Chapter 12 Lecture Notes

“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

Reaction Rate Reaction Rate Combustion of Fe(s) powder:

+ 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): (s) (mol / L) (mol L-1 s-1) 0.0 5.000 x 10-5 13.2 x 10-7 10.0 3.680 x 10-5 9.70 x 10-7 20.0 2.710 x 10-5 7.20 x 10-7 30.0 1.990 x 10-5 5.30 x 10-7 40.0 1.460 x 10-5 3.82 x 10-7 + - 50.0 1.078 x 10-5 Cv (aq) + OH (aq) → CvOH(aq) 2.85 x 10-7 60.0 0.793 x 10-5 -7 + + -5 1.82 x 10 change in concentration of Cv Δ [Cv ] 80.0 0.429 x 10 rate = = 0.99 x 10-7 elapsed time Δt 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 Δ[A] 1 Δ[B] 1 Δ[C] 1 Δ[D] Rate = − = − = + = + Another example: a Δt b Δt c Δt d Δt

2 N2O5(g) 4 NO2(g) + O2(g)

Reactants decrease with time. Products increase with time. Negative sign. 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

Δ[H ] Δ[I ] Δ[HI] (mol/L) ]

1 + Rate = − 2 = − 2 = + 2.0E-5

Δt Δt 2 Δt [Cv 1.0E-5

Δ[H2] 1 Δ[HI] Rate = − = − (-0.0040) = + 0 Δt 2 Δt 0 20 40 60 80 100 t (s) Δ[HI] So = +0.0080 mol L-1 s-1 Δt

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

2 Rate Law and Order of Reaction Determining Rate Laws from Initial Rates A general reaction will usually have a rate law: rate = k [A]m [B]n . . .

The orders are usually integers (-2, -1, 0, 1, 2…), but may also be fractions (½, ⅓…)

Determining Rate Laws from Initial Rates Determining Rate Laws from Initial Rates Data for the reaction of methyl acetate with base:

- - CH3COOCH3 + OH CH3COO + CH3OH

Dividing the first two data sets: 4.5 x 10-4 M/s = k (0.040 M)m(0.080 M)n 2.2 x 10-4 M/s = k (0.040 M)m(0.040 M)n

1 raised to any Rate law: power = 1 m - n rate = k [CH3COOCH3] [OH ]

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

Use experiments 2 & 3 to find m:

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

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-] Δ[A] 3 3 rate = – = k [A] Δt 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… But a graphical method is better.

The Integrated Rate Law Zeroth-order reaction The Integrated Rate Law slope = - Rate data for the decomposition of cyclopentene k The reaction: 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] 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:

ln[A]t = -kt + ln[A]0

st Half-lives are only useful for 1 -order reactions. When t = t1/2 [A]t = ½[A]0 Why?

Then: ln(½[A]0) = -kt1/2 + ln[A]0

ln(½[A]0/[A]0) = -kt1/2 {note: ln x – ln y = ln(x/y)}

ln(½) = -ln(2) = -kt1/2 {note: ln(1/y) = –ln y }

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): • ½ the cisplatin lost after 475 min. • (0.0100 M → 0.0050 M) 0.010 • [cisplatin] halves every 475 min 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/ t½ = 0.6931/(400. s) = 1.733x10 s 1 [reactant] 1 [reactant] t 2 0 4 0 1/2 and ln [A] = -kt + ln [A] 1 t 0 [reactant] 1 [reactant] t 4 0 8 0 1/2 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 (c) Calculate t for [reactant] to drop to 0.0500 mol/L ?

From part (a): k = 1.733 x 10-3 s-1

ln [A]t = -kt + ln [A]0 then ln (0.0500) = -(0.001733 s-1) t + ln(0.500) -2.996 = -(0.001733 s-1) t – 0.693

t = -2.303 -0.001733 s-1 t = 1.33 x 103 s

5 Elementary reactions Unimolecular Reactions 2-butene isomerization is unimolecular:

H3C CH3 H3C H unimolecular C=C (g) C=C (g) H H H CH3 bimolecular

unimolecular

Unimolecular Reactions Transition State transition state or activated complex

)

J 500 -21 Ea is the activation 400

energy, the minimum E to J

300 -21 go over the barrier.

200 = 435= x10 100 a E

0 -21 Potentialenergy (10 ΔE = -7 x 10 J Initial state Final state -30° 0 30° 60° 90° 120° 150° 180° 210° Reaction Progress (angle of twist) Exothermic overall 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.

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= x10 a -21 E ΔE = 63 x 10 J 25°C 0 Potentialenergy (10 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. numbermoleculesof 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 I + CH3Br reaction:

-Ea / RT k = A e T (K) k (L mol-1 s-1)

) 273 4.18 x 10-5 -1

s -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 0.00 0.10 0.20 0.30 250 300 350 400 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 3 8.314 J -Ea / RT 18 = -(-9.29 x10 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 + − RT ln e ln 3 K = 77.2 kJ/mol -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

+ -3 -5 When [H3O ] is between 10 M and 10 M, - rate = k [I ][H2O2]

Reaction Mechanisms Reaction Mechanisms

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

slow

Shows the bonding in H2O2 fast

fast

- + overall 2 I + H2O2 + 2 H3O I2 + 4 H2O

Reaction Mechanisms Reaction Mechanisms

A good analogy is supermarket shopping: • You run in for 1 item (~1 min = fast step), but… • 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:

2 NO (g) + Br2 (g) 2 NOBr (g)

The generally accepted reaction mechanism is:

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

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

The earlier rate law:

k 1 rate = k2 [NOBr2] [NO] NO + Br2 NOBr2 reversible, equilibrium k-1 becomes: k1[NO][Br2] rate = k [NO] 2 k rate forward = rate back -1 k1k2 2 rate = [Br2][NO] k-1 k1[NO][Br2] = k-1[NOBr2]

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. unimolecular rate = k[A] bimolecular rate = k[A]2 or rate = k[A][B]

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 e-. (shown by the dot) H H H CH3 No catalyst: 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 ]½ [cis-2-butene] 3 • 2 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 • I C–C C–C I Rotation around C-C • I• leaves; double bond H CH H H 3 reforms I dissociates to 2 Ea = 262 kJ/mol H3C H H3C H I• + I• I• + I• regenerates I I C–C C=C + I• 2 • Reactants (initial state) Ea = 115 kJ/mol Products CH3 H CH3 H ΔE = -4 kJ/mol (final state) Loss of I• and formation of C=C Reaction Progress I• adds to cis-2-butene,

I2 is regenerated (double → single bond)

Catalysts and Reaction Rate Enzymes: Biological Catalysts

10 Enzyme Activity and Specificity Enzyme Activity and Specificity

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

Enzyme Activity and Specificity Catalysis in Industry Enzyme catalyzed reactions:

RhI3 CH3OH(l) + CO(g) CH3COOH(l)

. auto exhausts are cleaned by catalytic converters: Pt-NiO 2 CO(g) + O2(g) 2 CO2 Pt-NiO 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) + ½ 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 like CO(g) + 2 H2(g) → CH3OH(l) atoms… A Pt-coated ceramic catalyst allows the 1st reaction to occur at low T. …they form N2 and O2 and leave the surface.