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Chamras

CHEMISTRY 110 LECTURE NOTES

s: Chp’s. 12, 13, 14, 15, 16 EXAM 4 Material

Chapter 12 Liquids & Solids

Remember: Energetics of States:

S L G P

Rel. E.:

Remember: Phase Changes:

Liquid

Solid Gas Boiling: L à G Boiling Point: Normal Boiling Point: Freezing: L à S Freezing Point: What happens at the Molecular Level, at the Freezing and Boiling points?

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Intermolecular Attractive Forces: (Intramolecular Attractive Forces:) Heating-Cooling Diagram: (Thermodynamics) Temp.

Heat (Thermal Energy) Content

***Two Types of Energetic Changes:

a) Temperature Change Within A State:

b) State Change:

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Temperature Change Within A State:

Q = m . s . ΔT State Change:

Q = Kf . m Q = Kv . m

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Sample Problem: A 12.50mg sample of iron is heated from 10.0oC to 2,323.3oC. Below is a list of some physical properties of iron you may find useful: MP=1,538.0oC, BP=2,861.5oC, kf=2.01kJ.g–1, kv=0.78kJ.g–1, s(s)=1.02J.g–1.oC–1, s(l)=0.87 J.g–1.oC–1 s(g)=0.445 J.g–1.oC–1.

a) Construct a complete and correct heating-cooling diagram reflecting the energetic changes described above.

b) Calculate the amount of heat exchanged.

c) Determine whether this thermal energy exchange is exothermic or endothermic for the iron sample.

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Ener

gy

Reaction Progress (time)

Ener

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Reaction Progress (time)

Ener

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Reaction Progress (time)

Ener

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Reaction Progress (time)

Ener

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Reaction Progress (time)

Ener

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Reaction Progress (time)

Essential Parts from Chapter 15: Thermodynamics of Chemical Reactions Energy Diagrams for Chemical Reactions: A + B C Exothermic Vs. Endothermis Vs. Isothermic Chemical reactions:

A + B C Endothermic Exothermic Isothermic

Energetically “Reversible” Vs. “Irreversible” Chemical Reactions: Reversible Irreversible

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Ener

gy

Reaction Progress (time)

For reversible cases: A + B C C A + B A + B C Forward Rate Vs. Reverse Rate

RF > RR RF < RR RF = RR

Dynamic Equilibrium:

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Liquids More on Intermolecular Attractive Forces: Types:

a) Dipole-Dipole:

***Hydrogen Bonding: A specific type of Dipole-Dipole*** Criteria for H-bonding:

b) Induced Dipole-Induced Dipole (London Dispersion):

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Vapor Pressure: A B *What is keeping the liquid molecules from complete evaporation? * How is the vapor pressure related to the boiling point of a liquid?

BP ßà P(vap) How is boiling point related to the strength of intermolecular forces?

BP ßà Intermol. Forces

How is vapor pressure related to the strength of intermolecular forces?

P(vap) ßà Intermol. Forces

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Solids The least energetic state Types:

Crystalline Solids:

i) Ionic Solids: Ions ii) Molecular Solids: Molecules

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iii) Atomic Solids: Elements Metallic Bond:

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Chapter 13 Solutions

*Remember:

a) Solution: b) Components of Solutions: c) Solubility: d) Types of Solutions: e) The process of Dissolution: For soluble ionic compounds, strong acids & bases, weak

acids & bases, polar molecular compounds. Types of Solutions: Based on the relative amounts of Solute & Solvent:

a) Unsaturated:

b) Saturated:

c) Supersaturated: Concentration:

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Concentrated Vs. Dilute Solutions: A Quantitative Method for Measuring Concentration:

Molarity Meaning: Unit: Symbol: Usage of the term: Solution examples:

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Sample Problems: Type A: Given information on the amount of the solute & the volume of solution, simply plug in, and then convert, if necessary, using dimensional analysis to determine the molarity. Examples: a) Determine the concentration of a 44.6mL of an aqueous solution of calcium chloride with 3.39 moles of calcium chloride dissolved: b) Determine the concentration of a 44.6mL of an aqueous solution of sodium chloride with 3.39 grams of sodium chloride dissolved: Type B: Given the volume of solution and the molarity, start your set up with volume, convert if necessary into liters, and then use the value for molarity (as a wild card) to determine the moles of solute. In case the problem asks for the mass of the solute, then link one more conversion factor to determine the amount of the solute in mass units rather than in moles. Example: a) Determine the moles of potassium nitrate present in 250.0 mL of 0.20M solution of potassium nitrate. b) Determine the mass of potassium nitrate in milligrams present in 250.0 mL of 0.20M solution of potassium nitrate.

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Type C: Given the moles or the mass of the solute and the molarity, start your set up with mass or moles of solute, convert if necessary into moles, and then use the value for molarity (as a wild card) to determine the volume of solution. In case the problem asks for the volume of the solute in any unit other than in liters, link one more conversion factor to convert the calculated volume into the desired unit. Example: a) Determine the volume of a 4.5M solution of lithium hydroxide in milliliters with 2.2 moles of lithium hydroxide present. b) Determine the volume of a 4.5M solution of lithium hydroxide in milliliters with 33.3 milligrams of lithium hydroxide present.

Dilution Definition: A decrease of the concentration of a solution through the addition of more solvent (water). Equation:

M1.V1 = M2.V2 M1 = V1 = M2 = V2 =

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Sample dilution problems: a) Determine the final concentration of 55.6 mL solution of 2.2M of hydrochloric acid, which was diluted to a final volume of 1.00L. b) Determine the final volume of 750.0mL solution of potassium permanganate which was diluted from 2.40M to 1.8M concentration.

Solution Stoichiometry Volume of sol’n. of A Volume of sol’n. of B Molarity of Molarity of mass of A of sol’n. of A sol’n. of A mass of B molar mass molar mass of A of B PV=nRT mol:mol PV=nRT Volume of moles of A moles of B Volume of A(g) ratio B(g) avogadro’s # avogadro’s # particles of A particles of B

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Sample Problems: 1. The balanced equation for the reaction of zinc with a solution of hydrochloric acid is shown below:

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) If 3.33 grams of zinc reacts with 25.0mL of 6.0M HCl solution, determine the mass of hydrogen gas expected to be produced: 2. The equation below describes the double displacement reaction between solutions of 2.00M aluminum bromide and 4.20M silver nitrate: AlBr3(aq) + AgNO3(aq)

a) Complete and balance this equation. b) If 100.0 mL of each solution is mixed in the reaction, determine the milligrams of

silver bromide expected to be produced:

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Chapter 14 Acids & Bases

Definitions: 1. Arrhenius Definition: Acid: Base: Example: HCl(aq) NaOH(aq) 2. Bronsted-Lowry: Acid: Base: Example: HCl(aq) NaOH(aq) HCl(aq) + NaOH(aq)

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Acids in water: (Hydronium Ion Formation) General Equation: Specific Example: Bases in water: (Hydroxide ion Formation) General Equation: Specific Example:

Water: An Acid or A Base? Conjugate Base: Example:

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Acid Strength Determined by the extent of dissociation (i.e. the extent of Hydronium Ion Formation) How to measure the extent of the Hydronium Ion Formation?

HA(aq) H2O(l) H3O+(aq) + A–(aq)

Ionization of Water Water: a polar molecular compound [ ] = concentration [H3O]+ = [OH]– = Kw = Ionization Constant for Water @ 25oC = = = Kw = A very small number! Check your observation of Electrolytes Demo on water. 2. Quantitative Assessment Method: An acidic Solution: A Basic Solution: A Neutral Solution:

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The pH Scale for Acidity The Scale in Water (for aqueous solutions of acids and bases): p (lower case) = a function Meaning = p… = –log … px = –logx à pH = –log [H+] Examples: Solution A Solution B [H]+ = 4.4 M [H]+ = 0.010M pH(A) = pH(B) =

pOH pH + pOH = 14 Example:

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pH Test: ***In-Class Demonstration: Solution Sample pH pH

Ranking pOH Comments

Results: 1. pH and acidic strength are ________________ proportional. 2. pH and basic strength are ________________ proportional.. 3. pOH and acidic strength are ________________ proportional. 4. pOH and basic strength are ________________ proportional.

Buffer Solutions Definition: Examples:

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Chapter 16 Oxidation–Reduction (Redox)

Definitions: Oxidation (half-reaction): Example: Reduction (half-reaction): Example: Oxidation State: Change in Oxidation State: Examples of Oxidation-Reduction Reactions:

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Examples of Commonly Encountered Redox Reactions: 1. Bleach:

2. Photo-chromic Lens:

Many people who wear eye glasses prefer those made with photochromic lenses or glass lenses which darken when exposed to bright light. These eyeglasses eliminate the need for sunglasses as they can reduce up to 80% of transmitted light. The basis of this change in color in response to light can be explained in terms of oxidation-reduction reactions. Glass consists of a complex matrix of silicates that is ordinarily transparent to visible light. In photochromic lenses, silver chloride (AgCl) and copper (I) chloride (CuCl) crystals are added during the manufacturing of the glass while it is in the molten state and these crystals become uniformly embedded in the glass as it solidifies. One characteristic of silver chloride is its susceptibility to oxidation and reduction by light as described below.

Cl- -----------> Cl + e- oxidation Ag+ + e- -----------> Ag reduction The chloride ions are oxidized to produce chlorine atoms and an electron. The electron is then transferred to silver ions to produce silver atoms. These atoms cluster together and block the transmittance of light, causing the lenses to darken. This process occurs almost instantaneously. As the degree of "darkening" is dependent on the intensity of the light,

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these photochromic lenses are quite convenient and all but eliminate the need for an extra pair of sunglasses.

The photochromic process would not be useful unless it were reversible. The presence of copper (I) chloride reverses the darkening process in the following way. When the lenses are removed from light, the following reactions occur:

Cl + Cu+ ------> Cu+2 + Cl- oxidizing agent reducing agent oxidized species reduced species The chlorine atoms formed by the exposure to light are reduced by the copper ions, preventing their escape as gaseous atoms from the matrix. The copper (+1) ion is oxidized to produce copper (+2) ions, which then reacts with the silver atoms as shown.

Cu+2 + Ag ------> Cu+1 + Ag+ oxidizing agent reducing agent reduced species oxidized species The net effect of these reactions is that the lenses become transparent again as the silver and chloride atoms are converted to their original oxidized and reduced states. 3. Metabolism: Metabolism is a general term used to refer to all of the chemical reactions which occur in a living system. Metabolism can be divided into two parts; anabolism, or reactions involving the synthesis of compounds; and catabolism, or reactions involving the breakdown of compounds. In terms of oxidation-reduction principles, anabolic reactions are primarily characterized by reduction reactions, such as the dark reaction in photosynthesis where carbon dioxide is reduced to form glucose. Catabolic reactions are primarily oxidation reactions. Although catabolism involves many separate reactions, an example of such as process can be described by the oxidation of glucose as shown below. Note that this equation is the reverse of the photosynthetic equation.

C6H12O6 + 6 O2 ----------> 6 CO2 + 6 H2O + Energy Note also, that in this reaction, the carbon atoms in glucose are oxidized, undergoing an increase in oxidation state (each carbon loses 2 electrons) as they are converted to carbon dioxide. At the same time, each oxygen atom is reduced by gaining 2 electrons when it is converted to water. Part of the energy is released as heat and the remainder is stored in the chemical bonds of "energetic" compounds such as adenosine triphosphate (ATP) and nicotinamide adenine dinucleotide (NADH).

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Catabolic reactions can be divided into many different groups of reactions called, catabolic pathways. In these pathways (referred to as Glycolysis, the Citric Acid Cycle, and Electron Transport) the carbon atoms are slowly oxidized by a series of reactions which gradually modify the carbon skeleton of the compound as well as the oxidation state of carbon. Coupled to these reactions are other reversible oxidation-reduction reactions designed to capture the energy released and temporarily store it within the chemical bonds of compounds called adenosine triphosphate (ATP) and nicotamide dinucleotide (NADH) . These compounds are then utilized to provide energy for driving the cellular machinery.

Assigning Oxidation States Examples: CO2 NO3

– CH4

Oxidizing & Reducing Agents

Examples:

Assigning Oxidized & Reduced Species

Examples:

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Balancing Oxidation-Reduction Reactions Given a unbalanced oxidation-reduction reaction: 1. Separate the reaction into oxidation and reaction half-reactions 2. Balance the half-reactions separately:

• Balance the charges as well as the atoms. • Balance charges by adding electron/s to the appropriate side of each half-reaction.

3. Equalize the number of electrons gained and lost, if necessary: Use multipliers for the half reactions. 4. Add the half-reactions together and cancel the electrons to acquire the overall balanced equation: 5. Check to make sure the elements and the charges balance.

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Example: Balance the following Redox Equation using the half-reaction method: Note: The complete (unbalanced) molecular equation is: Al(s) + NiCl2(aq)àAlCl3(aq) + Ni(s)

Al(s) + Ni2+(aq) Al3+(aq) + Ni(s)

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Balancing Oxidation-Reduction Reactions Occurring in Acidic Solutions Given a unbalanced oxidation-reduction reaction:

MnO4–(aq) + Fe2+(aq) Acid Fe3+(aq) + Mn2+(aq)

1. Identify and write the equations for the oxidation and the reduction half-reactions: 2. Balance each half-reaction:

1. Balance all elements except for H and O. 2. Balance O using H2O. 3. Balance H using H+. 4. Balance the charge using electron/s.

3.Equalize the number of electrons gained and lost, if necessary: Use multipliers for the half reactions.

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4. Add the half-reactions together and cancel identical species including the electrons to acquire the overall balanced equation: 5. Check to make sure the elements and the charges balance.