21-2-2011

1. 21-2-2011

2. Clausius – Clapeyron Equation • This equation is a relation between DHvap and pressure at a certain Temperature.

4. Example 1: Water has a vapor pressure of 24 mmHg at 25oC and a heat of vaporization of 40.7 kJ/mol. What is the vapor pressure of water at 67oC? • Solution: Simply use the Clausius-Clapeyron Equation to figure out the vapor pressure. We have to be a bit careful about the units of R: the units we're using are kJ, so R = 8.31x10-3 kJ/mol K.ln(P2/P1) = -DHvap/R * (1/T2- 1/T1) ln(P2/24) = - 40.7 kJ/8.31x10-3 kJ/mol K *(1/340- 1/298)ln(P2/24) = 2.03P2/24 = 7.62P2 = 182 mmHg

5. Example 2: An unknown liquid has a vapor pressure of 88 mmHg at 45oC and 39 mm Hg at 25oC. What is its heat of vaporization? • Solution: use the Clausius-Clapeyron Equation. Here, the only thing we don't know isDHvap ln(88/39) = (DHvap/8.31x10-3){(1/318) – (1/298)}DHvap = 32.0 kJ

6. Liquid-solid equilibrium • A solid can be transformed into a liquid at a specific temperature called the melting point. • Melting occurs as temperature increases the kinetic energy of the molecules and thus make them move around. • The melting point (or freezing point) is the temperature at which solid and liquid phases coexist in equilibrium.

7. Normal melting or freezing points • The temperature at which both solid and liquid phases coexist at equilibrium at 1 atm is called the normal melting point (normal freezing point). Ice Water

8. Molar heat of fusion (DHfus) • The energy required to melt 1 mole of a solid is called the molar heat of fusion (DHfus). • The molar heat of fusion is definitely smaller than the molar heat of vaporization since vaporization requires complete removal of the molecules from the surface while fusion only requires rearrangement from solid to liquid.

11. Solid-vapor equilibrium • Solids can undergo direct evaporation, and thereby solids are said to have a vapor pressure. Ice vapor Sublimation: is a process in which molecules can go directly from the solid to the vapor phase. The reverse process is called “deposition”. Naphthalene (mothballs) and I2 are examples.

12. Molar heat of sublimation (DHsub) • The energy required to sublime 1 mole of a solid is called the molar heat of sublimation. DHsub = DHfus + DHvap • This is, in fact, a manifestation of Hess’s law, where the amount of energy needed to transform a solid to vapor is the same whether we go directly or in steps by first transforming the solid to a liquid and then transforming the liquid to vapor.

13. Example Find the energy in kJ necessary to melt 1.00 g of ice. DHfus of ice = 5.98 kJ/mol Solution Melting of one mole of ice requires 5.98 kJ, therefore, find moles of ice present in 1.00 g and find energy required. Energy needed = DHfus * Number of moles Energy needed = 5.98 kJ/mol * {1.00 g/(18.0 g/mol)} = 0.332 kJ

14. Heating and Cooling Curves When heat is added to a solid its temperature will rise till the melting point is reached where the temperature stays constants till the entire solid is converted to liquid. When extra heat is added to the system, the temperature of the liquid starts to rise till the boiling point is reached where the temperature stays constant till all the liquid is converted to vapor. This can be represented by a heating curve as below:

16. Example Find the energy needed to convert 36.0 g of ice at -23 oC into vapor at 120 oC. DHfus = 5.98 kJ/mol and DHvap = 44.0 kJ/mol, specific heat of water is 4.184 J g-1oC-1, specific heat of ice is 2.06 J g-1oC-1, and specific heat of steam is 1.99 J g-1oC-1. Solution Number of moles = 36.0/18.0 = 2.00 moles

17. Following the heating curve above, ice should first be • It will require energy to be converted to ice at 0 oC (heat capacity 1) • It will require the heat of fusion to convert to liquid at 0 oC • It will need energy to be converted to liquid at 100 oC, boiling temperature for water (heat capacity 2) • It will require the heat of vaporization to be converted to water vapor at 100 oC • It will require energy to raise the temperature of the vapor to 120 oC (heat capacity 3)

18. Therefore, five energy terms should be summed together. Steps 1, 3, and five can be summed together as heat capacity term: Energy required = (heat capacity term) +  number of moles *(DHfus + DHvap) Energy required = {36.0 g * 0.00206 kJ g-1oC-1 {0 – (-23)} oC + 36.0 g * 0.004184 kJ g-1oC-1 {100 – (0)} oC + 36.0 g * 0.00199 kJ g-1oC-1 {120 – (100)} oC} + 2.00 mol * (5.98 + 44.0)kJ/mol = 118.2 kJ

19. Phase Diagrams • A phase diagram is a graph that represents and summarizes conditions under which a substance exist as liquid, solid, or gas. • We will only look at the phase diagrams of water and carbon dioxide

20. A phase diagram

21. Triple point • The triple point is the point at which all three phases coexist at equilibrium. • For water, the triple point occurs at 0.01 oC, and 0.006 atm.

22. Water phase diagram

24. Example Describe any changes in the phases present when water is: • Kept at 0 oC while pressure is increased from that at point 1 to that at point 5. • Kept at 1 atm while the temperature is increased from that at point 6 to that at point 9.

27. CO2 Phase diagram

28. Answer the following problems: • 1-3, 7, 9, 11-20, 22, 27-29, 31, 54, 56-60, 62, 65, 66, 69, 71-75, 80, 86, 99.