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Chemsheets AS006 (Electron arrangement)
10/06/2019 BORN-HABER CYCLES
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Lattice enthalpy For an ionic compound the lattice enthalpy is the heat energy released when one mole of solid in its standard state is formed from its ions in the gaseous state. Eg Na+(g) Cl-(g) NaCl(s) - L.E cannot be determined directly enthalpy cycle (Hess’s Law) links the data. MOLAR HEATS OF FORMATION - enthalpy cycle is based on the formation of the compound from its elements in standard states.
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Hformation = sum of all other H’s
Metal ions, e-’s, non-metal atoms (g) H H ionisation energy/ies H electron affinity/ies Gas atoms (g) Gas ions (g) Hatomisation(s) H lattice energy of formation Elements (std states) H formation Ionic compound (s) Hformation = sum of all other H’s
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kJ H= +107kJmol-1 Atomisation of sodium +800 +700 +600 +500 +400 +300
+200 Na(g) + 1/2 Cl2(g) +100 H= +107kJmol-1 Na(s) + 1/2 Cl2(g) -400 -300 -200 -100 Atomisation of sodium
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Atomisation of chlorine
kJ +800 +700 +600 +500 +400 +300 Na(g) + Cl(g) +200 H = +121kJmol-1 Na(g) + 1/2 Cl2(g) +100 Na(s) + 1/2 Cl2(g) -400 -300 -200 -100 Atomisation of chlorine
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e- leaving e- e- e- leaving e- e- e- + kJ H = +502kJmol-1
+800 Na+(g) + Cl(g) +700 e- leaving e- e- +600 e- leaving +500 e- e- +400 H = +502kJmol-1 e- +300 Na(g) + Cl(g) +200 + Na(g) + 1/2 Cl2(g) +100 Na(s) + 1/2 Cl2(g) -400 -300 -200 -100 First Ionisation of sodium
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First electron affinity of chlorine
kJ kJ kJ +800 +800 +800 Na+(g) + Cl(g) +700 +700 +700 +600 +600 +600 H = -355kJmol-1 +500 +500 +500 e- +400 +400 +400 Na+(g) + Cl-(g) +300 +300 +300 Na(g) + Cl(g) Na(g) + Cl(g) Na(g) + Cl(g) +200 +200 +200 - e- Na(g) + 1/2 Cl2(g) Na(g) + 1/2 Cl2(g) Na(g) + 1/2 Cl2(g) +100 +100 +100 Na(s) + 1/2 Cl2(g) Na(s) + 1/2 Cl2(g) Na(s) + 1/2 Cl2(g) -400 -300 -200 -100 -400 -300 -200 -100 -400 -300 -200 -100 First electron affinity of chlorine
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- - - - - kJ kJ kJ + + + + H = -786 kJmol-1 +800 +800 +800 +700 +700
Na+(g) + Cl(g) +700 +700 +700 +600 +600 +600 +500 +500 +500 +400 +400 +400 Na+(g) + Cl-(g) +300 +300 +300 Na(g) + Cl(g) Na(g) + Cl(g) Na(g) + Cl(g) +200 +200 +200 - - Na(g) + 1/2 Cl2(g) Na(g) + 1/2 Cl2(g) Na(g) + 1/2 Cl2(g) + +100 +100 +100 + - + Na(s) + 1/2 Cl2(g) Na(s) + 1/2 Cl2(g) Na(s) + 1/2 Cl2(g) - + - -400 -300 -200 -100 -400 -300 -200 -100 -400 -300 -200 -100 H = -786 kJmol-1 Lattice enthalpy for sodium chloride NaCl(s)
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CaO Ca2+(g) + O2– (g) Ca2+(g) + 2e– + O (g) Ca+(g) + e– + O(g)
2nd electron affinity of O 2nd ionisation of Ca 1st electron affinity of O Ca+(g) + e– + O(g) 1st ionisation of Ca Ca2+(g) + e– + O – (g) Ca(g) + O(g) atomisation of O Ca(g) + ½O2(g) Lattice formation of CaO atomisation of Ca Ca(s) + ½O2(g) formation of CaO CaO (s)
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CaO Ca2+(g) + O2– (g) Ca2+(g) + 2e– + O (g) Ca+(g) + e– + O(g)
2nd electron affinity of O 2nd ionisation of Ca 1st electron affinity of O –142 1150 844 Ca+(g) + e– + O(g) 1st ionisation of Ca Ca2+(g) + e– + O – (g) 590 Ca(g) + O(g) atomisation of O 248 Ca(g) + ½O2(g) Lattice formation of CaO atomisation of Ca X 193 Ca(s) + ½O2(g) –635 formation of CaO CaO (s) Hf CaO= –635; Atomisation calcium = +193; Atomisation oxygen = +248 1st IE Ca = +590; 2nd IE Ca = +1150 1st e- affinity of oxygen = –142 kJ mol-1 2nd e- affinity= +844
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CaO Hlattice = – 635 – 193 – 248 – 590 – 1150 + 142 – 844
Ca2+(g) + O2– (g) Ca2+(g) + 2e– + O (g) 2nd electron affinity of O 2nd ionisation of Ca 1st electron affinity of O –142 1150 844 Ca+(g) + e– + O(g) 1st ionisation of Ca Ca2+(g) + e– + O – (g) 590 Ca(g) + O(g) atomisation of O 248 Ca(g) + ½O2(g) Lattice formation of CaO atomisation of Ca X 193 Ca(s) + ½O2(g) –635 formation of CaO CaO (s) – 635 = – Hlattice – 635 = – Hlattice Hlattice = – 635 – 193 – 248 – 590 – – 844 = – 3518 kJ mol-1 © A May-2016
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MgBr2 Mg2+(g) + 2e– + 2Br (g) Mg+(g) + e– + 2Br(g) Mg2+(g) 2Br – (g)
2nd ionisation of Mg 2 x electron affinity of Br Mg+(g) + e– + 2Br(g) Mg2+(g) 2Br – (g) 1st ionisation of Mg Mg(g) + 2Br(g) 2 x atomisation of Br Mg(g) + Br2(l) Lattice formation of MgBr2 atomisation of Mg Mg(s) + Br2(l) formation of MgBr2 MgBr2 (s)
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MgBr2 Hformation = – 518 kJ mol-1 Mg2+(g) + 2e– + 2Br (g)
2nd ionisation of Mg 2 x electron affinity of Br 1450 2(–342) Mg+(g) + e– + 2Br(g) Mg2+(g) 2Br – (g) 1st ionisation of Mg 736 Mg(g) + 2Br(g) 2 x atomisation of Br 2(112) Mg(g) + Br2(l) Lattice formation of MgBr2 atomisation of Mg 150 –2394 Mg(s) + Br2(l) X formation of MgBr2 MgBr2 (s) Hformation = (112) (–342) –2394 Hformation = (112) (–342) –2394 Hformation = – 518 kJ mol-1 © A May-2016
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KCl Hformation = 90 + 121 + 418 – 364 – 710 = – 445 kJ mol-1
K+ (g) + e- + Cl (g) H 1st ionisation of K –364 1st electron affinity of Cl 418 K (g) + Cl (g) K+ (g) + Cl- (g) Atomisation of Cl 121 K (s) + ½ Cl2 (g) Lattice enthalpy of formation of KCl –710 Atomisation of K 90 K (s) + ½ Cl2 (g) ? Formation of KCl KCl (s) Hformation = – 364 – 710 = – 445 kJ mol-1
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Na2S Hlattice = – 370 – 2(107) – 279 – 2(496) + 200 – 649
2Na+(g) + S2–(g) 2nd electron affinity of S 2Na+(g) + 2e– + S(g) 1st electron affinity of S 2 x 1st ionisation of Na 649 –200 2(496) 2Na+(g) + e– +S –(g) 2Na(g) + S(g) atomisation of S 279 2Na(g) + S(s) Lattice formation of Na2S Na2S 2 x atomisation of Na ? 2(107) 2Na(s) + S(s) formation of Na2S –370 Na2S(s) – 370 = 2(107) (496) – Hlattice Hlattice = – 370 – 2(107) – 279 – 2(496) – 649 = – 2304 kJ mol-1
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CaBr2 Hformation = 193 + 2(112) + 590 + 1150 – 2(342) – 2125
Ca2+(g) + 2e– + 2Br (g) 2nd ionisation of Ca 2 x electron affinity of Br H 1150 2(–364) Ca+(g) + e– + 2Br(g) Ca2+(g) + 2Br – (g) 1st ionisation of Ca 590 Ca(g) + 2Br(g) 2 x atomisation of Ca 2(121) Ca(g) + Br2(l) Lattice formation of CaBr2 atomisation of Ca 193 –2125 Ca(s) + Br2(l) ? formation of CaBr2 CaBr2 (s) Hformation = (112) – 2(342) – 2125 = – 652 kJ mol-1
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Al2O3 2Al3+(g) + 6e– + 6O(g) 2Al3+(g) + 6O2–(g) 2Al2+(g) + 4e– + 3O(g)
2 x 3rd ionisation of Al 2(2740) 3 x 1st electron affinity of O 2Al2+(g) + 4e– + 3O(g) 3 x 2nd electron affinity of O 2 x 2nd ionisation of Al 3(–142) 3(844) 2(1820) 2Al+(g) + 2e– + 3O(g) 2Al3+(g) + 3e– + 3O –(g) 2 x 1st ionisation of Al 2(577) 2Al(g) + 3O(g) Lattice formation of Al2O3 3 x atomisation of O 3(248) 2Al(g) + 3/2O2(g) ? Al2O3 2 atomisation of Al 2(314) 2Al(s) + 3/2O2(g) formation of Al2O3 –1669 Al2O3(s) –1669 = 2(314) + 3(248) + 2(577) + 2(1820) + 2(2740) + 3(-142) + 3(844) + H H = –1669 – 2(314) – 3(248) – 2(577) – 2(1820) – 2(2740) + 3(142) – 3(844) = – kJ mol-1
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CaI2 Ca2+(g) + 2e– + 2I (g) 2nd ionisation of Ca 2 x electron affinity of I H 1150 2 X Ca+(g) + e– + 2I(g) Ca2+(g) + 2I – (g) 1st ionisation of Ca 590 Ca(g) + 2I(g) 2 x atomisation of I 2(107) Ca(g) + I2(s) Lattice formation of CaI2 atomisation of Ca 193 –2054 Ca(s) + I2(s) –535 formation of CaI2 CaI2 (s) – 535 = (107) X – 2054 2X = – 535 – 193 – 2(107) – 590 – 1150 – 2054 2X = X = – 314 kJ mol-1
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CuO X = – 155 – 248 – 745 – 1960 + 142 – 844 + 4149 = + 339 kJ mol-1
Cu2+(g) + O2–(g) Cu2+(g) + 2e– + O(g) 2nd electron affinity of O 2nd ionisation of Cu 1st electron affinity of O 1960 844 Cu+(g) + e– + O(g) –142 1st ionisation of Cu Cu2+(g) + e– +O –(g) 745 Cu(s) + O(g) Lattice formation of CuO atomisation of O 248 Cu(g) + ½O2(g) –4149 CuO atomisation of Cu X Cu(s) + ½O2(g) formation of CuO –155 CuO(s) – 155 = X – – 4149 X = – 155 – 248 – 745 – – = kJ mol-1
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AgI Ag+ (g) + e- + I (g) H 1st ionisation of Ag –314 1st electron affinity of I 732 Ag (g) + I (g) Ag+ (g) + I- (g) Atomisation of I 107 Ag (s) + ½ I2 (s) Lattice enthalpy of formation of AgI ? Atomisation of Ag 289 Ag (s) + ½ I2 (s) –62 Formation of AgI AgI (s) –62 = – HLE HLE = –289 – 107 – – 62 = –876 kJ mol-1
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