P. J. Grandinetti
Chem. 4300
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 1 / 60 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?
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 2 / 60 Chemical Kinetics Consider generic reaction: ◦ 휉 nA = nA − a n = n◦ − b휉 → B B aA + bB cC + dD where ◦ 휉 nC = nC + c ◦ 휉 nD = nD + d for each substance we have where 휉 is extent (or advancement) of chemical reaction. ◦ 휈 휉 ni = ni + i ni is number of moles of ith species, or n◦ is initial number of moles of ith species, 1 ◦ i 휉 = (n − n ) 휈 휈 i i i is stoichiometric coefficient for ith species in i reaction, ▶ 휈 > for products d휉 1 dni i 0 rate = = ▶ 휈 < 휈 i 0 for reactants dt i dt
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 3 / 60 Reaction Rates We define the reaction rate as d휉 1 dni rate = = 휈 dt i dt 1 dnNO dnO 1 dnN O e.g., 4 NO (g) + O (g) ←←←←←←←←←←←←←←←←←→ 2 N O (g), rate = − 2 = − 2 = 2 5 2 2 2 5 4 dt dt 2 dt In definition so far reaction rate is extensive property. To make it intensive divide by system volume ( ) rate 1 d휉 1 1 dni 1 d[i] R = = = 휈 = 휈 V V dt V i dt 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.
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 4 / 60 Definition Rate Law relates rates to concentration of reactants and products
Often we can define rate law in the form:
R = k[A]훼[B]훽 ⋯
k is the rate constant Overall reaction order is 훼 + 훽. 훼 and 훽 are partial reaction orders (usually integers but not necessarily). Reaction is 훼 order with respect to A Reaction is 훽 order with respect to B Determining 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 / 60 Zeroth order reaction
d[A] Reaction: A ←←←←←←←←←←←←←←←←←→ P has rate law: R = − = k[A]훼 dt if 훼 = 0 then reaction is zeroth order in A and has rate law: d[A] = −k dt
Rearranging and integrating [A]t t ∫ d[A] = −k ∫ dt [A]0 0 gives [A]t − [A]0 = −kt or [A]t = [A]0 − kt The unit of k would be mol/(L⋅s).
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 6 / 60 Zeroth order reaction 100
80
0th order 60 [A] 40
20
0 0 200 400 600 800 1000 time/seconds 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. P. J. Grandinetti (Chem. 4300) Chemical Kinetics 7 / 60 First order reaction
Consider this simple example A ←←←←←←←←←←←←←←←←←→ Products where d[A] rate = − = k[A]1 dt
Integrated rate law is ′ [A]t′ d[A] t = −k dt ∫ [A] ∫ [A]0 0
[A]t −kt ln = −kt or [A]t = [A]0e [A]0 units of k would be 1/s.
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 8 / 60 First order reaction
100
80
0th order 60 [A] 40
20 1st order
0 0 200 400 600 800 1000 time/seconds
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 9 / 60 First order reaction
100
0th order 80
60 [A] 40
20 1st order
0 0 20 40 60 80 100 time/seconds
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 10 / 60 First order reaction – Log concentration plot
5
0th order 4
3
2 log([A]/M) 1st order 1
0 0 200 400 600 800 1000 time/seconds
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 11 / 60 First order reaction – Log concentration plot
5 0th order
4
3
1st order 2 log([A]/M)
1 2nd order
0 0 20 40 60 80 100 time/seconds
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 12 / 60 First order reaction - Half Life
Sometimes it’s convenient to define a quantity called the half life, t1∕2, particularly for 1st order reactions.
t1∕2 is the time needed for the reaction to be half complete, i.e.,
[A] 1 [A] 1 t1∕2 0 1 when [A] = [A] then ln = ln 2 = ln = −kt t1∕2 0 1∕2 2 [A]0 [A0] 2 which gives t 1 = ln 2∕k ≈ 0.693∕k (only for 1st order reaction) 2
Half life of 1st-order reaction depends only on reaction rate constant and is independent of [A]0. This is not true for 2nd and 3rd order reactions.
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 13 / 60 First order reaction
Example ◦ Benzene diazonium chloride undergoes 1st-order thermal decomposition in H2O at 50 C with . . a rate constant of k = 0 071/min. If the initial concentration is [A]0 = 0 01 M, how long must it be heated at 50 ◦C to reduce its concentration by half?
. . t1∕2 = ln 2∕k = ln 2∕(0 071∕min) = 9 9 min
Note that result is independent of initial concentration, [A]0.
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 14 / 60 Second order reaction Here we have d[A] A ←←←←←←←←←←←←←←←←←→ Products with − = k[A]2 dt Integrating rate expression gives
′ [A]t′ d[A] t = −k dt ∫ 2 ∫ [A]0 [A] 0 1 1 = + kt [A]t [A]0 units of k would be L/(mol⋅s). Rearranging gives [A]0 [A]t = 1 + [A]0kt
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 15 / 60 Second order reaction
100
0th order 80
60 [A] 40
1st order 20 2nd order 0 0 20 40 60 80 100 time/seconds
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 16 / 60 Second order reaction – inverse concentration plot
2.0 2nd order
1.5
1/[A] 1.0
0.5 0th order 1st order 0.0 0 200 400 600 800 1000 time/seconds
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 17 / 60 Second order reaction – inverse concentration plot
2.0 2nd order
1.5
1/[A] 1.0
0.5 1st order
0.0 0 10 20 30 40 50 60 time/seconds
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 18 / 60 Another kind of second order reaction
d[A] d[B] A + B ←←←←←←←←←←←←←←←←←→ Products where − = − = k[A][B] dt dt
after some clever math we obtain 1 [B] [A] ln 0 t = kt [A]0 − [B]0 [A]0[B]t
If [A]0 = [B]0 then solution is like previous 2nd-order example, i.e. 1 1 = + kt [A]t [A]0
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 19 / 60 Another kind of second order reaction
30
25
20
15
10
5
0 0 5 10 15 20 25 30 time/seconds
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 20 / 60 Measuring Reaction Rate Order
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 21 / 60 Measuring Reaction Rate Order
rate = k[A]훼[B]훽 ⋯ [L]휆 need to determine orders, 훼, 훽,..., 휆 before measuring k.
Half-life method
Differential (initial rate) method
Isolation method
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 22 / 60 Half-Life Method When rate law has the form rate = k[A]n then one can show that
2n−1 − 1 log t = log + (1 − n) log [A] 10 1∕2 10 (n − 1)k 10 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.
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 23 / 60 Half-Life Method
0.0 100 -0.2 90 t1/2= 0.100 min for [A] = 100 to 50 80 -0.4 70 t1/2= 0.125 min for [A] = 80 to 40 /min) 60 1/2 -0.6 [A] 50
t1/2= 0.167 min for [A] = 60 to 30 Log(t 40 -0.8 30 20 -1.0 10 0 -1.2 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 time/minutes Log([A]0/M) In this example, a 2nd-order rate (n = 2) gives slope of 1 − n = −1.
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 24 / 60 Differential (Initial Rate) Method n Since rate = k[A] then ln(rate) = ln k + n ln[A]t, that is, slope is n.
100 3.00 d[A=100]/dt = 1000 mol/(L•s) 2.75 80 d[A=80]/dt = 640 mol/(L•s) 2.50 2.25 d[A=60]/dt = 360 mol/(L•s) 2.00 60 d[A=40]/dt = 160 mol/(L•s) 1.75 [A] 1.50 40 1.25 1.00 0.75 20 log(rate/ mol/(L•s)) 0.50 0.25 0.00 0 0.0 0.1 0.2 0.3 0.0 0.5 1.0 1.5 2.0 2.5 3.0 time/seconds log([A]0/M) For rate = k[A]n[B]m ⋯, vary [A] while holding other reactant concentrations constant. Then repeat with other reactants.
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 25 / 60 Isolation Method
Start with rate = k[A]훼[B]훽[C]훾 ⋯ [L]휆
Make the initial concentration of A much less than all other reactants, B, C, …, so you can assume other reactant concentrations remain constant in time. With this assumption the rate becomes
훼 훽 훾 ⋯ 휆 rate = k[A] [B]0[C]0 [L]0 rearranging becomes [ ] 훽 훾 ⋯ 휆 훼 rate = k[B]0[C]0 [L]0 [A] ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ k′ ≈ constant.
Combine this approach with the initial rate method to get partial reaction orders 훼, 훽, 훾, ….
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 26 / 60 Predicting Reaction Order
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 27 / 60 Predicting Reaction Order Most reactions can be broken down into sequence of steps that involve either Unimolecular Reaction where single molecule shakes itself apart or into new configuration: k d[P] A ←←←←←→ P rate = = k[A] dt Bimolecular Reaction where pair of molecules collide and exchange energy, atoms, or groups of atoms. k d[P] A + B ←←←←←→ P rate = = k[A][B] dt Reaction of some specific order can often be accounted for in terms of sequence of several unimolecular or bimolecular steps. Generally, order of reaction is distinct from molecularity of individual steps. Order is empirically determined. Molecularity is characteristic of underlying reaction mechanism.
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 28 / 60 Predicting Reaction Order If one step in overall mechanism is rate limiting then order of overall rate will be simply related to 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...
Example Unimolecular (decomposition reaction)
←←←←←←←←←←←←←←←←←→ N2O4(g) 2 NO2(g)
←←←←←←←←←←←←←←←←←→ CH3CH−CH2
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 29 / 60 Predicting Reaction Order Example Bimolecular
←←←←←←←←←←←←←←←←←→ D + H2 DH + H
←←←←←←←←←←←←←←←←←→ NO2 + CO NO + CO2
Example Termolecular
←←←←←←←←←←←←←←←←←→ D + H + H2 DH + H2 These type of reactions are rare, particularly in the gas phase. Higher molecularities than 3 have not been observed.
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 30 / 60 Mechanisms Consecutive elementary steps may be reversible Elementary forward and reverse rate constants related to concentration-based equilibrium constant, K
k1 k ←←←←←←←←←←←←←⇀ , 1 d[A] A ↽←←←←←←←←←←←←← B where K = and − = k1[A] − k−1[B] k−1 k−1 dt 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 0 d[A]> ( ) ( ) ( ) ¨¨* 0 d[A] dx eq ¨ − = − − = k1 x + [A]eq −k−1 −x + [B]eq = k1 + k−1 x + k1[A]eq − k−¨1[B]eq dt dt dt ⏟⏞⏞⏞⏟⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟¨¨ ¨ [A] [B] ¨2¨ rates equal at eq. dx ( ) ( ) − = k + k x ⟶ x = x e−(k1+k−1)t or [A] = [A] + [A] − [A] e−(k1+k−1)t dt 1 −1 0 t eq 0 eq x “relaxes” to zero with a time constant 휏 when system is displaced from equilibrium 1 휏 = k1 + k−1
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 31 / 60 Mechanisms Consecutive elementary steps may be reversible Near equilibrium the same equation −t∕휏 x = x0e holds for any forward and reverse molecularity reaction, but definition of 휏 depends on molecularity. Homework Show for k1 A + B ↽←←←←←←←←←←←←←←←←←←←←←←←←←←⇀ C k−1 that exponential time decay describing relaxation back to equilibrium is given by 1 휏 = ( ) k1 [A]eq + [B]eq + k−1
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 32 / 60 Series Reaction with Reversible Step Step between reactant, A, and an intermediate, I, is reversible:
k1 k A ↽←←←←←←←←←←←←←←←←←←←←←←←←←←⇀ I ←←←←←←←←←→2 P k−1 Rate law for each species is d[A] d[I] d[P] = −k [A] + k [I], = k [A] − k [I] − k [I], = k [I] dt 1 −1 dt 1 −1 2 dt 2
After a tricky derivation, for case when [I]0 = [P]0 = 0, solution is { } { } k [A] 휆 − k 휆 휆 − k 휆 k [A] 휆 휆 1 0 2 2 − 2t 3 2 − 3t , 1 0 − 3t − 2t [A]t = 휆 휆 휆 e − 휆 e [I]t = 휆 휆 e − e 2 − 3 2 3 2 − 3 { } 휆 휆 휆 휆 3 − 2t 2 − 3t [P]t = [A]0 1 + 휆 휆 e − 휆 휆 e 2 − 3 2 − 3 ( ) 1∕2 where 휆 = 1 (p + q), 휆 = 1 (p − q), p = k + k + k , and q = p2 − 4k k 2 2 3 2 1 −1 2 1 2 Solution requires integration by partial fractions. P. J. Grandinetti (Chem. 4300) Chemical Kinetics 33 / 60 Series Reaction with Reversible Step Step between reactant, A, and an intermediate, I, is reversible:
k1 k A ↽←←←←←←←←←←←←←←←←←←←←←←←←⇀ I ←←←←←←←←←→2 P k−1
100 [A] [P] 80
60 [A],[I],[P] 40
20 [I] 0 0 200 400 600 800 1000 time/seconds
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 34 / 60 Series Reaction with Reversible Step – Steady State Approximation
k1 k A ↽←←←←←←←←←←←←←←←←←←←←←←←←←←⇀ I ←←←←←←←←←→2 P k−1 If intermediate is so reactive that it remains at a low concentration compared to A or P, then one can ≫ make the steady-state approximation. It requires k−1 + k2 k1.
d[I] k1[A] = k1[A] − k−1[I] − k2[I] ≈ 0 which gives [I]ss ≈ dt k−1 + k2 Substituting back into rate expressions for A and P gives k k [A] k k k k d[A] −1 1 1 2 , d[P] 1 2 − = k1[A] − k−1[I] ≈ k1[A] − = [A] and = k2[I] ≈ [A] dt k−1 + k2 k−1 + k2 dt k−1 + k2
d[P] If 1st step is slow, k ≪ k then ≈ k [A] −1 2 dt 1 k ≪ d[P] 1 ← If 2nd step is slow, k2 k−1 then ≈ k2[A] = K k2[A] pre-equilibrium approximation dt k−1 K is equilibrium constant, given by K = k1∕k−1
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 35 / 60 Lindemann Mechanism Pre-equilibrium approximation example Many gas phase reactions follow 1st order kinetics
kr [P] A(g) ←←←←←←←←→ P(g) with = k [A] dt r For A to get enough energy to react it needs to collide with other molecules. We expect mechanism like The Lindemann mechanism uses the steady-state k1 A(g) + M(g) ↽←←←←←←←←←←←←←←←←←←←←←←←←⇀ A∗(g) + M(g) approximation, i.e., treat A∗ like I in last case. k−1 ∗ k2 [A ] A∗(g) ←←←←←←←←←→ P(g) = k [A][M] − k [A∗][M] − k [A∗] ≈ 0 dt 1 1 2 A(g) ⟶ P(g) so ∗ k1[A][M] d[P] ∗ k1k2[A][M] [A ] ≈ and = k2[A ] ≈ k−1[M] + k2 dt k−1[M] + k2
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 36 / 60 Lindemann Mechanism k [M][A] [A∗] ≈ 1 k−1[M] + k2 isomerization reaction CH NC ←←←←←←←←←←←←←←←←←→ CH CN at 472.5 K d[P] ∗ k1k2[M][A] 3 3 = k2[A ] ≈ dt k−1[M] + k2
≫ at high pressures we have k−1[M] k2 and then d[P] k k ≈ 1 2 [A] dt k−1
reaction is 1st order in A with kobs = k1k2∕k−1 ≫ at low pressures we have k2 k−1[M] then d[P] ≈ k [M][A] dt 1
reaction is 1st order in A with kobs = k1[M]
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 37 / 60 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] k[H ][Br ]3∕2 = 2 2 ′ dt [Br2] + k [HBr] Let’s examine this one in more detail...
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 38 / 60 ←←←←←←←←←←←←←←←←←→ Chain Reaction Mechanism: H2(g) + Br2(g) 2 HBr(g)
Initiation: k ←←←←←←←←→i ⋅ , Br2 + M 2 Br + M rate = ki[Br2][M]
M is either Br2 or H2. Collisions with M give Br2 enough energy to break apart. Propagation: k ⋅ ←←←←←←←←←→p ⋅ , ⋅ Br + H2 HBr + H rate = kp[Br ][H2]
k′ ⋅ ←←←←←←←←←→p ⋅ , ′ ⋅ H + Br2 HBr + Br rate = kp[H ][Br2] Retardation: k ⋅ ←←←←←←←←←→r ⋅ , ⋅ H + HBr H2 + Br rate = kr[H ][HBr] Termination: k ⋅ ←←←←←←←←→t , ⋅ 2 2 Br + M Br2 + M rate = kr[Br ] [M] Next we assemble the rate law...
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 39 / 60 ←←←←←←←←←←←←←←←←←→ Chain Reaction Mechanism: H2(g) + Br2(g) 2 HBr(g) k ←←←←←←←←→i ⋅ Br2 + M 2 Br + M rate = ki[Br2][M] k ⋅ ←←←←←←←←←→p ⋅ ⋅ Br + H2 HBr + H rate = kp[Br ][H2] k′ ⋅ ←←←←←←←←←→p ⋅ ′ ⋅ H + Br2 HBr + Br rate = kp[H ][Br2] k ⋅ ←←←←←←←←←→r ⋅ ⋅ H + HBr H2 + Br rate = kr[H ][HBr] k ⋅ ←←←←←←←←→t ⋅ 2 2 Br + M Br2 + M rate = kr[Br ] [M]
d[HBr] = k [Br ⋅ ][H ] + k′ [H ⋅ ][Br ] − k [H ⋅ ][HBr] dt p 2 p 2 r Make the steady state approximation for intermediates H ⋅ and Br ⋅ d[H ⋅ ] = k [Br ⋅ ][H ] − k′ [H ⋅ ][Br ] − k [H ⋅ ][HBr] ≈ 0 dt p 2 p 2 r d[Br ⋅ ] = 2k [Br ][M] − k [Br ⋅ ][H ] + k′ [H ⋅ ][Br ] + k [H ⋅ ][HBr] − 2k [Br ⋅ ]2[M] ≈ 0 dt i 2 p 2 p 2 r t ⋅ ⋅ Two equations, two unknowns, [H ]ss and [Br ]ss ... P. J. Grandinetti (Chem. 4300) Chemical Kinetics 40 / 60 ←←←←←←←←←←←←←←←←←→ Chain Reaction Mechanism: H2(g) + Br2(g) 2 HBr(g)
d[H ⋅ ] = k [Br ⋅ ][H ] − k′ [H ⋅ ][Br ] − k [H ⋅ ][HBr] ≈ 0 dt p 2 p 2 r d[Br ⋅ ] = 2k [Br ][M] − k [Br ⋅ ][H ] + k′ [H ⋅ ][Br ] + k [H ⋅ ][HBr] − 2k [Br ⋅ ]2[M] ≈ 0 dt i 2 p 2 p 2 r t ⋅ ⋅ Two equations, two unknowns, [H ]ss and [Br ]ss. You can show that ( ) ( ) k 1∕2 1∕2 k i [H ][Br ]1∕2 k p k 2 2 ⋅ i 1∕2 and ⋅ t [Br ]ss = [Br2] [H ]ss = ′ kt kp[Br2] + kr[HBr] Note [M] cancelled out. Finally we get ( ) 1∕2 ki 3∕2 2k [H ][Br ] 3∕2 d[HBr] p k 2 2 d[HBr] k[H ][Br ] t matching observed rate law: 2 2 = ′ = ′ dt [Br2] + (kr∕kp)[HBr] dt [Br2] + k [HBr]
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 41 / 60 Svante August Arrhenius 1859–1927
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 42 / 60 Temperature Dependence of Rate Constants Experimentally, a plot of ln k versus 1∕T for many reactions are linear or very close to linear.
In 1889 Arrhenius proposed the 23,5 relationship 23
−Ea∕(RT) 4 k = Ae ]) 3,5 -1 s
-1 3 A is pre-exponential factor mol
3 2,5
cm Ea is activation 2 -4 energy—corresponds to energy 1,5 needed for reaction to occur.
ln(k [10 1 As T increases so does fraction of
0,5 molecules with translational energy greater than E . 0 a 0 0.0015 0.00155 0.0016 0.00165 0.0017 1/T [1/K] P. J. Grandinetti (Chem. 4300) Chemical Kinetics 43 / 60 Temperature Dependence of Rate Constants
Arrhenius relationship: k = Ae−Ea∕(RT) A is pre-exponential factor
Ea is activation energy—corresponds to 1000 K energy needed for reaction to occur. As T increases so does fraction of molecules with translational energy greater than Ea. Energy Not all reactions follow Arrhenius behavior. It assumes Ea and A are temperature independent.
300 K More general expression gives Ea as slope Ea at specific temperature: reactants 100 K ( ) products 2 d ln k 0 Ea = RT 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 reaction coordinate dT p(E) P. J. Grandinetti (Chem. 4300) Chemical Kinetics 44 / 60 Transition State Theory and the Eyring Equation
k T ◦‡ ◦‡ k = B eΔS ∕Re−ΔH ∕(RT) c◦h
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 45 / 60 Potential Energy Surfaces Consider 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)?
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 46 / 60 Potential Energy Surfaces What 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 A-B + C What is reaction path on 2D energy surface? Could follow path of minimum potential energy from AB + C to BC + A.
(A-B-C)‡
r(BC)
Ea A-B + C B-C + A Potential Energy
re(BC) A + B-C reaction coordinate
0 (ABC)‡ is the transition state at top of saddle 0 r(AB) point in 2D contour plot. re(AB) P. J. Grandinetti (Chem. 4300) Chemical Kinetics 47 / 60 Transition State Theory aka Activated Complex Theory Assume reactants are in equilibrium with an activated unstable intermediate X‡.
k1 휈‡ A + B ↽←←←←←←←←←←←←←←←←←←←←←←←←←←⇀ X‡ ←←←←←←←←←←→ P k−1 k1 Assume A + B ↽←←←←←←←←←←←←←←←←←←←←←←←←←←⇀ X‡ is described by equilibrium expression: k−1 k ‡ c◦ ‡ 1 [X ]∕ , ‡ ◦ Kc = = ◦ ◦ where Kc is equilibrium constant, and c = 1M k−1 ([A]∕c )([B]∕c ) 휈‡ 휈‡ is frequency that activated complex crosses over barrier. Statistical mechanics says = kBT∕h. Overall rate is d[P] K‡ k T K‡ = 휈‡[X‡] = 휈‡ c [A][B] = B c [A][B] = k[A][B] dt c◦ h c◦ ⏟⏟⏟ k
‡ ‡ Kc cannot be measured (can’t observe X ) but can be inferred from k (assuming theory is correct).
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 48 / 60 Transition State Theory aka Activated Complex Theory
‡ k1 휈‡ k T K ←←←←←←←←←←←←←⇀ ‡ ←←←←←←←←←←→ , d[P] 휈‡ ‡ B c A + B ↽←←←←←←←←←←←←← X P where = [X ] = k[A][B] and k = ◦ k−1 dt h c
‡ From Kc we can examine kinetics in terms of thermodynamics. ◦‡ ‡ ‡ −ΔG◦‡∕(RT) ΔG = −RT ln Kc or Kc = e Since ΔG◦‡ = ΔH◦‡ − TΔS◦‡ we have ‡ −ΔH◦‡∕(RT) ΔS◦‡∕R Kc = e e
k1 Define ΔH◦‡ and ΔS◦‡ for A + B ↽←←←←←←←←←←←←←←←←←←←←←←←←←←⇀ X‡ k−1 Finally, we obtain the Eyring Equation:
k T ◦‡ ◦‡ k = B eΔS ∕Re−ΔH ∕(RT) c◦h
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 49 / 60 Transition State Theory Eyring Equation
k T ◦‡ ◦‡ k = B eΔS ∕Re−ΔH ∕(RT) c◦h ◦‡ ◦‡ 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 ( ) d ln k E = RT2 a dT Start with ( ) k T k ln k = ln B K‡ = ln B + ln T + ln K‡ c◦h c c◦h c Next calculate ‡ ‡ d ln k d ln T d ln K 1 d ln K = + c = + c dT dT dT T dT Then obtain ( ) ( ) ‡ ‡ d ln k 1 d ln K d ln K E = RT2 = RT2 + c = RT + RT2 c a dT T dT dT
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 50 / 60 Thermodynamics Review
Homework (A) For ideal gas phase reaction a A + b B ↽←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←⇀ d D + f F ( )( ) ◦ ◦ ◦ ◦ pD∕p pF∕p ([D]∕c )([F]∕c ) K ( )( ) and K where p◦ 1 bar, c◦ 1 M p = ◦ ◦ c = ◦ ◦ = = pA∕p pB∕p ([A]∕c )([B]∕c ) ( ) Δn c◦RT show that K = K where Δn = d + f − a − b c p p◦
◦ ◦ d ln Kp ΔH d ln K ΔU (B) From van’t Hoff Equation: = prove that c = dT RT2 dT RT2
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 51 / 60 Transition State Theory Back to Eyring Equation
d ln K‡ E = RT + RT2 c a dT For reaction of ideal gas molecules we know ◦ d ln K ΔU d ln K ◦ c = or RT2 c = ΔU dT RT2 dT With this relation we obtain ◦‡ Ea = 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
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 52 / 60 Transition State Theory ( ) ◦‡ ‡ Ea = ΔH + 1 − Δn RT
→ ‡ ‡ ‡ ‡ For unimolecular reaction: A X we have nf = 1, ni = 1, so Δn = 0,
◦ kBT ◦‡ ◦‡ kBT ◦‡ ΔH ‡ = E − RT and k = eΔS ∕Re−ΔH ∕(RT) = eΔS ∕Re−(Ea−RT)∕(RT) a c◦h c◦h [ ] kBT ◦‡ k = e eΔS ∕R e−Ea∕(RT) unimolecular gas phase reaction c◦h
→ ‡ ‡ ‡ ‡ For bimolecular reaction: A + B X we have nf = 1, ni = 2, so Δn = −1, and similarly obtain [ ] kBT ◦‡ k = e2 eΔS ∕R e−Ea∕(RT) bimolecular gas phase reaction c◦h
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 53 / 60 Transition State Theory
Problem The 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◦‡ and ΔS◦‡ 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 { } ◦ 11 ◦ hAc h(1.09 × 10 L/(mol⋅s))(1mol/L) ΔS ‡ = R ln = R ln = −60.3 J/(mol⋅K) 2 2 e kBT e kB(1000K)
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 54 / 60 Reactions in Solution
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 55 / 60 Reactions in Solution In solution reactants are surround by solvent molecules and need to “encounter” each other before they can react.
A “cage” of solvent molecules trap reactants together, allowing them to collide many times 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 to diffuse together. This is slower than free motion of gas molecules.
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 56 / 60 Reactions in Solution Consider the following mechanism: kd A + B ←←←←←←←←←→ AB ⋅ Diffusion of A and B into solvent cage and formation of activated complex. kr AB ←←←←←←←←←→ A + B ⋅ Dissociation back to reactants. kp ⋅ AB ←←←←←←←←←→ P Formation of products. dP Reaction rate is = k [AB]. With [AB] as intermediate make steady state approximation dt p k [A][B] d[AB] , d = kd[A][B] − kr[AB] − kp[AB] ≈ 0 gives [AB]ss ≈ dt kr + kp
Using [AB]ss in the rate expression gives
dP kpkd = kp[AB] ≈ [A][B] dt kr + kp ≫ If product formation is faster than dissociation back to reactants, i.e., kp kr, then we obtain a diffusion controlled reaction rate: dP = k [A][B] dt d
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 57 / 60 Reactions in Solution diffusion controlled reaction rate dP = k [A][B] dt d
Rate constants for diffusion can be related to diffusion coefficients ( ) 휋 kd = 4 NA rA + rB (DA + DB)
rA and rB are molecular radii, and DA and DB are diffusion coefficients. Relationship between D and viscosity of solution, 휂, is k T D = B Stokes-Einstein equation 6휋휂r Note: mass ∝ r3, so D ∝ m−1∕3
If we assume rA = rB = r and DA = DB = D then 8 RT k = d 3 휂
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 58 / 60 Reactions in Solution
Problem Water viscosity is 휂 = 8.9 × 10−4 N⋅s/m2. What is rate constant in water at room temperature?
8 RT 8 R(298K) . 9 ⋅ kd = = = 7 4 × 10 L/(mol s) 3 휂 3 8.9 × 10−4N⋅s/m2
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 59 / 60 Homework from McQuarrie
Chapter 28: 1, 2*, 4, 6*, 8, 11, 12, 15*, 17*, 18, 22, 23, 24*, 31, 34 Chapter 29: 1, 3*, 7*, 8, 12, 16, 21, 24 Chapter 28: 41, 42*, 43*,46, 49*
P. J. Grandinetti (Chem. 4300) Chemical Kinetics 60 / 60