1 / 6

Mg(s) + C l 2 (g)

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

reese-walter
Télécharger la présentation

Mg(s) + C l 2 (g)

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. Starter: write the corresponding equations for each enthalpy change. 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) 1 Mg2+(g) + 2Cl(g) 2 5 6 Mg+(g) + 2Cl(g) 3 4 4 Mg2+(g) + 2Cl–(g) Mg(g) + 2Cl(g) 3 5 Mg(g) + Cl2(g) 2 7 6 Mg(s) + Cl2(g) 1 7 MgCl2(s)

  2. Born-Haber cycle calculations L.O.: Carry out calculations related to the Born-Haber cycle.

  3. Indentify the trend in the enthalpy values. Explain the reason for this trend. Lattice Enthalpy Values Cl¯ Br¯ F¯ O2- Na+-780 -742 -918 -2478 K+ -711 -679 -817 -2232 Rb+ -685 -656 -783 Mg2+ -2256 -3791 Ca2+ -2259 Units: kJ mol-1 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.

  4. Na+ Cl¯ Cl¯ K+ Lattice Enthalpy Values Cl¯ Br¯ F¯ O2- Na+-780 -742 -918 -2478 K+ -711 -679 -817 -2232 Rb+ -685 -656 -783 Mg2+ -2256 -3791 Ca2+ -2259 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. 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.

  5. Task • Construct a Born-Haber cycle for sodium oxide. • Calculate the lattice enthalpy for sodium oxide • DHa(Na) = +108 KJ mol-1 • DHa(O) = + 249 KJ mol-1 • DHI1 =+249 KJ mol-1 • DHEA1 = - 141 KJ mol-1 • DHEA2 = + 790 KJ mol-1 • DHf(Na2O) = -414 KJ mol-1

More Related