AP* Chemistry Chemical Kinetics = How fast will the ...

AP* Chemistry (modified from NMSI resources by René McCormick) Chemical Kinetics

Reactions & Stoichiometry = What are we making and how much? Spontaneity & Thermodynamics = Will the reaction occur?

Chemical Kinetics = How fast will the process occur?

Reaction Rate: the speed at which a chemical reaction occurs

Factors Affecting Reaction Rates 1. Nature of Reactants- some chemicals react more quickly than others

a. State- molecules must be in a state in which they are able to mix: • aqueous (freely-moving ions) … gas (rapidly moving)… liquid (freely moving)… solid (little motion)

b. Chemical identity- How reactive are the reactants? • Strength & number of bonds to be broken/made • High/low electronegativity & ionization energy • Molecular complexity: monatomic vs. polyatomic ions • Strong vs. weak acid/base

2. Concentration of Reactants- more molecules… more collisions… faster reaction • For dissolved or gaseous reactants

o Solids & liquids do NOT have molarities

3. Surface Area of Reactants- more exposed molecules… more collisions… faster reaction • Heterogeneous reactions: reactants in different states • Only react at the surface of the solid

o More surface = more opportunity to react

4. Temperature- hotter/faster moving particles… more frequent & more powerful collisions… faster reaction

• Temperature: measure of average kinetic energy of particles in a sample

• Faster = more frequent AND more forceful collisions o In general, increasing 10°C doubles reaction rate

(near room temperature)

5. Catalysts- chemicals that speed up chemical reactions without changing themselves • Catalysts are used in intermediate steps and regenerated later

o Creates a new reaction pathway • Lowers the activation energy

o Less energetic collisions are successful (doesn’t speed up particles) • Enzymes = biological catalysts • Can be homogenous or heterogeneous

Collision Theory of Reaction Rates In order to react… 1. Particles must collide

• Only two particles may collide at a time

2. Must collide with the proper orientation • Use drawings when explaining

AP* is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product. Portions of this document are modified from NMSI AP Chemistry notes © 2008 by René McCormick.

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3. Must collide with enough energy Activation Energy (Ea): minimum kinetic energy

that reacting species must have in order to react

• Energy overcomes repulsion so that atoms can get close enough to rearrange electrons, forming new bonds

Activated Complex: the temporary transition state at the peak of the energy diagram

• Changing the temperature does NOT change the Ea, it causes a larger % of the molecules to have the minimum KE to get over the energy barrier

Reaction Rates Rate = measurable change in any quantity per unit time

Reaction rate: decrease in concentration of a reactant or increase in concentration of a product per unit time • Reaction rate is always reported as a positive number • Concentration can be determined by measuring: gas volume, gas pressure, mass, solution pH, solution color

Δ[Chemical species] Rate =

time

Because reaction rate is based upon how many reactants particles are colliding at any time, the reaction rate will change (slow down) over time.

Rate can be: 1. Average rate: use initial and final values for a time interval

Exercise #1 Use the data to calculate the average reaction rate during the first 100 seconds.

SO2Cl2 SO2 + Cl2 [SO2Cl2] (mol ⋅ L-1) Time (sec) 0.200 0 0.160 100 0.127 200 0.100 300 0.080 400 0.067 500

Rate = 0.0004 M/s

2. Instantaneous rate: rate at a specific time • Found by determining slope of the curve at any point

Initial rate: instantaneous rate at t = 0


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3. Relative rate: rate of one species as compared to another • Use the stoichiometric coefficients to determine the rate of an unknown species compared to a known one

a A + b B c C + d D

1 Δ[A] 1 Δ[B] 1 Δ[C] 1 Δ[D] Rate = –

a Δt = – b Δt =

c Δt = d Δt

Exercise #2 The rate of disappearance of dinitrogen pentoxide is 1.2 x 10-3 M⋅s-1 at a given time. What are the rates of appearance for each product given the following balanced equation?

2 N2O5 4 NO2 + O2

NO2 = 2.4 x 10-3 M⋅s-1; O2 = 6.0 x 10-4 M⋅s-1

Rate Laws & Reaction Order Rate Law: equation showing how the rate of a reaction depends on reactant concentrations

• Dependent on the reaction process, not the coefficients of the balanced reaction • Use initial rates to avoid confusion caused by presence of products • Help us to understand the mechanism (step-by-step process) for a reaction

Rate = k[A]m[B]n

k = rate constant m & n = reaction order (NOT the coefficients)

Rate constant (k)… • Changes with temperature and/or the addition of a catalyst • Magnitude gives indicator of how fast reaction will occur • Determined experimentally using known concentrations & initial rate • Unit varies based upon overall order, but will ALWAYS cancel units from rate law expression

Reaction order (m, n, etc)… • Order relative to each reactant is the exponent (0, 1, 2…) on the concentration in rate law expression • Normally whole numbers, but can be fractions • Overall order of the reaction is sum of the orders with respect to each reactant • MUST BE DETERMINED BY EXPERIMENT

Zero Order: concentration of products has no effect on the rate of reaction • Equation: Rate = k • Doubling concentration has no effect on the rate

First Order: reaction rate is directly proportional to the concentration of a reactant • Equation: Rate = k[A]1 • Doubling concentration double the rates, etc • Radioactive decays are first order (half-life)

Second Order: reaction rate is proportional to the concentration of a reactant squared • Equation: Rate = k[A]2 or Rate = k[A]1[B]1 • Doubling concentration quadruples the rate, etc


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Differential Rate Law Uses data in which initial RATES and reactant CONCENTRATIONS are given.

Experiment Number

Initial Rate mol/(L• hr)

Initial concentration [A]o

Initial concentration [B]o

1 5.0 × 10−3 0.50 0.20 2 5.0 × 10−3 0.75 0.20 3 5.0 × 10−3 1.00 0.20 4 1.00 × 10−2 0.50 0.40 5 1.50 × 10−2 0.50 0.60

Must determine how changing the concentration of each reactant changes the initial rate by looking at multiple trials of the same experiment.

By Inspection (“Table logic”)- 1. Look for trials where concentration of one species is constant 2. Look at how the concentration of the other reactant changes (doubled, halved, etc) 3. Look at how the change in concentration changes the initial rates (doubled, halved, etc) to determine order

Should be mostly SIMPLE math (2x 1, 2, 4, 8, 16… 3x 1, 3, 9, 27, 81… 4x 1, 4, 16, 64, 256…) 4. Repeat process with each reactant 5. After determining order of each reactant (and therefore overall order) plug concentration in to rate law

expression with data from one experiment to solve for rate constant

MUST SHOW YOUR WORK/LOGIC TO GET CREDIT!! !

Experiment 1 & 4- • A stays constant while B is doubled (0.20 0.40) • When B is doubled, initial rate is doubled (0.0050 0.0100) • Therefore B is first order: 21 = 2

Experiment 1 & 3- • A is doubled (0.50 1.00) while B is held constant • When A is doubled, initial rate remains constant (0.0050 0.0050) • Therefore A is zero order: 20 = 1

Rate law expression- • Rate = k[A]0[B]1 = [B]

Finding k- • 0.0050 = k[0.20] • k = 0.025 hr-1

Algebraically- 1. Write the rate law expression for each trial by plugging in values for concentrations and rate 2. Choose two equations in which a reactant concentration will cancel (along with k) 3. Solve for the exponent on the reactant that didn’t cancel 4. Repeat process with each reactant 5. After determining order of each reactant (and therefore overall order) plug concentration in to rate law

expression with data from one experiment to solve for rate constant

Experiments 1 & 4 Exp. 4 0.0100 = k(0.50)m(0.40)n 0.0100 = k(0.50)m(0.40)n 0.100 = (0.40)n 2 = 2n n = 1

Exp. 1 0.0050 = k(0.50)m(0.20)n 0.0050 = k(0.50)m(0.20)n 0.0050 = (0.20)n


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Experiments 1 & 3 Exp. 3 0.0050 = k(1.00)m(0.20)n 0.0050 = k(1.00)m(0.20)n 0.0050 = (1.00)m 1 = 2m m = 0

Exp. 1 0.0050 = k(0.50)m(0.20)n 0.0050 = k(0.50)m(0.20)n 0.0050 = (0.50)m

Rate law expression- • Rate = k[A]0[B]1 = [B]

Finding k- • 0.0050 = k[0.20] • k = 0.025 hr-1

Units in rate law expressions: • Concentration is always M (mol/L) • Initial rate is always M/time (mol/L⋅time) • Units on k will vary so that the units will cancel…

Zero Order First Order Second Order Nth Order

M 1 1 1 time time M⋅time M(n-1)⋅time

Example #3 Write the rate law expression and find the value for k for the chemical reaction given the following data.

NH4+ + NO2

– N2 + 2 H2O

Trial Number [NH4+]0 (M) [NO2

–]0 (M) Initial Rate (M/s) 1 0.0100 0.200 5.4 x 10-7

2 0.0200 0.200 10.8 x 10-7

3 0.0400 0.200 21.5 x 10-7

4 0.200 0.0202 10.8 x 10-7

5 0.200 0.0404 21.6 x 10-7

6 0.200 0.0808 43.3 x 10-7

Rate = k[NH4

+]1[ NO2–]1 k = 2.7 x 10-4 M-1s-1

Example #4 Write the rate law expression and find the value for k for the chemical reaction given the following data. What is the initial rate when [NO] = 0.050 M and [H2] = 0.150 M?

2 NO + 2 H2 N2 + 2 H2O

Trial Number [NO]0 (M) [H2]0 (M) Initial Rate (M/s) 1 0.10 0.10 1.23 x 10-3

2 0.10 0.20 2.46 x 10-3

3 0.20 0.10 4.92 x 10-3

Rate = k[NO]2[H2]1 k = 1.23 M-2s-1 rate = 4.5 x 10-4 M/s


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Example #5 The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation:

BrO3–(aq) + 5 Br– (aq) + 6 H+(aq) 3 Br2 (l) + 3 H2O (l)

The table below gives the results of four experiments. Using these data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant. What is the value of k? What are the units of k?

Trial Initial [BrO3–] Initial [Br –] Initial [H+] Measured initial rate (mol/L⋅s)

1 0.10 0.10 0.10 8.0 x 10-4

2 0.20 0.10 0.10 1.6 x 10-3 3 0.20 0.20 0.10 3.2 x 10-3 4 0.10 0.10 0.20 3.2 x 10-3

Rate = k[BrO3–]1[Br–]1[H+]2 k = 8.0 M-3 s-1

Integrated Rate Law Uses data in which reactant CONCENTRATIONS and TIME are given.

• Must graph data in order to determine the order • Rate constant, k, is related to the slope of the line (absolute value) • Will use graph(s) and y= mx +b to solve for unknown quantities

Graphs • X-axis is ALWAYS time • Y-axis should be plotted as:

o Concentration, [A] o ln(concentration), ln[A] o Reciprocal of concentration, 1/[A]

• The order of the reaction is based upon which graph is a straight line (or which is straightest)

[A] vs. time

Zero (0) Order

k = negative slope

[A] = -kt + [A]o

ln[A] vs. time

First (1st) Order

k = negative slope

ln[A] = -kt + ln[A]o

1/[A] vs. time

Second (2nd) Order

k = slope

1/[A] = kt + 1/[A]o


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Tips: • k is always a positive value (b/c rate can’t decrease with added reactant)

o k = |slope| • If graphs are not given, look at data to see which set ([A], ln[A], 1/[A]) describes linearly

o You can graph data with a graphing calculator to perform linear regressions • Graphs aren’t always presented in order in AP questions. Look at Y-axes ! • Partial pressure of a gas can replace concentration on x-axis

Zero Order: Rate = k

First Order: Rate = k[A]

Second Order: Rate = k[A]2


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Exercise #6 The decomposition of N2O5 in the gas phase was studied at constant temperature.

2 N2O5 4 NO2 + O2 The following results were collected.

[N2O5] 0.1000 0.0707 0.0500 0.0250 0.0125 0.00625 Time (s) 0 50 100 200 300 400

a) Determine the rate law and calculate the value of k (with unit).

b) What is the concentration of N2O5 at 600 seconds?

c) At what time is the concentration of N2O5 equal to 0.00150 M?

a) Rate = k[N2O5]1 k = 0.00693 s-1 b) 0.00156 M c) 606 sec

Exercise #7 For the reaction of (CH3)3CBr with OH–, the following data were obtained.

(CH3)3CBr + OH– (CH3)3COH + Br– The following results were collected.

Time (s) 0 30 60 90 [(CH3)3CBr] 0.100 0.074 0.055 0.041

ln[(CH3)3CBr] –2.30 –2.60 –2.90 –3.20 [(CH3)3CBr]-1 10. 13 18 24

a) Determine the rate law and calculate the value of k (with unit).

b) At what time is the concentration of (CH3)3CBr equal to 0.0500 M?

a) Rate = k[(CH3)3CBr] k = 0.0099 s-1 b) 70 sec Exercise #8 Hydrogen peroxide decomposes into water and oxygen gas. The graphs show the results of the study of the reaction.

a) Write the rate law for the reaction.

b) Calculate the value for k with units.

c) Determine [H2O2] after 2000 minutes.

a) Rate = k[H2O2]1 b) k = 0.000866 min-1 c) 0.142 M


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Half-Life Half-life is the time required for half of the reactants to disappear (turn into products)

• Often seen in nuclear reactions, but applies to first-order chemical reactions as well o The half-life changes over time for 0 and 2nd order reactions (initial half-life can be calculated)

• Nuclear decays are first order reactions with a constant half-life

ln[A] = -kt + ln[Ao]

ln[Ao/2] = -kt1/2 + ln[Ao]

kt1/2 = ln[Ao] - ln[Ao/2]

[Ao] kt1/2 = ln [Ao/2]

kt1/2 = ln[2]

0.693 t1/2 = k Exercise #9 The rate constant for the first order of transformation of cyclopropane to propene is 5.40 x 10-2/hr.

a) What is the half-life of the reaction? b) What percent of cyclopropane remains after 51.2 hours? c) How long until 1/9 of the original sample remains?

a) half-life = 12.8 hr b) 6.30% c) 40.7 hr Exercise #10

The radioactive sample of I-131 was measured. The data collected are plotted in the graph to the left.

a) Determine the half-life using the graph.

b) The data can be used to show that the decay is a first order reaction.

i) Label the vertical axis of the graph to the right

ii) What are the units on the rate constant, k, for the decay reaction?

a) 8 days b) i) ln[I-131] ii) days-1


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Integrated Rate Law with Multiple Reactants • Can’t graph concentration of multiple reactants at once • Must isolate the effect of one substance by “swamping” the reaction with an overwhelming amount of one

of the reactants o [Substance #1] = 1 x 10-3 M o [Substance #2] = 1 M (1000 times larger) o When all of Substance #1 is used, the concentration of Substance #2 barely changes

• Rate is now dependent on the [Substance #1]

o Rate = k’[Substance #1]m o Pseudo-rate law with pseudo-rate constant

Reaction Mechanisms Chemical reactions occur as a series of elementary steps (or sometimes just one elementary step)

• Elementary steps: individual collisions and simple changes that occur during the reaction process • VERY rare that multiple reactants collide at once to make products • Elementary steps are classified by their molecularity

Molecularity- number of molecules that participate in a collision • Unimolecular: involves one reactant molecule (not a collision, just an atomic rearrangement) • Bimolecular: involves a collision between two reactant molecules • Termolecular: involves a simultaneous collision between three molecules (very rare!)

The step-by-step process of turning reactants into products (with all the in-between details… elementary steps) is the reaction mechanism

• Breaks complex reaction into a series of elementary steps o Includes things that aren’t reactants or products, like catalysts and intermediates

Catalyst: substance that speeds up reaction without being consumed in the reaction; used in one step and regenerated in another

Intermediate: substance formed in one step of a reaction and consumed in another; does not appear in the overall reaction

• Mechanism is determined by experiment o Purpose of determining rate laws

Give data about the collisions that we can’t see o Rate laws eliminate incorrect mechanisms and narrow down the “possible paths” for a reaction

• Mechanism connects to rate law and to the stoichiometry (coefficients) of the balanced reaction o Connecting to stoichiometry: showing that the steps combine to give the overall reaction

Canceling out catalysts and intermediates Combining coefficients

o Connecting to rate law: showing that the exponents on the rate law connect to the coefficients on the elementary steps

Rate Expression for Elementary Steps- Elementary Step A Products 2 A Products A + B Products 2 A + B Products

Molecularity Unimolecular Bimolecular Bimolecular Termolecular Rate Expression Rate = k[A] Rate = k[A]2 Rate = k[A][B] Rate = k[A]2[B]

The order of an ELEMENTARY step is dependent on the coefficients of the reactants in the elementary step. • More molecules = more collisions = faster rate • NOT true of the overall reaction!


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Exercise #11 Nitrogen monoxide is reduced by hydrogen to give water and nitrogen.

2 H2 + 2 NO N2 + 2 H2O

Show that the stoichiometry of the following mechanism agrees with the overall reaction. Point out any catalysts or intermediates.

2 NO N2O2

N2O2 + H2 N2O + H2O

N2O + H2 N2 + H2O Exercise #12 Write the overall reaction for the following mechanism. Label catalysts and intermediates.

H2O2 + I– H2O + IO–

H2O2 + IO– H2O + O2 + I–

Mechanisms and Rate Expressions The rate law can be determined by the mechanism (or the mechanism by the rate law) by determining the rate determing step (RDS).

• RDS: the slowest elementary step in the reaction process o Obligatory food-related metaphor

• The rate law for the overall reaction includes all elementary steps up to and including the RDS o Order (exponents) on rate law are based upon how many times that reactant appears

Exercise #13 The balanced equation for the reaction of nitrogen dioxide and fluorine is:

2 NO2 + F2 2 NO2F

The experimentally determined rate law is: Rate = k[NO2][F2]. A suggested mechanism is:

NO2 + F2 NO2F + F SLOW

NO2 + F NO2F FAST

Explain is this mechanism if acceptable. Justify in terms of stoichiometry and the rate law.

Exercise #14 Write the overall reaction for the following mechanism:

HBr + O2 HOOBr

HBr + HOOBr 2 HOBr

2 HBr + 2 HOBr 2 Br2 + 2 H2O

If the rate law is experimentally determined to be: Rate = k[HBr][O2], label the slow step.

4 HBr + O2 2 Br2 + 2 H2O; Step 1 is slow step


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Potential Energy Diagrams Graphical representation of the energy changes in a chemical reaction

In a multi-step mechanism, each elementary step has its own activation energy

• RDS has highest activation energy


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Catalysts Substance that creates a mechanism with a lower activation energy & therefore speeds up the reaction

• Ea is lower but ΔH (ΔE) does NOT change • Enzymes are biological catalysts • Homogeneous catalysts: in the same phase as the rest

of the reaction system • Heterogeneous catalysts: solid on which reactant

takes place o Often platinum, palladium, or nickel o Adsorption: collection of one substance on the

surface of another o Absorption: penetration of one substance into

another • Many reactions are catalyzed by the presence of an

acid (H+ modifies structure making it more reactive) • Catalysts do NOT make molecules move faster or

“heat up” the reaction. o They allow the slow molecules to react

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