Three States of Matter

The state of matter depends on balance between kinetic energy and intermolecular forces:

Properties of Solids

Characteristics

1. Definite shape and volume

   • Particles in fixed positions
   • Strong intermolecular forces hold structure

2. High density

   • Particles closely packed
   • Little empty space

3. Incompressible

   • Cannot be compressed significantly
   • Particles already at minimum distance

4. Very low compressibility

   • Particles vibrate in place
   • Cannot flow or move past each other

Types of Solids

1. Crystalline Solids

• Ordered, repeating 3D structure
• Sharp melting point
• Examples: NaCl, diamond, ice

Types of crystalline solids:

a) Ionic Solids

• Ions held by electrostatic forces
• Hard, brittle, high melting points
• Conduct when molten/dissolved
• Example: NaCl, CaCO₃

b) Molecular Solids

• Molecules held by IMFs (LDF, dipole-dipole, H-bonding)
• Soft, low melting points
• Do not conduct electricity
• Example: Ice (H₂O), dry ice (CO₂), I₂

c) Covalent Network Solids

• Atoms covalently bonded throughout
• Very hard, very high melting points
• Do not conduct (except graphite)
• Example: Diamond, SiO₂ (quartz), SiC

d) Metallic Solids

• Metal cations in sea of electrons
• Malleable, ductile, lustrous
• Conduct electricity and heat
• Variable melting points
• Example: Na, Fe, Cu, Au

2. Amorphous Solids

• No long-range order
• Broad melting range
• Examples: Glass, rubber, plastics

Phase Diagram Features (Solids)

Triple point: Temperature and pressure where all 3 phases coexist

Normal melting point: Temperature at 1 atm where solid ↔ liquid

Properties of Liquids

Characteristics

1. Definite volume, no definite shape

   • Takes shape of container
   • Particles mobile but close together

2. High density (similar to solids)

   • Particles still close
   • Slightly more space than solids

3. Incompressible

   • Little empty space between particles

4. Can flow

   • Particles slide past each other
   • Viscosity depends on IMF strength

Key Liquid Properties

1. Viscosity

Definition: Resistance to flow

Factors affecting viscosity:

• Stronger IMFs → Higher viscosity
• Larger molecules → Higher viscosity
• Higher temperature → Lower viscosity (more KE overcomes IMFs)

Examples:

• Water: Low viscosity (H-bonding, but small molecule)
• Glycerol: High viscosity (3 OH groups, extensive H-bonding)
• Motor oil: High viscosity (large molecules, strong LDF)

2. Surface Tension

Definition: Energy required to increase surface area of liquid

Cause: Molecules at surface experience net inward force

• Bulk molecules: attracted equally in all directions
• Surface molecules: only attracted downward and sideways
• Creates "skin" on liquid surface

Factors:

• Stronger IMFs → Higher surface tension
• Water has high surface tension (H-bonding)
• Allows some insects to walk on water

3. Capillary Action

Definition: Spontaneous rise of liquid in narrow tube

Two competing forces:

Adhesion: Attraction between liquid and tube surface

• Liquid "sticks" to walls

Cohesion: Attraction between liquid molecules (IMFs)

• Liquid sticks to itself

Result:

• If adhesion > cohesion: Liquid rises (water in glass)
• If cohesion > adhesion: Liquid depressed (mercury in glass)

Meniscus:

• Water: Concave (curves up at edges)
• Mercury: Convex (curves down at edges)

4. Vapor Pressure

Definition: Pressure exerted by vapor in equilibrium with liquid

Dynamic equilibrium:

$\text{Liquid} \rightleftharpoons \text{Vapor}$

Rate of evaporation = Rate of condensation

Factors affecting vapor pressure:

a) Temperature:

• Higher T → Higher vapor pressure
• More molecules have enough KE to escape

b) IMF Strength:

• Stronger IMFs → Lower vapor pressure
• Harder for molecules to escape liquid
• Example: Water (H-bonding) has lower VP than ethanol at same T

Volatile liquids: High vapor pressure (weak IMFs)

• Examples: Acetone, diethyl ether

Nonvolatile liquids: Low vapor pressure (strong IMFs)

• Examples: Glycerol, motor oil

Normal Boiling Point: Temperature where vapor pressure = 1 atm (760 mmHg)

Properties of Gases

Characteristics

1. No definite shape or volume

   • Fills entire container
   • Expands to occupy all available space

2. Very low density

   • Particles far apart
   • Mostly empty space (~99.9%)

3. Highly compressible

   • Can squeeze particles closer
   • Volume decreases with pressure

4. Complete mixing

   • Particles move randomly
   • Rapid diffusion

Kinetic Molecular Theory (KMT)

Five postulates:

1. Gas particles are in constant, random motion

   • Move in straight lines until collision

2. Negligible particle volume

   • Volume of particles ≈ 0 compared to container
   • Mostly empty space

3. No intermolecular forces

   • Particles don't attract or repel (ideal gas assumption)

4. Elastic collisions

   • No energy lost in collisions
   • Total KE conserved

5. Average kinetic energy ∝ Temperature

   • Higher T → Higher average KE
   • $\overline{KE} = \frac{3}{2}RT$

Gas Pressure

Cause: Collisions of gas particles with container walls

Factors affecting pressure:

1. Temperature (↑T → ↑P at constant V, n)

   • Higher KE → more forceful, frequent collisions

2. Volume (↓V → ↑P at constant T, n)

   • Smaller space → more collisions per area

3. Number of moles (↑n → ↑P at constant T, V)

   • More particles → more collisions

4. Molar mass (indirect effect)

   • At same T, all gases have same average KE
   • Lighter gases move faster, heavier gases move slower

Phase Changes

Types of Phase Changes

Endothermic (absorb energy):

1. Melting (fusion): Solid → Liquid
2. Vaporization: Liquid → Gas
3. Sublimation: Solid → Gas (direct)

Exothermic (release energy): 4. Freezing: Liquid → Solid 5. Condensation: Gas → Liquid 6. Deposition: Gas → Solid (direct)

Energy of Phase Changes

Enthalpy of fusion (ΔHfus): Energy to melt 1 mole

• Breaking IMFs in solid structure
• Example: H₂O: ΔHfus = 6.01 kJ/mol

Enthalpy of vaporization (ΔHvap): Energy to vaporize 1 mole

• Breaking all remaining IMFs
• Example: H₂O: ΔHvap = 40.7 kJ/mol

Key relationship:

$\Delta H_{vap} > \Delta H_{fus}$

Why: More IMFs to break going liquid → gas than solid → liquid

Heating Curve

Temperature vs. Heat Added graph:

Features:

1. Solid heating: Temperature increases

   • Slope depends on specific heat of solid

2. Melting plateau: Temperature constant at melting point

   • Energy goes to breaking IMFs, not increasing KE
   • Length depends on ΔHfus

3. Liquid heating: Temperature increases

   • Slope depends on specific heat of liquid

4. Boiling plateau: Temperature constant at boiling point

   • Energy goes to breaking IMFs
   • Length depends on ΔHvap (longer than melting)

5. Gas heating: Temperature increases

   • Slope depends on specific heat of gas

Calculation:

$q = n \cdot \Delta H_{phase}$ (during phase change)

$q = m \cdot c \cdot \Delta T$ (within one phase)

Clausius-Clapeyron Equation

Relates vapor pressure to temperature:

$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Where:

• $P_1$, $P_2$ = vapor pressures
• $T_1$, $T_2$ = temperatures (K)
• $\Delta H_{vap}$ = enthalpy of vaporization
• $R$ = 8.314 J/(mol·K)

Linear form:

$\ln P = -\frac{\Delta H_{vap}}{R} \cdot \frac{1}{T} + C$

Plot: ln P vs. 1/T gives straight line with slope = $-\frac{\Delta H_{vap}}{R}$

Water's Unique Properties

Due to hydrogen bonding:

1. High boiling point (100°C)

   • Much higher than other hydrides (H₂S: -60°C)

2. High specific heat

   • Moderates temperature changes
   • Important for climate, organisms

3. High heat of vaporization

   • Cooling effect of sweating

4. Ice less dense than water

   • Ice floats
   • H-bonding creates open crystal structure
   • Unique among common substances

5. High surface tension

   • Capillary action in plants
   • Water transport in trees

Summary Table

Applications

1. Distillation: Separates based on different boiling points
2. Refrigeration: Uses phase changes to transfer heat
3. Weather: Water vapor pressure affects humidity, precipitation
4. Materials science: Solid structure determines material properties
5. Biological systems: Cell membranes, protein folding affected by IMFs

• Only C-C and C-H bonds
• C-H bonds slightly polar (ΔEN = 0.4)
• Molecular geometry: Nonpolar overall (symmetrical)
• No N-H, O-H, or F-H bonds

IMFs present:

• London Dispersion Forces (LDF) ONLY
• 42 total electrons (more polarizable than water)
• Linear structure → good surface contact

Water (H₂O):

Lewis structure analysis:

• O-H bonds (highly polar, ΔEN = 1.4)
• Bent geometry (2 lone pairs on O)
• Molecular geometry: Polar

IMFs present:

• Hydrogen bonding (strongest)
• Dipole-dipole forces
• London Dispersion Forces (weakest, only 10 electrons)

Step 2: Explain boiling point difference

Expected based on molar mass:

• Hexane: 86 g/mol → more electrons → stronger LDF
• Water: 18 g/mol → fewer electrons → weaker LDF
• Prediction: Hexane should have higher BP

Actual result: Water has higher BP (100°C vs. 69°C)

Explanation:

Type of IMF matters more than size!

Water:

• Hydrogen bonding strength: 10-40 kJ/mol per H-bond
• Each H₂O can form up to 4 H-bonds (2 donors, 2 acceptors)
• Extensive H-bonding network
• Total IMF energy per molecule: ~40 kJ/mol
• Requires significant energy to break H-bonds and boil

Hexane:

• LDF strength: ~30 kJ/mol total (for all LDF in molecule)
• Even with 42 electrons (strong for LDF)
• LDF still weaker than H-bonding
• Requires less energy to overcome IMFs and boil

$\text{H-bonding (H₂O)} > \text{LDF (hexane)}$

Result: Water boils at higher temperature

Step 3: Explain viscosity difference

Viscosity = resistance to flow

Factors:

1. Strength of IMFs
2. Molecular size/shape
3. Temperature

Water:

• Higher viscosity (relatively)
• Strong H-bonding between molecules
• Molecules must "unstick" from neighbors to flow
• H-bonds constantly breaking/reforming as molecules move

Hexane:

• Lower viscosity
• Only weak LDF
• Molecules slide past each other easily
• Less resistance to flow
• Despite being larger molecule, weaker IMFs dominate

However, note: Water actually has relatively LOW viscosity for a H-bonding liquid because it's so small. Glycerol (3 OH groups) has much higher viscosity.

Step 4: Explain surface tension difference

Surface tension = energy to increase surface area

Cause: Molecules at surface experience net inward pull

Water:

• High surface tension (~73 mN/m at 25°C)
• Strong H-bonding pulls surface molecules inward
• Surface molecules "miss" the H-bonds above them
• Strong cohesive forces
• Creates strong "skin" effect
• Allows insects to walk on water, water drops to bead up

Hexane:

• Low surface tension (~18 mN/m at 25°C)
• Only weak LDF
• Less "pulling" of surface molecules inward
• Weaker cohesive forces
• Spreads out easily (wets surfaces better)

Ratio: Water has ~4× higher surface tension than hexane

Step 5: Summary of IMF effects

Property relationships:

All these properties depend on IMF strength:

Step 2: Identify known and unknown values

Known:

• P₁ = 23.8 mmHg
• P₂ = 92.5 mmHg
• T₁ = 298 K
• T₂ = 323 K
• R = 8.314 J/(mol·K)

Unknown:

• ΔHvap = ?

Step 3: Calculate pressure ratio

$\frac{P_2}{P_1} = \frac{92.5}{23.8} = 3.887$

$\ln\left(\frac{P_2}{P_1}\right) = \ln(3.887) = 1.358$

Step 4: Calculate temperature term

$\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{323} - \frac{1}{298}$

$= 0.003096 - 0.003356$

$= -0.000260 \text{ K}^{-1}$

Or more precisely: $= -2.60 \times 10^{-4} \text{ K}^{-1}$

Step 5: Substitute into Clausius-Clapeyron equation

$1.358 = -\frac{\Delta H_{vap}}{8.314} \times (-2.60 \times 10^{-4})$

$1.358 = \frac{\Delta H_{vap}}{8.314} \times 2.60 \times 10^{-4}$

Step 6: Solve for ΔHvap

$\Delta H_{vap} = \frac{1.358 \times 8.314}{2.60 \times 10^{-4}}$

$\Delta H_{vap} = \frac{11.29}{2.60 \times 10^{-4}}$

$\Delta H_{vap} = 43,400 \text{ J/mol}$

$\Delta H_{vap} = 43.4 \text{ kJ/mol}$

Step 7: Check reasonableness

Literature value: ΔHvap for water ≈ 40.7 kJ/mol at 100°C

Our calculated value: 43.4 kJ/mol

Why slightly higher?

• ΔHvap decreases slightly with temperature
• At 25-50°C (lower than boiling point), ΔHvap is slightly higher
• Our value is reasonable!

Step 8: Verify with unit analysis

$\Delta H_{vap} = \frac{\ln(P_2/P_1) \times R}{(1/T_2 - 1/T_1)}$

Units: $= \frac{\text{(dimensionless)} \times \text{J/(mol·K)}}{\text{K}^{-1}}$

$= \frac{\text{J/(mol·K)}}{\text{K}^{-1}} = \text{J/mol}$ ✓

Answer:

$\boxed{\Delta H_{vap} = 43.4 \text{ kJ/mol}}$

Step-by-step summary:

1. Converted temperatures to Kelvin
2. Calculated ln(P₂/P₁) = 1.358
3. Calculated (1/T₂ - 1/T₁) = -2.60 × 10⁻⁴ K⁻¹
4. Substituted into Clausius-Clapeyron equation
5. Solved for ΔHvap = 43.4 kJ/mol

Physical interpretation:

43.4 kJ/mol is the energy required to convert 1 mole of liquid water to water vapor at constant temperature (around 25-50°C).

This energy is needed to:

• Break hydrogen bonds between water molecules
• Separate molecules from liquid to gas phase
• Overcome intermolecular attractions

Comparison:

• ΔHvap of water: 43.4 kJ/mol (calculated)
• ΔHfus of water: 6.01 kJ/mol (literature)

Ratio: ΔHvap/ΔHfus ≈ 7.2

Much more energy required for vaporization than melting because:

• Melting: Only partially breaks IMFs (molecules still close)
• Vaporization: Completely breaks all IMFs (molecules far apart)

Alternative approach - Linear form:

Could also use: ln P = -ΔHvap/R × (1/T) + C

Plot ln P vs. 1/T → slope = -ΔHvap/R

This problem uses the two-point form, which is more common for single calculations.