Changes of State H2O (g) H2O (s) H2O ()

Phase Changes

Phase Changes Evaporation molecules at the surface gain enough energy to overcome IMF Volatility measure of evaporation rate depends on temp & IMF

Phase Changes Equilibrium trapped molecules reach a balance between evaporation & condensation

Phase Changes temp v.p. IMF v.p. Vapor Pressure pressure of vapor above a liquid at equilibrium v.p. depends on temp & IMF directly related to volatility temp temp v.p. IMF v.p.

Phase Changes Patm b.p. IMF b.p. Boiling Point temp at which v.p. of liquid equals external pressure depends on Patm & IMF Normal B.P. - b.p. at 1 atm Patm b.p. IMF b.p.

Phase Changes IMF m.p. Melting Point equal to freezing point Which has a higher m.p.? polar or nonpolar? covalent or ionic? polar ionic

Phase Changes Sublimation solid  gas v.p. of solid equals external pressure EX: dry ice, mothballs, solid air fresheners

Phase Diagrams Show the phases of a substance at different temps and pressures.

Phase Changes From To …is Called Energy is: Solid Liquid Melting Absorbed Freezing Released Vapor Boiling or Vaporization Condensation Sublimation Deposition The temperature at which a substance melts/freezes at standard pressure (1 atm) is known as the Normal melting point or Normal freezing point. For water, this is 0 °C. The temperature at which a substance boils/condenses at standard pressure is known as the Normal boiling point or Normal condensation point. For water, this is 100 °C.

The Triple point is the condition of temperature and pressure in at which all three phases exist together at equilibrium. For water, this is 0.0099 °C and 0.006 atmospheres. The Critical Temperature, Tc is the temperature beyond which the solid and liquid phases of the substance cannot exist. Put another way, above the critical temperature, the substance can only be found as a gas. For water, this temperature is 373.99 °C. The Critical Pressure, Pc is the pressure above which the substance cannot exist as a gas. For water, this pressure is 217.75 atmospheres. The Critical Point is the point defined by the critical temperature and the critical pressure. The slope of the line between the solid and liquid phase provides important information about the substance: If the slope is negative (as it is for water), then the substance is more dense as a liquid than it is as a solid. If the slope is positive (as it is for most substances), then the substance is more dense as a solid than it is as a liquid.

Heating Curves Gas - KE  Boiling - PE  Liquid - KE  Melting - PE  Solid - KE 

Every Substance has a unique heating curve Shape is the same Melting and Boiling points are different Each phase has its own specific heat capacity

Heating Curves Temperature Change change in KE (molecular motion) depends on heat capacity Heat Capacity energy required to raise the temp of 1 gram of a substance by 1°C “Volcano” clip - water has a very high heat capacity

Heating Curves Phase Change change in PE (molecular arrangement) temp remains constant Heat of Fusion (Hfus) energy required to melt 1 gram of a substance at its m.p.

Heating Curves Heat of Vaporization (Hvap) energy required to boil 1 gram of a substance at its b.p. usually larger than Hfus…why? EX: sweating, steam burns, the drinking bird

Problem Solving Draw heating curve Mark starting and ending points Calculate EACH step individually Add energies for each step You must know the following information: Cp (ice) = 2.06 J/goC Cp (water) = 4.184 J/goC Cp (steam) = 2.02 J/goC This is ONLY for H2O. Every compound has their own numbers!

Heating and cooling curve for water heated at a constant rates. A-B = Solid ice, temperature is increasing as HEAT energy is absorbed Particles gain kinetic energy, vibration of particles increases. Ice

H2O (s)  H2O () energy required  6 kJ/mol B-C = 2. At 0 °C a phase change begins: Moving from left to right along LEG ‘B’, ice is melting to form liquid water Moving from right to left along LEG ‘B’, liquid water is freezing to form ice The distance of LEG ‘B’ along the Heat axis (x-axis) is known as the Heat of Fusion d) Note that temperature remains constant during a phase change as energy is used to break inter-molecular bonds. 0ºC

C-D = temperature starts to rise once all the solid has melted C-D = temperature starts to rise once all the solid has melted. Particles gain kinetic energy as heat absorbed by water is no longer going toward changing the phase of the substance. Liquid water

a) Moving from left to right along LEG ‘D’, water is boiling to form water vapor b) Moving from right to left along LEG ‘D’, water vapor is undergoing condensation to form liquid water The distance of LEG ‘D’ along the Heat axis (x-axis) is known as the Heat of Vaporization D-E = Liquid starts to vaporize, turning from liquid to gas at 100oC. The temperature remains constant as energy is used to break inter-molecular forces. H2O ()  H2O (g) energy required  41 kJ/mol 100ºC

E-F = temperature starts to rise once all liquid is vaporized E-F = temperature starts to rise once all liquid is vaporized. Gas particles gain kinetic energy. steam

Energy Requirements for changing state: In ice the water molecules are held together by strong intermolecular forces. The energy required to melt 1 gram of a substance is called the heat of fusion ( H fus) For ice it is 334J/g The energy required to change 1 gram of a liquid to its vapor is called the heat of vaporization (Hvap ) For water it is 2260 J/g H (delta H) is the change in energy or heat content.

What is specific heat capacity? The amount of energy required to change the temperature of one gram of a substance by 1C . 10 C 11 C Another name for specific heat is a calorie (1 calorie = 4.184 Joules) Specific heat capacity of liquid water (H2O (L) ) is 4.18 J /gC. Water (s) = 2.03 J/ g C Water (g) = 2.0 J/ g C

Calculating Energy Requirements using the equation: Q = m x T x Cp Q = energy (heat) required Cp = specific heat capacity m = mass of the sample T = change in temperature in C EXAMPLE: How much energy does it take to heat 10g of water from 50 to 100 C ? Q = m  T x Cp Q = (10g)  (4.184 J /g C )  (50 C) = 2.1  10 3 J Specific heat capacity of water = 4.184 J/ gC

Problem How much energy is required to heat 25 g of liquid water from 25C to 100C and change it to steam?

Step 1: Calculate the energy needed to heat the water from 25C to 100C Q = m  Cp  T Q = 25g  4.184 J/ g C  75 C = 7.8  10 3 J

Step 2: Vaporization: Use the vap H to calculate the energy required to vaporize 25g of water at 100C Q=m x  H vap .25g  1mol H2O / 18g mol-1 H2O = 1.4 mol H2O vap H (H2O) = 1.4 mol H2O  40.6kJ/mol = 57 kJ

 H vap (H2O) = 2260 J/g Q = 25g  2260J/g= 57000J

Total energy change is: 7800J + 57000J = 64800 J