REACTION KINETICS

S E C T I O N 1 – T H E R E A C T I O N P R O C E S S

B Y S TUD Y ING M A NY TY PE S OF E X PE R IM E NTS ,C HE M I S TS HA VE F OUND THA T C HE M IC A LR E A C T IONS OC C UR A T W ID EL Y D I F F ER ING R A TE S .F OR E XA M PLE , I N THE PR ES E NC E OF A IR , IR ONR US TS VE R Y S L OWL Y , WHE R EA S M E THA NE INNA TUR A L GA S B UR NS RA P ID L Y . THE S PEE D OF AC HE M IC A L R EA C T ION D E PE ND S ON THE E NE R GYPA THWA Y THA T A R E A C T ION F OLL OWS A ND THEC HA NGE S THA T TA K E PL A CE ON THE M OL E C ULA RL E VE L WHE N S UB S TA NC ES INTE R A C T . I N TH I SC HA PTE R , YOU W I L L S TUD Y THE F A C TORS THA TA F F E C T HOW F A S T C HE M IC A L R EA C T IONS TA K EPL A C E .

REACTION MECHANISMS• If you mix solutions of HCl and NaOH, a quick neutralization reaction happens

H3O+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) → 2H2O(l) + Na+(aq) + Cl−(aq)

•Reaction is nearly instantaneous – the rate is only limited by the speed H3O+ and OH− can diffuse through water to meet each other

•Reactions between ions of same charge and between molecules are not immediate•Positive and negative ions repel each other

•Electron clouds of molecules strongly repel each other at short distances•Only ions or molecules with very high kinetic energy overcome repulsive forces enough to react

• Colorless hydrogen gas is made of pairs of hydrogen atoms bonded together as diatomic molecules, H2• Violet-colored iodine vapor is also diatomic, I2• Chemical reaction between two at high temps makes hydrogen iodide, HI, a colorless gas

• HI molecules decompose and re-form H2 and I2molecules• Produces violet gas

H2(g) + I2(g) → 2HI(g)2HI(g) → H2(g) + I2(g)

• These equations only show which molecular species disappear as a result of the reactions and which species are made • They do not show

reaction mechanism àthe step-by-step sequence of reactions by which the overall chemical change occurs

H2(g) + I2(g) → 2HI(g)2HI(g) → H2(g) + I2(g)

•Experiments can often be designed that suggest most likely steps in a reaction mechanism

•Each reaction step is usually simple

•Equation for each step represents the actual atoms, ions or molecules that participate

•It was thought the formation of HI was a simple one-step process

•Involves two molecules, H2 and I2 in forward reaction and 2 HI molecules in reverse reaction

•Experiments showed direct reaction between H2 and I2 never takes place

•Alternative mechanisms for reaction suggested based on experimental results

•Steps in each reaction had to add up to give overall equation

•Some species appear in steps but not in overall net equation

• Intermediates à species that appear in some steps but not in net equation

2 POSSIBLE MECHANISMS FOR FORMATION OF HI

•This reaction is example of homongeneous reaction à a reaction whose reactants and products exist in a single phase

•In this case, gas

COLLISION THEORY

•For reactions to occur, the atoms/molecules must collide•The collisions must also result in interaction between particles•Collision theory à set of assumptions regarding collisions and reactions•Chemists use theory to understand observations about chemical reactions

• Let’s consider what happens on molecular scale in one step of homogeneous reaction system

AB + AB ⇌A2 + 2B

•According to collisions theory, the two AB molecules must collide in order to react•Also must collide in the right position with enough energy to combine valence electrons and disturb the bonds

• If they do, collision results in recombination (A2and 2B)

•If collision too gentle, two molecules rebound unchanged (a)

• If molecules aren’t in right position, they rebound without interacting (b)

•An effective collision must be in right position and with enough energy (c)•NOT EVERY COLLISION CREATES A REACTION

•Chemical reaction produces new bonds between atoms in colliding molecules•Unless collision brings correct atoms close together in proper orientation, molecules do not react

•Ex. If Cl2 collides with oxygen end of NO, the following reaction may occur:

NO (g) + Cl2 (g) à NOCl (g) + Cl

•This won’t happen unless Cl2collides with O of NO

ACTIVATION ENERGY• Let’s consider reaction for formation of water from H2 and O2

2H2(g) + O2(g) → 2H2O(l)

•Heat of formation ΔH˚f = −285.8 kJ/mol at 298.15 K• The free-energy change is also large: ΔG˚ = −237.1 kJ/mol•Why don’t oxygen and hydrogen combine spontaneously and immediately to form water?

•When diatomic H2 and O2 approach each other, the electron clouds repel each other•Molecules bounce off and never actually meet•For reaction to happen, molecules must have enough kinetic energy to actually mix together the valence electrons

•Bonds of species must be broken in order for new ones to be formed•Bond breaking is endothermic

•Bond forming is exothermic•Initial input of energy needed to overcome repulsion •This initial energy activates the reaction

•Once an exothermic reaction is started, the energy released is enough to continue the reaction by activating other molecules•Result à reaction rate keeps increasing•Limited by time required for reactant particles to get energy and make contact

•Activation energy (Ea) à minimum energy required to transform the reactants into an activated complex

•Reverse reaction (decomposition of water molecules) is endothermic•Water molecules lie at energy level lower than H2 and O2 levels•Require higher activation energy before they decompose•Represented by Eaʹ

THE ACTIVATED COMPLEX•When molecules collide, some kinetic energy converted into internal potential energy within colliding molecules• If enough energy converted, molecules with correct orientation become activated•New bonds can now form

•During interval of breaking and reforming bonds, collision complex is in transition state

•Some kind of partial bonding exists

•Activated complex à a transitional structure that results from an effective collision and that remains while old bonds are breaking and new bonds are forming

•Both forward and reverse reactions go through same activated complex•Bond broken in activated complex for forward reaction must be re-formed in reverse reaction

•Remember speeds (therefore kinetic energies) of molecules increases as temperature increases• Increase in speed à increase in collisions à increase in reactions•Still depends on enough energy to make activated complex no matter how many collisions•Raising temperature gives more molecules this activation energy, so reactions increase

•Activated complex is characteristic of both product and reactant•May re-form original bonds and separate back to reactant particles•May form new bonds and separate into product particles•Either are just as likely

SEC TIO N 4 –REAC TIO N RATE

THE C HA NGE IN C ONC E NTRA T ION OFR E A C TA NTS PE R UN I T T IM E A S A RE A C T IONPR OC E E D S I S C AL LE D THE R EA C T ION R A TE .THE S TUD Y OF R E A C T ION R A TE S I SC ONC E R NE D W ITH THE F A C TOR S THA TA F F E C T THE RA TE A ND W ITH THEM A THE M A T I C A L E X PR E SS IONS THA T R E VEA LTHE S PE C I F I C D E PE NDE NC I E S O F THE R A TEON C ONC E NTR A T ION .THE A R EA OF C HE M I S TR Y THA T I SC ONC E R NE D W ITH R EA C T ION R A TES A NDR E A C T ION M E C HA N I S M S I S C A LL E DC HE M IC A L K INE T I C S .

RATE-INFLUENCING FACTORS

•Other than decompositions, particles must come into contact in the right position and with enough energy for activation•Rate of reaction depends on collision frequency of reactants and on collision effectiveness

•5 important factors influence rate of chemical reaction

1.Nature of reactants2.Surface area3.Temperature4.Concentration5.Presence of catalysts

1. NATURE OF REACTANTS

•Substances vary greatly in their ability to react•Ex. H2 combines strongly with chlorine under certain conditions–Under same conditions, hardly reacts at all with nitrogen

HYDROGENAND

CHLORINE

•Na and O2 combine more rapidly than Fe and O2 under similar conditions

2. SURFACE AREA•Gaseous mixtures and dissolved particles can mix and collide freely•Reactions with them happen more often and faster•Heterogeneous reactions à involve reactants in two different phases•Can occur only when two phases are in contact•Surface area of solid reactant important in determining rate•Higher surface area, higher rate of reaction

EXAMPLE – ZN AND HCL•Solid zinc reacts with hydrochloric acid to make zinc chloride and hydrogen gas

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

•Reaction occurs on surface of zinc•Piece of zinc measuring 1cm on each edge only give 6 cm2 of contact area•Powdered form – 1,000 times the surface area

EXAMPLE – BURNING COAL

•Big lump of coal burns slowly in air•Rate of burning increased when you break up coal into smaller pieces•If piece of coal is powdered and ignited while suspended in air à BOOM!•This is cause of explosions in coal mines

Coal mine explosion in the Ukraine

3. TEMPERATURE

•Increase in temperature increases kinetic energy•Can result in greater number of effective collisions•IF number of effective collisions increase, reaction rate will increase

•To be effective, energy of collisions must be equal to or greater than (≥) the activation energy•At higher temperatures, more particles have enough energy to form activated complex when collisions occur•So, rise in temperature produces an increase in collision energy as well as in collision frequency

•Decreasing temperature has opposite effect•Average kinetic energy decreases•They collide less frequently with less energy•Produces fewer effective collisions•Beginning near room temp, reaction rates of many common reactions roughly double with each 10K rise in temperature

4. CONCENTRATION•Pure oxygen has 5 times the concentration of oxygen molecules in air•Substance that reacts with oxygen in air reacts more strongly with pure oxygen

Carbon burns faster in pure oxygen (a) than in air (b) because the concentration of the reacting species, O2, is greater.

(a) (b)

•Light produced in pure oxygen from charcoal is brighter than in air•Oxidation of charcoal is heterogeneous reaction system (gas and solid)•Reaction rate depends on amount of exposed charcoal AND concentration of O2

•In homogeneous reaction systems, reaction rates depend on concentration of reactants•If number of effective collisions increases, rate increases as well•In general, increase in rate expected if concentration of one or more reactants is increased

•In system with only two molecules, only one collision can occur•When there are four molecules, there can be four possible collisions

•Under constant conditions, as number of molecules in system increases, so does the total number of possible collisions•Lowering concentration should have the opposite effect

4. PRESENCE OF CATALYSTS•Some chemical reactions are slow•Can be increased by catalyst à a substance that changes the rate of a chemical reaction without itself being permanently used•Catalysis à the action of a catalyst

•Catalyst provides alternative energy pathway or reaction mechanism •Potential-energy barrier between reactants and products is lowered

• The reaction rate of the decomposition of hydrogen peroxide, H2O2, can be increased by using a catalyst• The catalyst used here is manganese dioxide, MnO2, a black solid• A 30% H2O2 solution is added dropwise onto the MnO2 in the beaker and rapidly decomposes to O2 and H2O• Both the oxygen and water appear as gases because the high heat of reaction causes the water to vaporize

•Catalysts do not appear among final products•May participate in one step along a reaction pathway and be regenerated in later step•Homogeneous catalyst à a catalyst that is in the same phase as all the reactants and products in a reaction system•Heterogeneous catalyst à a catalyst that is in a different phase from the reactants

RATE LAWS FOR REACTIONS

•Relationship between rate of reaction and concentration of one reactant is determined experimentally by keeping concentrations of other reactants and temperature constant•Reaction rate measured for various concentrations of reactant in question

•Hydrogen gas reacts with nitrogen monoxide gas at constant volume and at elevated constant temperature

2H2(g) + 2NO(g) → N2(g) + 2H2O(g)

•4 mol reactant gases produce 3 mol product gases•Result: pressure of system decreases as reaction proceeds•Rate of reaction can be determined by measuring pressure

•Suppose a series of experiments is done using same initial concentration of NO but different initial concentrations of H2•Initial reaction rate varies directly with H2 concentration

2H2(g) + 2NO(g) → N2(g) + 2H2O(g)

•Double H2 à double rate• If R represents reaction rate and [H2] is concentration of H2 in mol/L, relationship is shown as follows•R α [H2]where α means “is proportional to”

2H2(g) + 2NO(g) → N2(g) + 2H2O(g)

2H2(G) + 2NO(G) → N2(G) + 2H2O(G)

•Suppose same initial [H2] is used BUT initial [NO] is varied• Initial rate increase fourfold when [NO] is doubled • Increased 9-fold when [NO] tripled•Reaction rate varies with the square of concentration of NO

R α [NO]²

•Because R is proportional to [H2] and [NO]² it is proportional to their product

R ∝ [H2][NO]²

•By introduction of a proportionality constant, k, the expression becomes an equality

R = k[H2][NO]²

•Rate law à an equation that relates reaction rate and concentrations of reactants •Can use for specific reaction at a given temperature•Rise in temp increases rates of most reactions•Value k usually increases as temp increases•Relationship between rate and concentration almost always remains unchanged

USING THE RATE LAW

•General form for rate law isR = k[A]n[B]m•R = reaction rate•K = specific rate constant•[A] and [B] are concentration of reactants•n and m are powers to which concentrations are raised

•Order à the power to which a reactant concentration is raised•Value of “n” is order of reaction with respect to [A]•Reaction is “nth order in A”•Or “mth order in B”•n and m usually small integers or zero

•Order of 1 means reaction rate (R) is directly proportional to concentration of reactant•Order of 2 means R is directly proportional to the square of reactant•Zero means rate does not depend on concentration, as long as some reactant is present

•Sum of all reactant orders is called order of the reaction or overall order•Equal to n + m•Orders in rate law may or may not match coefficients in balanced equation•These must be determined from experimental data

SPECIFIC RATE CONSTANT

•k•Proportionality constant relating rate of reaction to concentrations of reactants

1. Once reaction orders are known, value of k must be determined from experimental data

2. Value of k is for specific reaction3. Units of k depend on overall order of

reaction4. Value of k does not change for different

concentrations of reactants or products5. Value of k is for reaction at specific

temperature – increasing temp increases k6. Value of k becomes larger with catalyst

SAMPLE PROBLEM

Three experiments that have identical conditions were performed to measure the initial rate of the reaction2HI (g) à H2(g) + I2 (g)The results for the three experiments in which only the HI concentration was varied are as follows:

Experiment [HI] (M) Rate (M/s)

1 0.015 1.1 x 10-3

2 0.030 4.4 x 10-3

3 0.045 9.9 x 10-3

•Write the rate law for the reaction•Find the value and units of the specific rate constant

2HI (g) à H2(g) + I2 (g)

•The general rate law for this reaction is R= k[HI]n•we can find the ratio of the reactant concentrations between experiments 1 and 2, [HI]2÷[HI]1•Then see how ratio of concentration affects ratio of rates, R2÷R3

SAMPLE PROBLEM C3 experiments were performed to measure the initial rate of reaction A + B à CConditions were identical except that the concentration of reactants varied.Write the rate law for the reaction.Find the value and units of the specific rate constant

RATE LAWS AND REACTION PATHWAY•The form of the rate law depends on the reaction mechanism

•For reaction with single step, reaction rate of that step is proportional to the product of the reactant concentrations, each of which is raised to its stoichiometriccoefficient

(wait a minute, you will understand)

EXAMPLE•1 molecule of gas A collides with 1 molecule of gas B to form molecules of C

A + B à 2C

•One particle of each reactant is involved in each collision•Doubling concentration of either reactant will double collision frequency•Also doubles reaction rate for this step

A + B à 2C

•The rate for this step is directly proportional to the concentration of A and B

R = k[A][B]

•Now suppose reaction is reversible• In reverse, 2 molecules of C must decompose to form 1 molecule of A and 1 molecule of B

2C à A + B

•Reaction rate for the reverse is directly proportional to [C] x [C]

R = k[C]²

•Power of the molar concentration of each reactant is same as the coefficient in balanced equation•Relationship is true ONLY if reaction is one-step•If reaction is in sequence of steps, rate law is determined from slowest step because it has the lowest rate•Rate-determining step à the slowest-rate step for the chemical reaction

•Consider reaction of nitrogen dioxide and carbon monoxide

NO2(g) + CO(g) → NO(g) + CO2(g)

•Reaction is believed to be in two steps

•Step 1: NO2 + NO2 → NO3 + NO slow•Step 2: NO3 + CO → NO2 + CO2 fast

STEP 1: NO2 + NO2 → NO3 + NO SLOWSTEP 2: NO3 + CO → NO2 + CO2 FAST

•Step 1 – two molecules NO2 collide, forming intermediate NO3•NO3 then collides with one molecule CO and reacts quickly to make one molecule NO2 and one CO2•First step is slower, so it is the rate-determining step

•Rate law from rate-determining stepR = k[NO2]²

•Tells us 2 molecules NO2 are reactants in slower, rate-determining step•CO not in rate law because it reacts AFTER the rate-determining step

GENERAL FORM OF RATE LAWR = k[A]n[B]m….

•R à reaction rate• k à rate constant• [A], [B] à molar concentrations of reactants•n, m, à respective powers reactants are raised

•Must be determined from experimental data

SAMPLE PROBLEM D•Nitrogen dioxide and fluorine react in the gas phase according to the following equation.

2NO2(g) + F2(g) → 2NO2F(g)•A proposed mechanism for this reaction follows.•Step 1: NO2 + F2 → NO2F + F slow•Step 2: F + NO2 → NO2F fast• Identify the rate-determining step and write an acceptable rate law.

• If we combine these two steps, the intermediate, F, cancels out and we are left with the original equation.•The first step is the slower step, and is considered the rate-determining step.•We can write the rate law from this rate-determining step.

R = k [NO2][F2]

SAMPLE PROBLEM E

•A reaction involving reactants X and Y was found to occur by a one-step mechanism:

X + 2Y → XY2•Write the rate law for this reaction, and then determine the effect of each of the following on the reaction rate:• a. doubling the concentration of X• b. doubling the concentration of Y• c. using one-third the concentration of Y

• Because the equation represents a single-step mechanism, the rate law can be written from the equation (otherwise, it could not be)• The rate will vary directly with the concentration of X, which has an implied coefficient of 1 in the equation, and will vary directly with the square of the concentration of Y, which has the coefficient of 2: R = k[X][Y]²• a. Doubling the concentration of X will double the rate (R = k[2X][Y]²)• b. Doubling the concentration of Y will increase the rate fourfold (R = k[X][2Y]²)• c. Using one-third the concentration of Y will reduce the rate to one-ninth of its original value (R = k[X][ 1Y]²)

PRACTICE PROBLEM 1

•The rate of a reaction involving L, M, and N is found to double if the concentration of L is doubled, to increase eightfold if the concentration of M is doubled, and to double if the concentration of N is doubled. Write the rate law for this reaction.•R = k[L][M]³[N]

PRACTICE PROBLEM 2•At temperatures below 498 K, the following reaction takes place.

NO2(g) + CO(g) → CO2(g) + NO(g)•Doubling the concentration of NO2 quadruples the rate of CO2 being formed if the CO concentration is held constant. However, doubling the concentration of CO has no effect on the rate of CO2 formation. Write a rate-law expression for this reaction.•R = k[NO2]²