Thermodynamics Solutions: #10
10.* (1996 3 4) Solubility depends on temperature. When solid potassium nitrate KNO3(s) is dissolved in water the process reaches equilibrium heat + KNO3 (s) + H2O(l) < -- > K+(aq) + NO3-(aq) (1) This dissolving process occurs with the absorption of head (endothermic process). When gaseous O2 is dissolved in water, the process reaches equilibrium O2 (gas) + H2O(l) < -- > O2 (aq) + heat (2) Answer the following questions. A. For reaction 1 what is the sign of DH? DH is positive (heat is absorbed while dissolving KNO3) B. For reaction 2 what is the sign of DH? DH is negative (heat is released while dissolving O2) C. If a saturated solution of patassium nitrate is in contact with solid KNO3 and the solution is heated, will the solubility of KNO3 increase, decrease, or stay the same? The solubility of KNO3 increases with the addition of heat. According to Le Chatêlier’s Principle, heat will drive the reaction to the right, making more KNO3 dissolve. D. If a saturated solution of oxygenated water is in contact with gaseous oxygen and the solution is heated, will the solubility of O2 increase, decrease, or stay the same? The solubility of O2 decreases with the addition of heat. According to Le Chatêlier’s Principle, heat will drive the reaction to the left, making less O2 dissolve. E. Explain in thermodynamic terms why KNO3 (s) and O2 (g) both dissolve spontaneously in pure water. DG = DH -TDS For KNO3: DH > 0 (the reaction is endothermic), but DS < 0 (dissolving a crystal causes more disorder) so |TDS| > |DH| (entropy wins and DG < 0) For O2: DH < 0 (the reaction is exothermic), but DS < 0 (dissolving a gas in water causes less disorder) so |H| > |TDS| (enthalpy wins and DG < 0 ) For both cases DGdissolution is less than zero, so dissolving is spontaneous for both KNO3(s) and O2(g). |