Molecular Bonds Slides Molecular Bonds
The Physics of Solvation and Solubility
12th Grade Physics • Thermodynamics
The "Clean-Up" Dilemma
Why does spilled motor oil float stubbornly on water, while spilled rubbing alcohol disappears completely?
"The physics of separation begins with understanding the physics of the bond."
How does molecular structure dictate our clean-up strategies?
The Toolkit: Intermolecular Forces (IMFs)
LDFs
London Dispersion Forces: Temporary dipoles in all molecules. Proportional to molar mass.
Dipole-Dipole
Attraction between permanent dipoles in polar molecules.
H-Bonding
Extreme dipole-dipole: H bonded to N, O, or F. High energy requirement to break.
The Thermodynamic Trade-off
Break Solute Bonds: \(\Delta H > 0\) (Endothermic)
Break Solvent Bonds: \(\Delta H > 0\) (Endothermic)
Form Solution Bonds: \(\Delta H < 0\) (Exothermic)
\[ \Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 \]
If the net energy released is greater than the energy required, solubility increases.
"Like Dissolves Like"
Polar Solutes
• Dissolve in Polar Solvents
• Example: NaCl in Water
• Interaction: Ion-Dipole
Non-Polar Solutes
• Dissolve in Non-Polar Solvents
• Example: Oil in Hexane
• Interaction: LDFs
Wait... what about soaps? They do both!
Measuring the Mixture
Molarity (M)
\[ M = \frac{n_{solute}}{V_{solution}(L)} \]
Dependent on temperature (volume expands).
Mole Fraction (\(\chi\))
\[ \chi_A = \frac{n_A}{n_{total}} \]
Essential for calculating Vapor Pressure later!
Solvation Physics Worksheet Solvation Physics
Topic 1.1: Intermolecular Forces & Concentration
Student Name:
Date:
Part 1: Predictive Solubility
Analyze the molecular structures below. Predict whether the solute will dissolve in the solvent and identify the primary intermolecular force (IMF) responsible for the interaction.
Solute: Iodine (I₂)
Non-polar diatomic molecule
Solvent: Carbon Tetrachloride (CCl₄)
Non-polar tetrahedral solvent
Solubility Prediction:
Select one: Soluble / Insoluble
Primary Interaction Force:
Solute: Potassium Chloride (KCl)
Ionic crystalline solid
Solvent: Water (H₂O)
Highly polar, H-bonding
Solubility Prediction:
Primary Interaction Force:
Part 2: The Enthalpy of Solution
Recall the Born-Haber style cycle for solvation:
\[ \Delta H_{soln} = \Delta H_{solute} + \Delta H_{solvent} + \Delta H_{mix} \]
1. Why is \(\Delta H_{solute}\) always endothermic (positive)? Explain in terms of electrostatic forces.
2. A certain mixture has a \(\Delta H_{soln}\) of -45.2 kJ/mol. Describe the temperature change of the container during the mixing process. Is this an ideal mixture?
Part 3: Quantitative Concentration
A solution is prepared by dissolving 25.0g of Glucose (C₆H₁₂O₆) in 200.0g of Water. The density of the resulting solution is 1.04 g/mL.
A) Calculate the Mole Fraction (\(\chi\)) of glucose.
B) Calculate the Molarity (M) of the solution.
Solvation Physics Answer Key Answer Key
Solvation Physics Worksheet • Teacher Reference
INTERNAL USE ONLY
Part 1: Predictive Solubility
1. I₂ in CCl₄
Prediction: Soluble
Reasoning: Both are non-polar. Dispersion forces (LDFs) allow them to mix readily.
2. KCl in Water
Prediction: Soluble
Reasoning: Ion-Dipole forces. The polar water molecules stabilize the K+ and Cl- ions, overcoming the lattice energy.
Part 2: The Enthalpy of Solution
1. Why is \(\Delta H_{solute}\) always endothermic?
Energy is required to overcome the electrostatic attractions (lattice energy in solids or IMFs in liquids) between the solute particles. Breaking any attractive force is an energy-consuming process.
2. \(\Delta H_{soln} = -45.2\) kJ/mol. Describe the temperature change.
The container will increase in temperature (Exothermic). This is NOT an ideal mixture; an ideal mixture has \(\Delta H_{soln} = 0\). The release of energy suggests that the new bonds formed (\(solute-solvent\)) are stronger than the original ones.
Part 3: Quantitative Concentration
Solution Setup:
Moles Glucose: 25.0g / 180.16 g/mol = 0.1388 mol
Moles Water: 200.0g / 18.02 g/mol = 11.099 mol
Total Moles: 0.1388 + 11.099 = 11.238 mol
Total Mass: 225.0g
Volume: 225.0g / 1.04 g/mL = 216.3 mL = 0.2163 L
A) Mole Fraction (\(\chi\))
\(\chi_{glu} = 0.1388 / 11.238\)
B) Molarity (M)
\(M = 0.1388 mol / 0.2163 L\)
Vapor Pressure Slides Pressure & Vapor
The Thermodynamics of Volatile Mixtures
Lesson 2 • Raoult's Law
The Moonshine Problem
In illegal distillation, the "danger zone" is the first few ounces. Methanol boils at 64.7°C, while Ethanol boils at 78.4°C.
How do the molecules "know" when to leave the liquid phase, and how does mixing them change their escape pressure?
Vapor Pressure Equilibrium
// Equilibrium State
Rate of Evaporation = Rate of Condensation
Raoult's Law: The Definition
The partial vapor pressure of a component in an ideal mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction.
\[ P_A = \chi_A \cdot P^\circ_A \]
\(P_A\)
Partial Pressure
\(\chi_A\)
Mole Fraction
\(P^\circ_A\)
Pure Vapor Pressure
Total Mixture Pressure
For a mixture of liquid A and liquid B:
\[ P_{total} = P_A + P_B \]
\[ P_{total} = \chi_A P^\circ_A + \chi_B P^\circ_B \]
As concentration of A increases, the contribution of B to the vapor phase decreases linearly.
Pure B
Pure A
Linear P-x Graph
Ideal Behavior
Reality vs. Ideality
Positive Deviation
A-B interactions are weaker than A-A or B-B. Molecules escape easier.
\(P_{exp} > P_{calc}\)
Negative Deviation
A-B interactions are stronger (e.g. H-bonding). Molecules "cling" together.
\(P_{exp} < P_{calc}\)
Temperature Dependency
Vapor pressure of pure components changes exponentially with temperature:
\[ \ln(P) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T}\right) + C \]
Higher Temp = Higher Vapor Pressure = Faster Separation.
Mixture Math Practice Mixture Math Workshop
Topic 2.1: Applying Raoult's Law to Volatile Solutions
Name:
Period:
Key Formula
\[ P_{tot} = \chi_A P^\circ_A + \chi_B P^\circ_B \]
• \(P^\circ\) is the vapor pressure of the pure component.
• \(\chi\) is the mole fraction in the liquid phase.
• R = 8.314 J/mol·K or 0.0821 L·atm/mol·K
1
The Binary Ideal Mixture
A solution is prepared by mixing 2.0 moles of Benzene (C₆H₆) and 3.0 moles of Toluene (C₇H₈) at 25°C. At this temperature, the vapor pressure of pure benzene is 95.1 mmHg and pure toluene is 28.4 mmHg.
A) Calculate \(\chi_{benzene}\) and \(\chi_{toluene}\)
B) Total Vapor Pressure of Solution (\(P_{tot}\))
C) Determine the composition (mole fraction) of the vapor above the liquid.
Hint: \(\chi_{A, vapor} = P_A / P_{tot}\)
2
Identifying Deviations
You are testing a mixture of Acetone and Chloroform. Calculated \(P_{tot}\) using Raoult's Law is 260 mmHg. However, experimental data shows a pressure of 235 mmHg.
A) Is this a positive or negative deviation from ideality? Justify your answer using the concept of intermolecular forces (A-B vs A-A/B-B).
B) Based on this deviation, would the boiling point of the real mixture be higher or lower than the ideal mixture? Explain the physics of why.
3
Vapor Pressure & Temperature
A sample of pure ethanol has a vapor pressure of 44 mmHg at 20°C. If the heat of vaporization (\(\Delta H_{vap}\)) is 38.6 kJ/mol, use the Clausius-Clapeyron equation to calculate the vapor pressure at 40°C.
Mixture Math Answer Key Answer Key
Mixture Math Practice • Teacher Reference
Physics Answer Key
Problem 1: Benzene/Toluene Mixture
A) Mole Fractions
Total moles = 2.0 + 3.0 = 5.0 mol
B) Total Pressure
\(P_{tot} = (0.40)(95.1) + (0.60)(28.4)\)
C) Vapor Composition
\(\chi_{benz, vap} = 38.04 / 55.08 = \mathbf{0.69}\)
Problem 2: Acetone & Chloroform
A) Deviation Type
Negative Deviation. The experimental pressure (235) is lower than the calculated ideal pressure (260). This indicates that Acetone-Chloroform (A-B) attractions are stronger than the attractions in the pure liquids. Specifically, they form a strong hydrogen bond.
B) Boiling Point Effect
The boiling point will be higher than ideal. Since the vapor pressure is lower (molecules are "held" in the liquid phase more tightly), more thermal energy is required to raise the vapor pressure to atmospheric pressure.
Problem 3: Clausius-Clapeyron
// Equation Setup
\(\ln(P_2/P_1) = (-\Delta H_{vap}/R) \cdot (1/T_2 - 1/T_1)\)
\(T_1 = 293.15 K, T_2 = 313.15 K\)
\(\ln(P_2/44) = (-38600/8.314) \cdot (1/313.15 - 1/293.15)\)
Distillation Dynamics Slides Fractional Flow
The Physics of Distillation & Theoretical Plates
Lesson 3 • Vapor-Liquid Equilibrium
The Crude Tower
The Giant Separator
Crude oil is a mess. It's a mixture of thousands of hydrocarbons. Yet, a single tower separates it into:
Gasoline (Top)
Jet Fuel
Diesel
Asphalt (Bottom)
How does height translate to purity?
The Fractionating Column
Simple Distillation
One evaporation/condensation cycle. Good for separating liquids with boiling point differences > 70°C.
RESULT
Moderate Purity
Fractional Distillation
Dozens of tiny "mini-distillations" happening at once on the surface area of packing material.
RESULT
High Purity (>99%)
Temperature-Composition Diagrams
Liquid + Vapor
Dew Point Curve
Bubble Point Curve
The Tie-Line
A horizontal line connecting the liquid phase to the vapor phase at a specific temperature.
The Enrichment
Moving along a tie-line shows you the composition of the vapor. Notice it is always richer in the lower-boiling component.
Calculating "Plates"
One Theoretical Plate represents one complete cycle of evaporation and condensation (one tie-line jump).
HETP
Height Equivalent to a Theoretical Plate
The Goal
Minimize HETP to maximize separation in a fixed-height column.
The Azeotrope: Physics Wins
Some mixtures reach a point where the liquid composition equals the vapor composition.
"Distillation cannot separate an azeotrope."
Example: Ethanol/Water at 95.6% ethanol. You can never reach 100% via boiling alone.
Engineering Solution
Breaking the Limit
• Change the Pressure (P-swing)
• Add a third component (Entrainer)
• Molecular Sieves (Filtering)
Column Calculations Lab Column Dynamics Lab
Topic 3.1: VLE Analysis & Plate Calculation
Lab Group:
Station:
Objective
In this activity, you will analyze a Temperature-Composition (T-x-y) phase diagram for a binary mixture of n-Hexane and n-Heptane to determine the theoretical number of plates required to achieve industrial purity.
1
Reading the Phase Curve
Temperature (°C)
Mole Fraction n-Hexane (\(\chi_{Hex}\))
0.0
0.25
0.50
0.75
1.0
Figure 1: T-x-y Diagram for Hexane/Heptane at 1 atm
1. At a liquid mole fraction of \(\chi_{Hex} = 0.25\), find the boiling point of the mixture and the composition of the vapor being released.
2. Draw a second "step" (tie-line) starting from the vapor composition you found in question 1. What is the new vapor composition?
2
Column Efficiency Calculation
You are designing a fractionating column that is 1.5 meters tall. Using the diagram from Figure 1, you find that to get from 20% purity to 95% purity, you need 8 theoretical plates.
Formula: HETP
\[ HETP = \frac{\text{Height}}{\text{N}_{plates}} \]
Calculate the HETP for this column in cm/plate.
3. If you replace the glass beads in the column with stainless steel mesh (which has more surface area), how will the HETP and the final purity change? Explain the physics of surface area in condensation/evaporation cycles.
4. Discuss the concept of "Reflux Ratio". Why must some of the purified liquid be sent back down the column? How does this affect the thermodynamics of the separation?
Column Calculations Teacher Guide Teacher Guide
Column Dynamics Lab • Facilitation & Key
PHYSICS 12
Tie-Lines
Horizontal lines connect liquid comp (bubble point) to vapor comp (dew point) at equilibrium.
Stepping
The process of moving from \(\chi_{liq}\) to \(\chi_{vap}\) then dropping down to the next liquid point at a lower temp.
Efficiency
HETP represents the vertical height needed for one such "step" to occur.
Lab Solution Key
Part 1, Q1: Initial Reading
At \(\chi_{Hex} = 0.25\): Boiling Point \(\approx 88^\circ\text{C}\). Vapor Composition \(\approx 0.52\) (enrichment in Hexane).
Part 1, Q2: Second Step
Starting from 0.52 on the liquid curve (lower temp \(\approx 78^\circ\text{C}\)), the next vapor composition is \(\approx 0.80\). Each step exponentially increases the concentration of the more volatile component.
Part 2: HETP Calculation
Height = 150 cm / 8 plates = 18.75 cm/plate.
Part 2, Q3: Surface Area Physics
Increasing surface area decreases HETP (more efficiency). Physics: More sites for condensation allow for faster heat exchange between rising vapor and falling liquid, increasing the likelihood of reaching phase equilibrium at every height increment.
Teaching Tips
Common Pitfalls
• Students often draw diagonal lines instead of horizontal tie-lines. Emphasize that evaporation happens at a constant temperature.
• Mixing up the liquid curve (lower) and vapor curve (upper). Remind them that at a given temp, vapor is always richer in the low-boiler.
Extension
Ask advanced students to predict what happens if the column is insulated versus uninsulated. (Heat loss can lead to premature condensation, actually helping separation but reducing yield/speed).
Crystal Kinetics Slides Crystal Kinetics
Thermodynamics of Solid-Liquid Phase Changes
Lesson 4 • Evaporation & Nucleation
The Garden of Glass
In a supersaturated solution, the physics of "nothing" becomes "everything."
What determines whether you get a fine powder or a single, massive sapphire-like geometric crystal?
Critical Question
"Is crystallization a victory of entropy or enthalpy?"
Lattice Formation in Progress...
Driving Separation: Evaporation
Heat Transfer
Energy required to overcome IMFs in the solvent.
\[ Q = m \cdot \Delta H_{vap} \]
Entropy Drive
Solvent moves from low-entropy liquid to high-entropy gas.
\[ \Delta S_{vap} = \frac{\Delta H_{vap}}{T_{bp}} \]
As the solvent leaves, the concentration of the solute rises, forcing it toward the solid state.
The Physics of Nucleation
The Energy Barrier
To form a crystal, you must overcome the surface energy of a tiny new cluster.
Gibbs Free Energy for a Cluster
\[ \Delta G = -V \Delta G_v + A \gamma \]
V = Volume, A = Surface Area, \(\gamma\) = Surface Tension
Slow Cooling
Fewer nuclei. Larger, perfect crystals.
Fast Evaporation
Many nuclei. Tiny, "dusty" crystals.
"Thermodynamic Control vs. Kinetic Control"
The 3 Pillars of Formation
Saturation
Solute reaches its maximum solubility limit for a given temperature.
Nucleation
Solute particles collide and stick, forming a stable "seed" cluster.
Growth
New particles add to the existing lattice, following the internal symmetry.
When Heat Fails
Some compounds undergo Thermal Decomposition before they can be separated by evaporation.
Example: Proteins
Heat causes denaturation, permanently altering the structure.
Solution: Lyophilization
Freeze-drying. Sublimation of solvent at low pressure and low temperature.
Phase Change Problem Set Phase Change Problems
Topic 4.1: Energetics of Evaporation and Crystal Growth
Student:
Date:
Reference Data
• Water \(\Delta H_{vap} = 40.7\) kJ/mol
• Ethanol \(\Delta H_{vap} = 38.6\) kJ/mol
• R = 8.314 J/mol·K
1
The Cost of Drying
An industrial processor needs to recover 5.0 kg of solid Copper Sulfate (CuSO₄) from a 15% solution (by mass) in water.
A) Calculate the total mass of water that must be evaporated to leave behind 5.0 kg of dry CuSO₄.
B) Calculate the total energy (in MJ) required to evaporate this mass of water, assuming the solution is already at 100°C.
2
Nucleation Barrier
Recall the equation for the Gibbs energy of a spherical nucleus: \(\Delta G = -\frac{4}{3} \pi r^3 \Delta G_v + 4 \pi r^2 \gamma\).
where \(\Delta G_v\) is the volume energy (driving force) and \(\gamma\) is the surface tension.
A) Derive the expression for the critical radius (\(r^*\)) where the nucleus becomes stable (i.e., when \(d\Delta G / dr = 0\)).
Physics Check
If you increase the degree of supercooling (making \(\Delta G_v\) larger), what happens to the critical radius? How does this explain why fast-cooled solutions produce smaller crystals?
3
The Entropy Drive
Calculate the molar entropy of vaporization (\(\Delta S_{vap}\)) for Ethanol at its boiling point of 78.37°C (\(351.52 K\)). Use \(\Delta H_{vap} = 38.6\) kJ/mol.
Compare this value to the "Trouton's Rule" approximation (\(\approx 85-88\) J/mol·K). Is ethanol an ideal liquid in this regard? Explain why or why not based on intermolecular forces.
Phase Change Answer Key Answer Key
Phase Change Problem Set • Teacher Reference
PHYSICS 12
Problem 1: CuSO₄ Recovery
A) Mass of Water
Total Mass = (5.0 kg solute) / 0.15 = 33.33 kg solution
B) Energy Required
Moles water = 28,333 g / 18.02 g/mol = 1,572.3 mol
Problem 2: Nucleation Barrier
A) Derivation
\(\Delta G = -\frac{4}{3} \pi r^3 \Delta G_v + 4 \pi r^2 \gamma\)
\(\frac{d\Delta G}{dr} = -4 \pi r^2 \Delta G_v + 8 \pi r \gamma = 0\)
\(4 \pi r^2 \Delta G_v = 8 \pi r \gamma\)
\(r^* = \frac{2 \gamma}{\Delta G_v}\)
B) Physics Reasoning
As supercooling increases, \(\Delta G_v\) (the driving force) increases. This makes the critical radius smaller. Smaller \(r^*\) means a lower energy barrier, so many more nuclei can form simultaneously. This results in many small crystals rather than few large ones.
Problem 3: Entropy Drive
\(\Delta S_{vap} = \Delta H_{vap} / T_{bp}\)
Reasoning:
Ethanol is NOT ideal (Trouton's rule \(\approx 88\)). The value (109.8) is much higher. This is because ethanol is highly ordered in the liquid phase due to Hydrogen Bonding. Breaking these bonds leads to a much larger increase in entropy upon transition to the gas phase than a non-polar liquid would experience.
Membrane Physics Slides The Filter Physics
Reverse Osmosis & Membrane Desalination
Lesson 5 • Osmotic Pressure
Boil or Push?
To separate salt from water, you have two choices:
Distillation
Heat the whole ocean to turn water to steam.
Reverse Osmosis
Use pressure to force molecules through a microscopic sieve.
One uses 10x less energy. Why?
Salt Water
Fresh Water
MEMBRANE
Applied Pressure > \(\Pi\)
The van't Hoff Equation
Osmotic pressure (\(\Pi\)) is the pressure required to stop the natural flow of solvent into a more concentrated solution.
\[ \Pi = i \cdot M \cdot R \cdot T \]
i
van't Hoff Factor
M
Molarity
R
Gas Constant
T
Abs Temperature
Selectivity & Flux
The Solution-Diffusion Model
Unlike a mechanical filter, membranes work by dissolving water into the polymer and diffusing it through.
High permeability for Solvent (H₂O)
Low permeability for Solute (Na⁺, Cl⁻)
Efficiency Factors
Concentration Polarization
Salt builds up on the membrane surface, increasing local \(\Pi\).
Fouling
Organic matter clogs the pores, decreasing flux.
Thermodynamic Limit
Phase Change (Thermal)
Requires overcoming the latent heat of vaporization (\(\approx 2260\) J/g).
~ 10-15 kWh/m³
Membrane (RO)
Requires only the work of compression against osmotic pressure.
~ 2-4 kWh/m³
The "Physics Winner" for modern global water security.
Membrane Osmosis Case Study The Desalination Dilemma
Topic 5.1: Osmotic Pressure and Global Solutions
Analyst:
Region:
1
Calculating the Pressure Floor
Average seawater has a salinity of approximately 35.0 g/L. Assume the salt is entirely NaCl (Molar Mass = 58.44 g/mol) and the temperature is 20°C.
A) Calculate the Osmotic Pressure (\(\Pi\)) of seawater in atmospheres.
Show work for Molarity (M), i, and final calculation...
B) Minimum Pressure for RO
If you want to produce fresh water, what is the absolute minimum pressure (in psi) the pump must generate? (1 atm = 14.7 psi)
2
Case Study: The Sorek Plant, Israel
Technical Specs
• Capacity: 624,000 m³/day
• Efficiency: 45% recovery
• Energy: 3.4 kWh/m³
As fresh water is removed, the remaining "brine" becomes twice as concentrated.
1. Based on the "Brine Concentration" problem, why does the energy requirement per liter of water increase as you extract more fresh water? Use the van't Hoff equation in your explanation.
2. Environmental Physics: Brine is denser than seawater. If a plant pumps its brine waste directly back into the ocean, explain what happens to the oxygen levels at the sea floor using the physics of fluid density and diffusion.
3
Critical Synthesis
Question: If Reverse Osmosis is so energy-efficient, why do we still use Multi-Stage Flash Distillation (MSF) in many Middle Eastern countries?
Consider: Waste heat from power plants, membrane fouling in high-salinity/high-temp water, and infrastructure costs.
Desalination Teacher Resource Teacher Resource
Lesson 5: Desalination & Membrane Physics • Answer Key & Facilitation
UNIT: THERMODYNAMICS
Part 1: Seawater Calculations
A) Osmotic Pressure (\(\Pi\))
Molarity (M) = (35.0 g/L) / (58.44 g/mol) = 0.599 M
van't Hoff (i) = 2 (for NaCl)
Temp (T) = 293.15 K
\(\Pi = 2 \cdot 0.599 \cdot 0.0821 \cdot 293.15\)
\(\Pi \approx 28.8 \text{ atm}\)
B) Minimum Pressure
Pressure must exceed \(\Pi\) to reverse flow.
28.8 atm \(\cdot\) 14.7 psi/atm = 423.4 psi
*Note: Industrial RO pumps typically operate at 800-1000 psi to maintain high flux.
Part 2: Case Study Insights
1. Why does energy requirement increase with extraction?
As water is removed, the remaining solution becomes more concentrated (Molarity \(M\) increases). According to \(\Pi = iMRT\), the osmotic pressure increases. Therefore, the pump must work against a higher "back-pressure" floor, consuming more energy per unit of water produced as the recovery percentage increases.
2. Brine Density & Diffusion Physics
Brine is significantly denser than ambient seawater. Upon discharge, it sinks to the sea floor (negative buoyancy). This "brine plume" acts as a physical barrier that prevents the vertical mixing of oxygen-rich surface water with bottom water. Furthermore, the high salinity can pull water out of local benthic organisms via osmosis, causing biological stress or "dead zones."
Debriefing the Dilemma
The final question (MSF vs RO) is designed for a Socratic discussion. Key points to highlight:
• Waste Heat: In the Middle East, desalination is often coupled with oil-fired power plants. MSF uses the "free" waste steam, making it economically viable despite the thermodynamic inefficiency.
• Durability: RO membranes are delicate. High-temp, high-silt water in the Persian Gulf can destroy a membrane in weeks. MSF is just a "big metal tank" and is much more robust.
Sequence Wrap-Up
Connect back to the Essential Question:
"Separation is never free. Whether we use heat to change phase (Lessons 2-4) or work to overcome osmosis (Lesson 5), the entropy of mixing must always be paid for in energy."
Sequence Complete: Thermodynamics of Mixtures
