Chemical Kinetics

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Chemical Kinetics Chemistry: The Molecular Science Moore, Stanitski and Jurs “The study of speeds of reactions and the nanoscale pathways or rearrangements by which atoms and molecules are transformed to products”

Chapter 13: Chemical Kinetics: Rates of Reactions Chemical kinetics is also called reaction kinetics or kinetics.

Reaction Rate Reaction Rate

Factors affecting the speed of a reaction: Combustion of Fe(s) powder: • Properties of reactants and products especially their structure and bonding. • Concentrations of reactants (and products). • Temperature • Catalysts and if present, their concentration.

Reactions are either: Homogeneous - reactants & products in one phase. Heterogeneous - species in multiple phases.

Reaction Rate Reaction Rate

+ Change in [reactant] or [product] per unit time. Average rate of the Cv reaction can be calculated: Time, t [Cv+] Average rate + Cresol violet (Cv ; a dye) decomposes in NaOH(aq): -1 -1 (s) (mol / L) (mol L s ) Sample calculation 0.0 5.000 x 10-5 Avg. rate = [Cv+] 13.2 x 10-7 t 10.0 3.680 x 10-5 9.70 x 10-7 = (5.00x10-5 – 3.68x10-5) mol/L 20.0 2.710 x 10-5 (10.0 – 0.0) s 7.20 x 10-7 30.0 1.990 x 10-5 5.30 x 10-7 = -1.32 x 10-6 mol L-1min-1 40.0 1.460 x 10-5 3.82 x 10-7 + - → 50.0 1.078 x 10-5 Table shows positive rates Cv (aq) + OH (aq) CvOH(aq) 2.85 x 10-7 60.0 0.793 x 10-5 – explained soon… 1.82 x 10-7 change in concentration of Cv+ [Cv+ ] 80.0 0.429 x 10-5 rate = = 0.99 x 10-7 elapsed time t 100.0 0.232 x 10-5

1 Reaction Rates and Stoichiometry Reaction Rates and Stoichiometry Cv+(aq) + OH-(aq) → CvOH(aq) For any general reaction: a A + b B c C + d D Stoichiometry: The overall rate of reaction is: Loss of 1 Cv+ Gain of 1 CvOH Rate of Cv+ loss = Rate of CvOH gain 1 [A] 1 [B] 1 [C] 1 [D] Rate = = =+ = + Another example: a t b t c t d t

2 N2O5(g) 4 NO2(g) + O2(g) Loss of 2 N O Gain of 1 O 2 5 2 Reactants decrease with time. Products increase with time. Negative sign. Rate of N2O5 loss = 2 x (rate of O2 gain) Positive sign

Negative rate Positive rate

Rate of loss of N2O5 divided by -2, equals rate of gain of O2

Reaction Rates and Stoichiometry Average Rate and Instantaneous Rate For: Graphical view of Cv+ reaction: 5.0E-5 H2 (g) + I2 (g) 2 HI (g) -1 -1 the rate of loss of I2 is 0.0040 mol L s . What is 4.0E-5 the rate of formation of HI ? 3.0E-5

[H ] [I ] 1 [HI] ] (mol/L) Rate = 2 = 2 = + + 2.0E-5

t t 2 t [Cv 1.0E-5

[H2] 1 [HI] Rate = = (-0.0040) = + 0 t 2 t 0 20 40 60 80 100 t (s) [HI] So = +0.0080 mol L-1 s-1 Average rate (from 0 to 80 s) = slope of the blue triangle t …but the avg. rate depends on interval chosen.

Average Rate and Instantaneous Rate Effect of Concentration on Reaction Rate Rate may change when [reactant] changes. •Cv+ example shows this. •For Cv+ the rate is proportional to concentration.

t [Cv+] Rate of Cv+ Rate/[Cv+] (s) (M) loss (M / s) (s-1)

0 5.00 x 10-5 1.54 x 10-6 0.0308 80 4.29 x 10-6 1.32 x 10-7 0.0308

Instantaneous rate = slope of a line tangent to the curve. rate = k [Cv+] • t = 0 s and t = 80 s have different instantaneous rates

2 Rate Law and Order of Reaction Determining Rate Laws from Initial Rates A general reaction will usually have a rate law: Rate laws must be measured. rate = k [A]m [B]n . . . They cannot be predicted from reaction stoichiometry.

Initial Rate Method where To find the order for a reactant: k rate constant • Run the experiment with known [reactant]0. m, n order for A & B, respectively • Measure the initial rate of reaction (slope at t = 0). m + n + … overall order of the reaction • Change [reactant]0 of 1 reactant; keep all others constant. • Remeasure the initial rate. The orders are usually integers (-2, -1, 0, 1, 2…), • The ratio of the two rates gives the order for the chosen but may also be fractions (, …) reactant.

Determining Rate Laws from Initial Rates Determining Rate Laws from Initial Rates

Initial concentration (M) Data for the reaction of methyl acetate with base: - Expt. [CH3COOCH3][OH] Initial rate (M/s) 1 0.040 0.040 2.2 x 10-4 - - -4 CH3COOCH3 + OH CH3COO + CH3OH 2 0.040 0.080 4.5 x 10 3 0.080 0.080 9.0 x 10-4

Initial concentration (M) Dividing the first two data sets:

- -4 m n Expt. [CH3COOCH3][OH ] Initial rate (M/s) 4.5 x 10 M/s = k (0.040 M) (0.080 M) 1 0.040 0.040 2.2 x 10-4 2.2 x 10-4 M/s = k (0.040 M)m(0.040 M)n 2 0.040 0.080 4.5 x 10-4 m n 3 0.080 0.080 9.0 x 10-4 Thus: 2.05 = (1) (2.00) n 2.05 = (2.00) 1 raised to any Rate law: and n = 1 power = 1 rate = k [CH COOCH ]m [OH-]n 3 3 It is 1st order with respect to OH-.

Determining Rate Laws from Initial Rates Determining Rate Laws from Initial Rates

Use experiments 2 & 3 to find m: The rate law is: - rate = k [CH3COOCH3][OH ] 9.0 x 10-4 M/s = k (0.080 M)m(0.080 M)n 4.5 x 10-4 M/s = k (0.040 M)m(0.080 M)n Overall order for the reaction is: m + n = 1 + 1 = 2 So: 2.00 = (2.00)m (1)n 2.00 = (2.00)m The reaction is: 2nd order overall. and m = 1 1st order in OH- st 1 order in CH3COOCH3 st Also 1 order with respect to CH3COOCH3.

3 Determining Rate Laws from Initial Rates The Integrated Rate Law If a rate law is known, k can be determined: Calculus is used to integrate a rate law. st rate Consider a 1 -order reaction: A products k = [CH COOCH ][OH-] [A] 3 3 rate = – = k [A] t 2.2 x 10-4 M/s Using run 1: k = d [A] (0.040 M)(0.040 M) = – = k [A] (as a differential equation) dt

-1 -1 -1 -1 k = 0.1375 M s = 0.1375 L mol s Integrates to: ln [A]t = k t + ln [A]0 y = m x + b (straight line) Could repeat for each run, take an average… If a reaction is 1st-order, a plot of ln [A] vs. t will be But a graphical method is better. linear.

The Integrated Rate Law Zeroth-order reaction The Integrated Rate Law slope = - Rate data for the decomposition of cyclopentene The reaction: k A products [A] C H (g) → C H (g) + H (g) doesn’t have to be 1st order. 5 8 5 6 2 time t were measured at 850°C. Determine the order of the Some common integrated rate laws: First-order reaction reaction from the following plots of those data: slope = -

Order Rate law Integrated rate law Slope k time t ln[A]

0 rate = k [A]t = -kt + [A]0 -k 1 rate = k[A] ln[A] = -kt + ln[A] -k t 0 Second-order reaction 2 1 1 2 rate = k[A] = kt + +k k [A]t [A]0 slope = 1/[A]

y x time t • The reaction is first order (the only linear plot) • k = -1 x (slope) of this plot. The most accurate k is obtained from the slope of a plot.

Half-Life Half-Life For a 1st-order reaction: t1/2 = Time for [reactant]0 to fall to [reactant]0. ln[A]t = -kt + ln[A]0

st Half-lives are only useful for 1 -order reactions. When t = t1/2 [A]t = [A]0 Why?

Then: ln([A]0) = -kt1/2 + ln[A]0 • t is independent of the starting concentration. 1/2 ln([A] /[A] ) = -kt {note: ln x – ln y = ln(x/y)} only true for 1st order reactions (not 0th, 2nd…) 0 0 1/2 ln() = -ln(2) = -kt1/2 {note: ln(1/y) = –ln y } st • t1/2 is constant for a given 1 -order reaction. t =ln 2 = 0.693 1/2 k k

4 Half Life Calculating [ ] or t from a Rate Law

st t1/2 of a 1 -order reaction can be used to find k. Use an integrated rate equation. For cisplatin (a chemotherapy agent): • the cisplatin lost after 475 min. Example → • (0.0100 M 0.0050 M) In a 1st-order reaction, [reactant] = 0.500 mol/L and 0.010 • [cisplatin] halves every 475 min 0 t1/2 = 400.s. Calculate: 0.008 k =ln 2 = 0.693 a) [reactant],1600.s after initiation. 0.006 t 475 min 1/2 b) t for [reactant] to drop to 1/16th of its initial value. -3 -1 0.004 = 1.46 x 10 min c) t for [reactant] to drop to 0.0500 mol/L.

[cisplatin] (mol/L) 0.002

0 0 400 800 1200 1600 2000 t (min)

Calculating [ ] or t from a Rate Law Calculating [ ] or t from a Rate Law

st st In a 1 -order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s In a 1 -order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s (a) Calculate [reactant] ,1600.s after initiation. (b) Calculate t for [reactant] to drop to 1/16th of its initial value.

st 1 order: [reactant] 1 [reactant] t 0 2 0 1/2 -3 -1 k = ln 2/ t = 0.6931/(400. s) = 1.733x10 s 1 1 [reactant]0 [reactant]0 t1/2 2 4 and ln [A]t = -kt + ln [A]0 1 1 4 [reactant]0 8 [reactant]0 t1/2 so ln[A] = -(0.001733 s-1)(1600 s) +ln(0.500) t 1 [reactant] 1 [reactant] t 8 0 16 0 1/2 ln[A]t = -2.773 + -0.693 = -3.466 4 t1/2 = 4 (400 s) = 1600 s -3.466 [A]t = e = 0.0312 mol/L Note: part (a) could be solved in a similar way. 1600 s = 4 t1/2 so 0.500 0.250 0.125 0.0625 0.0313 M.

Calculating [ ] or t from a Rate Law Nanoscale View: Elementary Reactions

st In a 1 -order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s Individual molecules undergo: (c) Calculate t for [reactant] to drop to 0.0500 mol/L ? • unimolecular reactions – a single particle (atom, ion, molecule) rearranges into 1 or 2 different particles. From part (a): k = 1.733 x 10-3 s-1 • bimolecular reactions – two particles collide and rearrange. ln [A] = -kt + ln [A] t 0 Both are elementary reactions – what actually then ln (0.0500) = -(0.001733 s-1) t + ln(0.500) occurs at the nanoscale. -2.996 = -(0.001733 s-1) t – 0.693 Observed reactions (macroscale) may be: t = -2.303 • elementary – directly occur by one of these two processes, -0.001733 s-1 or • complex – occur as a series of elementary steps. t = 1.33 x 103 s

5 Elementary reactions Unimolecular Reactions Example 2-butene isomerization is unimolecular: Label these elementary reactions as unimolecular or

bimolecular: H3C CH3 H3C H unimolecular C=C (g) C=C (g) H H H CH3 bimolecular cis-2-butene trans-2-butene

unimolecular

Unimolecular Reactions Transition State transition state or activated complex During isomerization, 2-butene passes through a )

J 500 transition state or activated complex. -21 Ea is the activation 400 energy, the minimum E

J • Occurs at the top of the activation barrier

300 -21 to go over the barrier. • Exists for very short time (few fs, 1 fs = 10-15 s). 200 = 435 x 10 100 a • Falls apart to form products or reactants. E

0 -21 • If an exothermic reaction Potential energy (10 Potential E = -7 x 10 J Initial state Final state products are at lower E than the reactants -30° 0 30° 60° 90° 120° 150° 180° 210° Reaction Progress (angle of twist) Exothermic overall the reverse reaction will be slower. cis-trans conversion twists the C=C bond. -19 • This requires a lot of energy (Ea= 4.35x10 J/molecule = 262 kJ/mol) • Even more (4.42x10-19J/molecule) to convert back.

Bimolecular Reactions Bimolecular Reactions e.g. Iodide ions reacting with methyl bromide:

- - I (aq) + CH3Br(aq) ICH3(aq) + Br (aq) I- must collide in the right location to cause the inversion. transition state

I- must collide with enough E and in the right location to cause the inversion. • unless the I- “hits” as shown it cannot drive out the Br- • only a small fraction of the collisions have this orientation. • the fractional factor is called the steric factor

6 Bimolecular Reactions Temperature and Reaction Rate

Increasing T will speed up most reactions. )

J 150 transition state -21 120 Higher T = higher average Ek for the reactants.

90 J -21 = larger fraction of the molecules can

60 Products overcome the activation barrier. (final state)

30 = 126 x 10 a E = 63 x 10-21 J E 25°C 0 Potential energy (10 Potential Reactants (initial state) 75°C Many more molecules have Reaction Progress (changing bond lengths and angles) E enough E to react at 75°C, • Also has an activation barrier (E ). a a so the reaction goes much

• Forward and back Ea are different. of moleculesnumber faster. • Here the forward reaction is endothermic. kinetic energy

Temperature and Reaction Rate Temperature and Reaction Rate

Reaction rates are strongly T-dependent. Data for The Arrhenius equation shows how k varies with T - the I + CH3Br reaction: -E / RT k = A e a T (K) k (L mol-1 K-1)

) -5

-1 273 4.18 x 10

K -4 Quantity Name Interpretation and/or comments

-1 290 2.00 x 10 A Frequency factor How often a collision occurs with -3 310 2.31 x 10 the correct orientation. -2

(L mol (L 330 1.39 x 10

k Ea Activation energy Barrier height. 350 6.80 x 10-2 e-Ea/RT Fraction of the molecules with 370 2.81 x 10-1 enough E to cross the barrier. T Temperature Must be in kelvins. 0.00 0.10 0.20 0.30 0.20 0.10 0.00 250 300 350 400 R Gas law constant 8.314 J K-1 mol-1. T (K)

Determining Activation Energy Determining Activation Energy Take the natural logarithms of both sides: The iodide-methyl bromide reaction data:

-Ea / RT 28 Ea = -(slope) x R ln k = ln A e intercept = 23.85 ln ab = ln a + ln b 8.314 J -Ea / RT 18 = -(-9.29 x103 K) ln k = ln A + ln e slope = -9.29 x 10 K mol ln e = 1 3

k = 77.2 x 10 J/mol Ea 8

ln k = ln A + ln e ln RT 3 = 77.2 kJ/mol K -2

Ea 1 ln k = + ln A A = eintercept = e23.85 R T -12 0 0.001 0.002 0.003 0.004 A = 2.28 x 1010 L mol-1 s-1 1/T (K-1) (A has the same units as k) A plot of ln k vs. 1/T is linear (slope = Ea/R).

7 Rate Laws for Elementary Reactions Reaction Mechanisms A complex reaction example: Elementary reactions Occur as written and their rate laws are predictable. 2 I-(aq) + H O (aq) + 2 H O+(aq) I (aq) + 4 H O(l) • Unimolecular reactions are always 1st -order. 2 2 3 2 2

• Bimolecular reactions are always 2nd-order. + -3 -5 When [H3O ] is between 10 M and 10 M, - rate = k [I ][H2O2]

Complex reactions It must be complex • Do not occur as written. • Exponents in the rate law do not match the stoichiometry. • They are carried out in a series of elementary steps. • Five reactants do not collide, form a transition state, and break into I2 and 4H2O.

Reaction Mechanisms Reaction Mechanisms

A reaction mechanism • lists the series of elementary steps that occur. • shows how reactants change into products.

- + 2 I (aq) + H2O2(aq) + 2 H3O (aq) I2 (aq) + 4 H2O(l)

Mechanism HOOH + I- slow OH- + HOI Shows the bonding in H O 2 2 - fast - HOI + I OH + I2

- + fast 2{ OH + H3O 2 H2O } - + overall 2 I + H2O2 + 2 H3O I2 + 4 H2O

Reaction Mechanisms Reaction Mechanisms Rate-limiting step Step one HOOH + I- OH- + HOI • The slowest step in the sequence Slow. It determines the overall rate. • Overall reaction rate is limited by, and equal to, the rate of the slowest step. - - Steps 2 & 3 HOI + I OH + I2 A good analogy is supermarket shopping: - + 2{ OH + H3O 2 H2O } • You run in for 1 item (~1 min = fast step), but… Fast. Will not affect the rate. • The checkout line is long (~10 min = slow step). • Time spent is dominated by the checkout-line wait. The overall rate is expected to be • In a reaction, a slow step may be thousands or - rate = k [H2O2][ I ] as observed! even millions of times slower than a fast step.

8 Mechanisms with a Fast Initial Step Mechanisms with a Fast Initial Step Consider: Step 2 is rate limiting: rate = k2 [NOBr2] [NO] 2 NO (g) + Br2 (g) 2 NOBr (g)

But NOBr2 is an intermediate. It is difficult (sometimes impossible) to measure its concentration. The generally accepted reaction mechanism is:

NOBr2 can form and fall apart many times (step 1) Step one NO + Br2 NOBr2 fast before it converts to products.

Step two NOBr2 + NO 2 NOBr slow Step1 is reversible (double arrows).

2 NO + Br2 2 NOBr NO + Br2 NOBr2 fast, reversible

Mechanisms with a Fast Initial Step Mechanisms with a Fast Initial Step

Equilibrium is established. The earlier rate law:

k 1 rate = k2 [NOBr2] [NO] NO + Br2 NOBr2 reversible, equilibrium k-1 becomes: k1[NO][Br2] rate = k2 [NO] At equilibrium: k-1 rate forward = rate back k1k2 2 rate = [Br2][NO] k-1 k1[NO][Br2] = k-1[NOBr2] 2 rate = k' [Br2][NO] k [NO][Br ] [NOBr ] = 1 2 Now only contains starting materials - can be 2 k -1 checked against experiment.

Summary Catalysts and Reaction Rate Elementary reactions: the rate law can be written down from the stoichiometry. A Catalyst is a substance that: unimolecular rate = k[A] • increases reaction rate without being consumed bimolecular rate = k[A]2 or rate = k[A][B] (reactants are consumed). • You can’t work backwards: • changes the mechanism for the reaction. Rate law matches stoichiometry: the reaction may be elementary or complex. • provides a lower Ea in the rate limiting step. • The reaction must be complex if: rate law doesn’t match the stoichiometry A catalyst does not change the products or their a species has a non-integer order. relative proportions. the overall order is greater than 2.

9 Catalysts and Reaction Rate Catalysts and Reaction Rate I is not in the overall equation, and is not used up. 2-butene isomerization is catalyzed by a trace of I2. 2 The mechanism changes! H C CH H C H 3 3 3 I2 splits into 2 atoms. C=C (g) C=C (g) Each has an unpaired Mechanism: e-. (shown by the dot) H H H CH 3 step 1: {I 2 I•} No catalyst: 2 rate = k [cis-2-butene] I• attaches and breaks one C-C bond A trace of I (g) speeds up the reaction, and: H C 2 H C CH3 3 CH3 rate = k [ I ] [cis-2-butene] 3 • 2 step 2: I• + C=C I C–C k uncatalyzed k H H H H

Catalysts and Reaction Rate Catalysts and Reaction Rate

Rotation around C-C Transition state for the H C uncatalyzed reaction 3 CH3 H3C H • step 3: I C–C C–C I Rotation around C-C • I• leaves; double bond H CH3 H H reforms I2 dissociates E = 262 kJ/mol H C a 3 H H3C H to I• + I• I• + I• regenerates I step 4: I C–C C=C + I• 2 • Reactants (initial state) Ea = 115 kJ/mol Products CH3 H CH3 H E = -4 kJ/mol (final state) Loss of I• and formation of C=C Reaction Progress I• adds to cis-2-butene, (double single bond) step 5: {2 I• I2} I2 is regenerated

Catalysts and Reaction Rate Enzymes: Biological Catalysts Key points: Enzymes are:

•I2 dissociates and reforms. • Usually very large proteins (often globular proteins) Not consumed, not in the overall reaction equation. • Very efficient catalysts for 1 or more chemical reactions • The activation energy is much lower. • can increase rates by factors of 109 - 1019 115 kJ/mol vs. 262 kJ/mol • can catalyze millions of reactions per minute. The catalyzed reaction is 1015 times faster at 500 K. • Highly specific • 5 step mechanism = 5 humps in energy diagram. • react with 1 or a small number of substrates (the molecule • Catalyst and reactant are both in the same phase. undergoing the reaction). Homogeneous catalysis • May require a cofactor to be present before they work • The initial and final energies are identical • Small organic or inorganic molecule or ion. • Many use NAD+ (nicotinamide adenine dinucleotide ion) H is the same whether catalyzed or not!

10 Enzyme Activity and Specificity Enzyme Activity and Specificity

Enzymes are often large globular proteins, but only a small part (the active site) interacts with the substrate.

The induced fit stretches and bends the substrate (and active site).

Enzyme Activity and Specificity Enzyme kinetics

Enzymes are effective catalysts because they: Transition state for the • Bring and hold substrates together while a reaction occurs. Formation of the uncatalyzed reaction enzyme-substrate • Hold substrates in the shape that is most effective for complex reaction. Transformation of Potential energy, the substrate to • Can donate or accept H+ from the substrate (act as acid or Ea products

base) Activation energy E'a is much smaller than E E' a • Stretch and bend substrate bonds in the induced fit so the a and so the enzyme reaction starts partway up the activation-energy hill. makes the reaction E Reactants E much faster (initial state) Products (final state) Reaction Progress

Enzyme Activity and Specificity Catalysis in Industry Enzyme catalyzed reactions: Catalysts are used extensively in industry. • have unusual T dependence. • Many are heterogeneous catalysts speed up as T increases… usually a solid catalyst with gas or liquid reactants & but will denature (change gross products. shape) and stop working. acetic acid is prepared using solid rhodium(III) iodide:

RhI3 • have a maximum reaction rate. CH3OH(l) + CO(g) CH3COOH(l) [enzyme] limiting. auto exhausts are cleaned by catalytic converters: • can be blocked by an inhibitor 2 CO(g) + O (g)Pt-NiO 2 CO binds, but doesn’t react and 2 2 Pt-NiO release. 2 C8H18(g) + 25 O2 16 CO2(g) + 18 H2O(g) catalyst 2 NO(g) N2(g) + O2(g)

11 Controlling Automobile Emissions Converting Methane to Liquid Fuel Methane is hard to transport. It can be converted …forms a bond with the Pt surface… to methanol:

…dissociates into N and O atoms (each CH4(g) + O2(g) CO(g) + 2 H2(g) bonded to Pt)… NO approaches the Pt surface… …N and O migrate on the surface until they get close to CO(g) + 2 H2(g) CH3OH(l) like atoms… A Pt-coated ceramic catalyst allows the 1st reaction to occur at low T. …they form N2 and O2 and leave the surface.