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Unit 3 - Energy Changes and Rates of Reaction

Chapter 5.1 Changes in Matter and Energy

Thermochemistry The study of energy changes that accompany physical or

chemical changes in matter All thermodynamics depends on the LAW OF CONSERVATION

OF ENERGY: the total energy of a system and its surroundings remains the same

Heat and Energy Changes Changes that occur in matter may be classified as physical,

chemical or nuclear Physical: a change in the form of a substance. No chemical bonds

are broken. Ex. hydrogen boils at -252°C

H2 (l) + heat H2 (g)

Chemical: a change in the chemical bonds between atoms resulting in new substances

Ex. hydrogen is burned as fuel in the space shuttle’s engines2H2 (g) + O2 (g) 2H2O (l) + heat

Nuclear: a change in the protons or neutrons in an atom resulting in a new atom.

Ex. hydrogen undergoes nuclear fusion in the Sun, producing helium

Chemical system – the substances undergoing a change Surroundings – the system’s environment Thermal Energy – energy available from a substance as a result of

the motion of its molecules Temperature (T) – the average kinetic energy of the molecules in

a sample, measured in °C or K Heat (q) – amount of energy transferred between substances,

measured in Joules (J)

Energy flows between substances because of their difference in temperature.

A. Exothermic: releasing thermal energy, heat (q) flows from the system to the surroundings, usually causing an increase in the temperature of the surroundings.

- q has a negative value

B. Endothermic: absorbing thermal energy, heat (q) flows into the system from the surroundings, usually causing a decrease in the temperature of the surroundings

- q has a positive value

Measuring Energy Changes Calorimetry : experimental technique used to measure energy

changes in chemical systems Different substances vary in their ability to absorb amounts of heat Specific heat capacity : amount of energy required to raise the

temperature of one gram of a substance one °C or one K (Table 5.2, p.234)

The amount of heat transferred (q) depends on – mass of sample, m measured in grams– change, ΔT measured in °C or K– specific heat capacity, c measured in J/g•°C or J/g•K

surroundings

system

surroundings

system

Q = mc ΔT

Sample Problem 1Many water heaters use the combustion of natural gas (assume methane) to heat the water in the tank. When 150.0 L of water at 10.0°C is heated to 65.0°C, how much heat flows into the water?

Sample Problem 2If 25.0 g of aluminum cools from 310°C to 37°C, how many joules of heat energy are lost by the sample?

Sample Problem 3Calculate the molar heat capacity (C) of water (J/mol•°C).

Heat Transfer and Enthalpy Change The internal energy of a system is equal to the sum of the

potential and kinetic energy of all species in the system Kinetic Energy

– moving electrons in atoms– vibration, rotation and translation of atoms and molecules

Potential Energy– nuclear potential energy of protons and neutrons– bond energy– intra and intermolecular forces

It is not possible to measure all of these energies for a system Instead, we study the energy absorbed or released to the

surroundings during a change in the system – the change in enthalpy, ΔH

This example of exothermic change shows

that the change in potential energy of the system (ΔH) equals the change in kinetic energy of the surroundings (q)

Reaction Progress

Changes in Kinetic and Potential Energyhigh potential

energy

low potential energy

low kinetic energy

high kinetic energy

ΔH

qEnergy

Molar Enthalpy, ΔHx

A thermochemical equation represents the energy change that accompanies a chemical reaction.

Ex. 2 H2 (g) + O2 (g) 2 H2O (g) + 483.6 kJ The equation representing the energy released when ONE mole of

H2 combusts would be:H2 (g) + ½ O2 (g) H2O (g) + 241.8 kJ

The enthalpy change per mole of a substance undergoing a change is called the molar enthalpy, ΔHx, where x represents the type of change that is occurring (see p. 227)

Hsol solution NaBr(s) Na+(aq) + Br-(aq)Hcomb combustion CH4(g) + 2O2(g) CO2(g) + H2O(l)Hvap vaporization CH3OH(l) CH3OH(g)Hfr freezing H2O(l) H2O (g)Hneut neutralization 2NaOH + H2SO4 2Na2SO4 + 2H2OHf formation C(s) + 2H2(g) + ½O2(g) CH3OH(l)

Thus the molar enthalpy for the combustion of hydrogen is ΔHcomb = – 241.8 kJ/mol

Enthalpy changes for exothermic reactions are given a negative sign (heat is a product)

Enthalpy changes for endothermic reactions are given a positive sign (heat is a reactant)

The molar enthalpy of a physical change may be expressed similarly:

H2O (l) + 40.8 kJ H2O (g)

Thus the molar enthalpy of vaporization for water is ΔHvap = 40.8 kJ/mol

The amount of energy involved in a change depends on the quantity of matter undergoing that change.

To calculate an enthalpy change, ΔH (kJ), for some amount other than a mole, use the formula:

ΔH = n ΔHx where n = # of moles and ΔHx = molar enthalpy (kJ/mol)

Sample Problem 1A common refrigerant, Freon-12 (molar mass 120.91 g/mol) is alternately vaporized in tubes inside a refrigerator, absorbing heat, and condensed in tubes outside the refrigerator, releasing heat. This results in heat being transferred from the inside to the outside of the refrigerator. The molar enthalpy of vaporization of Freon-12 is 34.99 kJ/mol. If 500.0g of Freon-12 is vaporized, what is the expected enthalpy change, ΔH?

Chapter 5.2 Determining Enthalpy by Experiment

Calorimetry of Physical Changes How are the values for molar enthalpies obtained? To study energy changes, an isolated system (a calorimeter) and

precise measurements are required There are three assumptions made when using calorimetry

– No heat is transferred between the calorimeter and the outside environment– Any heat absorbed or released by the calorimeter materials, such as the container, is negligible– A dilute aqueous solution is assumed to have a density and specific heat capacity equal to that of pure water (1.00 g/mL and 4.18 J/g•°C or J/g•K)

Calorimetry The analysis is based on the law of conservation of energy: the

total energy change of the chemical system is equal to the total energy change of the surroundings

ΔHsystem = ±| qsurroundings|

the chemical system is dissolved in the surrounding water

Sample Problem 2In a calorimetry experiment, 5.56 g of sodium chloride are dissolved in 75.0 mL of water at an initial temperature of 24.9°C. The final temperature of the solution is 23.7°C. What is the molar enthalpy of solution of sodium chloride?

Calorimetry of Chemical Changes Chemical reactions that occur in aqueous solutions can also be

studied using a simple calorimeter. Reactions that absorb or produce more heat must be conducted in

a more substantial calorimeter called a “bomb calorimeter”.

The reaction is carried out inside a steel “bomb”, and the heat evolved is transferred to the surrounding water, where the change in temperature is measured.

Sample Problem 3Calculate the molar heat of combustion of 1.00 g of octane in a “bomb” calorimeter containing 1.20 L of water, if the temperature rises from 25.0°C to 33.2°C and the heat capacity of the “bomb” is 837 J/ °C.

Representing Enthalpy Changes

Method OneTHERMOCHEMICAL EQUATIONS WITH ENERGY TERMS A balanced chemical equation that includes the heat transferred

to or from the surroundingsEx. H2 (g) + ½O2 (g) H2O(l) + 285.8 kJ

MgCO3 (s) + 117.3 kJ MgO(s) + CO2(g) To write a thermochemical equation, you must know

– the balanced chemical equation– the molar heat of reaction for one substance– if the reaction is endothermic or exothermic

Sample Problem 1The combustion of magnesium releases 24.7 kJ per gram of magnesium. Calculate the molar heat of combustion of magnesium and write the thermochemical equation.

Sample Problem 2Use the following thermochemical equation to calculate the molar heat of combustion of hexane.2C6H14 (l) + 19O2 (g) 12CO2 (g) + 14H2O (g) + 7086kJ

Method TwoTHERMOCHEMICAL EQUATIONS WITH ΔH VALUES The enthalpy of reaction is indicated by a separate expression

beside the chemical equationRemember: exothermic reactions have -ΔH

endothermic reactions have +ΔHExamples:

H2 (g) + ½O2 (g) H2O(l) Horxn = -285.8 kJ

MgCO3 (s) MgO(s) + CO2(g) Horxn = 117.3 kJ

Method ThreeMOLAR ENTHALPY OF REACTION

The quantity of heat transferred in a reaction per mole of a specified substance

– Hr (kJ/mol) To calculate the molar heat of reaction:

Hr = specific heat of reaction x molar mass standard molar enthalpy of reaction (Hr

o)

Sample Problem 4Ethane is the second largest component of natural gas. If its specific heat of combustion is -51.85 kJ/g, what is the molar heat of combustion of ethane?

Method FourPOTENTIAL ENERGY DIAGRAM a graphical representation of the energy transferred during a physical or chemical change exothermic reactions: products are lower in energy than reactants (energy lost to surroundings) endothermic reactions: products are higher in energy than reactants (energy absorbed from surroundings)

Ep

Reaction Progress

Reactants

Products

Exothermic Reactions

H Ep

Reaction Progress

Reactants

Products

Endothermic Reactions

H

Homework: Read Chapter 5.2 p. 236 – 239Answer p. 238 # 9-12

Chapter 5.3 Hess’s Law

Hess’s Law Law of Additivity of Reaction Enthalpies

– the enthalpy change of a physical or chemical process depends only on the beginning conditions (reactants) and the end conditions (products).

Enthalpy change is independent of the pathway of the process and the number of intermediate steps in the process

For any chemical change made in several steps, the net H is equal to the sum of the H values of the separate steps

For example, there are many ways to get from C(s) and O2(g) to CO2(g): (p. 243, fig. 5.13)

The direct route:C(s) + O2(g) CO2(g) H = -393 kJ/mol

Or a less direct route:

Same result!

Predicting ΔH using Hess’s Law Hess’s Law may be used when the molar enthalpy may not be

measured using calorimetry If 2 or more equations with known enthalpy changes can be added

together to form a new “target” equation, then their enthalpies may be added together to yield the enthalpy of the “target” equation

Two rules to remember1) When you reverse an equation, you need to change the sign

of Ho (multiply by –1)2) When you multiply the coefficients of an equation, you need

to multiply Ho by the same number

Sample Problem 1Calculate H for S(s) + 3/2 O2(g) --> SO3(g) using: S(s) + O2(g) --> SO2(g) H1 = -320.5 kJ

SO2(g) + 1/2 O2(g) --> SO3(g) H2 = -75.2 kJ

This can be represented by an enthalpy diagram:

enthalpy, H Ho=-395.7 kJ

S(s) + O2(g)

SO2(g), O2(g)

SO3(g)

Ho=-320.5 kJ

Ho=-75.2 kJ

Sample Problem 2Using Hess’s Law, find the enthalpy change for the reaction2 CH4 (g) + 3 O2 (g) 2 CO (g) + 4 H2O (l) using:

CH4 (g) + 2 O2 (g) 2 CO2 (g) + 2 H2O (l) H = -890 kJ 2 CO (g) + O2 (g) 2 CO2 (g) H = -566 kJ

Multistep Energy Calculations Several energy calculations may be required to solve multistep

problems:Heat flow q = mcΔTEnthalpy changes ΔH = nΔHx

Hess’s Law ΔHtarget = ΣΔHknown

Sample Problem 3

In the Solvay process for the production of sodium carbonate (or washing soda), one step is the endothermic decomposition of sodium hydrogen carbonate:

2 NaHCO3 (s) + 129.2 kJ Na2CO3(s) + CO2(g) + H2O(g)

What quantity of chemical energy, ΔH, is required to decompose 100.0 kg of sodium hydrogen carbonate?

1) Calculate the molar enthalpy (ΔHr) of NaHCO3:ΔH = n ΔHr

ΔHr = ΔH n= 129.2 kJ 2 mol= 64.6 kJ/mol

2) Convert 100.0 kg to an amount in moles:n(NaHCO3) = 100.0 kg x 1 mol

84.01g= 1.190 kmol

3) Find ΔH:ΔH = n ΔHr

= 1.190 kmol x 64.6 kJ1 mol

= 76.9 MJHomework: p. 247 # 13-16

Chapter 5.3 Standard Enthalpies of Formation

The quantity of energy associated with the formation of one mole of a substance from its elements in their standard states (at SATP)

Symbol: ΔH°fUnits: kJ/mol For example:

C(s) + ½ O2 (g) + 2 H2 (g) CH3OH (l) ΔH°f = -239.1 kJ/mol

2 Al (s) + 3/2 O2 (g) Al2O3 (s) ΔH°f = -1675.7 kJ/mol

All of the elements on the left sides of the equations are in their standard states (SATP)

Writing Formation EquationsStep 1: Write one mole of the product in the state that has been specified.

Example: FeSO4 (s)

Step 2: Write the reactant elements in their standard states (reference periodic table).

Fe (s) + S (s) + O2 (g) FeSO4 (s)

Step 3: Write coefficients for the reactants to give a balanced equation yielding one mole of product.

Fe (s) + S (s) + 2 O2 (g) FeSO4 (s)

Calculating ΔH A third way of calculating ΔH is possible using standard enthalpies

of formation. Table E.8, p.597 gives standard enthalpies of formation for

common molecules. According to Hess’s Law, the enthalpies of known equations may

be used to calculate the enthalpy of an unknown reaction. For example:

The formation of hydrogen and carbon monoxide gas from water and graphite:

H2O (g) + C (s) H2 (g) + CO (g) ΔH = ?Use the formation equations for each of the products and reactants to create the target equation.

H2O (g) + C (s) H2 (g) + CO (g)

ReactantsH2(g) + 1/2 O2(g) --> H2O(g) H0

f H2O = - 241.8 kJ/molC (s) – graphite in standard state H0

f = 0 kJ/molProductsC(s) + 1/2 O2(g) --> CO(g) H0

f CO = - 110.5 kJ/molH2 (g) – hydrogen in standard state H0

f = 0 kJ/mol

H2O(g) --> H2(g) + 1/2 O2(g) H0 = 241.8 kJ/molC(s) + 1/2 O2(g) --> CO(g) H 0 = - 110.5 kJ/mol H2O (g) + C (s) H2 (g) + CO (g) H0 = 131.3 kJ/mol

In general, when all enthalpies of formation are known, the enthalpy of a reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.

where n = the amount in moles of each product or reactant

Sample ProblemCalculate the heat of combustion of methanol.

CH3OH (l) + 3/2 O2 (g) --> CO2 (g) + 2 H2O (g) H0comb = ?

Multistep Energy Calculations Several energy calculations may be required to solve multistep problems

Heat flow q = mcΔTEnthalpy changes ΔH = n ΔHx

Hess’s Law ΔHtarget = ΣΔHknown

Enthalpies of Formation

Homework:Read Chapter 5.3 p. 250 – 254Answer p. 251 #17-20

p. 255 # 1-5,7

Chapter 7.2 Thermodynamics and Equilibrium:

The Laws of ThermodynamicsFirst Law of Thermodynamics

(also known as the Law of Conservation of Energy)The total amount of energy in the universe is constant. Energy can be neither created nor destroyed, but can be transferred from one object or place to another, or transformed from one form to another.

In a chemical reaction, a transfer of heat energy occurs resulting in an endothermic or exothermic change.

In both cases, the total energy of the universe (system plus surroundings) remains constant.

Entropy, S A measure of energy flow from being constricted or concentrated

to being more widely spread out.Second Law of Thermodynamics

All changes either directly or indirectly increase the entropy of the universe.

Examples of entropy increase: hot pan cools – the faster moving molecules in the pan disperse

their energy by making the slower air/counter molecules around them speed up

water flows down Niagara Falls – water molecules disperse their potential energy as they fall

Activation Energy Energy ALWAYS disperses if it is not hindered from doing so. BUT energy dispersal can be delayed for seconds, years or even

millennia by barriers. Examples:

– paint prevents oxygen from combining with iron to form rust

– paper and oxygen will not react unless a spark is introduced– mountain rock will not fall unless external energy breaks the rock into pieces

Chemical bonds are forces that hold atoms together in a molecule. Most chemical bonds are strong and require a great deal of energy to break them.

This is why most reactions require a “push” to start. The “push” is called activation energy. Most spontaneous reactions require an initial input of energy to aid

the first few molecules to react. The energy evolved from the reaction of these first few molecules serves as activation energy for the next few molecules and the cycle repeats.

Predicting Entropy Predict whether a reaction will cause the energy of the reactants

to become more spread out (increase entropy) or less spread out (decrease entropy)

Look at the state of the reactants and products. Gases have more entropy than liquids, and liquids have more than solids.

If the states are the same, an increase in the number of moles allows the energy to be more spread out and entropy increases.

N2O4 (g) 2 NO2 (g)

Entropy and Temperature The entropy of a substance varies with the temperature of a

substance. The lower the temperature, the lower the entropy. As we decrease the temperature of water vapour from 100°C, it

condenses to form a liquid and then freezes to become a solid. Even then, the molecules still have

some entropy as they vibrate in their spots.

As the temperature is cooled further, the entropy continues to decrease. At

absolute zero (0 K or –273.15 °C) the ice will have an entropy of zero

Third Law of ThermodynamicsAt absolute zero, the entropy of a pure crystal is also zero.

S = 0 J/K at T = 0 K The entropy of a substance measured at a temperature above

absolute zero will also be above zero. If the entropy of one mole of a substance is measured at SATP, it

is called the standard entropy, S°. (see table C6 p.799) With the standard entropies, we can calculate the ΔS° of a

reaction the same way we calculated ΔH°

Calculating Entropy Change, ΔS°

Example: Calculate the standard entropy change (ΔS°) when 1 mole of urea reacts with water.

CO(NH2)2 (aq) + H2O (l) CO2 (g) + 2 NH3 (g)

Gibbs Free Energy, G° Whether a reaction will occur spontaneously or not depends on

the enthalpy (ΔH) and the entropy (ΔS) of the reaction. Gibbs free energy (kJ) is the energy released from a chemical

reaction that is available to do work.ΔG = ΔH – T Δ where ΔH is the enthalpy change (kJ)

T is the temperature (K)ΔS is the entropy change (kJ/K)

A change can only be spontaneous if it is accompanied by a decrease in free energy, G must be negative.

In terms of enthalpy (ΔH) and entropy (ΔS):– If a change is exothermic (- ΔH) and entropy increases (+ΔS), both factors favor spontaneity

ΔG = ΔH – T ΔSΔG = (–) – T (+)ΔG = – , therefore spontaneous

– If a change is endothermic (+ ΔH) and entropy decreases (- ΔS), both factors work against spontaneity

ΔG = ΔH – T ΔSΔG = (+) – T (–)ΔG = + , therefore not spontaneous

When ΔH and ΔS have the same sign, the temperature at which the change occurs determines whether or not the change will be spontaneous.

– If ΔH and ΔS are both positive, only a high temperature will make ΔG negative

ΔG = ΔH – T ΔSΔG = (+) – T (+)

– If ΔH and ΔS are both negative, only a low temperature will make ΔG negative

ΔG = ΔH – T ΔSΔG = (–) – T (–)

Therefore:

 ENDOTHERMIC(ΔH > O)

EXOTHERMIC(ΔH < O)

ENTROPY INCREASES(ΔS > O)

Depends on Temperature(spontaneous only at higher temperatures)

Spontaneous(at all temperatures)

ENTROPY DECREASES(ΔS < O)

Nonspontaneous(at all temperatures)

Depends on Temperature(spontaneous only at lower temperatures)

Important Note:– Using the Laws of Thermodynamics, we can determine whether a change will be spontaneous or not at a certain temperature BUT this gives NO information about how quickly the reaction takes place!

Sample ProblemWill the following reaction proceed spontaneously at 25°C?4 KClO3 (s) 3 KClO4 (s) + KCl(s)

Chapter 6.1 Reaction Rates

Chemical Kinetics Thermodynamics gives NO information about HOW FAST a change will take place Chemical kinetics is the study of REACTION RATES and MECHANISMS (events at molecular level that control the speed and outcome of a reaction)

Decomposition of H2O2 (aq)

The reaction by itself would occur too slowly to be seen. A catalyst has been added and the reaction is visible.

Describing Reaction Rates Reaction rate (r) is determined by measuring the rate at which a product is formed or the rate at which a reactant is consumed over a series of time intervals Properties like mass, colour, conductivity, volume, pressure or concentration may be measured to determine the reaction rate Reaction rate is expressed mathematically in terms of a change in property of reactant or product per unit time UNITS: mol/L•s = change in concentration of reactant or product (Δc, mol/L) over time (Δt, s)

absolute value is taken (answer is always positive)

Consider the following reaction:2H2O2 (l) 2H2O (l) + O2 (g)

The rate of loss of hydrogen peroxide, the reactant, is not the same as the rate of formation of oxygen gas, the product. Only one mole of oxygen gas is produced for every two moles of hydrogen peroxide that decomposes.

Sample Problem Measurements taken during the following reaction showed a concentration of carbon monoxide (CO) of 0.019 mol/L at 27 min and of 0.013 mol/L at 45 min. Calculate the average rate of the loss of carbon monoxide (CO) AND the gain of carbon dioxide (CO2).

CO (g) + NO2 (g) CO2 (g) + NO (g)

Average Rate of Reaction Typically in a reaction, the reaction rate is not constant as the reaction progresses The average rate of reaction gives an overall idea of how quickly the reaction is progressing over a particular time intervalTo find the average rate1. Plot the concentration – time graph2. Draw a line between 2 points on

the curve (secant)3. Find the slope of the secant (equal to average rate)

Instantaneous Rate of Reaction The instantaneous rate of reaction is the rate of reaction at a particular time during the reactionTo find the average rate1. Plot the concentration – time graph2. Draw a tangent to the curve at the specified time3. Find the slope of the tangent (equal to instantaneous rate)

See Fig 6.2 p.269

Initial Rate and Graphing Initial Rate of Reaction: the speed of the reaction the instant the reactants are mixed (t = 0)

– like instantaneous rate of reaction but for t = 0 Concentration-Time Graphs

– The shape of the curve depends on what is measured – the concentration of the reactants or the concentration of the products

– Concentration of reactantdecreases over time, therefore slope will be negative

– Concentration of productincreases over time, thereforeslope will be positive

Rate of Reaction

time

conc

entr

atio

n of

reactant

Rate of Reaction

conc

entr

atio

n of

product

time

Measuring Reaction Rates Ideally, measurements are taken without disturbing the reaction itself. This is possible for the following:

Reactions That Produce a Gas– measure volume or pressure

of gas producedEx. Zn (s) + 2 HCl (aq)

H2 (g) + ZnCl2 (aq)

Reactions That Involve Ions– measure conductivity of solutionEx. (CH3)3CCl (aq) + H2O (l)

(CH3)3COH (aq) + H+(aq) + Cl-(aq)

Reactions That Change Colour– measure intensity of colour using a spectrophotometerEx. ClO-

(aq) + I-(aq) IO-

(aq) (yellow) + Cl-(aq)

Chapter 6.2 Rate Law and Order of Reaction

Quantitative Effects of Factors The mathematical relationship between the reaction rate and the factors that affect it is called the RATE LAW.

The rate law may not be predicted theoretically. It must be determined experimentally.

The oxidation of iodide by hydrogen peroxide in an acidic solution.

Determination of Rate Laws Consider the general reactionA + B AB As the reaction proceeds, [A] and [B] decreases, and [AB] increases The average rate =

The rate law is proportional to the product of the initial concentrations of the reactants to some exponential values

r ∞ [A]m[B]n

m and n describe the relationship between the rate and the initial concentration and can only be determined experimentally

The Rate Law Equation: r = k[A]m[B]n

where k is a rate constant, determined empirically, valid only for a specific reaction at a specific temperature

t

[AB]or

t

[B]or

][

t

A

Example 1: For the reaction 3X + 4Y Z, the following data was collected:Experiment [X] (mol/L) [Y] (mol/L) Initial Rate (mol/Ls)

1 0.1 0.1 102 0.2 0.1 203 0.3 0.1 304 0.4 0.1 405 0.1 0.1 106 0.1 0.2 407 0.1 0.3 908 0.1 0.4 160

For Experiment 1 to 4, rate ∞ [X]1

For Experiment 5 to 8, rate ∞ [Y]2

The reaction is first order with respect to X and second order with respect to Y. The overall order of the reaction is 3 (1 + 2) For this reaction, at this temperature, r = k[X] [Y]2

We can solve for k by using data from experiment 1:r = k[X] [Y]2

10 mol/Ls = k (0.1 mol/L)1 (0.1 mol/L)2

k = 10 mol/Ls . (0.1 mol/L)1 (0.1 mol/L)2

k = 1.0 x 104 L2/mol2s

* The value for k is the same no matter which data is used.

Chapter 6.4 Collision Theory and Rate of Reaction

Collision Theory Rates of reaction can be explained with collision theory.

Concepts of Collision Theory- system consists of particles in constant motion at speed proportional to temperature of sample- chemical reaction must involve collisions of particles with each other and with the walls of the containers- effective collision – has sufficient energy and correct orientation for bonds can be broken and new bonds form- ineffective collision – particles rebound from collision, essentially unchanged- rate of reaction depends on frequency of collisions and the fraction of collisions that are effective

Activation Energy The minimum increase in potential energy of a system required for molecules to rearrange their structure and result in an effective collision

Correct Orientation Particles must collide with correct orientation for the collision to be effective

HF + CH3Cl CH3F + HCl HF + CH3Cl NR

Transition State Theory Activation Energy (Ea): the minimum energy required for the reaction to occur.– The difference between the initial energy of the reactants and the energy of the activated complex at the transition state.

Activated Complex: an unstable molecule containing partially broken and partially formed bonds representing the maximum potential energy point in the change

Transition State: energy maximum where the activated complex is formed