Chemsheets AS006 (Electron arrangement) 27/04/2018 BORN-HABER CYCLES

At the end of this lesson you will know BORN-HABER CYCLES At the end of this lesson you will know The following terms: Lattice Enthalpy, 2. Ionization energy, 3. Electron affinity, 4. Atomization How to construct a Bohr-Haber Diagram How to calculate any aspect within the diagram 4 minutes on Salt production from the its elements (Formation) https://www.youtube.com/watch?v=wK9_DtwPaV0 Berkley salt production https://www.youtube.com/watch?v=d2geiGKFveE

Standard enthalpy of formation Definitions The change in enthalpy when one mole of a substance in its standard state is formed from its elements in their standard states. Standard enthalpy of formation  

Formation of Sodium Chloride kJmol-1 Formation of Sodium Chloride +800 +800 +800 +700 +700 +700 +600 +600 +600 +500 +500 +500 +400 +400 +400 - + +300 +300 +300 +200 +200 +200 +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 = -411kJmol-1 θ F NaCl(s)

I am sure you can see the PROBLEM here , if you look above Definitions The energy released when one mole of an ionic compound is formed from its constituent ions in their gaseous states.   Atomization Energy The energy required to break one mole of bonds between 2 atoms Is called Bond Energy for Cl2 it is 244 kJ/mol . Cl2  2 Cl I am sure you can see the PROBLEM here , if you look above

Atomization of Sodium Or Sublimation kJmol-1 Atomization of Sodium Or Sublimation +800 +700 +600 +500 +400 +300 +200 Na(g) + 1/2 Cl2(g) +100 H = +107kJmol-1 θ S Na(s) + 1/2 Cl2(g) -400 -300 -200 -100

Atomization of Chlorine Or Bond Dissociation kJmol-1 +800 +700 +600 +500 +400 +300 Na(g) + Cl(g) +200 H = +121kJmol-1 θ D ½ Na(g) + 1/2 Cl2(g) +100 Na(s) + 1/2 Cl2(g) -400 -300 -200 -100

First ionization energy Definitions The energy required to remove the outermost electron of each atom in one mole of an element in its gaseous state. X (g)  X+ (g) + e − First ionization energy  

First Ionization of Sodium kJmol-1 First Ionization of Sodium +800 Na+(g) + Cl(g) +700 e- +600 e- +500 e- H = +502kJmol-1 θ I +400 e- e- +300 Na(g) + Cl(g) +200 + Na(g) + 1/2 Cl2(g) +100 Na(s) + 1/2 Cl2(g) -400 -300 -200 -100

Definitions The energy released or absorbed when an electron is added to each atom in one mole of a substance in its gaseous state. X (g) + e − → X− (g) Electron affinity  

Electron Affinity of Chlorine kJmol-1 Electron Affinity of Chlorine +800 +800 +800 Na+(g) + Cl(g) +700 +700 +700 +600 +600 +600 H = -355kJmol-1 θ E +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) e- +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

Definition Lattice Energy: The heat energy (enthalpy) released when one mole of solid in its standard state is formed from its ions in the gaseous state. Ex: Na+(g) + Cl-(g)  NaCl(s) - + - + - - - + + + - - - + +

Born-Haber Cycle for Sodium Chloride kJmol-1 Born-Haber Cycle for Sodium Chloride +800 +800 +800 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) Lattice Enthalpy for Sodium Chloride +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 θ L + + - - + NaCl(s)

Hformation = sum of all other H’s Metal ions, e-’s, non-metal atoms (g) H H ionisation energy H electron affinity 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

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

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

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

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

H = -411kJ/mol Hformation = 107+121+502+-355+-786 Answer is -411 kJ/mol for heat of formation NaCl kJ kJ kJ +800 +800 +800 Na+(g) + Cl (g) +700 +700 +700 +600 +600 +600 +500 +500 +500 +400 +400 +400 Na+(g) + Cl-(g) + -355 +300 +300 +300 Na(g) + Cl(g) Na(g) + Cl(g) + 502 Na(g) + Cl(g) +200 +200 +200 - - Na(g) + 1/2 Cl2(g) + 121 Na(g) + 1/2 Cl2(g) Na(g) + 1/2 Cl2(g) + +100 +100 +100 + - + Na(s) + 1/2 Cl2(g) + 107 Na(s) + 1/2 Cl2(g) Na(s) + 1/2 Cl2(g) H = -786 kJmol-1 - + - -400 -300 -200 -100 -400 -300 -200 -100 -400 -300 -200 -100 H = -411kJ/mol Lattice enthalpy for sodium chloride NaCl(s)

Ok, so we just found the heat of formation for NaCl Using Born-Haber. Now lets try another (CaO) and find something different like Lattice Energy.

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)

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 2nd e- affinity= +844

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 = 193 + 248 + 590 + 1150 – 142 + 844 + Hlattice – 635 = 193 + 248 + 590 + 1150 – 142 + 844 + Hlattice Hlattice = – 635 – 193 – 248 – 590 – 1150 + 142 – 844 = – 3518 kJ mol-1 © www.chemsheets.co.uk A2 1015 19-May-2016

Ok, so we just found the heat of Lattice Energy For CaO Now try MgBr2 and Find it’s Lattice energy By drawing and Using Born-Haber Diagram.

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)

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 = 150 + 2(112) + 736 + 1450 + 2(–342) –2394 Hformation = 150 + 2(112) + 736 + 1450 + 2(–342) –2394 Hformation = – 518 kJ mol-1 © www.chemsheets.co.uk A2 1015 19-May-2016

Next, find Lattice energy for Na2S

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) + 279 + 2(496) – 200 + 649 + Hlattice Hlattice = – 370 – 2(107) – 279 – 2(496) + 200 – 649 = – 2304 kJ mol-1

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 = 193 + 2(112) + 590 + 1150 – 2(342) – 2125 = – 652 kJ mol-1

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) = – 15421 kJ mol-1

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 = 193 + 2(107) + 590 + 1150 + 2X – 2054 2X = – 535 – 193 – 2(107) – 590 – 1150 – 2054 2X = - 628 X = – 314 kJ mol-1

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 + 248 + 745 + 1960 – 142 + 844 – 4149 X = – 155 – 248 – 745 – 1960 + 142 – 844 + 4149 = + 339 kJ mol-1

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 = 289 + 107 + 732 – 314 + HLE HLE = –289 – 107 – 732 + 314 – 62 = –876 kJ mol-1

BORN-HABER CYCLES You should now know The following terms: A) Lattice Enthalpy, B) Ionization energy, C) Electron affinity, D) Atomization How to construct a Bohr-Haber Diagram How to calculate any aspect within the diagram