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19 Chemical Thermodynamics Solutions to Exercises 19.95 At the normal boiling point of a liquid, G = 0 and = T = By Trouton's rule, = The process of vaporization is: (a) Br₂(g) Tb = = 30.71kJ X kJ = 349 = 3.5 X K (b) According to WebElements™ 2005, the normal boiling point of is 332 K. Trouton's rule provides a good "ballpark" estimate. 19.96 (a) Formation reactions are the synthesis of 1 mole of compound from elements in their standard states. NH₃(g) C(s) KNO₃(s) In each of these formation reactions, there are fewer moles of gas in the products than the reactants, so we expect to be negative. If = - TAS° and is negative, - is positive and is more positive than (b) C(s) + CO(g) In this reaction, there are more moles of gas in products, is positive, - is negative and is more negative than 19.97 (a) (i) Ti(s) + 2Cl₂(g) TiCl4(g) = TiCl4(g) - Ti(s) Cl₂(g) = -763.2 2(0) = = 354.9 - 30.76 2(222.96) = -121.78 = -121.8J/K (ii) C₂H₆(g) + 7Cl₂(g) = + HCl(g) C₂H₆(g) - Cl₂(g) = 2(-106.7) +6(-92.30) - (-84.68) 7(0) = -682.52 = = 2(309.4) + 6(186.69) - 7(222.96) = -51.28 = -51.4J/K = 2(-64.0) + (-95.27) - (-32.89) - 7(0) = -666.73 kJ (iii) BaO(s) + BaCO₃(s) = BaCO₃(s) - BaO(s) = -1216.3 - (-553.5) - (-393.5) = = -1137.6 - (-525.1) - (-394.4) = -218.1 kJ 601