lecture 7 - Calculate the change in entropy of water...

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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 Clausius figured out that you can calculate the change in entropy by dividing the energy transferred, q, by the Kelvin temperature, T. Rudolf Clausius German Physicist 1822 - 1888 2 S = q/T and q = -n H fusion n = 2 mol H fusion = 6.01 kJ/mol T = 273 K 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 4 When 2 mol of water freezes, S water = –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. 5 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. More states in the surroundings are accessible. So the entropy of the surroundings increases. S total = S system + S surroundings S total also is called S universe
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3 6 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 7 Freezer (the surroundings) S freezer = q freezer /T freezer The freezer absorbs energy at its temp. ( 15°C) 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 8 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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