factors affecting the lattice enthalpy • The greater the charge of the ions, the stronger the electrostaticattraction http://www.chemhume.co.uk/A2CHEM/Unit%202b/9%20Lattice%20enthalpy/Ch9Latticec.htm

  9. Factors affecting the lattice enthalpy (2) The smaller the ionicradius, the shorter the distance, the strongerthe electrostaticattraction

  10. Electron affinity (electron gain enthalpy) • The enthalpy change when an atom gain one electron in gas phase e.g. Cl(g) + e-(g) Cl-(g) DHe.a. = -349 kJ/mol. • Electron affinity can be both exothermic and endothermic depending on element.

  11. Born-Haber cycle for the formation of NaCl (s) • Enthalpy of atomisation of Na Na (s) →Na (g) DHat= +108 kJ/mol • Enthalpy of atomisation of Cl ½ Cl2(s) →Cl(g) DHat= +121 kJ/mol (½ energy of Cl-Cl bond)

  12. Born-Haber cycle for the formation of NaCl (s)(2) • Electron affinity of Cl Cl(g) + e-→ Cl-(g)DHea= -349 kJ/mol • Ionisation energy of Na Na (g)→ Na+ (g) + e-DHie= + 496kJ/mol

  13. Born-Haber cycle for the formation of NaCl (s)(3) • Lattice enthalpy of NaCl Cl-(g)+ Na+ (g) →NaCl (s) DHlatt= -769 kJ/mol

  14. Theoretical value of Enthalpy of formation of NaCl = -411kJ/mol Using Hess Law: Enthalpy of formation of NaCl DHf(NaCl)= DHat(Na) + DHie(Na) + DHat(Cl) + DHea(Cl) + DHlatt(NaCl) 108+494+121+(-364) + (-771)= -393 kJ/mol

  15. Use of Born-Haber cycles • In the Chemistry Data Booklet the lattice enthalpies is given booth as: • Experimental values (obtained by Born-Haber cycle) • Theoretical values (calculated from electrostatic principles) • If the value differ in a significant way => indicate more covalent character of the salt

  16. 15.3 Entropy- disorder • Entropy, S Unit: J/K*mol • DS = change in disorder • DS = Sproducts- Sreactants • It’s possible to measure absolute values of S

  17. Increasing entropy- DS positive • Solid Liquid  Gas increase in S Ice  Water  Steam 48.0 69.9 188.7 JK-1mol-1 • Mixing different types of particles- dissolving NaCl in water • Increasing no of particles- N2O4 (g) →2 NO2 (g)

  18. Decreasing entropy- DS negative • System becomesmoreordered • Formation of solid ammonium chloride from hydrogen chloride and ammonia gas NH3(g) + HCl(g) NH4Cl(s)DS = - 285 J/K*mol

  19. In any conversion there is both a change in DH and DS. DS is probably positive if number of mol of gas increases and number of mol of solid/liquid decrease. NH4Cl(s) NH3(g) + HCl(g) DS = + 285J/K*mol Pb2+(aq) + 2 I- PbI2(s)DS = - 70 J/K*mol

  20. 15.4 Spontaneity of a reaction- DG • Nature likes low internal energy (DH to decrease) and high disorder (DS to increase) • Spontaneity: is a reaction going to occur • A reaction will occur if the final state is more probable than the initial state. => Decrease in DH => Increase in DS

  21. DGq= DHq -TDSq • Standard free energy, DGq (or Gibbs free energy) • Temperature is importantfor spontaneity • A reaction will be spontaneous if DG has a negative value (DGq < 0) • A positive DGq is a non-spontaneous reaction. • If DGq = 0 then its a equilibrium

  22. DGq= DHq -TDSq

  23. Just the fact that a reaction is spontaneous doesn’t mean that it will occur at once- it might need activation energy (topic 6)

More Related