013125 CHEM 102 SP25 Unit 02 pre

Unit 2: Phase Changes

Page 1

• Topic introduction: Phase Changes

Page 2: Recall Phase Changes

• Melting is also known as fusion.

• Enthalpy Changes:

  • ΔHvap and ΔHfus are endothermic processes.

  • ΔHcondensing and ΔHfreezing are exothermic processes.

  • Relationships:

    • ΔHcondensing = -ΔHvap

    • ΔHfreezing = -ΔHfus

    • ΔHvap + ΔHfus = ΔHsublimation

    • ΔHcond + ΔHfrz = ΔHdeposition

Page 3: Identifying Heating and Cooling Points

1. Determine if the process involves heating or cooling.

2. Mark melting/freezing point on the graph.

3. Mark boiling/condensation point on the graph.

4. Label points Ti (initial temperature) and Tf (final temperature).

5. Identify the relevant segments of the curve:

   • Heating or Cooling (sloped): q = msΔT

   • Phase Change (flat): q = nΔH

     • Condensing Gas <=> Freezing Liquid <=> Melting Solid <=> Vaporizing Liquid

Page 4: Using a Heating Curve

• Process: Heat is added (endothermic).

• Graph Points:

  • Point A = Ti

  • Point F = Tf

• Melting and boiling points are indicated on the curve. The location of Ti and Tf can vary along the curve.

Page 5: Using a Cooling Curve

• Process: Heat is removed (exothermic).

• Just like the heating curve, the positions of Ti and Tf may vary.

Page 6: Complete Vaporization of a Liquid

• When temperature exceeds the melting point, the location of Point A is positioned between C and D on the curve (boiling process).

Page 7: Calculation of Heat (q)

• Formulas:

  • q = msΔT indicates heat during heating/cooling processes:

    • Units: Must match when adding segments (J or kJ).

  • q = nΔH indicates heat during phase change.

• Multiple variations of these equations exist, where you need to pay attention to units used (J or kJ).

Page 8: General Information and Example Problem

• Key principles:

  • Tf > Ti ➔ heating (endothermic)

  • Tf < Ti ➔ cooling (exothermic)

• Given parameters for calculations:

  • Melting Point, Boiling Point, ΔH values, Specific heat (s) for all phases.

• Common calculation prompts:

  • Finding heat for temperature conversion of a substance.

  • Heat required for complete phase transition.

  • Enthalpy change calculations.

Page 9: Example Calculation

• Scenario: Convert 1.00 mol of ice at -25°C to water vapor at 125°C under constant pressure.

• Values Needed:

  • ssolid = 2.09 J/g°C

  • sliquid = 4.18 J/g°C

  • sgas = 1.84 J/g°C

  • ΔHfus = 6.01 kJ/mol

  • ΔHvap = 40.67 kJ/mol

Page 10: Phase Diagrams

• Key Components:

  • Triple Point: Unique temperature and pressure where all three phases exist.

  • Critical Point: Temperature and pressure beyond which separate phases of liquid and gas do not exist; a supercritical fluid emerges.

Page 11: Origin of Equilibrium Vapor Pressure

• Definition: Pressure in the headspace above the liquid is the equilibrium vapor pressure.

Page 12: Understanding Vapor Pressure

• The initial volume (V) and number of moles (n) remain constant at equilibrium.

• Applied Ideal Gas Law in context: P = nRT/V, explaining where equilibrium vapor pressure arises.

Page 13: Boiling Point

• Definition: Boiling point of a substance is when its vapor pressure equals atmospheric pressure.

• Normal Boiling Point: Boiling point at 1 atm pressure.

Page 14: Vapor Pressure and Intermolecular Forces

• Relationship: Increased intermolecular forces lead to decreased vapor pressure and increased boiling points.

Page 15: Factors Influencing Vapor Pressure

• Factors include:

  • Temperature

  • Enthalpy of vaporization (ΔHvap)

  • Strength of intermolecular forces affects how easily molecules escape into the gas phase.

Page 16: Clausius-Clapeyron Equation

• Equation Form:

  • ln(P2/P1) = -ΔHvap/R (1/T2 - 1/T1)

• This equation will be available for use on the exam.

Page 17: Vapor Pressure Example Problem

• Given:

  • Vapor pressure of benzene at 25.0°C is 95.1 mm Hg.

  • Find vapor pressure at 62.0°C where ΔHvap for benzene is 31.0 kJ/mol.