75 g of acetone ch 3 coch 3 δh vap 310 kjmol at 25c

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How much energy is required to evaporate 1.75 g of acetone (CH 3 COCH 3 , ΔH vap = 31.0 kJ/mol) at 25°C? Answer: (31.0 kJ/mol)(1.75 g)(1 mol / 58.08 g) = 0.934 kJ 1) A chemical reaction produces 0.132 g of water which is maintained at 50.0°C in a closed 525-mL flask. Will the water be present as vapor only or as a liquid↔vapor equilibrium? (the vapor pressure of water at 50.0°C = 92.5 torr) Answer: If the water is present as vapor only, then: P = nRT/V = [(0.132 g / 18.02 g/mol)(0.0821)(323K)] (0.525 L) = 0.370 atm or (0.370 atm x 760 torr/atm) = 281 torr Since 281 torr > 92.5 torr, some of the water must be liquid, and the water is present in the flask as a liquid↔vapor equilibrium at 92.5 torr and 50.0°C
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1) What mass of water from Problem 2 above is vapor? What mass is liquid? Answer: n = PV/RT = (92.5 torr / 760 torr/atm)(0.525 L) (0.0821)(323K) = 0.002 mol H 2 O x 18.02 g/mol = 0.036 g H 2 O vapor So, 0.132 g total H 2 O – 0.036 g H 2 O vapor = 0.096 g H 2 O liquid 1) Isooctane has a normal BP of 99°C and ΔH vap = 35.76 kJ/mol. What is the vapor pressure of isooctane at 25°C? Answer: ln (P 1 /P 2 ) = [(ΔH vap / R)(1/T 2 – 1/T 1 )] ln (760/P 2 ) = [(35,760 J / 8.314)(1/298 – 1/372)] ln (760/P 2 ) = 2.87 760/P 2 = e 2.87 = 17.6, so P 2 = 43.2 torr
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Melting Freezing H 2 O ( s ) H 2 O ( l ) The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
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Molar heat of fusion ( H fus ) is the energy required to melt 1 mole of a solid substance at its freezing point.
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Again, during a dynamic phase equilibrium, as in the case of a liquid freezing to a solid and vice-versa, where there are two phases present simultaneously, there is no ΔT during this process. The temperature remains constant until the one of the phases is gone, at which time the temperature may rise or fall again. When a hard freeze is predicted, citrus farmers will sprinkle their fruit trees with water, so that when both the liquid and solid phases of the water are present, the temperature will remain unchanged at 0°C around the fruit. This temperature stability is only maintained if BOTH phases are present, however. (While external pressure influences BP, it does not influence MP because liquids have very low compressibilities) Now let’s look at these phase-change stages visually in the form of a heating-cooling curve……
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Heating Curve
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There are 5 stages on the preceding slide’s heating-cooling curve: 1) Vapor cools – during this stage, the heat transfer is: q = m•s•ΔT 2) Condensation (vapor-liquid eq.) – during this stage, the heat transfer is: q = n•ΔH vap 3) Liquid cools – during this stage, the heat transfer is: q = m•s•ΔT 4) Freezing (liquid-solid eq.) – during this stage, the heat transfer is: q = n•ΔH fus 5) Solid cools – during this stage, the heat transfer is: q = m•s•ΔT Add all of the above up (as per Hess’s Law), and you’ll have the total heat transfer from the system (the substance) to the surroundings during this cooling process. The ΔH associated with this heat transfer will be negative since the cooling process is exothermic.
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