Phase Changes and Heat Calculations

Obj. 1…Vapor Pressure o Vapor pressure (VP) is the P exerted at the surface of a liquid by particles trying to escape the liquid.

Obj. 2…VP and Temperature o As T, KE will. (direct relationship) (direct relationship) o If liquid molecules gain enough KE, they will overcome the intermolecular bonds that hold them together.  become a gas

Obj. 3…Boiling/Melting Points o Boiling point (BP) = the temp. at which the VP of a liquid is equal to the external pressure. o BP is directly related to atmospheric pressure.  a pot of water in Denver (mountains…low pressure) will boil at a lower temp. than a pot of water in Houston (sea level). o normal BP is always measured at sea level. o Melting Point (MP) = the temp. at which a solid turns into a liquid.  KE increases pressure enough to break intermolecular bonds.  as KE of solid increases, molecules begin to vibrate  if vibrations are strong enough, molecules will break away from their fixed positions liquid

Obj. 3…Boiling/Melting Points

Obj. 4… Freezing/Melting and Boiling/Condensation Points o The freezing point (FP) and melting point (MP) of a substance occur at the same temp.  FP (liquid solid) is used as a substance loses KE (heat) molecules get slower and lock into place. molecules get slower and lock into place.  MP (solid liquid)is used as a substance gains KE (heat)  MP (solid liquid) is used as a substance gains KE (heat) molecules break away from solid bonds. molecules break away from solid bonds. o The boiling point (BP) and condensation point (CP) of a substance occur at the same temp.  BP (liquid gas) is used as a substance gains KE (heat) molecules break away from liquid bonds. molecules break away from liquid bonds.  CP (gas liquid)is used as a substance loses KE (heat)  CP (gas liquid) is used as a substance loses KE (heat) molecules get slower and more attracted to each other. molecules get slower and more attracted to each other.

Obj. 5…Sublimation o sublimation = a solid changing directly into a vapor (gas) w/out passing through the liquid stage. o only occurs in certain solids with high VP. o Ex…naphthalene (moth balls), CO 2 (dry ice) etc…

Obj. 6…Boiling vs. Evaporation o for a liquid to boil, the VP of the liquid MUST = the atmospheric pressure. o to accomplish this, we can…  increase temp. of liquid ( KE = VP)  reduce atmospheric pressure ** entire pot of water boils at the same time!!! o evaporation occurs w/out changing temp. or pressure.  surface molecules exposed to more KE (sun/atmosphere) than particles below surface.  this is a cooling process (high KE molecules leave, low KE molecules stay). ** only occurs at the SURFACE of a liquid!!!

Obj. 7…Volatile vs. Non-Volatile o volatile substances evaporate very easily and boil at low temps. o vapors are typically very strong and distinct. o Ex…ammonia, gasoline, rubbing alcohol, acetone o non-volatile substances contain stronger bonds and do not evaporate easily. o Ex…molasses, glue, paint

Obj. 8…KE and Intermolecular Bonds o As KE, the strength of intermolecular bonds will. (inverse relationship) (inverse relationship)  heat causes KE to  enough movement eventually breaks intermolecular bonds.  heat causes KE to  molecules get slower, move less.  eventually lock into place.  bond strength increases.

Obj. 9…Heating/Cooling Curves Time Temperature (KE) Solid Melting Boiling MP BP Heating Curve: KE is KE is (melting and boiling) Liquid Gas (vapor) Plateaus = phase changes! Plateaus = phase changes!  temp. remains constant until EVERY molecule changes phase.

Obj. 9 cont… Cooling Curve: KE is KE is (condensation and freezing) Time Temperature (KE) Solid Gas (vapor) Liquid Condensation Freezing CP FP

Obj. 10…Vocabulary Obj. 11…Heat Calculations o as a substance changes phases, temp. remains constant  plateaus on heating/cooling curves. until all molecules have completed the change! o to calculate heat gained/lost during a phase change … total heat (q) = mass x H (f or v) total heat (q) = mass x H (f or v) heat of fusion… use when melting! heat of fusion… use when melting! heat of vaporization… use when boiling! heat of vaporization… use when boiling! ** Both H f and H v will be given to you!

Obj. 11 cont… o Ex… How many kilojoules (kJ) of heat are required to melt a How many kilojoules (kJ) of heat are required to melt a 10.0 gram ice cube at 0°C and kPa? (Hf° = kJ/g) total heat (q) = mass x H f total heat (q) = mass x H f total heat (q) = 10.0 x total heat (q) = 10.0 x kJ/g = 3.34 kJ = 3.34 kJ o This can be used for any phase change, as long as temp. remains constant (plateaus).

Obj. 12 and 14…Temp. Changes o To calculate a temp. change (slope)… heat (q) = m x C p x Δ T mass specific heat capacity **given…changes w/ phases!** change in temp…(T f – T i ) o Ex… The temp. of a 64.0g sample of H 2 O is raised from 20.0°C The temp. of a 64.0g sample of H 2 O is raised from 20.0°C to 40.0°C. How much heat is required? (C p water = J/g°C) heat (q) = m x C p x ΔT = 20 q = 64 x x 20° = 5360 joules = 5360 joules

Obj. 12 and 14 cont… o We can combine the phase change eq. and the Δ temp. eq. to find the total heat absorbed on a heating curve. o Ex… How much heat is needed to change 32.0 grams of H 2 O at How much heat is needed to change 32.0 grams of H 2 O at -30.0°C to 45.0°C? time temp. -30°C 45°C ** must do 3 separate equations… ** must do 3 separate equations… 1) Δ temp. from -30°C to 0°C 2) Phase change (melting) 1) 32.0 x 3) Δ temp. from 0°C to 45°C (m x C p x Δ T) (m x H f ) 2016 J 2) 32.0 x J 3) 32.0 x J add together = J x 30 = = x 45 = (H f water = J/g, C p ice = 2.1 J/g°C, C p water = J/g°C)

o +q (heat) = endothermic reaction Obj. 12 and 14 cont… o -q (heat) = exothermic reaction (heat absorbed) (heat released) o graphically… R P Enthalpy (ΔH) Time Time R P * R have less heat than P = * R have less heat than P = endothermic! * R have more heat than P = lost heat = lost heat = exothermic! gained heat =

Obj. 12 and 14 cont… o to calculate heat of reaction (H r ) from a graph… H r = ΔH products – ΔH reactants H r = ΔH products – ΔH reactants (ΔH) Time R P Time R P 154 kJ 561 kJ H r = 154 – 561 = H r = 154 – 561 = -407kJ -407kJ -q = exothermic! -q = exothermic! 45.2 kJ kJ H r = – 45.2 = H r = – 45.2 = 68.3 kJ 68.3 kJ +q = endothermic! +q = endothermic!

Obj. 13 and 16…Phase Diagrams o a phase diagram represents the P-T relationships b/n the different phases of the same substance. o each point on the curves shows the T and P at which two phases are in equilibrium. (conditions for phase changes to occur!)

Obj. 13 and 16 cont… o the point at which all 3 curves intersect = triple point.  represents T and P at which all 3 phases exist simultaneously!  triple point for water is 0.016°C and 0.61 kPa  every substance has its own triple point.

Obj. 15…Liquefying Gases o two ways to liquefy (condense) a gas…  atmospheric pressure. VP of gas would be lower than atmospheric pressure VP of gas would be lower than atmospheric pressure  temperature of gas. KE of gas decreases causing bond strength to increase KE of gas decreases causing bond strength to increase