Understanding Phase Change for Water in General Chemistry

VI. Phases of matter: solid, liquid, and gas

Key focus of this chapter: phase change for water

This chapter focuses on the phase change for water and gives concise summaries of the important things about the phase diagram of H₂O and CO₂, heating curve of H₂O, colligative properties, Raoult’s law, osmotic pressure, and Hery’s law in more detail.

A. Solid, liquid, and gas

Classification

Features

Solid

• Particles are closely packed and vibrate about fixed positions.

• Crystalline solids have long-range order (ionic, metallic, covalent-network, or molecular solids).

• Amorphous solids lack long-range order (e.g., glass).

Liquid

• Particles remain close together but can move past one another.

• Cohesion: attraction between like liquid particles; it contributes to surface tension.

- Surface tension: resistance to increasing surface area.

- Viscosity: resistance to flow; it generally increases with stronger intermolecular attractions.

• Adhesion: attraction between a liquid and another surface.

Gas

• Gas particles may be atoms or molecules; in an ideal-gas model, intermolecular attractions are negligible.

• Particles move randomly between collisions.

• Vapor pressure: equilibrium pressure of vapor above its liquid (or solid) at a given temperature.

- At equilibrium, the rate of vaporization equals the rate of condensation.

Stronger intermolecular forces → lower vapor pressure → higher boiling point

Boiling occurs when a liquid’s vapor pressure equals the external pressure.

B. Phase change

: A phase change is the conversion of a substance from one physical state to another in response to changes in temperature, pressure, or both.

Fig. 1 Phase change

Deposition H < 0, S < 0 Sublimation H>0, S>0

Phase changes

Features

Phase changes

Features

Condensation

• Gas → liquid

ΔH < 0

ΔS < 0

Vaporization

• Liquid → gas

ΔH > 0

ΔS > 0

Freezing

• Liquid → solid

Fusion

(melting)

• Solid → liquid

Deposition

• Gas → solid directly.

Sublimation

• Solid → gas directly.

• Examples: iodine and dry ice

C. Phase diagram of H₂O and CO₂

Fig. 2 Phase diagram of H₂O and CO₂

Phase of CO 2 Phase of H 2 O

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H₂O

CO₂

• The solid–liquid boundary has a negative slope: increasing pressure lowers the melting point.

• The liquid–gas boundary has a positive slope.

• The solid–liquid boundary has a positive slope: increasing pressure raises the melting point.

• The liquid–gas boundary has a positive slope.

• Triple point

- unique temperature and pressure at which solid, liquid, and gas coexist in equilibrium

• Critical point

- endpoint of the liquid–gas coexistence curve

- above the critical temperature, a gas cannot be liquefied by pressure alone

D. Heating curve of H₂O

Fig. 3 Heating curve of H₂O

Lines of increasing molecular motions Temp Boiling point Vapor 100 o C Melting point Water 0 o C Lines of breaking bonds of in termolecular Ice Heat - 5 0 o C 36.6 J/(mol • o C) 75.4 J/(mol • o C) H vap = 40.7 kJ/mol H fusion = 6.0 kJ/mol

	Specific heat J/(g·°C)	Molar heat capacity J/(mol·°C)	DH_fusion kJ/mol	DH_vap kJ/mol
Ice	2.0	36.6
Ice → water			6.0
Water	4.2	75.4
Water → vapor				40.7

Heat required to convert ice to vapor

qtotal = nCp,iceΔTice + nΔHfusion + nCp,waterΔTwater + nΔHvap

Question: How much energy is required to convert 2.00 mol of ice at −10 °C to liquid water at 50 °C?

Solution: q1 (warming ice) = 36.6 J/(mol·°C)

= 36.6 J/(mol·°C) × 2.00 mol × 10 °C

= 732 J

q2 (melting ice) = ΔHfusion = 6.0 kJ/mol

= 6,000 J/mol × 2.00 mol

= 12,000 J

q3 (warming liquid water) = 75.4 J/(mol·°C)

= 75.4 J/(mol·°C) × 2.00 mol × 50 °C

= 7,540 J

Therefore, qtotal = q1 + q2 + q3 = 732 J + 12,000 J + 7,540 J

= 20,272 J ≈ 20.3 kJ

E. Colligative properties

: Colligative properties depend on the number of dissolved solute particles, not on their chemical identity.

For a nonvolatile solute, the major effects are:

1. Major colligative properties

• Vapor-pressure lowering and boiling-point elevation

• Freezing-point depression

• Osmotic pressure

Osmotic pressure Solution (m ixing solute & pure solvent ) Boiling point • Attraction b/w solute & pure solvent • Surface area of vaporizable solvent V apor pressure Freezing point

A solution containing a dissolved solute generally freezes at a lower temperature than the pure solvent.

See Fig. 4.

Melting point Freezing point

2. Boiling-point elevation and freezing-point depression

Pure solvent line Fig. 4 Effects of a nonvolatile solute on Δ Tb, Δ Tf, and Psoln P solv : V apor pressure of pure solvent P soln : V apor pressure of a solution Solution line

Classification

Features

Boiling-point elevation (ΔTb)

• Higher boiling point than the pure solvent

DT_b = K_b·m·i

Kb: boiling-point-elevation constant

m: molality

i: van ’t Hoff factor

Freezing-point depression (ΔTf)

• Lower freezing point than the pure solvent

DT_f = K_f·m·i

Kf: freezing-point-depression constant

van ’t Hoff factor, i: effective dissolved particles per formula unit (ideal limit)

• CH3OH: nonelectrolyte; i ≈ 1

LiCl → Li⁺ + Cl⁻; i ≈ 2

CaCl2 → Ca²⁺ + 2Cl⁻; i ≈ 3

T b = K b • m • i T f = K f • m • i Using van ’t Hoff factor P soln = P solv • X solv • i (Raoult’ s law) = M • R • T • i (osmotic pressure)

Question: Calculate the freezing point of 21.0 g LiCl in 1.000 kg water. Assume ideal complete dissociation; Kf = 1.86 °C·kg/mol.

ΔTf = iKfm

Molar mass of LiCl = 6.94 + 35.45 = 42.39 g/mol

Moles of LiCl = 21.0 g/(42.39 g/mol) = 0.495 mol

Molality, m = 0.495 mol/1.000 kg H2O = 0.495 mol/kg

For ideal complete dissociation, i ≈ 2 (LiCl → Li⁺ + Cl⁻).

ΔTf = (1.86 °C·kg/mol)(0.495 mol/kg)(2) = 1.84 °C

Freezing point = 0.00 °C − 1.84 °C = −1.84 °C

F. Raoult’s law

: For an ideal solution, the partial pressure of a volatile component equals its mole fraction in the solution

multiplied by the vapor pressure of the pure component.

Psoln = Xsolv · P°solv

Psoln: vapor pressure of the solution

P°solv: vapor pressure of the pure solvent

_{Xsolv: mole
fraction of solvent = nsolvent/ntotal}

For electrolytes, use i to count effective dissolved particles in ntotal.

Derived form of Raoult’s law

Psoln × total moles in solution = P°solv × moles of solvent

Psoln = P°solv × (moles of solvent/total moles in solution)

Psoln = P°solv × Xsolv

• A nonvolatile solute lowers the solvent mole fraction, so the solution’s vapor pressure is lower

than that of the pure solvent.

Fig. 5 Vapor pressure of a pure solvent and a solution

Molecule of vaporizable pure solvent Molecule of non - volatile solute

Pure solvent Solution (solvent + nonvolatile solute)

See Fig. 4 for the effects on P°solv and Psoln.

Question: Calculate the vapor pressure when 4.0 mol K2SO4 is dissolved in 12.0 mol H2O at 45 °C.

Assume ideal behavior and complete dissociation; P°H2O = 70 mmHg.

Psoln = P°solv × Xsolv

P°solv = 70 mmHg

Xsolv = nsolvent/(nsolvent + i·nsolute)

= 12/(12 + 3 × 4)

= 0.50

For ideal complete dissociation, i ≈ 3 (K2SO4 → 2K⁺ + SO4²⁻).

Therefore, the vapor pressure of the solution is:

Psoln = P°solv × Xsolv

= 70 mmHg × 0.50

= 35.0 mmHg

G. Osmotic pressure (P)

: Osmotic pressure is the external pressure required to stop the net flow of solvent into a solution

through a semipermeable membrane.

Pressure Fig. 6 Osmotic pressure between a pure solvent and a solution Pressure Solvent Solution Porous membrane

Osmosis: net movement of solvent through a semipermeable membrane from lower solute concentration to higher solute concentration.

P = M•R•T•i

P: Osmotic pressure (atm)

M: Molarity

R=0.082(L•atm)/(mol•K)

T= Kelvin temperature

i= van ’t Hoff factor

Relation to the ideal gas law

PV = nRT → P = (n/V)RT = MRT; for dissolved particles, Π = iMRT

Question: 116 g NaCl is dissolved to make 4.00 L of solution at 27 °C. Calculate the osmotic pressure.

Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol

Moles of NaCl = 116 g/(58.44 g/mol) = 1.985 mol

Molarity of NaCl = 1.985 mol/4.00 L = 0.496 M

R = 0.08206 L·atm/(mol·K)

T = 273.15 + 27 = 300.15 K

For ideal complete dissociation, i ≈ 2 (NaCl → Na⁺ + Cl⁻).

Π = iMRT

= (2)(0.496 mol/L)(0.08206 L·atm/(mol·K))(300.15 K)

= 24.4 atm

H. Henry’s law

: At constant temperature, dissolved-gas concentration is proportional to the gas’s partial pressure above the solution.

Fig. 7 Gas solubility as a function of partial pressure

0 atm of pressure 5 atm of pressure Gas Liquid

Solubility of gas (M) = k•P Solubility of gas: M ( Authenticated embedded content preview )

k: Henry’s law constant ( Authenticated embedded content preview )

P: partial pressure of the gas above the solution (atm)

Question: Find the gas solubility at 23 °C when Pgas = 3.0 atm and

k = 5.0 × 10⁻² mol/(L·atm).

Cgas = kP

= (5.0 × 10⁻² mol/(L·atm))(3.0 atm)

= 0.15 mol/L

VI. Phases of matter: solid, liquid, and gas

Key focus of this chapter: phase change for water

A. Solid, liquid, and gas

B. Phase change

Features

C. Phase diagram of H2O and CO2

D. Heating curve of H2O

E. Colligative properties

1. Major colligative properties

2. Boiling-point elevation and freezing-point depression

F. Raoult’s law

G. Osmotic pressure (P)

H. Henry’s law

C. Phase diagram of H₂O and CO₂

D. Heating curve of H₂O