07 Gibbs Free Energy

07 Gibbs Free Energy - Water Ice Problem Calculate the...

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1 1 Water – Ice Problem Calculate the change in entropy of water, S system , when 2 mol of water freezes at 0 o C (273 K). For ice, H fusion = 6.01 kJ/mol S = S final – S initial = q/T A) + 44.0 J/K B) – 44.0 J/K C) + 0.022 J/K D) – 0.022 J/K 2 Water (the system) Heat flow: system (water) to surroundings (freezer) During the freezing process T water is constant (0°C) S = q/T and q = -n H fus H fus = 6.01 kJ/mol q H2O is negative since energy flows out of the system. q H2O = -2 x 6.01 x 10 3 J = -1.202 x 10 4 J S H2O = q H2O /T H2O = -1.202x10 4 / 273 = -44.0 J K -1
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2 3 When 2 mol of water freezes, S system = - 44 J/K. The entropy decreases! Makes sense liquid solid. Water freezes spontaneously, but we said S > 0 for a spontaneous process. What gives here? Need to consider the total entropy not just the entropy of the system. 4 Total Entropy Change When water freezes, energy is transferred to the surroundings. Which means that the energy is dispersed over more states in the surroundings. So the entropy of the surroundings increases. S total = S system + S surroundings S total also called S universe
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3 5 Calculate the change in entropy of the freezer (surroundings), S surroundings , when 2 mol of water freezes at 0 o C (273 K). The temperature of the freezer is – 15 o C (258 K) For ice, H fusion = 6.01 kJ/mol S = S final – S initial = q/T A) + 46.6 J/K B) – 46.6 J/K C) + 44.0 J/K D) – 44.0 J/K 6 Freezer (the surroundings) The freezer absorbs energy at its temp. (-15°C). Absorbs energy so q freezer > 0. q freezer = -q H2O so q freezer = 2mol (6.01 kJ/mol) q freezer = 1.202 x 10 4 J S freezer = q freezer /T freezer = 1.202 x 10 4 / 258 = 46.6 J K -1
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4 7 The Universe (total) S universe = S system + S surroundings S universe = - 44.0 J/K + 46.6 J/K = + 2.6 J/K
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07 Gibbs Free Energy - Water Ice Problem Calculate the...

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