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OK STATE / Engineering / CHEM 1515 / What is the composition of solutions?

What is the composition of solutions?

What is the composition of solutions?

Description

School: Oklahoma State University
Department: Engineering
Course: General Chemistry 2
Professor: John gelder
Term: Fall 2015
Tags: chemistry2, Chemistry, and General Chemistry
Cost: 50
Name: Chemistry 2 Exam 2 Study Guide
Description: All of these notes cover what will be included on Exam 2
Uploaded: 03/01/2020
45 Pages 68 Views 2 Unlocks
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Chapter 14, Lecture 1


What is the composition of solutions?



SOLUTIONS

KEY:

Yellow Highlight = key words to know/memorize

Red Underline = definition of key words

Green Highlight = general concepts to remember 

Blue Highlight = Katy’s notes

1. COMPOSITION OF SOLUTIONS

A. In general, a solution consists of a solute (minority component) dissolved in a solvent

i. The solvent is the most abundant component of a given solution ii. The solubility (s) of a solute is by definition “the maximum amount that dissolves in a fixed quantity of a given solvent at a specific temperature” 1. Dif erent solutes have dif erent solubilities

iii. Solutions can be either dilute or concentrated.


What are the homogeneous mixtures?



1. These are qualitative terms that refer to the relative amount of solute: a. A dilute solution contains much less dissolved solute that a concentrated one. 

2. THIRSTY SOLUTIONS: SEAWATER

A. Drinking seawater can cause dehydration

B. Seawater:

i. Is a homogeneous mixture of salts with water

ii. Contains higher concentrations of salts than salt content of human body cells. 1. This means that there is more salt in sea water than in your body’s cells, causing a form of osmosis to occur. The sea water draws the water out of your body cells to balance out its own concentration levels, causing dehydration. 2. This is due to nature’s tendency toward spontaneous mixing. 


What are the common types of solutions?



If you want to learn more check out What is an action potential?

3. Usually results in diarrhea as the extra fluid flows out with the seawater

3. HOMOGENEOUS MIXTURES

A. A mixture of two or more substances 

B. Appear to be one substance, through really contains multiple materials. C. Most homogeneous materials we encounter are actually solutions i. i.e. Air, Seawater

D. Nature has a tendency towards spontaneous mixing

i. Generally, uniform mixing is more energetically favorable Don't forget about the age old question of What is the meaning of stressor?

4. COMMON TYPES OF SOLUTIONS

A. A solution may be composed of a solid and a liquid, a gas and a liquid, or other combinations.

B. In aqueous solutions, water is the solvent. 

Solution Phase Solute Phase Solvent Phase Examples

Gaseous Solution

Gas

Gas

Air (mainly

oxygen and

nitrogen)

Liquid Solution

Gas

Liquid

Club Soda (CO2

+ water)

Liquid

Liquid

Vodka (ethanol

+ water)

Solid

Liquid

Seawater (salt

+ water)

Solid Solution

Solid

Solid

Brass (copper +

zinc)

If you want to learn more check out What is the meaning of string literal?

5. SOLUBILITY

A. When one substance (solute) dissolves in another (solvent) is it said to be soluble. i. Salt is soluble in water

ii. Bromine is a liquid soluble in methylene chloride

B. When one substance does not dissolve in another, it is said to be insoluble i. Oil is insoluble in water

C. The solubility of one substance in another depends on:

i. Nature’s tendency towards mixing

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ii. The types of intermolecular attractive forces

6. NATURE’S TENDENCY TOWARD MIXING: ENTROPY

A. Many physical systems tend toward lower potential energy

B. But formation of a solution does not necessarily lower the potential energy of the system

C. A typical example is when two ideal gasses are put into the same container, they spontaneously mix, even though there are no significant intermolecular forces between their constituent particles.

i. The two gasses mix but their potential energy remains unchanged. D. The tendency to mix is related to a concept called entropy.

7. MIXING AND THE SOLUTION PROCESS We also discuss several other topics like What is the framework for financial accounting?

A. Entropy is a measure of energy randomization or energy dispersal throughout the system.

B. When mixing, each gas along with its kinetic energy becomes spread out or dispersed over as large a volume it is allowed

i. I.E. as large as its container allows it to become 

C. By each gas expanding to fill the container, the mixture of the two gases has a greater energy dispersal, or greater entropy, than the separated components.

8. SOLUTIONS: EFFECT OF INTERMOLECULAR FORCES

A. Energy changes in the formation of most solutions also involve differences in attractive forces between the particles. If you want to learn more check out What is the difference between life span and life expectancy?

B. For the solvent and solute to mix, the following must be overcome: i. All of the solute-to-solute attractive forces

ii. Or some of the solvent-to-solvent attractive forces

1. Both processes are endothermic (i.e. absorb heat) 

C. Solution is formed when new solute-to-solvent attractions are established. The process of solute solvent attraction is exothermic (i.e. releases heat) 

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D. When the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent-to-solvent attractions, the solution will only form if the energy difference is small enough to be overcome by the increase in entropy form mixing.

9. SOLUBILITY

A. The maximum amount of solute that can be dissolved in a given amount of solvent is called the solubility.

B. There is usually a limit to the solubility of one substance in another. i. Gases are always soluble in each other If you want to learn more check out What orbit the atom?

ii. To liquids that are mutually soluble are said to be miscible.

1. Alcohol and water are miscible

2. Oil and water are immiscible

C. The solubility of one substance in another varies with temperature and pressure.

10.WILL IT DISSOLVE?

A. Chemist’s rule of thumb- “like dissolves like”. 

B. A chemical will dissolve in a solvent if it has a similar structure to the solvent. C. Polar molecules and ionic compounds will be more soluble in polar solvents. D. Non polar molecules will be more soluble in non polar solvents 

11.HEAT OF SOLUTION

A. When some compounds, such as NaOH, dissolve in water, a lot of heat is released i. The container gets hot (exothermic)

B. When other compounds, such as NH4NO3, dissolve in water, heat is absorbed from the surroundings

i. The container gets cold (endothermic)

12.ENERGETICS OF SOLUTION FORMATION: THE ENTHALPY OF SOLUTION

A. To make a solution you must

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i. Overcome all attractions between the solute particles; therefore ∆H(solute) is endothermic

ii. Overcome some attractions between solvent molecules; therefore, ∆H(solvent) is endothermic

iii. Form new attractions between solute particles and solvent molecules; therefore ∆H(mix) is exothermic.

B. The overall ∆H for making a solution depends on the relative sizes of the ∆H for these three processes. ∆H(solution)=∆H(solute)+∆H(solvent)+∆H(mix) 

13.SOLUTION PROCESS

A. Step 1: Separating the solute into its constituent particles. ∆H(solute)>0

B. Step 2: Separating the solvent particles from each other to make room for the incoming solute particles. ∆H(solvent)>0

C. Step 3: Mixing the solute particles with the solvent particles. ∆H(mix)<0

14.ENERGETICS OF SOLUTION FORMATION

A. If the total energy cost for breaking attractions between particles in the pure solute and pure solvent is less than the energy released in making the new attractions between the solute and solvent, the overall process will be exothermic.

B. If the total energy cost for breaking attractions between particles in the pure solute and pure solvent is greater than the energy released in making the new attractions between the solute and solvent, the overall process will be endothermic.

15.HEATS OF HYDRATION

A. For aqueous solutions of ionic compounds, the energy added to overcome the attractions between water molecules and the energy released in forming attractions between the water molecules and ions are combined into a term called Heat of Hydration. 

i. Attractive forces between ions=lattice energy

1. ∆H(solute)=-∆H(lattice energy)

ii. Attractive forces in water=H bonds

iii. Attractive forces between ion and water=ion-dipole

iv. ∆H(hydration)=heat released when 1 mole of gaseous ions dissolves in water=∆H(solvent)+∆H(mix)

5

16.ION-DIPOLE INTERACTIONS

A. When ions dissolve in water they become hydrated

i. Each ion is surrounded by water molecules

B. The formation of these ion-dipole attractions causes the heat of hydration to be very exothermic.

17.HEATS OF SOLUTION FOR IONIC COMPOUNDS

A. For an aqueous solution of an ionic compound, the ∆H(solution) is the difference between the heat of hydration and the lattice energy

∆H(SOLUTION)=∆H(SOLUTE)+∆(SOLVENT)+∆H(MIX)

∆H(SOLUTION)=-∆H(LATTICE ENERGY)+∆H(SOLVENT+∆H(MIX)

∆H(SOLUTION)=∆H(HYDRATION)-∆H(LATTICE ENERGY)

18.COMPARING HEAT OF SOLUTION TO HEAT OF HYDRATION

A. Because the lattice energy is always exothermic, the size and sign on the

∆H(solution) tells us something about ∆H(hydration).

B. If the heat of solution is large and endothermic, then the amount of energy it costs to separate the ions is more than the energy released from hydrating the ions

i. ∆H(hydration)<∆H(lattice) when ∆H(solution) is (+) positive

C. If the heat of solution is large and exothermic, then the amount of energy it costs to separate the ions is less than the energy released from hydrating the ions.

i. ∆H(hydration)>∆H(lattice) when ∆H(solution) is (-) negative.

6

Chapter 14 Lecture 2

SOLUTIONS

1. SOLUTION EQUILIBRIUM

A. The dissolution of a solute in a solvent is an equilibrium process

B. Initially, when there is no dissolved solute, the only process possible is dissolution.

C. Shortly after some solute is dissolved, solute particles can start to recombine to reform molecules, but the rate of dissolution is much greater than the rate of deposition and the solute continues to dissolve.

D. Eventually the rate of dissolution = the rate of deposition

i. The solution is saturated and no more solute will dissolve.

2. SOLUBILITY LIMIT

A. A solution that has the solute and solvent in dynamic equilibrium is said to be saturated.

i. If you add more solute, it will not dissolve.

ii. The saturation concentration depends on the temp. and pressure. B. A solution that has less solute than saturation is said to be unsaturated. C. A solution that has more solute than saturation is said to be supersaturated.

3. HOW CAN WE MAKE A SOLVENT HOLD MORE SOLUTE THAN IT IS ABLE TO? A. Solutions can be made saturated at non-room conditions, and then can be allowed to come to room conditions slowly.

B. For some solutes, instead of coming out of solution when the conditions change, they get stuck between the solvent molecules and the solution becomes supersaturated.

C. Supersaturated solutions are unstable and lose all the solute above saturation when disturbed.

i. For example, shaking a carbonated beverage.

4. TEMPERATURE DEPENDENCE OF SOLUBILITY OF SOLIDS IN WATER

A. Solubility is generally given in grams of solute that will dissolve in 100g of water. B. For most solids, the solubility of the solid increases as the temp. increases. i. When ∆H(solution) is endothermic

C. Solubility curves can be used to predict whether a solution with a particular amount of solute dissolved in water is saturated (on the line), unsaturated (below the line), or supersaturated (above the line).

5. PURIFICATION BY RECRYSTALLIZATION

A. One of the common operations performed by a chemist is removing impurities from a solid compound.

B. One method of purification involves dissolving a solid in a hot solvent until the solution is saturated.

C. As the solution slowly cools, the solid crystallizes out, leaving impurities behind.

6. TEMPERATURE DEPENDENCE OF SOLUBILITY OF GASES IN WATER

A. Gases generally have lower solubility in water than ionic or polar covalent solids because most are non polar molecules

i. Gases with high solubility usually are actually reacting with water. B. For all gases, the solubility of the gas decreases as the temp. increases.

i. The ∆H(solution) is exothermic because you do not need to overcome solute-solute attractions.

7. PRESSURE DEPENDENCE OF SOLUBILITY OF GASES IN WATER

A. The larger the partial pressure of a gas in contact with a liquid, the more soluble the gas is in the liquid.

8. HENRY’S LAW

A. The solubility of a gas (S gas) is directly proportional to its partial pressure (P gas). i. S(gas) = K(h)P(gas)

B. K(h) is the Henry’s Law Constant

2

9. CONCENTRATIONS

A. Solutions have variable composition

B. To describe a solution, you need to describe the components and their relative amounts.

C. The terms dilute and concentrated can be used as qualitative descriptions of the amount of solute in solution.

D. Concentration= amount of solute in a given amount of solution. i. Occasionally amount of solvent

10.SOLUTION CONCENTRATION: MOLARITY/MOLALITY

A. Molarity= moles of solute per 1 liter of solution

i. Describes how many moles of solute in each liter of solution

ii. If a sugar solution concentration contains 2.0 M

1. 1 liter of solution contains 2.0 moles of sugar

2. 2 liters = 4.0 moles of sugar

3. 0.5 liters = 1.0 moles of sugar

iii. Molarity (M) = moles of solute/liters of solution

B. Molality (m) = moles of solute per 1 kilogram of solvent

i. Defined in terms of amount of solvent, not solution like the others ii. Does not vary with temperature

iii. Based on masses, not volume

iv. Molality (m) = moles of solute/kg of solvent

11.PARTS SOLUTE IN PARTS SOLUTION\

A. Parts can be measured by mass or volume

B. Parts are generally measured in the same units

i. By mass: grams, kilograms, pound, etc.

ii. By volume: mL, L, gallons, etc.

iii. Mass and volume combined in grams and mL.

C. Percentage = parts of solute in every 100 parts solution

i. If a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution (or 0.9 kg solute in every 100 kg of solution).

3

D. Parts per million = parts of solute in every 1 million parts solution

i. If a solution is 36 ppm by volume, then there are 36 mL of solute in 1 million mL of solution.

12.PARTS PER MILLION (PPM) AND PARTS PER BILLION (PPB)

A. PPM = grams of solute per 1,000,000g of solution or mg of solute per 1 kg of solution.

B. 1 liter of water = 1 kg of water

i. For aqueous solutions, we often approximate the kg of the solution as the kg or L of water

1. For dilute solutions, the dif erence in density between the solution and pure water is usually negligible.

13.SOLUTION CONCENTRATIONS: MOLE FRACTION

A. The mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution

B. Total of all the mole fractions in a solution = 1

C. Unit-less

D. The mole percentage is the percentage of the moles of one component in the total moles of all the components of the solution.

i. The mole % = mole fraction • 100%

4

14.CONCENTRATIONS: A SUMMARY

UNIT 

DEFINITION 

UNITS

MOLARITY (M)

AMOUNT SOLUTE/VOLUME SOLUTION

MOL/L

MOLALITY

AMOUNT SOLUTE/MASS SOLVENT

MOL/KG

MOLE FRACTION

AMOUNT SOLUTE/TOTAL (SOLUTE + SOLVENT)

NONE

MOLE PERCENT (MOL %)

AMOUNT SOLUTE/TOTAL (SOLUTE + SOLVENT) • 100 %

%

PARTS BY MASS

(MASS SOLUTE/MASS SOLUTION) • MULTIPLICATION FACTOR

PERCENT BY MASS

MULTIPLICATION FACTOR = 100

%

PARTS PER MILLION BY MASS (PPM)

MULTIPLICATION FACTOR = 10ˆ6

PPM

PARTS PER BILLION BY MASS (PPB)

MULTIPLICATION FACTOR = 10ˆ9

PPB

PARTS BY VOLUME

(VOLUME SOLUTE/VOLUME SOLUTION) • MULTIPLICATION FACTOR

5

Chapter 14, Lecture 3

SOLUTIONS

1. COLLIGATIVE PROPERTIES OF SOLUTIONS

A. There are four properties known as colligative properties (or collective properties): i. Vapor pressure lowering

ii. Boiling point elevation

iii. Freezing point depression

iv. Osmotic pressure

B. Usually, the number of solute particles dissolved in a given solution makes the  difference in the four solution properties, and not their chemical identity.  

C. We have two kinds of solutions:

i. Electrolyte solutions:

1. Strong electrolytes such as soluble salts (e.g. NaCl), strong acids (e.g.  HCl), and strong bases (e.g. NaOH). They dissolve completely and their  solutions conduct electricity well.  

2. Weak electrolytes such as weak acids (e.g. CH3COOH) and weak bases  (e.g. NH3). They dissociate very little and consequently their solutions  conduct electricity poorly.  

ii. Non-electrolyte solutions are those of compounds that do not dissociate into  ions (e.g. sucrose solution) at all and their solutions do not conduct electricity.  D. In assessing the magnitude of colligative properties, we use the solute formula to  find the number of particles in solution:  

i. Each mole of non-electrolyte yields 1 mole of particles in solution. 1. 0.60 M solution of sucrose contains 0.60 mol of solute particles per liter.  ii. Each mole of strong electrolyte dissociates into the number of moles of ions in  the formula unit.

1. 0.30 M solution of FeCl3 yields 0.30 mol of Fe(3+) and 0.90 mol of Cl(-)  ions per liter:  

a. FeCl3(aq)—> Fe(3+)(aq)+3Cl(-)(aq)

2. VAPOR PRESSURE LOWERING

A. The vapor pressure of a solvent in a solution is always lower than the vapor  pressure of the pure solvent.

B. The vapor pressure of the solution is directly proportional to the amount of the  solvent in the solution.  

C. The difference between the vapor pressure of the pure solvent and the vapor  pressure of the solvent in solution is called the vapor pressure lowering.

i. ∆P=P˚(solvent)-P(solution)=P˚(solvent)-X(solvent)•P˚(solvent)=

P˚(solvent)-[1-X(solute)]•P˚(solvent)

∆P=X(solute)•P˚(solvent)

D. Thus the lowering of the vapor pressure is directly proportion to the mole fraction  of the solute [X(solute)].

3. RAOULT’S LAW FOR VOLATILE SOLUTE

A. When both the solvent and the solute can evaporate, both molecules will be found in  the vapor phase.

B. The total vapor pressure above the solution will be the sum of the vapor pressures  of the solute and solvent.

i. For an ideal solution

1. P(total)=P(solute)+P(solvent)

C. The solvent decreases the solute vapor pressure in the same way the solute  decreases the solvent’s vapor pressure.

i. P(solute)=X(solute)•P˚(solute)

ii. P(solvent)=X(solvent)•P˚(solvent)

4. IDEAL VS. NON-IDEAL SOLUTION

A. In ideal solutions, the made solute-solvent interactions are equal to the sum of the  broken solute-solute and solvent-solvent interactions.

i. Ideal solutions follow Raoult’s Law.

B. Effectively, the solute is diluting the solvent.

C. If the solute-solvent interactions are stronger or weaker than the broken  interactions, the solution is non-ideal.

2

5. VAPOR PRESSURE OF A NON-IDEAL SOLUTION

A. When the solute-solute interactions are stronger than the solute-solute + solvent solvent interactions, the total vapor pressure of the solution will be less than  predicted by Raoult’s Law, because the vapor pressures of the solute and solvent are  lower than ideal.

B. When the solute-solvent interactions are weaker than the solute-solute + solvent solvent interactions, the total vapor pressure of the solution will be more than  predicted by Raoult’s Law.

6. OTHER COLLIGATIVE PROPERTIES RELATED TO VAPOR  PRESSURE LOWERING

A. Vapor pressure lowing occurs at all temperatures.

B. This results in the temperature required to boil the solution being higher than the  boiling point of the pure solvent = boiling point elevation

C. This also results in the temperature required to freeze the solution being lower than  the freezing point of the pure solvent = freezing point depression.  

7. FREEZING POINT DEPRESSION

A. The freezing point of a solution is lower than that of a pure solvent. i. Therefore, the melting point of the solution is also lower.

B. The difference between the freezing point of the solution and freezing point of the  pure solvent is directly proportional to the molal concentration (m) of solute  particles.

i. T(f)(solvent)-T(f)(solution)=∆T= m•K(f)

C. The proportionality constant is called the freezing point depression constant, K(f). i. The value of K(f) depends on the solvent:

1. Water: K(f)=1.86 ˚C/m

2. The units of K(f) are degrees Celsius per molal unit (˚C/m).

D. ∆T(f) is considered as a positive value

i. ∆T(f)=T(f)(solvent)-T(f)(solution)

E. In aqueous solutions the effect is quite small because the K(f) value for water is  small, only 1.86 ˚C/m.

i. Thus, 1m solution with 1 mol of particles per kg of water freezes at -1.86˚C at 1  atm instead of at 0.00˚C.

ii. Values of K(f) for several solvents are listed in Table 14.8 (page 609)

3

F. Plane de-icer and car antifreeze, ethylene glycol (C2H6O2 or HOCH2CH2OH), lowers  the freezing point of water in winter and raises its boiling point in the summer.  

G. Glycerol [HOCH2CH(OH)CH2OH] is produced by some fish and insects to lower the  freezing point of their blood, thus allowing them to survive the winter.  

i. Glycerol is a biological antifreeze

H. De-icing sidewalks and roads: NaCl or a mixture of NaCl+CaCl2 is used to melt ice on  roads and sidewalks

SOLVENT

NORMAL  FREEZING  POINT (˚C)

K(F)

NORMAL  BOILING  POINT  

(˚C)

K(B)

BENZENE 5.5 5.12 80.1 2.53

CARBON  

TETRACHLORIDE  (CCL4)

CHLOROFORM  

-22.9 29.9

76.7 5.03

(CHCL3) -63.5 4.70 61.2 3.63 (C2H5OH) -114.1 1.99 78.3 1.22 ETHANOL  

DIETHYL ETHER  

(C4H10O) -116.3 1.79 34.6 2.02 WATER (H2O) 0.0 1.86 100.0 0.512

4

8. BOILING POINT ELEVATION

A. The boiling point of a solution is higher than that of the pure solvent.

i. For a non-volatile solute

B. The difference between the boiling point of the solution and boiling point of the  pure solvent is directly proportional to the molal concentration of solute particles.  

i. T(b)(solution)-T(b)(solvent)=∆T(b)=m•K(b)

C. The proportionality constant is called the boiling point elevation constant, K(b). i. The value of K(b) depends on the solvent

ii. The units of K(b) are degrees Celsius per molal unit (˚C/m)

iii. K(b) for water is only 0.512 ˚C/m, so the changes in boiling point are quite small.  D. See Table on Page 609 of Text Book

9. OSMOSIS AND OSMOTIC PRESSURE

A. Osmosis is the flow of solvent from a solution of lower solute concentration into a  solution of higher solute concentration.  

B. The solutions may be separated by a semi-permeable membrane, yielding an  osmotic pressure across the semi-permeable membrane.

C. A semi-permeable membrane that lies at the curve of a U-shaped tube separating an  aqueous solution of sugar from pure water. The membrane allows water to pass in  both directions, but not the larger molecules of sucrose.

D. Because the sugar molecules are present, fewer water molecules touch the  membrane on the solution side.  

i. Therefore, more water molecules will enter the solution than those that would  leave.

ii. Thus there will be a net flow of water into the solution that will increase the  volume of the solution and consequently decreasing its concentration. iii. As the height of the solution rises and that of the pure solvent falls, a pressure  difference will be established and will push some water molecules from the  solution back through the membrane.  

1. An equilibrium will be established whereby water will be pushed out of  the solution at the same rate it will enter.  

2. The pressure difference at equilibrium is the osmotic pressure (I do not  have a symbol on my keyboard for osmotic pressure. It looks a lot like the  symbol for pi, π, only bigger).  

3. Osmotic pressure is, by definition, the applied pressure required to  prevent the net movement of water from solvent to solution.

5

E. The osmotic pressure is proportional to the number of moles of solute particles in a  given volume of solution, that is, to the Molarity (M):

Osmotic Pressure = n(solute)/V(solution) • RT = MRT

i. This equation is very similar to that of the ideal gas law (P=nRT/V). The  similarity is obvious since both equations relate the pressure of the system to its  concentration and temperature.

10. USE OF OSMOTIC PRESSURE IN MEDICINE

A. An isosmotic solution has the same osmotic pressure as the solution inside the cell;  as a result, there is no net flow of water in or out of the cell.

B. A hyperosmotic solution has a higher osmotic pressure than the solution inside the  cell; as a result there is a net flow of water out of the cell, causing it to shrivel

C. A hyposmotic solution has a lower osmotic pressure than the solution inside the  cell; as a result there is a net flow of water into the cell, causing it to swell. This may  cause the cells to burst.  

D. Thus intravenous solutions that are used in patient’s veins must have osmotic  pressures equal to those of body fluids. These solutions are called isosmotic or  isotonic, containing 0.9 g NaCl/100 mL of solution

6

11. COLLIGATIVE PROPERTIES OF SOLUTIONS- A SUMMARY A. Each of the 4 colligative properties rests on the inability of solute particles to cross  between two phases:

i. Solute particles cannot enter the gas phase, thus leading to vapor pressure  lowering and boiling point elevation.

ii. Solute particles cannot enter the solid phase, thus leading to freezing point  depression.  

iii. Solute particles cannot cross a semi-permeable membrane thus leading to the  generation of osmotic pressure. .

12. SOLUTE MOLAR MASS DETERMINATION USING COLLIGATIVE  PROPERTIES

A. Using colligative properties to find solute molar mass

i. Since all the four colligative properties relate concentration to some measurable  quantity, we can determine the amount (mol) of solute particles and, for a  known mass of solute, the molar mass of the solute as well.  

ii. Of the 4 colligative properties, osmotic pressure causes the largest changes, and  consequently the most precise measurement.

Osmotic pressure (atm)—>(M=osmotic pressure/RT)—> M(mol/L)—>(• L of  solution)—> amount (mol) of solute—>(÷ by g of solute)—>molar mass (M) in  g/mol

13. COLLIGATIVE PROPERTIES OF STRONG ELECTROLYTE  SOLUTIONS

A. A strong electrolyte dissociates completely to form ions. Each mole of solute gives  more than 1 mol of dissolved particles.

B. The formula of the compound indicates the expected number of particles in the  solution.  

i. Each mol of NaCl is expected to give 2 moles of dissolved ions

C. The van’t Hoff factor takes into account the dissociation of a strong electrolyte to  predict the effect on the solution.

i. i= measured value for electrolyte solution/expected value for non-electrolyte  solution

ii. For vapor pressure lowering: ∆P=i[X(solute)•P˚(solvent)]

iii. For boiling point elevation: ∆T(b)=i[K(b)m]

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iv. For freezing point depression: ∆T(f)=i[K(f)m]

v. For osmotic pressure: osmotic pressure=i(MRT)

D. For instance, the ∆T(b) of 0.050 m NaCl should be 2 • ∆T(b) of 0.050 m glucose  because NaCl dissociates into two particles per formula unit

E. For most ionic compounds dissolved in water, the van’t Hoff factor is equal to the  number of discrete ions in a formula unit of the substance. For instance, for NaCl  i=2; for MgCl2 i=3, for FeCl3 i=4.

SOLUTE  I EXPECTED I MEASURED NON-ELECTROLYTE 1 1 NACL 2 1.9

MG(SO4) 2 1.3 MGCL2 3 2.7

K2SO4 3 2.6

FECL3 4 3.4

14. MIXTURES

A. Solutions = homogeneous

B. Suspensions = heterogeneous, separate on standing

C. Colloids = heterogeneous, do not separate on standing

i. Particles can coagulate

ii. Cannot pass through semipermeable membrane

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15. STRUCTURE AND PROPERTIES OF COLLOIDS

A. Unlike large solid particles such as sand that form suspension (heterogeneous  mixture) when mixed with water and settle gradually to the bottom of the flask,  colloids are very fine particles the form dispersions (colloidal dispersions) when  mixed with a solvent, and do not settle to the bottom of the flask.

i. Colloidal particles are larger than simple molecules but small enough to remain  distributed and not settle out

ii. Colloids range in diameter from 1 to 1000 nm (10ˆ-9 to 10ˆ-6 m) B. A colloidal particle may consist of a single macromolecule such as a protein or  synthetic polymer or an aggregate of many atoms, ions or molecules.

C. Colloids have an enormous total surface area as a result of their small size i. Familiar colloidal products and natural objects:  

1. Milk (emulsion)

2. Butte (solid emulsion)

3. Whipped cream (foam)

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16. TYPES OF COLLOIDAL SUSPENSIONS

CLASSIFICATION

DISPERSING  SUBSTANCE  (SOLUTE

LIKE)

DISPERSING  MEDIUM  

(SOLVENT LIKT)

EXAMPLE

FOG (WATER  

AEROSOL LIQUID GAS

DROPLETS  IN AIR)

SOLID AEROSOL SOLID GAS SMOKE (ASH  IN AIR)

WHIPPED  

FOAM GAS LIQUID EMULSION LIQUID LIQUID SOLID EMULSION LIQUID SOLID

CREAM (AIR  BUBBLES IN  BUTTERFAT)

MILK (MILK  FAR  

GLOBULES  IN WATER)

OPAL  

(WATER IN  SILICA  

GLASS)

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17. CALCULATION EXAMPLE: USING HENRY’S LAW TO CALCULATE  GAS SOLUBILITY

A. The partial pressure of carbon dioxide gas inside a bottle of cola is 4 atm at 25˚C.  What is the solubility of CO2? The Henry’s Law constant for CO2 dissolved in water  is 3.3 • 10ˆ-2 mol/L•atm at 25˚C.

i. Solution: Substituting P for CO2 (4atm) and the value of K(H) into the Henry’s  Law equation S(CO2)= (3.4 • 10ˆ-2 mol/L •atm)(4 atm)= 0.14 mol/L

18. CALCULATION EXAMPLE: USING RAOULT’S LAW TO FIND ∆P A. Find the vapor pressure lowering, ∆P, when 10.0 mL of glycerol (C3H8O3) is added  to 500 mL of water at 50˚C. At this temperature, the vapor pressure of pure water is  92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.  

i. Solution: Applying the following steps, we can calculate ∆P. We just need the  mole fraction of glycerol, X(glycerol), since we know the vapor pressure of  water.

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19. CALCULATION EXAMPLE: DETERMINING BOILING AND  FREEZING POINTS OF A SOLUTION

A. You add 1.00 kg of ethylene glycol (C2H6O2) antifreeze to 4450 g of water in you  car’s radiator. What are the boiling and freezing points of the solution?

i. Solution: Applying the following steps, we can calculate the boiling point  elevation and freezing point depression. First, we need to find the Molality of the  solution.

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20. CALCULATION EXAMPLE: DETERMINING MOLAR MASS FROM  OSMOTIC PRESSURE

A. Biochemists have discovered more then 400 mutant varieties of hemoglobin, the  blood protein that carries O2. A physician dissolves 21.5 mg of one variety in water  to make 1.50 mL of solution at 5.0˚C. She measures an osmotic pressure of 3.61 torr.  What is the molar mass of this protein?

i. Solution: Applying the following steps, we can determine the molar mass using the number of moles and the known mass. First, we need to convert osmotic  pressure to atm and T to degrees K and then calculate Molarity from osmotic  pressure.

13

Chapter 15 Lecture 1  

CHEMICAL KINETICS

1. WHAT IS KINETICS IN CHEMISTRY?

A. Chemical kinetics deals with the speed of a reaction and its mechanism including the  stepwise changes that reactants undergo in their conversation to products.

B. Chemical kinetics isn’t he study of reaction rates, which reflect the changes in  concentrations of reactants as a function of time.

2. DEFINING RATE

A. Rate is how much a quantity changes in a given period of time.

B. The speed you drive your car is a rate- the distance your car travels (miles) in a  given period of time (1 hour)

i. So, the rate of your car has units of miles/hour.

3. FACTORS THAT INFLUENCE REACTION RATE

A. Under a given set of conditions, each reaction has its own characteristic rate, which  is determined by the chemical nature of the reactants

i. Some reactions are very fast while some others are very slow:  

H2(g) + F2(g) —> 2HF(g). [ very fast]

3H2(g) + N2(g) —> 2NH3(g). [very slow]

B. Four major factors can control the rate of the reactionL

i. Concentration of the reactants

ii. Physical state of the reactants

iii. Temperature of the reaction

iv. The use of a catalyst

C. Concentrations affect the rate of collision among reactant molecules. Molecules  must collide to react

i. The more molecules present in a reaction vessel, the more frequently they  collide, and the more often a reaction occurs.  

D. Reactants must mix to react: the frequency of collisions between molecules depends  on the physical states of the reactants.

i. When the molecules of reactants are in the same phase, random thermal motion  brings them into contact

ii. When the molecules are in different phase (liquid/solid for example), contact  occurs only at the interface. This limits the reaction rate.  

E. Temperature: molecules must collide with enough energy to react

i. At a higher temperature, more collisions occur in a given time. Also,  temperature affects the kinetic energy of the molecules and, in turn, the energy  of collisions.  

1. Increasing the temperature increases the reaction rate by increasing the  number and the energy of collisions

4. COLLISION THEORY OF KINETICS

A. For most reactions, for a reaction to take place, the reacting molecules must collide  with each other.

i. On average, about 10˚9 collisions per second.  

B. Once molecules collide, they may react together of they may not, depending on two  factors:

i. Whether the collision has enough energy to “break the bonds holding reactant  molecules together”

ii. Whether the reacting molecules collide in the proper orientation for new bonds  to form.

5. EFFECTIVE COLLISIONS: KINETIC ENERGY FACTOR  

A. For a collision to lead to overcoming the energy barrier, the reacting molecules must  have sufficient kinetic energy to that when they collide, the activated complex can  form.  

6. EFFECTIVE COLLISIONS: ORIENTATION EFFECT

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7. REACTION RATE EXPRESSION

A. Expressing quantitatively the effects of temperature and concentration on reaction  rate

i. Reaction rate reflects the changes in concentrations of reactants of products per  unit time. Usually, in a chemical reaction, reactant concentrations decrease  while product concentrations increase with time.

1. Consider the hypothetical reaction A—>B. If we measure the starting  reactant concentrations at time T(1) (conc. A(1)), and allow the reaction  to proceed, and then quickly measure the reactant concentration at time  T(2) (conc. A(2)).

B. The change in concentration divided by the change in time gives the average rate:  

C. Reaction rate is a positive number. The negative sign is used to convert the change  in concentration to a positive value since (conc. A(1)) is greater than (conc A(2).

D. In a chemical reaction, we usually use molar concentration (mol/L). The above  expression will become

i. The square bracket [] is to express concentration in moles per liter. Thus, the  rate has units mol Lˆ-1, sˆ-1, or Msˆ-1

E. If we chose to use the product concentration to measure the rate of a reaction, then  (conc. B(2)) is always greater than (conc. B(1)).

i. Thus, the change in product concentration ∆B is positive and the reaction rate  for A—>B expressed in terms of B is:

F. Average, Instantaneous, and Initial Reaction Rates

i. In most reactions, not only the concentration changes, but the rate itself varies  with time as the reaction progresses

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1. The expression Rate=-∆[A]/∆t gives us the average rate for the specified  period ∆t. This average does not tell us that the rate in changing nor does  it tell us how fast the reactant concentration is decreasing at any instant.

ii. Common sense: as the molecules of the reactant are used up, fewer of them are  present to collide with each other, and consequently the rate which is the  change in their concentration over time will decrease

iii. The change in rate can be seen by plotting the concentrations of the reactant  versus the times at which they were measured. A curve is usually obtained  which means that the rate changes.  

G. The slope of the straight line (∆y/∆x) joining any two points gives the average rate  over that period.

H. The shorter the time period chosen the closer we come to the instantaneous rate,  which is the rate at a particular instant during the reactions

i. The slope of a line tangent to the curve at a particular point gives the  instantaneous rate at that time

ii. WE USE THE TERM REACTION RATE TO MEAN THE INSTANTANEOUS  REACTION RATE

I. As a given reaction continues, the product concentration increase, and as a result,  the reverse reaction (reactants<— products) proceeds more quickly. To find the  overall or net rate, both the forward and reverse reactions must be taken into  account and calculate the difference between their rates

J. To avoid this complication, it is recommended to measure the initial rate, which is  the instantaneous rate at the moment the reactants are mixed. The product  concentrations are negligible so that the reverse rate is negligible.

K. The initial rate is measured by determining the slope of the line tangent to the curve  at t=0 s.

L. In the rest of this chapter, the initial rate data will be used to determine other  kinetic parameters.  

M. Expressing rates in terms of reactant and product concentration

i. In the following reaction: C2h4(g)+O3(g)<—>C2H4O(g)+O2(g) ii. The rate of disappearance of C2H4 is equal that for the disappearance of O3.  These rates are exactly the opposite in terms of the products because their  concentrations are increasing

1. From the balanced reaction: 1 mol of C2H4O and 1 mol of O2 appear for  every mole of C2H4 and O3 that disappear. We can write the following:

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8. THE RATE LAW AND ITS COMPONENTS

A. The rate law or rate equation expresses the rate as a function of reactant  concentrations, product concentrations and temperature.

B. Our focus here will be on reactions for which the products do not appear in the rate  law. That is the reaction rate will depend on reactant concentrations and  temperature.

C. First, we will look at the effect of concentration on rate of reactions occurring at a  fixed temperature

D. For the general reaction: aA+bB+…—>cC+dD+…

i. The rate law has the form: Rate= k[A]ˆm•[B]ˆn…

1. Square brackets [ ] are molar concentration

2. The proportionality constant k, called the rate constant, is specific for a  given substance at a given temperature.

3. K does not change as the reaction proceeds; k changes with the  temperature and therefor determine how temperature affects the rate. 4. The exponents m and n, called the reaction orders, define how the rate is  affected by reactant concentrations. Thus:

a. If the rate doubles when [A] doubles, the rate depends of [A] raised to  the first power [A]ˆ1, so m=1

b. If the rate quadruples when [B] doubles, the rate depends on [B]  raised to the second power [B]ˆ2, so n=2

c. In other cases, if the rate does not change when [A] doubles then the  rate does not depend on [A]. Thus, the rate depends on [A] raised to  the zero power [A]ˆ0, so m=0

d. The coefficients a, b, c, d in the balanced equations are not necessarily  related in any way to the reaction orders m and n.

5. The components of the rate law including rate, reaction orders and rate  constant, must be found by experiments; they cannot be deduced from the  reaction stoichiometry.

E. Reaction order terminology: the meaning of reaction order

i. Differentiation between individual order with respect to each reactant and  overall order:  

ii. For a reaction with a single reactant A:

1. The reaction is a first order overall if the rate is directly proportional to  [A]: Rate=k[A]

2. It is second order overall if the rate is directly proportional to the square  of [A]: Rate=k[A]ˆ2

3. It is zero order overall if the rate is not dependent on [A] at all:  Rate=k[A]ˆ0=k(1)=k

5

F. Typical real examples:

i. For the following reaction: NO(g)+O3(g)—>NO2(g)+O2(g)

1. The rate law has been experimentally determined to be: Rate=k[NO][O3] 2. This reaction is second order overall: 1+1=2

ii. For the following reaction: 2NO(g)+2H2(g)—>N2(g)+2H2O(g) 1. The rate law has been determined to be : Rate=k[NO]ˆ2[H2]

2. This shows that the reaction is second order in NO and first order in H2. It  is third order overall: 2+1=3

iii. For the following reaction: (CH3)3C- Br(l) + H2O(l)—> (CH3)3C-OH(l) + H(aq)  +Br (aq)

1. The rate law is as follows: Rate=k[(CH3)3C-Br]

2. This means that the rate is first order in (CH3)3C-Br and zero order in  H2). This means that the rate does not depend on water, and it is first  order overall

a. Reaction orders cannot be deduced from the balanced chemical  equation: reaction orders must be determined experimentally from  raw data

iv. Reaction orders are usually positive integers or zero, but they can be fractional  or negative. For instance, writing: Rate=k[A][B]ˆ1/2 means that the rate  depends on the square root of [B]. If increasing the concentration of B by a  factor of 4, while [A] is kept constant, the rate increases bu a factor of 2. 1. A negative exponent means that the rate decreases as the concentration  of that component increases: Rate=k[A][B]ˆ-1=k[A]/[B]

6

Chapter 15 Lecture 2

CHEMICAL KINETICS

1. THE RATE LAW AND ITS COMPONENTS (CONT.) A. Determining Reaction Orders Experimentally

i. For the reaction: 2NO(g)+O2(g)—>2NO2(g)

1. The rate law expressed in general form is: Rate=k[O2]ˆm[NO]ˆn 2. To find m and n (reaction orders), a series of experiments were run,  starting each one with a different set of reactant concentrations and  obtaining an initial rate in each case.

Experiment No.

Initial reactant  concentration: O2  (mol/L)

Initial reactant  concentration: NO  (mol/L)

Initial Rate  (mol/L.s)

1 1.10•10ˆ-2 1.30•10ˆ-2 3.21•10ˆ-3 2 2.20•10ˆ-2 1.30•10ˆ-2 6.40•10ˆ-3 3 1.10•10ˆ-2 2.60•10ˆ-2 12.8•10ˆ-3 4 3.30•10ˆ-2 1.30•10ˆ-2 9.60•10ˆ-3 5 1.10•10ˆ-2 3.90•10ˆ-2 28.8•10ˆ-3

3. In experiments 1 and 2, we can see the effect of doubling [O2] on rate:

4. [NO] and k is the same in experiments 1 and 2. We can write: 5. Using the values from the table, we can write:  

6. Rounding to one significant figure: 2=2m

a. Therefore, m=1 and the reaction order in O2 is first order: when O2  doubles, the rate doubles.

7. To determine the order with respect to NO, we compare experiments 3  and 1 in which [O2] is held constant and [NO] is doubled:

8. Since [O2] does not change in experiments 3 and 1 and k is a constant,  then:  

9. The values in table give:  

10. Dividing, we get: 3.99=(2.00)ˆn

a. Solving for n: log3.99=n log 2.00

b. n log3.99/log2.00=2.00

c. Therefore, n=2

11. The reaction is second order in NO. When [NO] doubles, the rate  quadruples

12. Finally, the rate law is: Rate=k[O2][NO]ˆ2

B. Determining the Rate Constant k

i. Once the rate, reactant concentrations, and reaction orders are known, then k is  readily determined.

ii. From experiment 2 in the table from above, we can write:  

iii. Using the other experiments, we can generate 5 values for k that differ slightly  from each other, and when averaged, we get the value 1.72 • 10ˆ-3 Lˆ2/molˆ2.s

2

2. INTEGRATED RATE LAWS: CONCENTRATION CHANGES OVER  TIME

A. The rate law tells us the rate or concentration at a given instant. It allows the  determination of how fast a reaction is proceeding at the moment when a certain  moles per liter of A are reacting with given moles per liter of B.

B. Using the integrated rate law, we can consider the time factor, and in turn, be able to  determine the time it will take to x moles per liter of A to be used up as well as  figuring out the concentration of A after a certain time or reaction

C. Integrated Rate Laws for First, Second, and Zero-Order Reactions

i. For a simple first order reaction such as A—>B, we can write: Rate=-∆[A]/∆t 1. It can also be expressed in terms of rate law: Rate=k[A]

2. Combining both equations and we get: -∆[A]/∆t=k[A]

3. Rearranging and integrating between time 0 and time t and between  [A](0) and [A](t)

4. First order integrated law is then: ln[A](0)-ln[A](t)=kt

ii. For a second order reaction, we have: Rate=-∆[A]/∆t = k[A]ˆ2

1. Rearranging and integrating:  

iii. For a zero order reaction, we can write: Rate=∆[A]/∆t = k[A]ˆ0= k  1. Rearranging and integrating:

3

D. Determining the Reaction Order Through Graphical Representation of the  Integrated Rate Law

i. The first order integrated rate law: ln[A](0)-ln[A](t)=kt can be rearranged as  an equation for a straight line y=mx+b as follows:

1. Ln[A](t)= -kt+ln[A](0)

2.           Y=   Mx+ b

3. Therefore, a plot of ln[A](t) versus time gives a straight line with a slope =  -k and a y-intercept = ln[A](0)

ii. For a simple second order reaction we have: 1/[A](t) – 1/[A](0) = kt  1. Rearranging gives: 1/[A](t) = kt + 1/[A](0)

2.                                               Y = mx. +  b

3. A plot of 1[A](t) versus time gives a straight line with a slope=k and a y intercept = 1/[A](0)

iii. For a zero-order reaction, we have: [A](t) – [A](0) = -kt

1. Rearranging we get: [A](t) = -kt + [A](0)

2.                                             Y = mx   +  b

3. A plot of [A](t) versus time gives a straight line with a slope= -k and a y intercept = [A](0)

E. In summary:  

i. Is is convenient to use graphical representation to determine the order of  reaction

1. The reaction is a first order with respect to a given reactant when the plot  of ln[reactant] versus time is a straight line

2. When a straight line is obtained upon plotting 1/[reactant] versus time,  the reaction order is second order with respect to that reactant

3. When a straight line is obtained upon plotting [reactant] versus time, the  reaction order is zero order with respect to that reactant.

3. DECOMPOSITION OF N2O5

A. The reaction is as follows: 2N2O5(soln)—>4NO2(soln)+O2(g)

Time (min) [N2O5] Ln[N2O5] 1/[N2O5] 0 0.0165 -4.104 60.0 10 0.0124 -4.390 80.6 20 0.0093 -4.68 1.1•10ˆ2 30 0.0071 -4.95 1.4•10ˆ2 40 0.0053 -5.24 1.9•10ˆ2 •• 0.0039 -5.55 2.6•10ˆ2 60 0.0029 -5.84 3.4•10ˆ2

4

4. GRAPHICAL DETERMINATION OF THE REACTION ORDER FOR THE  DECOMPOSITION OF N2O5

5. GRAPHICAL DETERMINATION OF THE RATE LAW FOR A —>  PRODUCT

A. Plots of [A] versus time, ln[A] versus time, and 1/[A] versus time allow  determination of whether a reaction is zero, first, or second order.

B. Whichever plot gives a straight line determines the order with respect to [A].

i. If linear is [A] versus time, Rate=k[A](0)

ii. If linear is ln[A] versus time, Rate=k[A](1)

iii. If linear is 1/[A] versus time, Rate=k[A](2)

6. REACTION HALF-LIFE: T(SUBSCRIPT ½)

A. T(subscript ½) of a reaction is the time required for the reactant concentration to  reach half its initial value

B. A half-life is expressed in time units for a given reaction and it is characteristic of  that reaction at a given temperature.

C. Under a given set of conditions, the half-life or a first order reaction is a constant  independent of reactant concentration.

5

D. From the integrated rate law for first order: ln([A](0)/[A](t)) = kt i. And considering that after t=t(subscript ½) , [A](t)=1/2[A](0), we can write:

1. the time required to reach one-half life the starting concentration in a  first-order reaction doers not depend on what that starting concentration  is

2. The radioactive decay of unstable nucleus is an example of first order  process: the half-life for U(235) is 7.1•10ˆ8 years; that is after 710 million  years a sample of 2kg U(235) will contain 1kg of U(235), and a sample of  1g of U(235) will contain 0.5g of U(235).

E. For a second-order reaction the half-life depends on the reactant concentration.  

i. The half-life for a second-order reaction is inversely proportional to the initial  reactant concentration. That is, increasing the initial concentration shortens the  half-life. As a second-order reaction proceeds, the half-life increases.

F. For a zero order reaction, the half-life is directly proportional to the initial reactant  concentration as follows:

6

7. RELATIONSHIP BETWEEN ORDER AND HALF-LIFE A. For a zero order reaction, the lower the initial concentration of the reactants, the  shorter the half-life: t(subscript ½) = [A](0)/2k  

B. For a first order reaction, the half-life is independent of the concentration:  t(subscript ½) = ln(2)/k=0.693/k

C. For a second order reaction, the half-life is inversely proportional to the initial  concentration, and therefore increasing the initial concentration shortens the half life: t(subscript ½) = 1/(k[A](0))

7

Chapter 15 Lecture 3  

CHEMICAL KINETICS

1. THE EFFECT OF TEMPERATURE ON REACTION RATE

A. For many reactions near room temperature, an increase of 10˚C causes a doubling  or tripling of the rate.

B. In 1889 the Swedish scientist Svante Arrhenius discovered a key relationship  between T and k:  

i. Where k is rate constant, e is the base of natural logarithms, T is the absolute  temperature and R is the universal gas constant in energy units, 8.314  J/(mol•K).

ii. A is a constant related to the orientation of the colliding molecules (A is also  called frequency factor)

iii. E(subscript a) is the activation energy of the reaction, which Arrhenius  considered as the minimum energy the molecules must have to react. iv. Increasing T leads to decreasing the negative exponent, and consequently the  value of k becomes larger, which means that the rate increases.

C. If we take the natural logarithms of both side of Arrhenius equation, we will get the  equation of a straight line for the plot of ln k versus 1/T:  

i. The slope of the line= -E(subscript a)/R, and the y-intercept is equal to ln A.  Knowing the value of the slope from the graph and using the known value of the  constant R we can estimate graphically the value of E(subscript a).  

ii. Since the relationship between ln k and 1/T is linear, we can use a simpler  method to estimate E(subscript a) if we know k at two temperatures T(subscript  1) and T(subscript 2):  

iii. Subtracting ln k(subscript 1) from ln k(subscript 2), the term ln A drops, and we  get the following relationship:

2

2. REACTION MECHANISMS: STEPS IN THE OVERALL REACTION A. The aim here is to get to know how a reaction works at the molecular level

i. Reactions occur through a reaction mechanism, a sequence of single reaction  steps that add up to yield the overall reaction

ii. Consider the hypothetical reaction: 2A+B—>E+F

iii. It might be arranged in three steps as follows:  

iv. C is a product in step 1 and a reactant in step 2, and D is a product in 2 and a  reactant in 3. C and D are therefore reaction intermediates

1. A reaction intermediate is a substance that is formed and used up during the overall reaction

2. Reaction intermediates do not appear in the overall reaction but they are  essential for the reaction to occur.

B. Elementary Reactions and Molecularity

i. A reaction mechanism is made up of elementary reactions or elementary steps,  each describing a single molecular even such as a one particle decomposition or  two particle collision.

1. An elementary step is characterized by its molecularity, which means the  number of reactant particles involved in the step.

ii. Let us consider the proposed mechanism for the breakdown of O3: 2O3(g)— >3O2(g) whereby a two step mechanism has been advanced:  

1. Step 1 is unimolecular: it involves the decomposition of a single particle 2. Step 2 is bimolecular involving the reaction between two particles  3. Conversely, to the rate of an overall reaction, the rate law for an  

elementary reaction can be deduced from the reaction stoichiometry.  That is, we use the equation coefficients as the reaction orders in the rate  law for an elementary step: reaction order equal molecularity.

3

Elementary step Molecularity Rate law A—> product Unimolecular Rate=k[A] 2A—>product Bimolecular  Rate=k[A]squared

A+B—>product Bimolecular Rate=k[A][B] 2A+B—>product Termolecular  Rate=k[A]squared[B]

C. The Rate-Determining Step of a Reaction Mechanism

i. In a reaction mechanism, one of the elementary steps is much slower than the  others, and consequently it influences how fast the overall reaction occurs 1. Thus the slower step is called the rate-determining step or rate limiting  step

2. Because the rate-determining step limits the rate of the overall reaction,  its rate law represents the rate law for the overall reaction

ii. For the following reaction: NO2(g)+CO(g)—>NO(g)+CO2(g) experiment shows  that the actual rate law is given by: Rate=k[NO2]squared

iii. The two step mechanism is proposed as follows:  

iv. NO3 is a reaction intermediate. The rate laws for (1) and (2) are:  Rate(1)=k(1)[NO2]squared

1. Rate(2)=k(2)[NO3][CO]

a. If k(1)=k, the rate law for the rate determining step (step 1) is  

identical to the experimental rate law given above.  

2. Since the first step is slow, [NO3] is very low. Thus, the fast second step  cannot increase the overall rate, and the reaction takes essentially as long  as the first step.

3. The reactant CO has a reaction order of zero because it takes part in the  reaction after the rate determining step.  

3. RATE DETERMINING-A SUMMARY

A. In most mechanisms, one step occurs slower than the other steps

B. The result is that the product production cannot occur any faster than the slowest  step; the step determines the rate of the overall reaction.  

C. We call the slowest step in the mechanism the rate determining step i. The slowest step has the largest activation energy E(subscript a)

4

D. The rate law of the rate determining step determines the rate law of the overall  reaction.

4. CATALYSIS: SPEEDING UP A CHEMICAL REACTION A. Raising the temperature generally speeds up a reaction, sometimes dramatically.

i. However, raising the temperature in a living system to speed up some  biochemical reactions can kill the living organism

ii. Therefore, living things rely on catalysis via enzymes (i.e., biocatalysts) rather  than temperature increases to speed up reactions  

B. Many industrial and laboratory processes require catalysis as well:

i. The decomposition of potassium chlordane is sometimes used to produce small  quantities of oxygen in the laboratory: 2KClO3(s)—>2KCl(s)+3O2(g) ii. Without a catalyst, the KClO3(s) must be heated to above 400˚C in order to  produce O2(g) at a useful rate

iii. However, if we add a small quantity of MnO2(s), we can get the same rate of  oxygen evolution at just 250˚C

iv. After the reaction is completed, we can recover essentially all the manganese  dioxide unchanged

C. MnO2 is a catalyst: a catalyst increases the reaction rate without itself being  changed in a chemical reaction

i. In general, a catalyst works by changing the mechanism of a chemical reaction ii. The pathway of a catalyzed reaction has a lower activation energy than that of  an un-catalyzed reaction

1. If the activation energy is lowered, then more molecules have sufficient  energies to engage in effective collisions

D. Homogeneous Catalysis:  

i. A catalyzed reaction that occurs in a homogeneous (one-phase) mixture involves  homogeneous catalysis. One simple mechanism for homogeneous catalysis in a  reaction in which two reactants A and B yield two products C and D involves two  steps:  

ii. As expected, neither the intermediate nor the catalyst appears in the net  equation

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iii. A practical example of this mechanism has O3 and O as the reactants and O2 as  the product with CL the catalyst and ClO the intermediate:  

1. The E(subscript a) for the uncatalyzed reaction is about 17.1 kJ, much  higher than that of the catalyzed reaction, which is 2.5 kJ. As a  

consequence, the rate constant for the catalyzed reaction is thousands of  times greater than that of the uncatalyzed reaction

E. Heterogeneous Catalysis:  

i. Many reactions can be catalyzed by the surfaces of appropriate solids. An  essential aspect of this catalytic activity is the ability of surfaces to adsorb (bind)  reactant molecules from the gaseous or liquid state (or phase)

1. Because a surface-catalyzed reaction occurs in a heterogeneous mixture  (solid catalyst-liquid reactant or gaseous reactant), the catalytic action is  called heterogeneous catalysis.  

2. A simplified mechanism of the surface-catalyzed conversion of oleic acid  to stearin acid:  

a. This is a surface reaction-hydrogenation (conversion of vegetable oil  into margarine)

ii. Another example of heterogeneous catalysts

1. Solid catalytic converter in a car’s exhaust system. It eliminates pollutants  in the car’s exhaust, and converts fuel fragments into environmentally  friendly species.

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CHAPTER 16 LECTURE 1

CHEMICAL EQUILIBRIUM

1. THE EQUILIBRIUM STATE

A. Equilibrium involves opposing processes occurring at equal rates:  

i. In vapor-pressure equilibrium, the rate of evaporation of a liquid is equal to the  rate of condensation of its vapor.

ii. In solubility equilibrium, the rate of dissolution of a solid is equal to its rate of  crystallization from solution

iii. In chemical reactions, and after sufficient time, the concentration of reactants  and products no longer change. This is an indication of a state of equilibrium,  and the reactions are reversible.

2. ARROW CONVENTIONS FOR CHEMICAL REACTIONS A. Chemists commonly use two kinds of arrows in reactions to indicate the degree of  completion of the reactions.

B. A single arrow indicates all the reactant molecules are converted to product  molecules at the end. We said reaction goes to completion

C. A double arrow indicates the reaction stops when only some of the reactant  molecules have been converted into products. We have a reversible reaction.

3. REACTION DYNAMICS

A. When a reaction starts, the reactants are consumed and products are made

i. The reactant concentrations decrease and the product concentrations increase ii. As reactant concentration decreases, the forward reaction rate decreases B. Eventually, the products can react to re-form some of the reactants, assuming the  products are not allowed to escape

i. As product concentration increases, the reverse reaction rate increases C. Processes that proceed in both the forward and reverse direction are said to be  reversible

i. Reactants<—-—>products

4. DYNAMIC EQUILIBRIUM

A. As the forward reaction slows and the reverse reaction accelerates, eventually the  reach the same rate.

B. Dynamic equilibrium is the condition wherein the rebates of the forward and  reverse reactions are equal

C. Once the reaction reaches equilibrium, the concentrations of all the chemicals  remain constant because the chemicals are being consumed and made at the same  rate

H2(G)+I2(G)<—-—>2HI(G)

i. At time 0, there are only reactants in the mixture, so only the forward reaction can take place

1. [H2]=8, [I2]=8, [HI]=0

ii. At time 16, there are both reactants and products in the mixture, so both the  forward reaction and reverse reaction can take place

1. [H2]=6, [I2]=6, [HI]=4

iii. At time 32, there are now more products that reactants in the mixture, the  forward reaction has slowed down as the reactants run out, and the reverse  reaction accelerated.

1. [H2]=4, [I2]=4, [HI]=8

iv. At time 48, the amounts of products and reactants in the mixture haven’t  changed; the forward and reverse reactions are proceeding at the same rate. It  has reached equilibrium.

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