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1 6 5 4 3 2 Mg(s) + Cl 2 (g) MgCl 2 (s) Mg(g) + Cl 2 (g) Mg(g) + 2Cl(g) Mg 2+ (g) + 2Cl (g) 7 Mg + (g) + 2Cl(g) Mg 2+ (g) + 2Cl(g) Enthalpy of formation of MgCl 2 Mg(s) + Cl 2 (g) ——> MgCl 2 (s) Enthalpy of sublimation of magnesium Mg(s) ——> Mg(g) Enthalpy of atomisation of chlorine ½Cl 2 (g) ——> Cl(g) x2 Ist Ionisation Energy of magnesium Mg(g) ——> Mg + (g) + 2nd Ionisation Energy of magnesium Mg + (g) ——> Mg 2+ (g) + Electron Affinity of chlorine Cl(g) + ——> Cl¯(g) x2 Lattice Enthalpy of MgCl 2 Mg 2+ (g) + 2Cl¯(g) ——> MgCl 2 (s) 1 2 3 4 5 7 6 Starter: write the corresponding equations for each enthalpy change.

Mg(s) + C l 2 (g)

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Starter: write the corresponding equations for each enthalpy change. Enthalpy of formation of MgC l 2 Mg(s) + C l 2 (g) ——> MgC l 2 (s) Enthalpy of sublimation of magnesium Mg(s) ——> Mg(g) Enthalpy of atomisation of chlorine ½C l 2 (g) ——> C l (g) x2 - PowerPoint PPT Presentation

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Page 1: Mg(s)  + C l 2 (g)

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Mg(s) + Cl2(g)

MgCl2(s)

Mg(g) + Cl2(g)

Mg(g) + 2Cl(g) Mg2+(g) + 2Cl–(g)

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Mg+(g) + 2Cl(g)

Mg2+(g) + 2Cl(g)

Enthalpy of formation of MgCl2

Mg(s) + Cl2(g) ——> MgCl2(s)

Enthalpy of sublimation of magnesium

Mg(s) ——> Mg(g)

Enthalpy of atomisation of chlorine

½Cl2(g) ——> Cl(g) x2

Ist Ionisation Energy of magnesium

Mg(g) ——> Mg+(g) + e¯

2nd Ionisation Energy of magnesium

Mg+(g) ——> Mg2+(g) + e¯

Electron Affinity of chlorine

Cl(g) + e¯ ——> Cl¯(g) x2

Lattice Enthalpy of MgCl2

Mg2+(g) + 2Cl¯(g) ——> MgCl2(s)

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Starter: write the corresponding equations for each enthalpy change.

Page 2: Mg(s)  + C l 2 (g)

Born-Haber cycle calculations

L.O.:

Carry out calculations related to the Born-Haber cycle.

Page 3: Mg(s)  + C l 2 (g)

Cl¯ Br¯ F¯ O2-

Na+ -780 -742 -918 -2478

K+ -711 -679 -817 -2232

Rb+ -685 -656 -783

Mg2+ -2256 -3791

Ca2+ -2259

Lattice Enthalpy ValuesLattice Enthalpy Values

Smaller ions will have a greater attraction for each other because of their higher charge density. They will have larger Lattice Enthalpies and larger melting points because of the extra energy which must be put in to separate the oppositely charged ions.

Units: kJ mol-1

Indentify the trend in the enthalpy values. Explain the reason for this trend.

Page 4: Mg(s)  + C l 2 (g)

Cl¯ Br¯ F¯ O2-

Na+ -780 -742 -918 -2478

K+ -711 -679 -817 -2232

Rb+ -685 -656 -783

Mg2+ -2256 -3791

Ca2+ -2259

Lattice Enthalpy ValuesLattice Enthalpy Values

Smaller ions will have a greater attraction for each other because of their higher charge density. They will have larger Lattice Enthalpies and larger melting points because of the extra energy which must be put in to separate the oppositely charged ions.

Cl¯Na+ Cl¯

The sodium ion has the same charge as a potassium ion but is smaller. It has a higher charge density so will have a more effective attraction for the chloride ion. More energy will be released when they come together.

K+

Page 5: Mg(s)  + C l 2 (g)

Task

1)Construct a Born-Haber cycle for sodium oxide.

2)Calculate the lattice enthalpy for sodium oxide

Ha(Na) = +108 KJ mol-1

Ha(O) = + 249 KJ mol-1

HI1 =+249 KJ mol-1

HEA1 = - 141 KJ mol-1

HEA2 = + 790 KJ mol-1

Hf(Na2O) = -414 KJ mol-1

Page 6: Mg(s)  + C l 2 (g)