分子间作用力愈大.PPT

Insert Fig. 13-11. Have it appear (wipe left) after the last sentence. Insert Fig. 13-11. Have it appear (wipe left) after the last sentence. Insert Fig. 13-11. Have it appear (wipe left) after the last sentence. * Example 13-6: Heat of Fusion The molar heat of fusion, ?Hfus, of Na is 2.6kJ/mol at its melting point, 97.5oC, How much neat must be absorbed by 5.0g of sodium Na at 97.5oC to melt it? ?J = 5.0g x 1mol 23g x = 0.57 kJ 2.6kJ mol * Example 13-2: Heat of Fusion Calculate the amount of heat that must be absorbed by 50.0g of ice at -12.0oC to convert to water at 20.0oC. -10oC H2O(s) 0oC H2O(s) 0oC H2O(l) 20oC H2O(l) ?J = 50.0g x 2.09J goC x (0.0-(-12).0oC) = 1.25x103 J ?J = 50.0g x 334J g = 16.7x103 J ?J = 50.0g x 4.18J goC x (20.0-0.0oC) = 4.18x103 J Total J = 1.25x103 + 16.7x103 + 4.18x103 = 22.1 kJ * Sublimation and the Vapor Pressure of Solids Sublimation昇華 In the sublimation process the solid transforms directly to the vapor phase without passing through the liquid phase. Solid CO2 or “dry” ice does this well. iodine solid gas sublimation deposition Sublimation can be used to purify volatile solids * Endothermic 吸熱 Exothermic 放熱 Transitions among the three states of matter 凝結 蒸發 昇華 * Phase Diagrams (P versus T)相圖 Phase diagrams are a convenient way to display all of the different phase transitions of a substance. The equilibrium pressure-temperature (壓力-溫度) relationships among the different phases of a given pure substance in a closed system (密閉系統). * Phase Diagrams (P versus T)相圖 This is the phase diagram for water. 三相點 AB Melting curve AD Sublimation curve Negative slope of line AB Ice is less dense than liquid The network of hydrogen bonding in ice is more extensive than that in a liquid water * Phase Diagrams (P versus T) Compare water’s phase diagram to carbon dioxide’s phase diagram. 三相點 Liquid CO2 cannot exist at atmospheric pressure Critical point For CO2: critical point is at 31oC and 73 atm For H2O: critical point is a