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Phase Changes Courtesy www.lab-initio.com The BIG Idea • Students know energy is released when a material condenses or freezes and is absorbed when a material evaporates or melts. constant Temperature remains __________ during a phase change. Water phase changes Phase Changes Energy must be absorbed to change from a solid to a liquid; heat of fusion or enthalpy of fusion, DHfus Energy must be absorbed to change from a liquid to a gas; heat of vaporization or enthalpy of vaporization, DHvap Phase Changes For water, the DHfus = 6.02 kJ/mol the DHvap = 40.7 kJ/mol Why is the DHvap much higher than DHfus? Example #1 How much energy is needed to convert 25g of ice at -15C to 200C steam? Plan: Follow the heating curve to determine where the energy goes at each step in the process Can you identify the FIVE steps in the process? Example #1 Continued Step 1: heat ice from -15C to 0C Cice = 2.03J/gC q = mCDT q = 25g (2.03J/gC)(15C) = 761.25J = .761kJ Step 2: phase change (s) (l) 25g (1mol/18.02g) (6.02kJ/mol) = 8.35kJ Step 3: heat water from 0C to 100C Cwater = 4.18J/gC q=mCDT q=25g (4.18J/gC)(100C)= 10450J = 10.4kJ Example #1 Continued Step #4: phase change (l) (g) 25g (1mol/18.02g)( 40.7kJ/mol) = 56.5kJ Step #5: heat steam from 100C to 200C q=mCDT q = 25g (2.0J/gC)(100C) = 5000J = 5kJ Csteam = 2.0J/gC Total heat required: 0.761kJ + 8.35kJ + 10.4kJ + 56.5kJ + 5kJ = 81kJ Now let’s try one: The molar heat of fusion for benzene, C6H6, is 9.92kJ/mol. The molar heat of vaporization is 30.7kJ/mol. Calculate the heat required to melt 8.25g of benzene at its normal melting point. 1.05kJ Calculate the heat required to vaporize 8.25g of benzene at its normal boiling point. 3.24kJ Why is the heat of vaporization more than 3 times the heat of fusion? Another Example You have a 75g sample of water at 125C. What phase(s) are present when 215kJ of energy is removed from the sample? Hint: Identify the steps on the cooling curve that may occur: Energy to cool from 125C to 100C Energy to phase change from a vapor to a liquid Energy to cool from 100C to 0C Energy to phase change from a liquid to a solid Energy to cool the solid A mixture of H2O(l) and H2O(s) will be present Vapor Pressure Pressure of the vapor present when equilibrium is achieved between the rate of vaporization and the rate of condensation. At the boiling point, the Patm = Pvapor As the vapor pressure on a pot of water is reduced, the energy needed to boil that water is also reduced. Pressure Cooker: By increasing the vapor pressure, additional energy is needed for the water to boil, therefore the water can boil at temperatures above 100C. Effect of Pressure on Boiling Point Water Phase Diagram Represents phases as a function of temperature and pressure. Critical temperature: temperature above which the vapor can not be liquefied. Critical pressure: pressure required to liquefy AT the critical temperature. Critical point: critical temperature and pressure (for water, Tc = 374°C and 218 atm). Phase changes by Name Carbon dioxide Phase Diagram for Carbon dioxide Carbon Phase Diagram for Carbon Vapor Pressure vs. Temperature As the temperature increases, a greater number of molecules have sufficient kinetic energy to convert from the liquid to the vapor phase. There is a nonlinear relationship between the vapor pressure of a liquid and temperature. Vapor Pressure vs. Temperature The Clausius – Clapeyron Equation • A mathematical expression which relates the variation of vapor pressure to temperature • ln P = (-DHvap/RT) + C where C is a constant • IMPORTANCE: • When the ln P is plotted vs (1/T) you create a line where the slope is equal to the –DHvap/R • Which means you can calculate the heat of vaporization from the slope of the line. • R = 8.314 J/Kmol Convert all Temps to Kelvin Example #1 Calculate the heat of vaporization of a compound that has a vapor pressure of 560 torr at 25C and a normal boiling point of 45C. Hint: normal BP occurs at 1atm (760 torr) Variation of the C-C Equation is used: ln (P1/P2) = (DHvap/R )(1/T2 – 1/T1) ln(560torr/760torr) = (DH/8.314J/Kmol) (1/318 – 1/298) 12033J, 12kJ = DH Example #2 In Breckenridge, CO, the typical atmospheric pressure is 520 torr. What is the boiling point of water (DH=40.7kJ/mol) in Breckenridge? Hint: 760torr BP=100C / 373K ln(520torr/760torr)= (40,700J mol-1 /8.314)(1/373 - 1/T) -0.379 = 4896(1/373 - 1/T) -7.75 E -5 = (1/373 - 1/T) 1/T = 2.76 E -3 T = 363K Example #3 The temperature inside a pressure cooker is 115C. Calculate the vapor pressure of water inside the pressure cooker. ln (P/1atm) = (40,700kJ mol-1 / 8.314)(1/373 – 1/388) lnP = 0.51 P = e.51 = 1.7atm What would be the temperature inside the pressure cooker if the vapor pressure of water was 3.50atm? ln(3.50/1.00) = (40.7kJ mol-1 / 8.314)(1/373 – 1/T) 413K = T Example #4 The normal boiling point for acetone is 56.5C . At an elevation of 5600ft, the atmospheric pressure is 630 torr. What would be the boiling point of acetone at this elevation? (DHvap = 32.0kJ/mol). ln(760/630) = (32kJ mol-1/8.314)(1/T – 1/330) 325K = T What would be the vapor pressure of acetone at 25C at this elevation? ln(630/P) = (32kJ mol-1/8.314)(1/298 – 1/325) P = 221 torr