chemical kinetics - grandinettichemical kinetics consider generic reaction: aa+bb → cc+dd where na...

Chemical Kinetics

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti (Chem. 4300) Chemical Kinetics 1 / 59

Chemical Kinetics

How do reactants become products?

▶ What is the mechanism?

What is the time scale of the reaction?

▶ How fast are reactants consumed and products formed?


Chemical KineticsConsider generic reaction:

aA + bB → cC + dD where

nA = n◦A − a𝜉nB = n◦B − b𝜉nC = n◦C + c𝜉nD = n◦D + d𝜉

for each substance we have

𝜂i = 𝜂◦i + 𝜈i𝜉

where𝜉 is extent (or advancement) of chemical reaction.𝜂i is the number of moles of the ith species,𝜂◦i is the initial number of moles of the ith species,𝜈i is the stoichiometric coefficient for the ithespecies in the reaction,

▶ 𝜈i > 0 for products▶ 𝜈i < 0 for reactants


Reaction RatesWe define the reaction rate as rate =

d𝜉dt

e.g., 4 NO2(g) + O2(g) ←←←←←←←←←←←←←←←←←→ 2 N2O5(g), rate = −14

dnNO2

dt= −

dnO2

dt= 1

2dnN2O5

dtIn definition so far reaction rate is extensive property. To make it intensive divide by system volume

R = rateV

= 1V

d𝜉dt

= 1V

(1𝜈i

dnidt

)= 1

𝜈i

d[i]dt

R is an intensive reaction rate. It has dimensionality of concentration per time.[i] is molar concentration of ith species. More accurate to use activities instead of concentration,but concentration is good approximation in ideal systems.Reaction rates will depend on p, T, and concentration of reaction species.Rates also depend on phase or phases in which reaction occurs.

▶ Homogeneous reactions occur in a single phase▶ Heterogeneous reactions involve multiple phases, e.g. reactions on surfaces.


DefinitionRate Law relates rates to concentration of reactants and products

Often we can define rate law in the form:

R = k[A]𝛼[B]𝛽 ⋯

k is the rate constantOverall reaction order is 𝛼 + 𝛽.𝛼 and 𝛽 are partial reaction orders (usually integers but not necessarily).Reaction is 𝛼 order with respect to AReaction is 𝛽 order with respect to BDetermining the rate law for reaction or order of reaction must be done experimentally.Later, we will examine on a deeper level why reactions have a given order.P. J. Grandinetti (Chem. 4300) Chemical Kinetics 5 / 59

Zeroth order reaction

Reaction: A ←←←←←←←←←←←←←←←←←→ P has rate law: R = −d[A]dt

= k[A]𝛼

if 𝛼 = 0 then reaction is zeroth order in A and has rate law:

d[A]dt

= −k

Rearranging and integrating

∫[A]t

[A]0d[A] = −k ∫

t

0dt

gives[A]t − [A]0 = −kt or [A]t = [A]0 − kt

The unit of k would be mol/(L⋅s).


Zeroth order reaction

0 200 400 600 800 10000

20

40

60

80

100

[A]

time/seconds

0th order

Zeroth order rate laws typically occurs when rate is limited by the concentration of a catalyst, e.g.,2 N2O(g) ←←←←←←←←←←←←←←←←←→ 2 N2(g) + O2(g) in presence of Pt surface catalyst.


First order reaction

Consider this simple exampleA ←←←←←←←←←←←←←←←←←→ Products

whererate = −d[A]

dt= k[A]1

Integrated rate law is

∫[A]t′

[A]0

d[A][A]

= −k ∫t′

0dt

ln[A]t[A]0

= −kt or [A]t = [A]0e−kt

units of k would be 1/s.



0 200 400 600 800 10000

20

40

60

80

100

[A]

time/seconds

0th order

1st order



0 20 40 60 80 1000

20

40

60

80

100

[A]

time/seconds

0th order

1st order


First order reaction – Log concentration plot

0 200 400 600 800 10000

1

2

3

4

5

log(

[A]/M

)

time/seconds

0th order

1st order


First order reaction – Log concentration plot

0 20 40 60 80 1000

1

2

3

4

5

log(

[A]/M

)

time/seconds

0th order

1st order

2nd order


First order reaction - Half Life

Sometimes it’s convenient to define a quantity called the half life, t1∕2, particularly for 1storder reactions.

t1∕2 is the time needed for the reaction to be half complete, i.e.,

when [A]t1∕2= 1

2[A]0 then ln

[A]t1∕2

[A]0= ln

12 [A]0[A0]

= ln 12= −kt1∕2

which gives t 12= ln 2∕k ≈ 0.693∕k (only for 1st order reaction)

Half life of 1st-order reaction depends only on reaction rate constant and is independentof [A]0.

This is not true for 2nd and 3rd order reactions.



ExampleBenzene diazonium chloride undergoes 1st-order thermal decomposition in H2O at 50 ◦C witha rate constant of k = 0.071/min. If the initial concentration is [A]0 = 0.01 M, how long mustit be heated at 50 ◦C to reduce its concentration by half?

t1∕2 = ln 2∕(0.071∕min) = 9.9 min

Note that result is independent of initial concentration, [A]0.


Second order reactionHere we have

A ←←←←←←←←←←←←←←←←←→ Products with − d[A]dt

= k[A]2

Integrating rate expression gives

∫[A]t′

[A]0

d[A][A]2

= −k ∫t′

0dt

ln[A]t[A]0

= −kt or 1[A]t

= 1[A]0

+ kt

units of k would be L/(mol⋅s).Rearranging gives

[A]t =[A]0

1 + [A]0kt


Second order reaction

0 20 40 60 80 1000

20

40

60

80

100

[A]

time/seconds

0th order

1st order

2nd order


Second order reaction – inverse concentration plot

0 200 400 600 800 10000.0

0.5

1.0

1.5

2.0

1/[A]

time/seconds

0th order1st order

2nd order


Second order reaction – inverse concentration plot

0 10 20 30 40 50 600.0

0.5

1.0

1.5

2.0

1/[A]

time/seconds

1st order

2nd order


Another kind of second order reaction

A + B ←←←←←←←←←←←←←←←←←→ Products where − d[A]dt

= −d[B]dt

= k[A][B]

after some clever math we obtain

1[A]0 − [B]0

ln[B]0[A]t[A]0[B]t

= kt

If [A]0 = [B]0 then solution is like previous 2nd-order example, i.e.

1[A]t

= 1[A]0

+ kt


Another kind of second order reaction

0 5 10 15 20 25 300

5

10

15

20

25

30

time/seconds


Measuring Reaction Rate Order


Measuring Reaction Rate Order

rate = k[A]𝛼[B]𝛽 ⋯ [L]𝜆

need to determine orders, 𝛼, 𝛽, . . . , 𝜆 before measuring k.

Half-life method

Differential (initial rate) method

Isolation method


Half-Life MethodWhen rate law has the form rate = k[A]n then one can show that

log10 t1∕2 = log102n−1 − 1(n − 1)k

+ (1 − n) log10[A]0

and a plot of log10 t1∕2 versus log10[A]0 gives a straight line with a slope of 1 − n.

1 Plot [A] versus t

2 Pick any [A] value and find the time interval, t1∕2, for it to fall to half its value.

3 Pick another [A] value and repeat step 2.

4 Repeat step 3 several times and then plot log10 t1∕2 versus log10[A]0

5 The slope will be 1 − n.


Half-Life Method

0.0 0.2 0.30.1 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.40

20

10

40

30

60

70

50

80

90

100

[A]

time/minutes

t1/2= 0.100 s for [A] = 100 to 50

t1/2= 0.125 s for [A] = 80 to 40

t1/2= 0.167 s for [A] = 60 to 30

1.5 1.6 1.7 1.8 1.9 2.0 2.1-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

Log([A]0/M)

Log(

t 1/2

/s)

In this example, a 2nd-order rate (n = 2) gives slope of 1 − n = −1.


Differential (Initial Rate) MethodSince rate = k[A]n then ln(rate) = ln k + n ln[A]t, that is, slope is n.

0.0 0.1 0.2 0.30

20

40

60

80

100

[A]

time/seconds

d[A=100]/dt = 1000 d[A=80]/dt = 640d[A=60]/dt = 360d[A=40]/dt = 160

0.0 0.5 1.0 1.5 2.0 2.5 3.00.000.250.500.751.001.251.501.752.002.252.502.753.00

log(

rate

)

log([A]0/M)For rate = k[A]n[B]m ⋯, vary [A] while holding other reactant concentrations constant.Then repeat with other reactants.


Isolation Method

Start with rate = k[A]𝛼[B]𝛽[C]𝛾 ⋯ [L]𝜆

Make the initial concentration of A much less than all other reactants, B, C, …, so youcan assume other reactant concentrations remain constant in time. With this assumptionthe rate becomes

rate = k[A]𝛼[B]𝛽0[C]𝛾0 ⋯ [L]𝜆0rearranging becomes

rate =[k[B]𝛽0[C]𝛾0 ⋯ [L]𝜆0

]⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

k′ ≈ constant.

[A]𝛼

Combine this approach with the initial rate method to get partial reaction orders 𝛼, 𝛽, 𝛾,….


Predicting Reaction Order


Predicting Reaction Order

Most reactions can be broken down into a sequence of steps that involve either aUnimolecular Reaction where a single molecule shakes itself apart or into a newconfiguration:

A → P

Bimolecular Reaction where a pair of molecules collide and exchange energy, atoms, orgroups of atoms.

A + B → P

A reaction of some specific order can often be accounted for in terms of a sequence of severalunimolecular or bimolecular steps.Generally, the order of a reaction is distinct from the molecularity of the individual steps.

Order is empirically determined.Molecularity is characteristic of the underlying reaction mechanism.


Predicting Reaction OrderIf one step in overall mechanism is rate limiting then order of overall rate will be simply relatedto molecularity.

Unimolecular rate limiting step leads to k[A].Bimolecular rate limiting step leads to k[A]2 or k[A][B]Termolecular rate limiting step leads to k[A]3 or k[A]2[B] or k[A][B]2 or k[A][B][C].and so on...

ExampleUnimolecular (decomposition reaction)

N2O4(g) ←←←←←←←←←←←←←←←←←→ 2 NO2(g)

←←←←←←←←←←←←←←←←←→ CH3CH−−CH2


Predicting Reaction OrderExampleBimolecular

D + H2 ←←←←←←←←←←←←←←←←←→ DH + H

NO2 + CO ←←←←←←←←←←←←←←←←←→ NO + CO2

ExampleTermolecular

D + H + H2 ←←←←←←←←←←←←←←←←←→ DH + H2

These type of reactions are rate, particularly in the gas phase. Higher molecularities than 3have not been observed.


MechanismsConsecutive elementary steps may be reversible

Elementary forward and reverse rate constants related to concentration-based equilibrium constant, K

Ak1←←←←←←←←←←←←←⇀↽←←←←←←←←←←←←←k−1

B, where K =k1

k−1and − d[A]

dt= k1[A] − k−1[B]

Integrate rate law by rewriting concentration in terms of displacement from equilibrium.Let x = [A] − [A]eq = [B] − [B]eq or [A] = x + [A]eq and [B] = −x + [B]eq then

−d[A]dt

= −dxdt

−��>

0d[A]eq

dt= k1

(x + [A]eq

)⏟⏞⏞⏞⏟⏞⏞⏞⏟

[A]

−k−1(−x + [B]eq

)⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟

[B]

=(k1 + k−1

)x +

��*0k1[A]eq − k−1[B]eq⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

2 rates equal at eq.

−dxdt

=(k1 + k−1

)x ⟶ x = x0e−(k1+k−1)t or [A]t = [A]eq +

([A]0 − [A]eq

)e−(k1+k−1)t

x “relaxes” to zero with a time constant 𝜏 when system is displaced from equilibrium

𝜏 = 1k1 + k−1


MechanismsConsecutive elementary steps may be reversible

Near equilibrium the same equationx = x0e−(k1+k−1)t

holds for any forward and reverse molecularity reaction, but definition of 𝜏 depends on molecularity.

HomeworkShow for

A + Bk1←←←←←←←←←←←←←⇀↽←←←←←←←←←←←←←k−1

C

that exponential time decay describing relaxation back to equilibrium is given by

𝜏 = 1k1

([A]eq + [B]eq

)+ k−1


Series Reaction with Reversible StepStep between reactant, A, and an intermediate, I, is reversible:

Ak1←←←←←←←←←←←←←⇀↽←←←←←←←←←←←←←k−1

Ik2←←←←←←←←←→ P

Rate law for each species is

d[A]dt

= −k1[A] + k−1[I],d[I]dt

= k1[A] − k−1[I] − k2[I],d[P]

dt= k2[I]

After a tricky derivation, for case when [I]0 = [P]0 = 0, solution is

[A]t =k1[A]0𝜆2 − 𝜆3

{𝜆2 − k2𝜆2

e−𝜆2t −𝜆3 − k2𝜆3

e−𝜆3t}

, [I]t =k1[A]0𝜆2 − 𝜆3

{e−𝜆3t − e−𝜆2t}

[P]t = [A]0

{1 +

𝜆3𝜆2 − 𝜆3

e−𝜆2t −𝜆2

𝜆2 − 𝜆3e−𝜆3t

}where 𝜆2 = 1

2 (p + q), 𝜆3 = 12 (p − q), p = k1 + k−1 + k2, and q =

(p2 − 4k1k2

)1∕2.Solution requires integration by partial fractions.


Series Reaction with Reversible StepStep between reactant, A, and an intermediate, I, is reversible:

Ak1←←←←←←←←←←←←⇀↽←←←←←←←←←←←←k−1

Ik2←←←←←←←←←→ P

20000 400 600 800 1000

20

40

60

80

100

[A],[I],[P]

time/seconds

[A][P]

[I]


Series Reaction with Reversible Step – Steady State Approximation

Ak1←←←←←←←←←←←←←⇀↽←←←←←←←←←←←←←k−1

Ik2←←←←←←←←←→ P

If intermediate is so reactive that it remains at a low concentration compared to A or P, then one canmake the steady-state approximation. It requires k−1 + k2 ≫ k1.

d[I]dt

= k1[A] − k−1[I] − k2[I] ≈ 0 which gives [I]ss ≈k1[A]

k−1 + k2

Substituting back into rate expressions for A and P gives

−d[A]dt

= k1[A] − k−1[I] ≈ k1[A] −k−1k1[A]k−1 + k2

=k1k2

k−1 + k2[A], and d[P]

dt= k2[I] ≈

k1k2k−1 + k2

[A]

If 1st step is slow, k−1 ≪ k2 then d[P]dt

≈ k1[A]

If 2nd step is slow, k2 ≪ k−1 then d[P]dt

≈k1

k−1k2[A] = K k2[A] ← pre-equilibrium approximation

K is equilibrium constant, given by K = k1∕k−1


Lindemann MechanismPre-equilibrium approximation example

Many gas phase reactions follow 1st order kinetics

A(g)kr←←←←←←←←→ P(g) with [P]

dt= kr[A]

For A to get enough energy to react it needs to collide with other molecules.We expect mechanism like

A(g) + M(g)k1←←←←←←←←←←←←⇀↽←←←←←←←←←←←←k−1

A∗(g) + M(g)

A∗(g)k2←←←←←←←←←→ P(g)

A(g) ⟶ P(g)

The Lindemann mechanism uses the steady-stateapproximation, i.e., treat A∗ like I in last case.

[A∗]dt

= k1[A][M] − k1[A∗][M] − k2[A∗] ≈ 0

so[A∗] ≈

k1[A][M]k−1[M] + k2

and d[P]dt

= k2[A∗] ≈k1k2[A][M]k−1[M] + k2


Lindemann Mechanism[A∗] ≈

k1[M][A]k−1[M] + k2

d[P]dt

= k2[A∗] ≈k1k2[M][A]k−1[M] + k2

at high pressures we have k−1[M] ≫ k2 andthen

d[P]dt

≈k1k2k−1

[A]

reaction is 1st order in A with kobs = k1k2∕k−1

at low pressures we have k2 ≫ k−1[M] then

d[P]dt

≈ k1[M][A]

reaction is 1st order in A with kobs = k1[M]

isomerization reactionCH3NC ←←←←←←←←←←←←←←←←←→ CH3CN at 472.5 K


Chain Reaction Mechanism

Chain reaction consists of series of steps in which reactive intermediate is consumed,reactants are converted to products, and intermediate is regenerated.Cycle repeats, allowing small amount of intermediate to produce large amount of product.Most combustions, explosions, and addition polymerization reactions are chain reactions.Intermediates are often free radicals.

The classic example is H2(g) + Br2(g) ←←←←←←←←←←←←←←←←←→ 2 HBr(g)The observed rate law is

d[HBr]dt

=k[H2][Br2]3∕2

[Br2] + k′[HBr]Let’s examine this one in more detail...


Chain Reaction Mechanism: H2(g) + Br2(g) ←←←←←←←←←←←←←←←←←→ 2 HBr(g)

Initiation:Br2 + M

ki←←←←←←←←→ 2 Br ⋅ + M, rate = ki[Br2][M]

M is either Br2 or H2. Collisions with M give Br2 enough energy to break apart.

Propagation:

Br ⋅ + H2kp←←←←←←←←←→ HBr + H ⋅ , rate = kp[Br ⋅ ][H2]

H ⋅ + Br2

k′p←←←←←←←←←→ HBr + Br ⋅ , rate = k′p[H ⋅ ][Br2]

Retardation:H ⋅ + HBr

kr←←←←←←←←←→ H2 + Br ⋅ , rate = kr[H ⋅ ][HBr]

Termination:2 Br ⋅ + M

kt←←←←←←←←→ Br2 + M, rate = kr[Br ⋅ ]2[M]

Next we assemble the rate law...


Chain Reaction Mechanism: H2(g) + Br2(g) ←←←←←←←←←←←←←←←←←→ 2 HBr(g)Br2 + M

ki←←←←←←←←→ 2 Br ⋅ + M rate = ki[Br2][M]

Br ⋅ + H2kp←←←←←←←←←→ HBr + H ⋅ rate = kp[Br ⋅ ][H2]

H ⋅ + Br2

k′p←←←←←←←←←→ HBr + Br ⋅ rate = k′p[H ⋅ ][Br2]

H ⋅ + HBrkr←←←←←←←←←→ H2 + Br ⋅ rate = kr[H ⋅ ][HBr]

2 Br ⋅ + Mkt←←←←←←←←→ Br2 + M rate = kr[Br ⋅ ]2[M]

d[HBr]dt

= kp[Br ⋅ ][H2] + k′p[H ⋅ ][Br2] − kr[H ⋅ ][HBr]

Make the steady state approximation for intermediates H ⋅ and Br ⋅d[H ⋅ ]

dt= kp[Br ⋅ ][H2] − k′p[H ⋅ ][Br2] − kr[H ⋅ ][HBr] ≈ 0

d[Br ⋅ ]dt

= 2ki[Br2][M] − kp[Br ⋅ ][H2] + k′p[H ⋅ ][Br2] + kr[H ⋅ ][HBr] − 2kt[Br ⋅ ]2[M] ≈ 0

Two equations, two unknowns, [H ⋅ ]ss and [Br ⋅ ]ss ...P. J. Grandinetti (Chem. 4300) Chemical Kinetics 40 / 59

Chain Reaction Mechanism: H2(g) + Br2(g) ←←←←←←←←←←←←←←←←←→ 2 HBr(g)

d[H ⋅ ]dt

= kp[Br ⋅ ][H2] − k′p[H ⋅ ][Br2] − kr[H ⋅ ][HBr] ≈ 0

d[Br ⋅ ]dt

= 2ki[Br2][M] − kp[Br ⋅ ][H2] + k′p[H ⋅ ][Br2] + kr[H ⋅ ][HBr] − 2kt[Br ⋅ ]2[M] ≈ 0

Two equations, two unknowns, [H ⋅ ]ss and [Br ⋅ ]ss. You can show that

[Br ⋅ ]ss =(

kikt

)1∕2[Br2]1∕2 and [H ⋅ ]ss =

kp

(kikt

)1∕2[H2][Br2]1∕2

k′p[Br2] + kr[HBr]

Note [M] cancelled out. Finally we get

d[HBr]dt

=2kp

(kikt

)1∕2[H2][Br2]3∕2

[Br2] + (kr∕k′p)[HBr]matching observed rate law: d[HBr]

dt=

k[H2][Br2]3∕2

[Br2] + k′[HBr]


Svante August Arrhenius1859–1927


Temperature Dependence of Rate ConstantsExperimentally, a plot of ln k versus 1∕T for many reactions are linear or very close to linear.

ln(k

[10

-4cm

3 mol

-1s

-1 ])

0

0,5

1

1,5

2

2,5

3

3,5

4

23

23,5

1/T [1/K]0.0015 0.00155 0.0016 0.00165 0.00170

In 1889 Arrhenius proposed therelationship

k = Ae−Ea∕(RT)

A is pre-exponential factor

Ea is activationenergy—corresponds to energyneeded for reaction to occur.

As T increases so does fraction ofmolecules with translationalenergy greater than Ea.


Temperature Dependence of Rate Constants

00.050.00 0.10 0.15 0.20 0.25 0.30 0.35

1000 K

300 K

100 K

Energy

reaction coordinate

reac

tant

s

prod

uctsEa

p(E)

Arrhenius relationship: k = Ae−Ea∕(RT)

A is pre-exponential factor

Ea is activation energy—corresponds toenergy needed for reaction to occur.

As T increases so does fraction ofmolecules with translational energygreater than Ea.

Not all reactions follow Arrheniusbehavior. It assumes Ea and A aretemperature independent.

More general expression gives Ea as slopeat specific temperature:

Ea = RT2(d ln k

dT

)P. J. Grandinetti (Chem. 4300) Chemical Kinetics 44 / 59

Potential Energy SurfacesConsider two diatomic molecules A–B and B–C.How does their potential energies vary with bond length, r(AB) or r(BC)?

re(AB) r(AB) re(BC) r(BC)

Consider a collision between A–B and C which leads to the reaction AB + C ←←←←←←←←←←←←←←←←←→ A + BC.What is the 2D potential energy surface as a function of r(AB) and r(BC)?


Potential Energy SurfacesWhat is the 2D potential energy surface as a function of r(AB) and r(BC)?Assume the angle A–B ⋯ C is 180◦ for collision.

2D Potential Energy Surface

00

r(AB)re(AB)

r(BC)

re(BC)

A-B + C

A + B-C

What is reaction path on 2D energy surface?

Could follow path of minimum potentialenergy from AB + C to BC + A.

reaction coordinate

A-B

+ C

B-C

+ AEa

(A-B-C)‡

Pote

ntia

l Ene

rgy

(ABC)‡ is the transition state at top of saddlepoint in 2D contour plot.


Transition State Theoryaka Activated Complex TheoryAssume reactants are in equilibrium with an activated unstable intermediate X‡.

A + Bk1←←←←←←←←←←←←←⇀↽←←←←←←←←←←←←←k−1

X‡ 𝜈‡←←←←←←←←←←→ P

Assume A + Bk1←←←←←←←←←←←←←⇀↽←←←←←←←←←←←←←k−1

X‡ is described by equilibrium expression:

K‡c =

k1k−1

=[X‡]∕c◦

([A]∕c◦) ([B]∕c◦), where K‡

c is equilibrium constant, and c◦ = 1M

𝜈‡ is frequency that activated complex crosses over barrier. Statistical mechanics says 𝜈‡ = kBT∕h.Overall rate is

d[P]dt

= 𝜈‡[X‡] = 𝜈‡K‡

cc◦

[A][B] =kBT

hK‡

cc◦

⏟⏟⏟k

[A][B] = k[A][B]

K‡c cannot be measured (can’t observe X‡) but can be inferred from k (assuming theory is correct).


Transition State Theoryaka Activated Complex Theory

A + Bk1←←←←←←←←←←←←←⇀↽←←←←←←←←←←←←←k−1

X‡ 𝜈‡←←←←←←←←←←→ P, where d[P]

dt= 𝜈‡[X‡] = k[A][B] and k =

kBTh

K‡c

c◦

From K‡c we can examine kinetics in terms of thermodynamics.

ΔG◦‡ = −RT lnK‡c or K‡

c = e−ΔG◦‡∕(RT)

Since ΔG◦‡ = ΔH◦‡ − TΔS◦‡ we haveK‡

c = e−ΔH◦‡∕(RT)eΔS◦‡∕R

Remember ΔH◦‡ and ΔS◦‡ describe A + Bk1←←←←←←←←←←←←←⇀↽←←←←←←←←←←←←←k−1

X‡

Finally, we obtain the Eyring Equation:

k =kBTc◦h

eΔS◦‡∕Re−ΔH◦‡∕(RT)


Transition State TheoryEyring Equation

k =kBTc◦h

eΔS◦‡∕Re−ΔH◦‡∕(RT)

How do ΔH◦‡ and ΔS◦‡ in Eyring equation related to Arrhenius parameters, A and Ea?Substitute predicted rate constant into the general expression for Arrhenius behavior

Ea = RT2(d ln k

dT

)Start with

ln k = ln(

kBTc◦h

K‡c

)= ln

kBc◦h

+ ln T + lnK‡c

Next calculated ln kdT

= d lnTdT

+d lnK‡

cdT

= 1T+

d lnK‡c

dTThen obtain

Ea = RT2(d ln k

dT

)= RT2

(1T+

d lnK‡c

dT

)= RT + RT2 d lnK‡

cdT


Thermodynamics Review

Homework(A) For ideal gas phase reaction a A + b B ←←←←←←←←←←←←←←←←←⇀↽←←←←←←←←←←←←←←←←← d D + f F

Kp =(pD∕p◦

) (pF∕p◦

)(pA∕p◦

) (pB∕p◦

) and Kc =([D]∕c◦) ([F]∕c◦)([A]∕c◦) ([B]∕c◦)

where p◦ = 1 bar, c◦ = 1 M

show that Kc = Kp

(c◦RT

p◦

)Δn

where Δn = d + f − a − b

(B) From van’t Hoff Equation:d lnKp

dT= ΔH◦

RT2 prove thatd lnKc

dT= ΔU◦

RT2


Transition State TheoryBack to Eyring Equation

Ea = RT + RT2 d lnK‡c

dTFor reaction of ideal gas molecules we know

d lnKcdT

= ΔU◦

RT2 or RT2 d lnKcdT

= ΔU◦

With this relation we obtainEa = RT + ΔU◦‡

For ideal gases we also know that ΔH = ΔU + pΔV = ΔU + ΔnRT so

Ea = RT +(ΔH◦‡ − Δn‡RT

)and finally obtain

Ea = ΔH◦‡ +(1 − Δn‡

)RT


Transition State Theory

Ea = ΔH◦‡ +(1 − Δn‡

)RT

For unimolecular reaction: A → X‡ we have n‡f = 1, n‡i = 1, so Δn‡ = 0,

ΔH◦‡ = Ea − RT and k =kBTc◦h

eΔS◦‡∕Re−ΔH◦‡∕(RT) =kBTc◦h

eΔS◦‡∕Re−(Ea−RT)∕(RT)

k =[

ekBTc◦h

eΔS◦‡∕R]

e−Ea∕(RT) unimolecular gas phase reaction

For bimolecular reaction: A + B → X‡ we have n‡f = 1, n‡i = 2, so Δn‡ = −1, and similarlyobtain

k =[

e2 kBTc◦h

eΔS◦‡∕R]

e−Ea∕(RT) bimolecular gas phase reaction


Transition State Theory

ProblemThe Arrhenius activation energy and pre-exponential factor for the gas phase reaction:

H(g) + Br2(g) ←←←←←←←←←←←←←←←←←→ HBr(g) + Br(g)

are 15.5 kJ/mol and 1.09 × 1011 L/(mol⋅s), respectively. What are the values of ΔH◦‡ andΔS◦‡ at 1000 K based on standard state c◦ = 1 M. Assume ideal gas behavior.

ΔH◦‡ = Ea = 2RT = 15.5kJ/mol − 2R(1000K) = −1.13 kJ/mol

and

ΔS◦‡ = R ln hAc◦

e2kBT= R ln

{h(1.09 × 1011L/(mol⋅s))(1mol/L)

e2kB(1000K)

}= −60.3 J/(mol⋅K)


Reactions in Solution


Reactions in SolutionIn solution reactants are surround by solvent molecules and need to “encounter” eachother before they can react.

A “cage” of solvent molecules trap reactants together, allowing them to collide manytimes before they move away from each other.Chance that molecules will react is much higher in solution than in gas phase.If reaction is very fast then rate limiting step becomes how long it takes for molecules todiffuse together. This is slower than free motion of gas molecules.P. J. Grandinetti (Chem. 4300) Chemical Kinetics 55 / 59

Reactions in SolutionConsider the following mechanism:

A + Bkd←←←←←←←←←→ AB ⋅ Diffusion of A and B into solvent cage and formation of activated complex.

ABkr←←←←←←←←←→ A + B ⋅ Dissociation back to reactants.

ABkp←←←←←←←←←→ P ⋅ Formation of products.

Reaction rate is dPdt

= kp[AB]. With [AB] as intermediate make steady state approximation

[AB]dt

= kd[A][B] − kr[AB] − kp[AB] ≈ 0, gives [AB]ss ≈kd[A][B]kr + kp

Using [AB]ss in the rate expression gives

dPdt

= kp[AB] ≈kpkd

kr + kp[A][B]

If product formation is faster than dissociation back to reactants, i.e., kp ≫ kr, then we obtain adiffusion controlled reaction rate:

dPdt

= kd[A][B]


Reactions in Solutiondiffusion controlled reaction rate

dPdt

= kd[A][B]

Rate constants for diffusion can be related to diffusion coefficientskd = 4𝜋NA

(rA + rB

)(DA + DB)

rA and rB are molecular radii, and DA and DB are diffusion coefficients.Relationship between D and viscosity of solution, 𝜂, is

D =kBT6𝜋𝜂r

Stokes-Einstein equation

Note: mass ∝ r3, so D ∝ m−1∕3

If we assume rA = rB = r and DA = DB = D then

kd = 83

RT𝜂


Reactions in Solution

ProblemWater viscosity is 𝜂 = 8.9× 10−4 N⋅s/m2. What is rate constant in water at room temperature?

kd = 83

R(298K)8.9 × 10−4N⋅s/m2 = 7.4 × 109L/(mol⋅s)


Homework from McQuarrie

Chapter 28: 1, 2, 4, 6, 8, 11, 12, 15, 17, 18, 22, 23, 24, 31, 34, 41, 42, 43,46Chapter 29: 1, 3, 7, 8, 12, 16, 21, 24


chemical kinetics - grandinettichemical kinetics consider generic reaction: aa+bb → cc+dd where na...

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