Brdy 6Ed Ch14 ChemicalKinetics

Chapter 14: Chemical Kinetics

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Speeds at Which Reactions Occur Kinetics: Study of factors that govern How rapidly reactions occur and How reactants change into products

Rate of Reaction: Speed with which reaction occurs How quickly reactants disappear and products

form

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Important Questions in Kinetics

Practical value!! Chemical /pharmaceutical manufacturers Is it practical to make drug? Made on manageable time scale? Can we adjust conditions to improve rate and

yield? Mechanism of Reaction Series of individual steps leading to overall

observed reaction How do reactants change into products? Detailed sequence of events

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Factors that Affect Reaction Rates

1. Chemical nature of reactants What elements, compounds, salts are

involved? What bonds must be formed, broken? What are fundamental differences in

chemical reactivity?

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Factors that Affect Reaction Rates 2. Ability of reactants to come in contact If two or more reactants must meet in order to react Gas or solution phase facilitates this

Reactants mix and collide with each other easily Homogeneous reaction

All reactants in same phase Occurs rapidly

Heterogeneous reaction Reactants in different phases Reactants meet only at interface between phases Surface area determines reaction rate ↑ area, ↑ rate ↓ area, ↓ rate

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3. Concentrations of reactants Rates of both homogeneous and

heterogeneous reactions affected by [X] Collision rate between A and B ↑ if we ↑ [A] or

↑ [B]. ∴ Often (but not always) Reaction rate ↑ as [X]↑

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4. Temperature Rates are often very sensitive to T Cooking sugar

Raising T usually makes reaction faster for two reasons:a. Faster molecules collide more often and

collisions have more Eb. Most reactions, even exothermic reactions,

require E to get going Crude Rule of Thumb: Rate doubles if ↑ T by 10 °C (10 K)

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5. Presence of Catalysts Catalysts Substances that ↑ rates of chemical and

biochemical reactions without being used up Rate-accelerating agents Speed up rate dramatically

Rate enhancements of 106 not uncommon

Chemicals that participate in mechanism but are regenerated at the end

Ex. Enzymes

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Measuring Rate of Reaction Rate = ratio with time unit in denominator

Ex. Rate of pay =

Rate of Chemical Reaction ↓ in [X] of particular species per unit time.

Always with respect to (WRT) given reactant or product

[reactants] ↓ w/ time [products] ↑ w/ time

reaction rate=Δ [ reactant ]

Δ time

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10 dollarshr

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Rate of Reaction with Respect to Given Species X

Concentration in M Time in s Units on rate: Ex. [product] ↑ by 0.50 mol/L per second ⇒ rate =

0.50 M/s [reactant] ↓ by 0.20 mol/L per second ⇒ rate =

0.20 M/s10

Rate WRT X =[ X ]t2

−[ X ]t1

t2−t1

¿Δ[ X ]

Δt

mol/Ls

=molL⋅s

=Ms

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Rate of Reaction Always + Whether something is ↑ or ↓ in [X].

Reactants Need – sign to make rate + Reactant consumed So ∆[X] = –

Products Produced as reaction goes along So ∆[X] = + Thus Rate = +

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Rate=−Δ [ reactant ]

Δt

Rate=Δ [ product ]

Δt

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Rates and Coefficients Relative rates at which reactants are

consumed and products are formed Related by coefficients in balanced chemical

equation. Know rate with respect to one product or reactant Can use equation to determine rates WRT all other

products and reactants. C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)

Rate of Reaction

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=−Δ [C3 H 8 ]

Δt=−

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Δ [O2 ]

Δt=

13

Δ[ CO2 ]

Δt=

14

Δ[ H 2O ]

Δt

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Rates and Coefficients

O2 reacts 5 times as fast as C3H8

CO2 forms 3 times faster than C3H8 consumed

H2O forms 4/5 as fast as O2 consumed

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Rate=−Δ [O2 ]

Δt=−5

Δ [C 3 H8 ]

Δt

Rate=Δ [ CO2 ]

Δt=−3

Δ [C3 H8 ]

Δt

Δ[ H 2O ]

Δt=−

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Δ[O2 ]

Δt

C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)

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Rates and CoefficientsIn general

αA + βB → γC + δD

Which one do I measure? Doesn't matter. All interrelated. Often determined by which one is easily

measured

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Rate=−1α

ΔAΔt

=−1β

ΔBΔt

=1γΔCΔt

=1δ

ΔDΔt

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Your Turn!In the reaction 2CO(g) + O2(g) 2CO→ 2(g), the rate of the reaction of CO is measured to be 2.0 M/s. What would be the rate of the reaction of O2?

A.the sameB.twice as greatC.half as largeD.you cannot tell from the given information

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Change of Reaction Rate with Time Generally reaction rate changes during

reaction i.e. Not constant

Often initially fast when lots of reactant present Slower at end when reactant depleted

Why? Rate depends on [reactants] Reactants being used up, so [reactant] is ↓ [A] vs. time is curve A is reactant ∴[A] is ↓ w/ time

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Measuring Rates Can measure rate using change in

concentration of any substance in reactionEx. 2A 3B→

Rates based on each substance are related to one another by the stoichiometric coefficients of reaction

Measured in three ways: Instantaneous rate Average rate Initial rate

Δ [B ]Δtime

=−32

Δ [A ]Δtime

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Instantaneous Reaction Rates Instantaneous rate

Slope of tangent to curve at any specific time Initial rate

Determined at initial time

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Average Rate of Reaction

Slope of line connecting starting and ending coordinates for specified time frame

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Δ[ Product ]Δtime

=rate

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Concentration vs. Time Curve for HI Decomposition at 508°C

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Rate=−slope=ΔyΔx

=riserun

Rate at any time t = negative slope (or tangent line) of curve at that point

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Table 14.1 Data at 508 °C

[HI] (mol/L) Time (s)

0.100 00.0716 500.0558 1000.0457 1500.0387 2000.0336 2500.0296 3000.0265 350

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2 HI(g) → H2(g) + I2(g)

rate=−(0 . 0716−0 .100 )M(50−0 )s

=−(−0 . 0284 M )

50 s =5 .68×10−4 M / s

Initial rate rate between first two data

points

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 22

Rate=−slope=ΔyΔx

=riserun

=−(0 . 044−0 . 068)M(150−50) s

=−−0 .024 M100 s

=2 . 4×10−4 M / s

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Rate at 300 s 2 HI(g) → H2(g) + I2(g)

Rate = tangent of curve at 300 s

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Rate=−( 0. 0265−0 .0296)M(350−300 )s

=0 .0031 M50 s

=6 .20×10−5 M /s

[HI] (mol/L) Time (s)

0.100 00.0716 500.0558 1000.0457 1500.0387 2000.0336 2500.0296 3000.0265 350

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Your Turn! A reaction was of NO2 decomposition was studied.

The concentration of NO2 was found to be 0.0258M at 5 minutes and at 10 minutes the concentration was 0.0097M. What is the average rate of the reaction between 5 min and 10 min?

A. 310 M/minB. 3.2 x 10-3 M/minC. 2.7 x 10-3 M/minD. 7.1 x 10-3 M/min

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( ) 30.0258 0.00973.2 x 10 / min

10min 5min

M MM−−

=−

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Concentration and Rate Rate Laws αA + βB → γC + δD Homogeneous reaction

Rate = k[A]m[B]n

Rate Law or Rate expression m and n = exponents found experimentally No necessary connection between stoichiometric

coefficients (α, β) and rate exponents (m, n) Usually small integers Sometimes simple fractions (½, ¾) or zero

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Rate Laws

Rate = k[A]m[B]n

k = Rate Constant Dependence of rate on concentration goes as

some power (m) of concentration [A] All other factors (T, solvent) are included in k

Specific rate constant

k depends on T Must specify T at which you obtained k.

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Rate=−Δ [ A]

Δt= k[A]m[B]n

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Learning Check

The rate law for the reaction 2A +B 3C is →rate= 0.045M-1s-1 [A][B]

If the concentration of A is 0.2M and that of B is 0.3M, what will be the reaction rate?

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rate=0.0027 M/s ⇒ 0.003 M/s

rate=0.045 M-1 s-1 [0.2][0.3]

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Rate Laws

Rate = k[A]m[B]n

Exponents tell Order of Reaction with respect to (WRT) each reactant

Order of Reaction m = 1 [A]1 1st order WRT reactant m = 2 [A]2 2nd order WRT reactant m = 3 [A]3 3rd order WRT reactant m = 0 [A]0 0th order WRT reactant [A]0 = 1 ⇒ means A doesn't affect rate

Overall order of reaction = sum of orders (m and n) of each reactant in rate law

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Example 1

5 Br− + BrO3− + 6H+ → 3Br2 + 3H2O

x = 1 y = 1 z = 2 1st order WRT BrO3

−

1st order WRT Br− 2nd order WRT H+

Overall order = 1 + 1 + 2 = 4

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−Δ [ BrO3

−]

Δt=k [BrO 3

−]x[Br−

]y[H +

]z

rate=k [ BrO3−]1[ Br− ]

1[ H+

]2

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Example 2 Sometimes n and m are coincidentally the

same as stoichiometric coefficients2 HI (g) → H2 (g) + I2 (g)

2nd order WRT HI 2nd order overall

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rate=−Δ[ HI ]

Δt=k [ HI ]2

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Your Turn!The following rate law has been observed:Rate = k[H2SeO][I-]3[H+]2. The rate with

respect to I- and the overall reaction rate is:

A. 6, 2B. 2, 3C. 1, 6D. 3, 6

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Calculating k from Rate Law

If we know rate and concentrations, can use rate law to calculate k

Ex. 2 at 508 °C Rate= 2.5 x 10−4 M/s [HI] = 0.0558 M

rate=−Δ[ HI ]

Δt=k [ HI ]2

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k=rate

[HI ]2=

2. 5×10−4 M /s

(0 .0558 M )2=0 . 08029 M−1 s−1

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How To Determine Exponents in Rate Law

Experiments Method of Initial Rates If reaction is sufficiently slow or have very fast technique

Can measure [A] vs. time at very beginning of reaction before it curves up very much, then

Set up series of experiments, where initial concentrations vary

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initial rate=−([ A]1−[ A ]0

t1−t0 )Person

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Ex 1. Method of Initial Rates3A + 2 B → productsRate = k[A]m[B]n Expt. # [A]0, M [B]0, M Initial Rate, M/s

1 0.10 0.10 1.2 × 10−4 2 0.20 0.10 4.8 × 10−4 3 0.20 0.20 4.8 × 10−4

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Convenient to set up experiments so [X] of one species is doubled or tripled while [X] of all other species are held constant

Tells us effect of [varied species] on initial rate

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Reaction Order and Rate If reaction is 1st order WRT given species X, Doubling [X]1 → 21

Rate doubles If reaction is 2nd order WRT X, Doubling [X]2 → 22

Rate quadruples If reaction is 0th order WRT X, Doubling [X]0 → 20

Rate doesn't change If reaction is nth order WRT X Doubling [X]n → 2n

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Back to our Example Expt. # [A]0, M [B]0, M Initial Rate, M/s

1 0.10 0.10 1.2 × 10−4 2 0.20 0.10 4.8 × 10−4 3 0.20 0.20 4.8 × 10−4

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Comparing 1 and 2 Doubling [A] Quadruples rate Reaction 2nd order in A [A]2

Rate 2Rate 1

=4. 8×10−4

1 . 2×10−4=4

4=Rate 2Rate 1

=k [A2 ]

m[B2]

n

k [ A1]m[B1 ]

n=

k [0 . 20 ]m [0 . 10 ]n

k [0. 10 ]m [0 .10 ]n=

[0 . 20 ]m

[0 . 10 ]m=2m

2m = 4 or m = 2

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Back to our Example Expt. # [A]0, M [B]0, M Initial Rate, M/s

1 0.10 0.10 1.2 × 10−4 2 0.20 0.10 4.8 × 10−4 3 0.20 0.20 4.8 × 10−4

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Comparing 2 and 3 Doubling [B] Rate does not change Reaction 0th order in B [B]0

Rate 3Rate 2

=4. 8×10−4

4. 8×10−4=1

1=Rate 3Rate 2

=k [ A3]

m[B3]

n

k [ A2]m[B2]

n=

k [0. 20 ]m [0 . 20 ]n

k [0 . 20]m [0. 10]n=

[0 . 20 ]n

[0 .10 ]n=2n

2n = 1 or n = 0

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Ex. 1 Method of Initial Rates

Conclusion: rate = k[A]2

Can use data from any experiment to determine k Let’s choose experiment 1

Expt. # [A]0, M [B]0, M Initial Rate, M/s

1 0.10 0.10 1.2 × 10−4 2 0.20 0.10 4.8 × 10−4

3 0.20 0.20 4.8 × 10−4

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k=rate

[ A ]2=

1 . 2×10−4 M/s

(0. 10 M )2=1 . 2×10−2 M−1 s−1

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Ex. 2. Method of Initial Rates

2 SO2 + O2 → 2 SO3

Rate =

Expt #

[SO2], M

[O2], M

Initial Rate of SO3 formation, M·s−1

1 0.25 0.30 2.5 × 10−3

2 0.50 0.30 1.0 × 10−2

3 0.75 0.60 4.5 × 10−2

4 0.50 0.90 3.0 × 10−2

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2 SO2 + O2 → 2 SO3

Rate = k[SO2]m[O2]n

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Ex. 2 Compare 1 and 2

[SO2] doubles, [O2] constant, Rate quadruples, 22

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Rate 2Rate 1

=1 .0×10−2

2 . 5×10−3=4

4=Rate 2Rate 1

=k [SO2 ]2

m[O2 ]2

n

k [SO2 ]1m[O2 ]1

n=k [0 . 50 ]m [0. 30 ]n

k [0 . 25 ]m

[0 .30 ]n

=[0. 50 ]

m

[0. 25 ]m=2m

2m = 4 or m = 2

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[O2] triples, [SO2] constant Rate triples, 31

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Rate 4Rate 2

=3 . 0×10−2

1 . 0×10−2=3

3=Rate 4Rate 2

=k [SO2]4

m[O2]4

n

k [SO2]2m[O2]2

n=k [0.50 ]m [0 . 90 ]n

k [0.50 ]m

[0 . 30 ]n

=[0. 90 ]

n

[0. 30 ]n=3n

3n = 3 or n = 1

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Ex. 2

Rate = k[SO2]2[O2]1

1st order WRT O2

2nd order WRT SO2

3rd order overall Can use any experiment to find k

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k=rate

[ SO2 ]2 [O2 ]1

=3. 0×10−2 M / s

(0 . 50 M )2( 0. 90 M )=0 . 13 M−2 s−1

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Ex. 3. Method of Initial Rates

BrO3− + 5 Br− + 6H+ → 3Br2 + 3H2O

Rate=−Δ [BrO3

−]

Δt=k [BrO3

−]m[Br−

]n[H +

]p

Expt #

[BrO3−],

mol/L[Br−], mol/L

[H+], mol/L

Initial Rate, mol/(L·s)

1 0.10 0.10 0.10 8.0 × 10−4

2 0.20 0.10 0.10 1.6 × 10−3

3 0.20 0.20 0.10 3.2 × 10−3

4 0.10 0.10 0.20 3.2 × 10−3

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Rate 2Rate 1

=1 .6×10−3 M / s

8 .0×10−4 M / s=

k (0 . 20 M )m(0. 10 M )

n(0 .10 M )

p

k (0. 10 M )m(0 . 10 M )n (0. 10 M )p

2 . 0=(0 .20 M0 .10 M )

m

=(2 . 0 )m ∴m=1

Rate 3Rate 2

=3 . 2×10−3 M / s

1 .6×10−3 M / s=

k (0 . 20 M )m( 0.20 M )

n(0 .10 M )

p

k (0. 20 M )m(0 .10 M )n(0 .10 M )p

2 . 0=(0 .20 M0 .10 M )

n

=(2 .0)n ∴n=1

Compare 2 and 3

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First order in [BrO3−] and [Br−]

Second order in [H+] Overall order = m + n + p = 1 + 1 + 2 = 4 Rate Law is: Rate = k[BrO3

−][Br−][H+]2

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Rate 4Rate 1

=3 .2×10−3 M /s

8 .0×10−4 M /s=

k ( 0. 10 M )m(0 . 10 M )

n(0 .20 M )

p

k (0 .10 M )m(0 . 10 M )n (0.10 M )p

4 . 0=(0 .20 M0.10 M )

p

=(2 .0 ) p ∴ p=2

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Your Turn!Using the following experimental data, determine the

order with respect to NO and O2 .

A. 2, 0B. 3,1C. 2, 1D. 1, 1

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Expt #

[NO], M

[O2], M

Initial Rate of NO2 formation, M·s−1

1 0.12 0.25 1.5 × 10−3

2 0.24 0.25 6.0 × 10−3

3 0.50 0.50 5.2 × 10−2

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Your Turn! - Solution

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3 12

3 11

22 13

3 1 21

0.24 0.256.0 x 10 1.5 x 10 0.12 0.25

2

0.50 0.505.2 x 10 1.5 x 10 0.12 0.25

1

x y

x y

y

y

M MR M sR M s M M

x

M MR M sR M s M M

y

− −

− −

− −

− −

= =

=

= =

=Pers

onal

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Concentration and Time Rate law tells us how speed of reaction varies

with [X]'s. Sometimes want to know [reactants] and [products] at given time during

reaction How long for [reactants] to drop below some

minimum optimal value Need dependence of Rate on Time

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Concentration vs. Time for 1st Order Reactions

Corresponding to reactions A → products

Integrating we get

Rearranging gives

Equation of line y = mx + b 49

Rate=−Δ [ A ]

Δt=k [ A ]

ln[ A ]0[ A ]t

=kt

ln [ A ]t=−kt+ ln [ A ]0Person

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Plot ln[A]t (y axis) vs. t (x axis)

Yields straight line Indicative of 1st

order kinetics slope = − k intercept = ln[A]0

If we don't know already

ln [ A ]t=−kt+ ln [ A ]0

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Slope = − k

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First Order Kinetics Graph Plot of [A] vs. time gives an exponential decay

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[ A]t=[ A ]o e−kt

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Concentration vs. Time for 2nd Order Reactions

Corresponding to reactions 2B → products

Integrating we get

Rearranging gives

Equation of line y = mx + b52

1[ B ]t

−1

[B ]0=kt

1[ B ]t

=kt+1

[ B ]0

Rate=k [B ]2=−Δ [ B ]

Δt

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Plot 1/[B]t (y axis) vs. t (x axis)

Yields straight line indicative of 2nd

order kinetics slope = + k intercept = 1/[B]0

1[ B ]t

=kt+1

[ B ]0

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Graph of [B] vs. time is still ↓ function, which curves up

Not an exponential

Slope= +k

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How to Determine Reaction Order Using Graphs

Make 2 plots1.ln [A] vs. time2.1/[A] vs. time If ln [A] is linear and 1/[A] is curved, then

reaction is 1st order WRT [A] If 1/[A] plot is linear and ln [A] is curved,

then reaction is 2nd order WRT [A] If both horizontal lines, then 0th order WRT

[A]

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Ex 4. SO2Cl2 → SO2 + Cl2

Time, min [SO2Cl2], M ln[SO2Cl2] 1/[SO2Cl2] (L/mol)

0 0.1000 -2.3026 10.000

100 0.0876 -2.4350 11.416

200 0.0768 -2.5666 13.021

300 0.0673 -2.6986 14.859

400 0.0590 -2.8302 16.949

500 0.0517 -2.9623 19.342

600 0.0453 -3.0944 22.075

700 0.0397 -3.2264 25.189

800 0.0348 -3.3581 28.736

900 0.0305 -3.4900 32.787

1000 0.0267 -3.6231 37.453

1100 0.0234 -3.7550 42.73555

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Ex 4. SO2Cl2 → SO2 + Cl2

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Clearly reaction is 1st order in SO2Cl2.

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Ex. 2 HI (g) → H2 (g) + I2 (g)

Time (s)

[HI] (mol/L)

ln[HI] 1/[HI] (L/mol)

0 0.1000 -2.3026 10.000

50 0.0716 -2.6367 13.9665

100 0.0558 -2.8860 17.9211

150 0.0457 -3.0857 21.8818

200 0.0387 -3.2519 25.840

250 0.0336 -3.3932 29.7619

300 0.0296 -3.5200 33.7838

350 0.0265 -3.6306 37.735857

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Ex. 5 HI (g) → H2 (g) + I2 (g)

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Clearly reaction is 2nd order in HI.

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Your Turn! A plot for a zeroth order

reaction is shown. What is the proper label for the y-axis in the plot ?

A. ConcentrationB. ln of ConcentrationC. 1/ConcentrationD. 1/ ln Concentration

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0 200 400 600 800 1000 1200

Zeroth Order Plot

time (min)

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Half-lives for 1st Order Reactions Half-life = t½ First Order Reactions Set

Substituting into

Gives

Canceling gives ln 2 = kt½

Rearranging gives

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[ A]t=12[ A]0

ln[ A ]0[ A ]t

=kt

ln[ A ]0

12 [ A]0

=kt 12

t 12

=ln 2k 1

=0 . 693k 1

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Half-life for 1st Order ReactionsObserve: 1. t½ is independent of [A]o

For given reaction (and T) Takes same time for concentration to fall from

2 M to 1 M as from 5.0 × 10-3 M to 2.5 × 10-3 M

1. k1 has units (time)-1, so t½ has units (time) t½ called half-life

Time for ½ of sample to decay

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Half-life for 1st Order ReactionsDoes this mean that all of sample is gone

in two half-lives (2 x t½)? No! In 1st t½, it goes to ½[A]o

In 2nd t½, it goes to ½(½[A]o) = ¼[A]o

In 3rd t½, it goes to ½(¼[A]o) = ⅛[A]o

In nth t½, it goes to [A]o/2n

Existence of [X] independent half-life is property of exponential function Property of 1st order kinetics

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Half-Life Graph

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Ex. Using Half-Life 131I is used as a metabolic tracer in hospitals.

It has a half-life, t½ = 8.07 days. How long before the activity falls to 1% of the initial value?

64

N=N oe−kt

lnNN o

=−kt=−t ln 2

τ 12

t=−

τ 12

lnNN o

ln 2=−

(8. 07days ) ln( 1100 )

ln 2=53. 6 days

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Learning Check

The radioactive decay of a new atom occurs so that after 21 days, the original amount is reduced to 33%. What is the rate constant for the reaction in s-1?

ln(10033

)=k ( 21da )

65

ln( A0

A )=kt

k = 6.11×10-7 s-1

k = 0.0528 da-1

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Learning Check

66

The half-life of I-132 is 2.295h. What percentage remains after 24 hours?

ln( 2)

k=t 1

20.302 h–1 = k

A = .0711 % Ao

k=ln 22 . 295h

ln( Ao

A )=kt

ln( Ao

A )=0 . 302h−1×24h=7 . 248

A=A o e−kt

=Aoe−7. 248

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Your Turn!Which order has a half-life that is independent of the original amount?

A. ZeroB. FirstC. SecondD. None depend on the original quantity

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Half-lives of Second Order Reactions

How long before [A] = ½[A]o?

t½, depends on ½[A]o

t½, not useful quantity for 2nd order reaction

68

t 12

=1

k∗[ A ]0

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Learning Check

The rate constant for the second order reaction 2A B is 5.3×10→ -5 M-1s-1. What is the original amount present if, after 2 hours, there is 0.35M available?

1[ 0.35 ]

−1

[ A 0 ]=

5 .3×10−5

M s×7200 s

69

1[ A ]

−1

[ A 0 ]=kt

A0=0.40 M

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Your Turn! Plutomium-239 has a

half life of 24,100 yrs. How many years will it

take for 1.0 grams of Pu-239 to decay to 0.025 g ?

A. 9.6 x 105 yrsB. 2.4 x 105 yrsC. 1.3 x 105 yrsD. 4.8 x 105 yrs

70

5

The number of half-lifes if given by:1.00 g

0.025 g = 2 402

log 2 = log 40 = 5.32

5.32 x 24,100 yrs = 1.3 x 10 yrs

nn

n n

=

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Theories about Reaction Rates Reaction rate depends on [reactants] and T Collision Theory Based on Kinetic Molecular Theory Accounts for both effects on molecular level Central Idea Molecules must collide to react Greater number of collision/sec = greater reaction

rate

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Theories about Reaction Rates

Collision Theory As [reactants] ↑ number of Collisions ↑ Reaction rate ↑

As T ↑ Molecular speed ↑ Molecules collide with more force (energy) Reaction rate ↑

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Collision Theory Rate of reaction proportional to number of effective

collisions/sec among reactant molecules Effective collision

1 that gives rise to product

Ex. At Room Temperature and Pressure H2 and I2 molecules undergoing 1010 collisions/sec

Yet reaction takes a long time Not all collisions lead to reaction

Only very small % of all collisions lead to net change

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1. Molecular Orientation

Molecules must be oriented in a certain way during collisions for reaction to occur

Ex. NO2Cl + Cl → NO2 + Cl2 Cl must come in pointing directly at another Cl atom

for Cl2 to form

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1. Molecular Orientation

75

Wrong Orientation

Correct Orientation

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2. Temperature Greatly Affects Rates

Over moderate T range, Ea unchanged

As ↑ T, More molecules have Ea

So more molecules undergo reaction

Reaction rate ↑ as T↑ 76

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3. Activation Energy, Ea Molecules must possess certain amount of kinetic

energy (KE) in order to reactActivation Energy, Ea

Minimum KE needed for reaction to occur Get energy from collision with other molecules Upon collision, KE converted to potential energy (PE),

used to stretch, bend, and break bonds leading to chemical reaction

If molecules move too slowly, too little KE, they just bounce off each other

Without this minimum amount, reaction will not occur even when correctly oriented

Major reason all collisions do not lead to reaction77

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Transition State Theory Used to explain details of reactions What happens when reactant molecules collidePotential Energy Diagram To visualize what actually happens during

successful collision Relationship between Ea and developing Total

PE

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Potential Energy Diagram

79

Reaction Coordinate (progress of reaction)

Po

ten

tia

l E

ner

gy

Activation energy (Ea) = hill or barrier

between reactants and products

heat of reaction (∆H) = difference in PE between

products and reactants

∆Hreaction = Hproducts – Hreactants ProductsPers

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Potential Energy Diagram (Exothermic)

80

Reaction Coordinate (progress of reaction)

Po

ten

tia

l E

ner

gy Exothermic reaction

• Products lower PE than reactants

Exothermic Reaction

∆H = −

Products

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Exothermic Reaction

∆Hreaction < 0 (–) ↓ in PE of system Appears as ↑ in KE So T system ↑

Reaction gives off heat Can’t say anything about Ea from size of ∆H Ea could be high and reaction slow even if ∆Hrxn

large and negative Ea could be low and reaction rapid

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Potential Energy Diagram (Endothermic)

82

Endothermic Reaction

∆H = +

∆Hreaction = Hproducts – Hreactants

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Endothermic Reaction

∆Hreaction > 0 (+) ↑ in PE Appears as ↓ in KE So T system ↓

Have to add E to get reaction to go Ea ≥ ∆Hrxn as Ea includes ∆Hrxn If ∆Hrxn large and + Ea must be high Reaction very slow

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Transition State or Activated Complex

Arrangement of atoms at top of activation barrier

Brief moment during successful collision when bond to be broken is partially broken and bond to be formed is partially formed

Ex.

84

N CH3C C NH3CH3CN

C

Transition State (TS)

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Ex. NO2Cl + Cl → NO2 + Cl2 As NO2Cl and Cl come

together Start to form Cl····Cl bond Start to break N····Cl bond

Requires E, as must bring 2 things together

In TS N····Cl bond ½ broken Cl····Cl bond ½ formed

After TS Cl—Cl bond forms N····Cl breaks

Releases E as products more stable

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Your Turn!Examine the potential energy diagram. Which is the Slowest (Rate Determining) Step?A. Step 1B. Step 2C. Can’t tell from the given information

Reaction ProgressPo

ten

tial

En

erg

y

Has greatest Ea

1 2

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Measuring Activation Energy

Generally we find k ↑ as T ↑ Usually magnitude of effect ↓ as T ↑ Arrhenius Equation Equation expressing T dependence of k

A = Frequency factor—has same units as k R = gas constant in energy units = 8.314 J·mol−1·K−1

Ea = Activation Energy—has units of J/mol T = Temperature in K

87

k=Ae−E

a/RT

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How To Calculate Activation Energy

Method 1. Graphically Take natural logarithm of both sides

Rearranging

Equation for a line y = b + mx

Arrhenius Plot Plot ln k (y axis) vs. 1/T (x axis)

88

ln k= ln A−( Ea

R )∗( 1T )

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Ex. 2N2O5(g) → 4NO2(g) + O2(g)

T (°C) T (K) 1/T (K−1) k (s−1) ln k

20 293 3.41 × 10−3 2.00 × 10−5 −10.82

30 303 3.30 × 10−3 7.30 × 10−5 −9.53

40 313 3.19 × 10−3 2.70 × 10−4 −8.22

50 323 3.09 × 10−3 9.10 × 10−4 −7.00

60 333 3.00 × 10−3 2.90 × 10−3 −5.84

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Arrhenius Plot

3.0E-03 3.1E-03 3.2E-03 3.3E-03 3.4E-03 3.5E-03-11

-10

-9

-8

-7

-6

-5

1/T (1/K)

ln k

90

2 N2O5 (g) → 4 NO2 (g) + O2 (g)

Yields straight line slope = − Ea/R intercept = A

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How To Determine Ea w/ Arrhenius Plot

Pick 2 points on line Point 1 (3.030 × 10−3, −6.20) Point 2 (3.375 × 10−3, −10.4)

91

slope=Δ (ln k )

Δ(1T )

=(−10 . 4−(−6 . 20))(3 . 375−3 . 030)×10−3 K−1

=−4 . 20

3 . 45×10−4K

slope=−1 . 22×104 K=−E a

R

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Determine Ea from Arrhenius Plot

Ea = − (−1.22 × 105 K) * 8.314 J·mol−1K−1

Ea = 1.01 × 105 J/mol

Ea=−slope∗R

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Working With Arrhenius Equation

Given the following data, predict k at 75°C using the graphical approach

k (M/s) T °C T, K0.000886 25 2980.000894 50 3480.000908 100 3980.000918 150 448

? 75 348

ln (k) = -36.025/T – 6.908

ln k=−EaR

×1T

+lnA

ln (k) = – 36.025/(348) – 6.908 = – 7.01152

k=e−7 . 012=9 . 01×10−4

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Working with Arrhenius Equation

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Method 2. van't Hoff Equation

Sometimes a graph is not needed Only have 2 k s at 2 Ts

Here use van't Hoff Equation Derived from Arrhenius equation

95

ln( k 2

k 1)=−Ea

R ( 1T 2

−1T 1)Pers

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Using van't Hoff EquationEx. CH4 + 2 S2 → CS2 + 2 H2S

k (L/mol·s) T (°C) T (K)1.1 = k1 550 823 = T1

6.4 = k2 625 898 = T2

96

ln(6 . 41. 1)=

−E a

8 .3145 J /K⋅mol ( 1898 K

−1

823 K )

Ea=

−(8 . 314 J /K⋅mol ) ln(6 . 41. 1)

( 1898K

−1

823 )=1 . 4×105 J /mol

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Learning CheckGiven that k at 25°C is 4.61×10-1 M/s and that at 50°C it is 4.64×10-1 M/s, what is the activation energy for the reaction?

ln(k 2

k 1)=

−E a

R ( 1T 2

−1T 1)

Ea = 208 J/mol

ln(4 . 64×10−1 M/s

4 . 61×10−1 M/s)=

−E a

8. 314J/ ( mol⋅K )( 1323K

−1

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Your Turn!A reaction has an activation energy of 40 kJ/mol. What happens to the rate if you increase the temperature from 70 oC to 80 0C?

A. Rate increases approximately 1.5 timesB. Rate increases approximately 5000 timesC. Rate does not increaseD. Rate increases approximately 3 times

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Your Turn! - Solution

99

40000 JJ

8.314 x(80 273)Kmol K

2

1 40000 JJ

8.314 x(70 273)Kmol K

Rate is proportional to the rate constant

e 1.49

e

kk

÷

− ÷ ÷+ ÷

÷

− ÷ ÷+ ÷

= =

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Collision Theory and Reaction Mechanisms Sometimes rate law has simple form N2O5 → NO2 + NO3

NO2 + NO3 → N2O5

But others are complex H2 + Br2 → 2 HBr

100

Rate=−d [ N 2O5 ]

dt=k 1[ N 2O5 ]

Rate=−d [ NO2 ]

dt=k 2 [NO 2 ][ NO3 ]

Rate =−d [ H 2 ]

dt=

k [ H 2 ][ Br2 ]1

2

1+k ' [HBr ]

[ Br2 ]

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Why?

Some reactions occur in a single step, as written NO2 bumps into NO3

Bond forms Others involve a sequence of steps Reaction Mechanism Entire sequence of steps

Elementary Process Each individual step in sequence (mechanism) Single step that occurs as written

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What makes an elementary step elementary?

Exponents in rate law for elementary process are equal to coefficients of reactants in balanced chemical equation for that elementary process

Rate laws for elementary processes are directly related to stoichiometry

Number of molecules that participate in elementary process defines molecularity of step

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Unimolecular Process Only one molecule as reactant H3C—N≡C → H3C—C≡N

Rate = k[CH3NC]

1st order overall As number of molecules ↑, number that rearrange

in given time interval ↑'s proportionally

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Bimolecular Process Elementary step with 2 reactants NO (g) + O3 (g) → NO2 (g) + O2 (g)

Rate = k[NO][O3]

2nd order overall From collision theory: If [A] doubles, number of collisions between A and

B will double If [B] doubles, number of collisions between A and

B will double Thus, process is 1st order in A, 1st order in B, and

2nd order overall104

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Termolecular Process Elementary reaction with 3 molecules Very rare

Why? Very low probability that 3 molecules will collide

simultaneously 3rd order overall

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Summary of Elementary Processes

Molecularity Elementary Step Rate Law

Unimolecular A → products Rate = k[A]

Bimolecular A + A → products Rate = k[A]2

Bimolecular A + B → products Rate = k[A][B]Termolecular A + A + A → products Rate = k[A]3

Termolecular A + A + B → products Rate = k[A]2[B]

Termolecular A + B + C → products Rate = k[A][B][C]

106

Significance of elementary steps: • If we know that reaction is elementary step• Then we know its rate law

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Multi-Step Mechanisms Contains 2 or more steps to yield net reaction Elementary processes in multi-step mechanism

must always add up to give chemical equation of overall process

Any mechanism we propose must be consistent with experimentally observed rate law

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Multi-Step Mechanisms

Ex. Net reaction is:NO2(g) + CO(g) → NO(g) + CO2(g)

Proposed mechanism is:NO2(g) + NO2(g) → NO3(g) + NO(g)

NO3(g) + CO(g) → NO2(g) + CO2(g)

2NO2(g) + NO3(g) + CO(g) → NO2(g) + NO3(g) + NO(g) + CO2(g)

or NO2(g) + CO(g) → NO(g) + CO2(g)

108

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Intermediates Species which are formed in one step and used

up in subsequent steps Species which are neither reactant nor product

in overall reaction NO3 in this reaction

Product in one step Reactant in later step

Mechanisms may involve one or more intermediates

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How Observed Rate Law Relates to Mechanism

Can we deduce mechanism based on rate law?

Complicated problem Sometimes we can make simplifying

assumptions Most important: concept of Rate-determining

or Rate-limiting step Slowest step in mechanism

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How Observed Rate Law Relates to Mechanism

If process follows sequence of steps, slow step determines rate.

Think of an assembly line Fast earlier steps may cause intermediates to pile

up Fast later steps may have to wait for slower initial

steps Rate-determining step governs rate law

for overall reaction Can only measure up to rate determining step

—too fast to see111

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Mechanism Examples

(CH3)3CC l(aq) + OH− (aq) → (CH3)3COH (aq) + Cl− (aq)

chlorotrimethylmethane trimethylmethanol

Observed rate = k[(CH3)3CCl]

If reaction was elementary Rate would depend on both reactants Frequency of collisions depends on both

concentrations ∴Mechanism is more complex than single step What is mechanism? Evidence that it is a two step process

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Mechanisms with Slow Initial Step

Step 1: (CH3)3CCl(aq) → (CH3)3C+(aq) + Cl−(aq) (slow)

Step 2: (CH3)3C+(aq) + OH−(aq) → (CH3)3COH(aq) (fast)

Two steps Go at different rates

Each step in multiple step mechanism is elementary process, so Has its own rate constant Has its own rate law

Hence only for each step can we write rate law directly

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Mechanisms with Slow Initial Step

For above reaction Observed rate law says that step 1 is very slow

compared to step 2 i.e. Carbonium ion is formed very slowly

Once it forms, it reacts immediately Thus, rate determining step of overall

reaction is controlled by this slow step ≡ Rate-determining step (RDS)

In this case, RDS = step 1 Overall rate = k1[(CH3)3CCl] Consistent with observed rate

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Mechanisms with Fast Initial Step

1st step involves fast, reversible reactionEx. Decomposition of Ozone (No catalysts)Net reaction: 2 O3 (g) → 3 O2 (g)

Proposed mechanism:

O3 (g) O2 (g) + O (g) (fast)

O (g) + O3 (g) → 2 O2 (g) (slow)115

Observed Rate =k [O3 ]

2

[O2 ]

kf

k2

kr

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Your Turn! What is the activated complex in the reaction

3O2 (g) ⇌ 2O3(g) ? The * indicates the species is in a high energy (activated) state.

A. [O2]*

B [O3]*

C. [O]*

D. [O4]* The activated complex is formed during the

slow, rate determining step.

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Consistent with Observed Rate Law? Rate of formation of O2 = Rate of rxn 2

= k2[O][O3] But O is intermediate Need rate law in terms of reactants and

products and possibly catalysts

Rate (forward) = kf[O3] Rate (reverse) = kr[O2][O] When step 1 comes to equilibrium Rate (forward) = Rate (reverse)

kf[O3] = kr[O2][O]117

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Consistent with Observed Rate Law?

Solving this for intermediate O gives:

Substitution into rate law for step 2 gives:

Rate of RXN 2 = k2[O][O3] = where

This is observed rate law Yes, mechanism consistent

Observed Rate =k [O3 ]

2

[O2 ]

118

[O ]=k f [O3 ]

k r [O2 ]

k 2k f [O3 ]2

k r [O2 ]k obs=k 2k f

k r

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????? Questions ?????1. How do you determine if a proposed

mechanism is consistent with the observed rate law?

2. What do you do if the first step is rate determining?

3. What if the first step is a fast equilibrium and the second step is slow?

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Learning Check

The reaction mechanism that has been proposed for the decomposition of H2O2 is

1. H2O2 + I– → H2O + IO– (slow)

2. H2O2 + IO– → H2O + O2 + I– (fast)

What is the expected rate law?

120

First step is slow so RDSrate=k[H2O2][I–]Pers

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Learning CheckThe reaction: A + 3 B → D + F was studied and the following mechanism was finally determined:

1. A + B C (fast)2. C + B → D + E (slow)3. E + B → F (very fast)

What is the expected rate law?

121

Rate Step 2=k2[C][B] Rate forward = kf[A][B]Rate reverse = kr[C]kf[A][B] = kr[C][C]= kf[A][B]/kr

Rate = kobs[A][B]2

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Catalyst Substance that changes rate of chemical

reaction without itself being used up Speeds up reaction, but not consumed by

reaction Appears in mechanism, but not in overall

reaction Does not undergo permanent chemical change Regenerated at end of reaction mechanism May appear in rate law May be heterogeneous or homogeneous

122

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How Does A Catalyst Work?

By providing alternate mechanism One with lower Ea

Because Ea lower, more reactants and collisions have minimum KE, so reaction proceeds faster

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Homogeneous Catalyst Same phase as reactantsConsider : S (g) + O2 (g) + H2O (g) → H2SO4 (g)

S (g) + O2 (g) → SO2 (g) NO2 (g) + SO2 (g) → NO (g) + SO3 (g) Catalytic pathway

SO3 (g) + H2O (g) → H2SO4 (g) NO (g) + ½O2 (g) → NO2 (g) Regeneration of catalyst

Net: S (g) + O2 (g) + H2O (g) → H2SO4 (g)

What is Catalyst? Reactant (used up) in early step Product (regenerated) in later step

Which are Intermediates? 124

NO2 (g)

NO and SO2

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Heterogeneous Catalyst

Exists in separate phase from reactants Usually a solid Many industrial catalysts are heterogeneous Reaction takes place on solid catalyst

125

Ex. 3 H2 (g) + N2 (g) → 2 NH3 (g)

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Heterogeneous Catalyst

126

H2 & N2 approach Fe catalyst

H2 & N2 bind to Fe& bonds break

N—H bonds forming

N—H bonds forming

NH3 formation complete

NH3 dissociates

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Other Examples of Heterogeneous Catalysts

Catalytic converters in cars to remove CO, NO, N2O, and unburned hydrocarbons

Enzymes Proteins that catalyze reactions in living

systems. Insulin, digestive enzymes, nitrogenase, etc.

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????? Question ????? What is the difference between a catalyst and

an intermediate?

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Documents

Brdy 6Ed Ch14 ChemicalKinetics