Physical Chemistry Lecture Notes

Physical Chemistry Intensive Training Camp @ KAUST 2025

• Dr. Ahmad Alsaleh, Assistant Professor of Physical Chemistry, King Saud University.
• © McGraw Hill LLC 1

Table of Content

• Short Review for Chemical Stoichiometry (مراجعة ألهم المفاهيم في الحسابات الكيميائية)
• Gases (الغازات)
• Thermochemistry and Chemical Thermodynamics (الكيمياء الحرارية والثيرموديناميك)
• Free Energy and Equilibrium (الحرة الطاقة واإلتزان الكيميائي)
• Chemical Equilibrium (اإلتزان الكيميائي)
• Intermolecular Forces (القوى البينية)
• Electrochemistry (الكيمياء الكهربائية)
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Defining Chemistry (Introduction)

• Chemistry is the "study of matter and the changes it undergoes."
• Matter is anything that occupies space and has mass.
• Physical Chemistry is the study of the (physical) principles of chemistry.
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Introduction & A Quick Review of the Basics Mass Relationships in Chemical Reactions

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IUPAC Periodic Table of the Elements

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• For notes and updates to this table, see www.iupac.org. This version is dated 4 May 2022.
• Copyright ©2022 IUPAC, the International Union of Pure and Applied Chemistry.

Dimensional Analysis (تحليل الأبعاد أو تحويل الوحدات)

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International System of Units (SI)

• Lists base quantity, name of unit, and symbol for:
  • Length: meter (m)
  • Mass: kilogram (kg)
  • Time: second (s)
  • Electrical current: ampere (A)
  • Temperature: kelvin (K)
  • Amount of substance: mole (mol)
  • Luminous intensity: candela (cd)
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Prefixes Used with SI Units

• Lists prefixes, symbols, meaning, and examples for:
  • peta- (P), 1,000,000,000,000,000, or $10^{15}$ (1 petameter (Pm)=$1 \times 10^{15}$ m)
  • tera- (T), 1,000,000,000,000, or $10^{12}$ (1 terameter (Tm)=$1 \times 10^{12}$ m)
  • giga- (G), 1,000,000,000, or $10^9$ (1 gigameter (Gm)=$1 \times 10^9$ m)
  • mega- (M), 1,000,000, or $10^6$ (1 megameter (Mm)=$1 \times 10^6$ m)
  • kilo- (K), 1,000, or $10^3$ (1 kilometer (km)=$1 \times 10^3$ m)
  • deci- (d), 1/10, or $10^{-1}$ (1 decimeter (dm) = 0.1 m)
  • centi- (c), 1/100, or $10^{-2}$ (1 centimeter (cm) = 0.01 m)
  • milli- (m), 1/1,000, or $10^{-3}$ (1 millimeter (mm)=0.001 m)
  • micro- ($\mu$), 1/1,000,000, or $10^{-6}$ (1 micrometer ($\mu$m)=$1 \times 10^{-6}$ m)
  • nano- (n), 1/1,000,000,000 or $10^{-9}$ (1 nanometer (nm)=$1 \times 10^{-9}$ m)
  • pico- (p), 1/1,000,000,000,000, or $10^{-12}$ (1 picometer (pm)=$1 \times 10^{-12}$ m)
  • femto- (f), 1/1,000,000,000,000,000, or $10^{-15}$ (1 femtometer (fm)=$1 \times 10^{-15}$ m)
  • atto- (a), 1/1,000,000,000,000,000,000, or $10^{-18}$ (1 attometer (am)=$1 \times 10^{-18}$ m)
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Dimensional Analysis Method of Solving Problems

1. Determine which unit conversion factor(s) are needed.
2. Carry units through calculation.
3. If all units cancel except for the desired unit(s), then the problem was solved correctly.
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5. given quantity $\times$ conversion factor = desired quantity
6. $\frac{\text{desired unit}}{\text{given unit}}$
7. $\frac{\text{desired unit}}{\text{given unit}} \times \text{given unit} = \text{desired unit}$

EXAMPLE 1.6

• A person's average daily intake of glucose (a form of sugar) is 0.0833 pound (lb). What is this mass in milligrams (mg)? (1 lb = 453.6 g.)
• Strategy: The problem can be stated as ? mg = 0.0833 lb
  • The relationship between pounds and grams is given in the problem. This relationship will enable conversion from pounds to grams. A metric conversion is then needed to convert grams to milligrams (1 mg = 1 × 10-3 g). Arrange the appropriate conversion factors so that pounds and grams cancel and the unit milligrams is obtained in your answer.
• Solution: pounds $\rightarrow$ grams $\rightarrow$ milligrams
  • Using the following conversion factors: $\frac{453.6 \text{ g}}{1 \text{ lb}}$ and $\frac{1 \text{ mg}}{1 \times 10^{-3} \text{ g}}$
  • $? \text{ mg} = 0.0833 \text{ lb} \times \frac{453.6 \text{ g}}{1 \text{ lb}} \times \frac{1 \text{ mg}}{1 \times 10^{-3} \text{ g}} = 3.78 \times 10^4 \text{ mg}$
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EXAMPLE 1.7

• An average adult has 5.2 L of blood. What is the volume of blood in m³?
• Strategy: The problem can be stated as ? m³ = 5.2 L
  • 1 L = 1000 cm³ , 5.2 L = 5200 cm³
  • How many conversion factors are needed for this problem? Recall that 1 L = 1000 cm³ and 1 cm = 1 × 10-2 m.
• Solution: We need two conversion factors here: one to convert liters to cm³ and one to convert centimeters to meters:
  • $\frac{1000 \text{ cm}^3}{1 \text{ L}}$ and $\frac{1 \times 10^{-2} \text{ m}}{1 \text{ cm}}$
  • 1 m = 100 cm
  • Because the second conversion factor deals with length (cm and m) and we want volume here, it must therefore be cubed to give
  • $\frac{1 \times 10^{-2} \text{ m}}{1 \text{ cm}} \times \frac{1 \times 10^{-2} \text{ m}}{1 \text{ cm}} \times \frac{1 \times 10^{-2} \text{ m}}{1 \text{ cm}} = \frac{1 \times 10^{-6} \text{ m}^3}{1 \text{ cm}^3}$
  • This means that 1 cm³ = $1 \times 10^{-6}$ m³.
  • $? \text{ m}^3 = 5.2 \text{ L} \times \frac{1000 \text{ cm}^3}{1 \text{ L}} \times \frac{1 \times 10^{-6} \text{ m}^3}{1 \text{ cm}^3} = 5.2 \times 10^{-3} \text{ m}^3$
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Chemical Problems!

• How many nm³ equals one mm³?
• What is the mass in Kg of 55 ml of black coffee with a density of 0.87g/cm³? (Knowing the, density = $\frac{mass}{volume}$)
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The Mole (mol)

• The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of ¹²C
• 1 mol = NA = $6.0221367 \times 10^{23}$ Avogadro’s number (NA)
• Dozen = 12, Pair = 2
• The Mole (mol): A unit to count numbers of particles
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Molar Mass

• Molar mass is the mass of 1 mole of eggs, shoes, marbles, atoms in grams.
• 1 mole ¹²C atoms = $6.022 \times 10^{23}$ atoms = 12.00 g
• 1 ¹²C atom = 12.00 amu
• 1 mole ¹²C atoms = 12.00 g ¹²C
• 1 mole lithium atoms = 6.941 g of Li
• For any element atomic mass (amu) = molar mass (grams)
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One Mole Of:

• Images of one mole of:
  • C
  • S
  • Fe
  • Cu
  • Hg
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Conversion Between Mass and Atoms

• 1 amu = $1.66 \times 10^{-24}$ g or 1 g = $6.022 \times 10^{23}$ amu
• 1 ¹²C atom 12.00 amu $\times \frac{12.00 \text{ g}}{6.022 \times 10^{23} \text{ }^{12}C \text{ atoms}} = 1.66 \times 10^{-24} \text{ g}$
• $\frac{1 \text{ amu}}{NA} = Avogadro’s \text{ number}$
• $\text{M}_{wt} = \text{molar mass in (g/mol)}$
• $n = \frac{m}{M_{wt}}$
• $N = nN_A$
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Example Calculation

• How many atoms are in 0.551 g of potassium (K)?
• 1 mol K = 39.10 g K
• 1 mol K = $6.022 \times 10^{23}$ atoms K
• $0.551 \text{ g K} \times \frac{1 \text{ mol K}}{39.10 \text{ g K}} \times \frac{6.022 \times 10^{23} \text{ atoms K}}{1 \text{ mol K}} = 8.49 \times 10^{21} \text{ atoms of K}$
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EXAMPLE 3.7

• How many hydrogen atoms are present in 25.6 g of urea [(NH₂)₂CO], which is used as a fertilizer, in animal feed, and in the manufacture of polymers? The molar mass of urea is 60.06 g.
• Solution: To calculate the number of H atoms, we first must convert grams of urea to moles of urea using the molar mass of urea. This part is similar to Example 3.2. The molecular formula of urea shows there are four moles of H atoms in one mole of urea molecule, so the mole ratio is 4:1. Finally, knowing the number of moles of H atoms, we can calculate the number of H atoms using Avogadro's number. We need two conversion factors: molar mass and Avogadro's number. We can combine these conversions
  • grams of urea $\rightarrow$ moles of urea $\rightarrow$ moles of H $\rightarrow$ atoms of H
  • into one step:
  • $25.6 \text{ g (NH₂ )₂CO} \times \frac{1 \text{ mol (NH₂ )₂CO}}{60.06 \text{ g (NH₂ )₂CO}} \times \frac{4 \text{ mol H}}{1 \text{ mol (NH₂ )₂CO}} \times \frac{6.022 \times 10^{23} \text{ H atoms}}{1 \text{ mol H}} = 1.03 \times 10^{24} \text{ H atoms}$
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How to “Read” Chemical Equations

• 2 Mg + O₂ $\rightarrow$ 2 MgO
• 2 atoms Mg + 1 molecule O₂ makes 2 formula units MgO
• 2 moles Mg + 1 mole O₂ makes 2 moles MgO
• 48.6 grams Mg + 32.0 grams O₂ makes 80.6 g MgO
• NOT 2 grams Mg + 1 gram O₂ makes 2 g MgO
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Balancing Chemical Equations

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
2. Ethane reacts with oxygen to form carbon dioxide and water: C₂H₆ + O₂ $\rightarrow$ CO₂ + H₂O
3. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
4. NOT C₄H₁₂
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Balancing Chemical Equations Continued

3. Start by balancing those elements that appear in only one reactant and one product.
4. C₂H₆ + O₂ $\rightarrow$ CO₂ + H₂O
5. start with C or H but not O
6. 2 carbon on left 1 carbon on right multiply CO₂ by 2
7. C₂H₆ + O₂ $\rightarrow$ 2CO₂ + H₂O
8. 6 hydrogen on left 2 hydrogen on right multiply H₂O by 3
9. C₂H₆ + O₂ $\rightarrow$ 2CO₂ + 3H₂O
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Balancing Chemical Equations continued

4. Balance those elements that appear in two or more reactants or products.
5. C₂H₆ + O₂ $\rightarrow$ 2CO₂ + 3H₂O
6. 2 oxygen on left 4 oxygen (2x2) + 3 oxygen (3x1) = 7 oxygen on right
7. multiply O₂ by $\frac{7}{2}$
8. C₂H₆ + $\frac{7}{2}$O₂ $\rightarrow$ 2CO₂ + 3H₂O
9. remove fraction multiply both sides by 2
10. 2C₂H₆ + 7O₂ $\rightarrow$ 4CO₂ + 6H₂O
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Balancing Chemical Equations Continued

5. Check to make sure that you have the same number of each type of atom on both sides of the equation.
6. 2C₂H₆ + 7O₂ $\rightarrow$ 4CO₂ + 6H₂O
7. Reactants:
   • 4 C (2 x 2)
   • 12 H (2 x 6)
   • 14 O (7 x 2)
8. Products:
   • 4 C
   • 12 H (6 x 2)
   • 14 O (4 x 2 + 6)
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Chemical Problems!

• Write a balanced chemical equation for the complete combustion of gasoline (C₈H₁₈) in air
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Amounts of Reactants and Products

1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
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Example Problem

• Methanol burns in air according to the equation
  • 2CH₃OH + 3O₂ $\rightarrow$ 2CO₂ + 4H₂O
• If 209 g of methanol are used up in the combustion, what mass of water is produced?
• grams CH₃OH $\rightarrow$ moles CH₃OH $\rightarrow$ moles H₂O $\rightarrow$ grams H₂O
• molar mass CH₃OH ,coefficients chemical equation ,molar mass H₂O
• $209 \text{ g CH₃OH} \times \frac{1 \text{ mol CH₃OH}}{32.0 \text{ g CH₃OH}} \times \frac{4 \text{ mol H₂O}}{2 \text{ mol CH₃OH}} \times \frac{18.0 \text{ g H₂O}}{1 \text{ mol H₂O}} = 235 \text{ g H₂O}$
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Limiting Reagent

• 2NO + O₂ $\rightarrow$ 2NO₂
• Reactant used up first in the reaction
• NO is the limiting reagent
• O₂ is the excess reagent
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Example Using Limiting Reagent

• In one process, 124 g of Al are reacted with 601 g of Fe₂O₃
  • 2Al + Fe₂O₃ $\rightarrow$ Al₂O₃ + 2Fe
• Calculate the mass of Al₂O₃ formed.
• g Al $\rightarrow$ mol Al $\rightarrow$ mol Fe₂O₃ needed $\rightarrow$ g Fe₂O₃ needed
• OR
• g Fe₂O₃ $\rightarrow$ mol Fe₂O₃ $\rightarrow$ mol Al needed $\rightarrow$ g Al needed
• $124 \text{ g Al} \times \frac{1 \text{ mol Al}}{27.0 \text{ g Al}} \times \frac{1 \text{ mol Fe₂O₃}}{2 \text{ mol Al}} \times \frac{160. \text{ g Fe₂O₃}}{1 \text{ mol Fe₂O₃}} = 367 \text{ g Fe₂O₃}$
• Start with 124 g Al $\rightarrow$ need 367 g Fe₂O₃
• Have more Fe₂O₃ (601 g) so Al is the limiting reagent
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Continued Limiting Reagent Example

• Use limiting reagent (Al) to calculate amount of product that can be formed.
• g Al $\rightarrow$ mol Al $\rightarrow$ mol Al₂O₃ $\rightarrow$ g Al₂O₃
• $124 \text{ g Al} \times \frac{1 \text{ mol Al}}{26.98 \text{ g Al}} \times \frac{1 \text{ mol Al₂O₃}}{2 \text{ mol Al}} \times \frac{102.0 \text{ g Al₂O₃}}{1 \text{ mol Al₂O₃}} = 234.4 \text{ g Al₂O₃}$
• 2Al + Fe₂O₃ $\rightarrow$ Al₂O₃ + 2Fe
• At this point, all the Al is consumed and Fe₂O₃ remains in excess.
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Reaction Yield

• Theoretical Yield is the amount of product that would result if all the limiting reagent reacted.
• Actual Yield is the amount of product actually obtained from a reaction.
• $\% ext{ Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$
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EXAMPLE 3.16

• Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets, aircraft, jet engines, and bicycle frames. It is prepared by the reaction of titanium(IV) chloride with molten magnesium between 950°C and 1150°C:
  • TiCl₄(g) + 2Mg(l) $\rightarrow$ Ti(s) + 2MgCl₂(l)
• In a certain industrial operation $3.54 \times 10^7$ g of TiCl₄ are reacted with $1.13 \times 10^7$ g of Mg.
  • (a) Calculate the theoretical yield of Ti in grams.
  • (b) Calculate the percent yield if $7.91 \times 10^6$ g of Ti are actually obtained.
• Solution Carry out two separate calculations to see which of the two reactants is the limiting reagent. First, starting with $3.54 \times 10^7$ g of TiCl₄, calculate the number of moles of Ti that could be produced if all the TiCl reacted. conversions are
  • grams of TiCl₄ $\rightarrow$ moles of TiCl₄ $\rightarrow$ moles of Ti
  • $moles \text{ of Ti} = 3.54 \times 10^7 \text{ g TiCl₄} \times \frac{1 \text{ mol TiCl₄}}{189.7 \text{ g TiCl₄}} \times \frac{1 \text{ mol Ti}}{1 \text{ mol TiCl₄}} = 1.87 \times 10^5 \text{ mol Ti}$
  • Next, we calculate the number of moles of Ti formed from $1.13 \times 10^7$ g of Mg. The conversion steps are:
  • grams of Mg $\rightarrow$ moles of Mg $\rightarrow$ moles of Ti
  • $moles \text{ of Ti} = 1.13 \times 10^7 \text{ g Mg} \times \frac{1 \text{ mol Mg}}{24.31 \text{ g Mg}} \times \frac{1 \text{ mol Ti}}{2 \text{ mol Mg}} = 2.32 \times 10^5 \text{ mol Ti}$
  • Therefore, TiCl₄ is the limiting reagent because it produces a smaller amount of Ti. The mass of Ti formed is
  • $1.87 \times 10^5 \text{ mol Ti} \times \frac{47.88 \text{ g Ti}}{1 \text{ mol Ti}} = 8.95 \times 10^6 \text{ g Ti}$
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Solution to Example 3.16 Continued

• The percent yield is given by
  • $\% ext{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% = \frac{7.91 \times 10^6 \text{ g}}{8.95 \times 10^6 \text{ g}} = 88.4\%$
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Chemical Problems!

• How many oxygen atoms are present in 3.14 g Dideoxysucrose (C₁₂H₂₂O₉)?
• 650 g of NH₃ are treated with 1200 g of CO₂ to produce (NH₂)₂CO:
  • 2NH₃(g) + CO₂(g) $\rightarrow$ (NH₂)₂CO(aq) + H₂O(l)
  • A) Which reactants is the limiting reagent?
  • B) Calculate the percent yield if 325 g of (NH₂)₂CO are obtained
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Gases

• Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.
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Physical Characteristics of Gases

• Gases assume the volume and shape of their containers.
• Gases are the most compressible state of matter.
• Gases have much lower densities than liquids and solids.
• Gases will mix evenly and completely when confined to the same container.
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Boyle’s Law

• $P \propto \frac{1}{V}$
• $P \times V = constant$
• $P*1 \times V*1 = P*2 \times V*2$
• Constant temperature
• Constant amount of gas
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Boyle’s Law Example

• A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?
• $P*1 \times V*1 = P*2 \times V*2$
• $P_1 = 726 \text{ mmHg}$
• $V_1 = 946 \text{ mL}$
• $P_2 = ?$
• $V_2 = 154 \text{ mL}$
• $P*2 = \frac{P*1 \times V*1}{V*2} = \frac{726 \text{ mmHg} \times 946 \text{ mL}}{154 \text{ mL}} = 4460 \text{ mmHg}$
• $P \times V = constant$
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Charles’ Law

• As T increases V increases
• Variation in Gas Volume with Temperature at Constant Pressure
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Charles’ Law Continued

• Variation of Gas Volume with Temperature at Constant Pressure
• $V \propto T$
• $V = constant \times T$
• $\frac{V*1}{T*1} = \frac{V*2}{T*2}$
• T \text{ (K)} = t \text{ (^oC)} + 273.15
• Temperature must be in Kelvin
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Charles’ Law Example

• A sample of carbon monoxide gas occupies 3.20 L at 125 ^oC. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?
• $\frac{V*1}{T*1} = \frac{V*2}{T*2}$
• $V_1 = 3.20 \text{ L}$
• $T_1 = 398.15 \text{ K}$
• $V_2 = 1.54 \text{ L}$
• $T_2 = ?$
• $T*2 = \frac{V*2 \times T*1}{V*1} = \frac{1.54 \text{ L} \times 398.15 \text{ K}}{3.20 \text{ L}} = 192 \text{ K}$
• T_1 = 125 \text{ (^oC)} + 273.15 \text{ (K)} = 398.15 \text{ K}
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Avogadro’s Law

• $V \propto \text{ number of moles (n)}$
• $V = constant \times n$
• $\frac{V*1}{n*1} = \frac{V*2}{n*2}$
• Constant temperature
• Constant pressure
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Avogadro’s Law Example

• Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?
• 4NH₃ + 5O₂ $\rightarrow$ 4NO + 6H₂O
• 1 mole NH₃ $\rightarrow$ 1 mole NO
• At constant T and P
• 1 volume NH₃ $\rightarrow$ 1 volume NO
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Ideal Gas Equation

• Charles’ law: $V \propto T$ (at constant n and P)
• Avogadro’s law: $V \propto n$ (at constant P and T)
• Boyle’s law: $P \propto \frac{1}{V}$ (at constant n and T)
• $V \propto \frac{nT}{P}$
• $V = constant \times \frac{nT}{P} = R \frac{nT}{P}$
• R is the gas constant
• PV = nRT
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Standard Temperature and Pressure

• The conditions 0 ^oC and 1 atm are called standard temperature and pressure (STP).
• PV = nRT
• $R = \frac{PV}{nT} = \frac{(1 \text{ atm})(22.414 \text{ L})}{(1 \text{ mol})(273.15 \text{ K})} = 0.082057 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}}$
• Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.
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Ideal Gas Equation Example

• What is the volume (in liters) occupied by 49.8 g of HCl at STP?
• PV = nRT
• $\text{V} = \frac{nRT}{P}$
• $$T = 0 \text{ }