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CH 302 HW1 - Thompson Margaret Homework 1 Due 11:00 am Inst...

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Thompson, Margaret – Homework 1 – Due: Sep 11 2007, 11:00 am – Inst: James Holcombe 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The vapor pressure of CH 3 COCH 3 (propa- none or acetone), is 67 Torr at 0 C and 222 Torr at 25 C. What is the normal boiling point of propanone? The universal gas con- stant is 8 . 314 J K · mol . Correct answer: 55 . 941 C. Explanation: T 1 = 0 C + 273 . 15 = 273 . 15 K T 2 = 25 C + 273 . 15 = 298 . 15 K P 1 = 67 Torr P 2 = 222 Torr P 0 = 1 atm = 760 Torr R = 8 . 314 J K · mol Using the Clausius-Clapeyron equation, ln P 1 P 2 = Δ H vap R 1 T 2 - 1 T 1 R ln P 1 P 2 1 T 2 - 1 T 1 = Δ H vap Δ H vap = ( 8 . 314 J K · mol ) ln 67 Torr 222 Torr 1 298 . 15 K - 1 273 . 15 K = 32445 . 7 J / mol . Applying the Clausius-Clapeyron equation again, ln P 0 P 2 = Δ H vap R 1 T 2 - 1 T 0 R Δ H vap ln P 0 P 2 = 1 T 2 - 1 T 0 1 T 0 = 1 T 2 - R Δ H vap ln P 0 P 2 = 1 298 . 15 K - 8 . 314 J K · mol 32445 . 7 J / mol ln 760 Torr 222 Torr = 0 . 00303867 C T 0 = 1 0 . 00303867 C = 329 . 091 C - 273 . 15 = 55 . 941 K 002 (part 1 of 1) 10 points Arsine (AsH 3 ) is a highly toxic compound used in the electronic industry for the pro- duction of semiconductor. Its vapor pressure is 35 Torr at - 111 . 95 C and 253 Torr at - 83 . 6 C. What is the standard enthalpy of vaporization of arsine? Correct answer: 17 . 7248 kJ / mol. Explanation: T 1 = - 111 . 95 C + 273 . 15 = 161 . 2 K P 1 = 35 Torr P 2 = 253 Torr T 2 = - 83 . 6 C + 273 . 15 = 189 . 55 K P 0 = 1 atm = 760 Torr Using the Clausius-Clapeyron equation, ln P 1 P 2 = Δ H vap R 1 T 2 - 1 T 1 R ln P 1 P 2 1 T 2 - 1 T 1 = Δ H vap Δ H vap = ( 8 . 314 J K · mol ) ln 35 Torr 253 Torr 1 189 . 55 K - 1 161 . 2 K = 17724 . 8 J / mol = 17 . 7248 kJ / mol . 003 (part 1 of 1) 10 points The vapor pressure of CH 3 COCH 3 (propa- none or acetone), is 67 Torr at 0 C and 222 Torr at 25 C. What is the standard entropy of vaporization of propanone? Correct answer: 0 . 0985919 J K · mol . Explanation: T 1 = 0 C + 273 . 15 = 273 . 15 K
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Thompson, Margaret – Homework 1 – Due: Sep 11 2007, 11:00 am – Inst: James Holcombe 2 P 1 = 67 Torr P 2 = 222 Torr T 2 = 25 C + 273 . 15 = 298 . 15 K P 0 = 1 atm = 760 Torr Using the Clausius-Clapeyron equation, ln P 1 P 2 = Δ H vap R 1 T 2 - 1 T 1 R ln P 1 P 2 1 T 2 - 1 T 1 = Δ H vap Δ H vap = ( 8 . 314 J K · mol ) ln 67 Torr 222 Torr 1 298 . 15 K - 1 273 . 15 K = 32445 . 7 J / mol . The standard entropy of vaporization is ob- tained from R ln P 1 P 0 = - Δ H vap T 1 + Δ S vap Δ S vap = R ln P 1 P 0 + Δ H vap T 1 = 8 . 314 J K · mol ln 67 Torr 760 Torr + 32445 . 7 J / mol 298 . 15 K = 98 . 5919 J K · mol = 0 . 0985919 J K · mol . 004 (part 1 of 1) 10 points The vapor pressure of hydrogen azide (HN 3 ) is 58 Torr at - 22 . 75 C and 512 Torr at 25 C. What is the standard free energy of vaporiza- tion of hydrogen azide? Correct answer: 979 . 118 C. Explanation: T 1 = - 22 . 75 C + 273 . 15 = 250 . 4 K P 1 = 58 Torr T 2 = 25 C + 273 . 15 = 298 . 15 K P 2 = 512 Torr P 0 = 1 atm = 760 Torr Using the Clausius-Clapeyron equation, ln P 1 P 2 = Δ H vap R 1 T 2 - 1 T 1 R ln P 1 P 2 1 T 2 - 1 T 1 = Δ H vap Δ H vap = ( 8 . 314 J K · mol ) ln 58 Torr 512 Torr 1 298 . 15 K - 1 250 . 4 K = 28310 J / mol .
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