Phase Changes and Heat Calculations

1. Phase Changes and Heat Calculations

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

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

4. Obj. 3…Boiling/Melting Points • Boiling point (BP) = the temp. at which the VP of a liquid is equal to the external pressure. • KE increases pressure enough to break intermolecular bonds. • 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). • normal BP is always measured at sea level. • Melting Point (MP) = the temp. at which a solid turns into a liquid. • as KE of solid increases, molecules begin to vibrate • if vibrations are strong enough, molecules will break away from liquid their fixed positions

5. Obj. 3…Boiling/Melting Points

6. Obj. 4… Freezing/Melting and Boiling/Condensation Points • 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. • MP (solid liquid)is used as a substance gains KE (heat) • molecules break away from solid bonds. • 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. • CP (gas liquid)is used as a substance loses KE (heat) • molecules get slower and more attracted to each other.

7. Obj. 5…Sublimation • sublimation = a solid changing directly into a vapor (gas) w/out passing through the liquid stage. • only occurs in certain solids with high VP. • Ex…naphthalene (moth balls), CO2 (dry ice) etc…

8. Obj. 6…Boiling vs. Evaporation • for a liquid to boil, the VP of the liquid MUST = the atmospheric pressure. • to accomplish this, we can… • increase temp. of liquid ( KE = VP) • reduce atmospheric pressure ** entire pot of water boils at the same time!!! • 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!!!

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

10. Obj. 8…KE and Intermolecular Bonds • As KE , the strength of intermolecular bonds will . (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.

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

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

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

14. Obj. 11 cont… • Ex… How many kilojoules (kJ) of heat are required to melt a 10.0 gram ice cube at 0°C and 101.3 kPa? (Hf° = 0.334 kJ/g) total heat (q) = mass x Hf total heat (q) = 10.0 x 0.334 kJ/g = 3.34 kJ • This can be used for any phase change, as long as temp. remains constant (plateaus).

15. Obj. 12 and 14…Temp. Changes • To calculate a temp. change (slope)… heat (q) = m x Cp x Δ T mass change in temp…(Tf – Ti) specific heat capacity **given…changes w/ phases!** • Ex… The temp. of a 64.0g sample of H2O is raised from 20.0°C to 40.0°C. How much heat is required? (Cp water = 4.184 J/g°C) heat (q) = m x Cp x ΔT 40-20 = 20 4.184 x q = 64 x 20° = 5360 joules

16. add together = Obj. 12 and 14 cont… • We can combine the phase change eq. and the Δ temp. eq. to find the total heat absorbed on a heating curve. • Ex… How much heat is needed to change 32.0 grams of H2O at -30.0°C to 45.0°C? (Hf water = 333.9 J/g, Cp ice = 2.1 J/g°C, Cp water = 4.184 J/g°C) ** must do 3 separate equations… (m x Cp x ΔT) 1) Δ temp. from -30°C to 0°C (m x Hf) 2) Phase change (melting) (m x Cp x ΔT) 3) Δ temp. from 0°C to 45°C 45°C 3 temp. 1) 32.0 x 2.1 x 30 = 2016 J 2 1 2) 32.0 x 333.9 = 10684.8 J -30°C 6024.96 J 3) 32.0 x 4.184 x 45 = 18700 J time

17. Obj. 12 and 14 cont… • +q (heat) = endothermic reaction (heat absorbed) (heat released) • -q (heat) = exothermic reaction • graphically… *R havemore heat than P = *R haveless heat than P = lost heat = gained heat = exothermic! endothermic! R P Enthalpy (ΔH) Enthalpy (ΔH) R P Time Time

18. Obj. 12 and 14 cont… • to calculate heat of reaction (Hr) from a graph… Hr = ΔHproducts – ΔHreactants (ΔH) (ΔH) R P 561 kJ 113.5 kJ P R 154 kJ 45.2 kJ Time Time Hr = 113.5 – 45.2 = 68.3 kJ Hr = 154 – 561 = -407kJ -q = exothermic! +q = endothermic!

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

20. Obj. 13 and 16 cont… • 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.

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