\u0394G o RT lnK Slightly soluble K is a small number If K 1 but bigger than 0 \u0394G o

Δg o rt lnk slightly soluble k is a small number if

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ΔG o = -RT lnK Slightly soluble K is a small number If K <1 (but bigger than 0), ΔG o = -RT ln(<1) ln(<1) = negative numbr ΔG o = positive number
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Clicker Q10 Using the data provided estimate ΔH° vap for CCl 4 ? A. 2.86 kJ/mol B. -344 J/mol C. 35 kJ/mol D. 3.44 J/mol Temp (°C) Temp (°C) P P vap vap (atm) (atm) 4.3 4.3 0.0526 0.0526 23.0 23.0 0.131 0.131
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Clicker Q10: Solution Using the data provided estimate ΔH° vap for CCl 4 ? A. 2.86 kJ/mol B. -344 J/mol C. 35 kJ/mol D. 3.44 J/mol Temp (°C) Temp (°C) P P vap vap (atm) (atm) 4.3 4.3 0.0526 0.0526 23.0 23.0 0.131 0.131 ) 1 1 ( ln 1 2 1 2 T T R H K K o ) 3 . 277 1 295 1 ( 0526 . 0 131 . 0 ln R H o ΔH° = ln(2.49)(0.0083)/(2.16 x 10 -4 ) = 35 kJ/mol
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Clicker Q11 Using your answer in Clicker Q10 and the data provided estimate ΔS° vap for CCl 4 ? Temp (°C) Temp (°C) P P vap vap (atm) (atm) 4.3 4.3 0.0526 0.0526 23.0 23.0 0.131 0.131 A. 0.0102 J/K B. 101.7 J/K C. 150.6 J/K D. I am lost
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Clicker Q11: Solution Using your answer in Clicker Q10 and the data provided estimate ΔS° vap for CCl 4 ? A. 0.0102 J/K B. 101.7 J/K C. 150.6 J/K D. I am lost Temp (°C) Temp (°C) P P vap vap (atm) (atm) 4.3 4.3 0.0526 0.0526 23.0 23.0 0.131 0.131 R S T R H K o o ) 1 ( ) ln( ln(0.0526)   35 0.0083 ( 1 277.3 ) S o 0.0083 ΔS° = (ln(0.052) + 15.2)*(0.0083) ΔS° = 0.101 kJ/K
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For the fusion reaction of water: H 2 0(s) H 2 O(l) At what temperature does ΔG = 0? (ΔH° fus = 6.02 kJ/mol, ΔS° fus = 22 J/mol-K) A. 0.27°C B. 0.27 K C. 274 K D. Need more information Clicker Q12
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A. 0.27°C B. 0.27 K C. 274 K D. Need more information Clicker Q12: Solution T = ΔH° fus /ΔS° fus = 6.02/0.022 = 273.6 K ΔG o = -RT lnK = ΔH° – TΔS° =0 ΔG = 0 at equilibrium, so T = ΔH° /ΔS° For the fusion reaction of water: H 2 0(s) H 2 O(l) At what temperature does ΔG = 0? (ΔH° fus = 6.02 kJ/mol, ΔS° fus = 22 J/mol-K)
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A. ΔG > 0 B. ΔG = 0 C. ΔG < 0 Clicker Q13 For the fusion reaction of water: H 2 0(s) H 2 O(l) What would you expect ΔG to be at T = 10°C? (ΔH° fus = 6.02 kJ/mol, ΔS° fus = 22 J/mol-K)
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A. ΔG > 0 B. ΔG = 0 C. ΔG < 0 Clicker Q13: Solution For the fusion reaction of water: H 2 0(s) H 2 O(l) What would you expect ΔG to be at T = 10°C? (ΔH° fus = 6.02 kJ/mol, ΔS° fus = 22 J/mol-K) At 10 o C you are not at equilibrium: ΔG = ΔG o = ΔH° – TΔS° ΔG = ΔG o + RT ln(1) ΔG = ΔG o ΔG = ΔH° – TΔS° ΔG = 6.02 – (283)(0.022) = -0.206 kJ
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A. 100 °C B. 93 °C C. 85 °C D. 78 °C Clicker Q14: Difficult one! What is the boiling point of water in Denver, where P atm is 0.823 atm?
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Clicker Q14: Solution What is the boiling point of water in Denver, where P atm is 0.823 atm? A. 100 °C B. 93 °C C. 85 °C D. 78 °C H 2 O(l) H 2 O(g) NOTE: Pvap is not = 1, ΔG o is NOT = 0 We cannot use the normal bp equation: T = ΔH° /ΔS° R S T R H K o o 1 ln If we can find ΔH° and ΔS° we can find T
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Clicker Q14: Solution What is the boiling point of water in Denver, where P atm is 0.823 atm? ln(0.823)   44 0.0083 1 T 0.1189 0.0083 H 2 O(l) H 2 O(g) R S T R H K o o 1 ln ΔH o rxn = 44 kJ ΔS o rxn = 0.119 kJ/mol K (ln0.823) – 14.32 = -5301(1/T) T = -(5315)/(-14.5) T = 366 K – 273 = 93 °C
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A. The entropy of the system will increase under standard conditions at 25 C.
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