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Ch 13: Chemical Kinetics
Objectives

Rates
 meaning, determination, factors influencing
 Reaction mechanisms, i.e.- pathways of bond changing
 Role of catalyst
13.1 Reaction Rates
 how rapidly do chemical reactions occur, i.e.- what is the speed, or velocity, or rate, of
reaction?
 expressed as change in concentration of a reactant or product in a certain amount of time
 conventional unit: M/s
(recall: M = mol/L)
 example: gas-phase reaction
NO2 (g)  CO(g)  NO(g) + CO2 (g)
 certain initial conditions, measure one (say, [NO]), then calc. other conc’ns by
stoichiometry at various times; see Figure 13.3
 calculate average rate for any time interval :
increase in [NO]
time interval
[NO]
=t
Average rate =
 “delta” symbol for “change in”
 i.e.- (final - initial) conc’n or time
 note: positive sign above since NO is being produced
 if [NO2] measured, minus sign above since NO2 is being consumed, disappearing
 [ NO2] would be negative, hence rate is positive (rates always so)
 note that average rate continuously decreases as time progresses
 also in Fig. 13.3: determine instantaneous rates from tangents to curve at various time points
(done in text for [NO] = 0.0288 M)
 same trend in decreasing rates with time
Chem 110 (’02)
2
 average rate becomes same as instantaneous as time interval approaches zero
 in calculus notation, rate = d[NO]/dt
 at t = 0, instantaneous rate is called initial rate
Reaction Rates & Stoichiometry
 example: decomposition of N2O5 in CCl4
1
O 2 (g)
2
 rate of disappearance of N2O5 = half of rate of appearance of NO2 = twice rate of
appearance of O2, i.e.- related by stoichiometry in reaction equation
N 2O 5 (soln)  2 NO 2 (soln) +
[N 2O5 ]
1 [NO 2 ]
[O 2 ]
=
= 2
t
2
t
t
d [N 2O5 ]
1 d [NO 2 ]
d [O 2 ]
= =
= 2
dt
2
dt
dt
Rate = -
 stoichiometry can be 1:1, eg.:
C 4 H 9 Cl(l) + H 2 O  C 4 H 9 OH(aq) + HCl(aq)
[C4 H 9 Cl]
[C4 H 9OH]
Rate = =
t
t
d [C4 H 9 Cl]
d [C4 H 9OH]
= =
dt
dt
 in general:
aA + bB  cC + dD
Rate = -
1 [A]
1 [B]
1 [C]
1 [D]
= =
=
a t
b t
c t
d t
Reaction Conditions and Rate
 what factors control the rate?




concentrations of the reactants - more collisions if higher conc’ns
temperature - rates increase with temp; collisions more frequent and higher energy
presence of a catalyst - permit collisions at lower energy and with correct geometry
surface area of solid or liquid reactants or catalysts - more chance for collision with other
reactant(s)
13.2 Rate Laws
Chem 110 (’02), ch. 13, Chemical Kinetics
3
 easiest to measure initial rate at varying [reactant] values
 eg. reaction:
1
O 2 (g)
2
 rate when [N2O5] = 1.0 M is 3-times rate when [N2O5] = 0.34 M
 initial rate proportional to [N2O5]; overall:
N 2O 5 (soln)  2 NO 2 (soln) +
Rate = k [N 2 O 5 ]
 this expression is the rate law = rate expression = rate equation
 constant, k, is the rate constant (= specific rate, i.e.- when [N2O5] = 1.0 M)
Order of a Reaction
 rate laws in general:
C
aA + bB  xX
(where C is a catalyst)
Rate = k [A]m [B]n [C]p
or, Rate = k [reactant 1]m [reactant 2]n ...[catalyst]p
where m  a, necessarily, etc.
 exponents m and n, etc. are called reaction orders; sum is overall reaction order
 example: Rate = k [NO]2 [Cl2 ]
 said to be second order in NO and first order in Cl2 and to be third order overall
 note: reaction orders are not necessarily stoichiometry numbers; orders come from
experimentally determined rate law
 orders commonly 0, 1 or 2 (sometimes fractional)
Units of Rate Constants
 depend on overall reaction order
 eg., for second order:
Units of rate = (Units of rate constant)(Conc’n)2
Units of rate constant = (Units of rate)/(Conc’n)2
= (M/s)/M2 = M-1 s-1
Chem 110 (’02), ch. 13, Chemical Kinetics
4
 in general:
units of rate const =
=
units of rate
concentration overall order
M/s
M overall order
Determine Rate Law from Initial Rates
 usually looking for zero-, first- or second-order dependence of each reactant
 NB: do not confuse rate (reactant concentration dependent) with rate constant (concentration
independent)
 practice with class Example (then do Examples 13.1, .2 and .3)
 reaction: 2 NO(g) + O 2 (g)  2 NO 2 (g)
 Rate equation and Rate constant?
 data Table, 3 experiments:
 triple [O2], initial rate triples, i.e.- proportional to [O2]
 double [NO], initial rate quadruples, i.e.- proportional to [NO]2
 hence, Rate equation: Rate = k [NO]2 [O2]
 could calculate k from any of the experiments in Table, using above equation
 eg., expt 1:
2.8 x 10-6 M/s = k (1 x 10-4 M)2 (1 x 10-4 M)
k = 2.8 x 106 M-2s-1
 and with k, can calc. rate at any other concns; eg. 0.100 M each:
Rate = (2.8 x 106M-2s-1)(0.100M)2(0.100M)
= 2.8 x 103 M s-1
Integrated Rate Laws
 change of concentration with time
 convert rate law to a form that can be used to calculate concentrations of reactant(s) and
products during course of reaction
 done by integrating (i.e.- calculus) rate equation
 especially for first-order reaction, one special second-order
First-Order Reactions
 for reaction:
Chem 110 (’02), ch. 13, Chemical Kinetics
5
A  B
eg. CH 3NC:  CH 3CN:
or, cyclopropane  propene
[CH 3NC]
d [CH 3NC]
= = k[CH 3NC]
t
dt
 this derivative corresponds to the change over a small “slice” of time
 written, in general, rearranged, then integrated:
Rate = -
Rate = -
d [A]
= kd t
[A]
[A] t
t
d [A]
= k d t

[A]
[A]0
0
 this gives: ln [A] = - kt + c
 where c = constant of integration
 set t = 0, then [A] = [A]o and c = ln [A]o
 hence:
ln [A] - ln [A]0 = - kt
(an exponential function: [A]t = [A]o e-kt)
[A]t
= - kt
[A]0
 rearrange: ln [A]t = - kt + ln [A]0
 or:
ln
 recall: ln x = 2.30 log x
Half-Life and First-Order Reactions
 [A]t = ½ [A]o at t = t1/2
1
[A]0
ln 2
= - kt1/ 2
[A]0
1
ln
= - kt1/ 2
2
1
ln
2 = 0.693
t1/ 2 = k
k
 time for concentration of reactant to drop by one-half is independent of starting point, Fig.
13.6
 Example 13.4
Chem 110 (’02), ch. 13, Chemical Kinetics
6
Second-Order Reactions
 one special case:
Rate = k [A]2
 from rate law, integration gives:
1
[A]t
= kt +
1
[A]0
(plot: Fig. 13.7)
 calculations in Example 13.5
 note: for second-order reaction, half-life not independent of conc’n
1
1
1
= kt +
;
t1/ 2 =
[A]t
[A]0
k[A]0
Zero-Order reactions
 Rate = k [A]o ; integrated: [A]o - [A]t = kt
Graphical Methods for Determining Reaction Order and Rate Constant
 Fig. 13.8 summarizes for 1st- and 2nd-order reactions (2 C2F4  C4F8)
13.3 Reaction Mechanisms
 balanced equation gives the starting point and the finishing point of a reaction, ratios of
reactant(s) and product(s)
 gives no information on how the process of bond rearrangement occurs, i.e.- pathway taken
 process is the reaction mechanism, derived from experimental kinetic data
Elementary Steps
 some reactions occur in a single event - elementary step
 eg. methyl isonitrile rearrangement, isonitrile must collide with another molecule to get
enough energy for rearrangement:
CH 3 NC  CH 3CN
 in the following eg., two different molecules must collide with enough energy and with the
correct geometry:
NO(g) + O 3 (g)  NO 2 (g) + O 2 (g)
 both of these reactions take place as the result of a single event
 number of molecules participating in an elementary step defines molecularity of step;
unimolecular, bimolecular, etc.
 on the other hand, a balanced chemical equation may be the result of a multi-step mechanism
 eg., the reaction:
Chem 110 (’02), ch. 13, Chemical Kinetics
7
Br2 (g) + 2 NO(g)
 2 BrNO(g)
 has 2 elementary steps, both bimolecular:
Br2 (g) + NO(g)  Br2 NO(g)
Br2 NO(g) + NO(g)  2 BrNO(g)
 elementary steps in a multi-step mechanism must always add together to give the overall,
balanced equation (verify)
 note: component Br2NO not in overall, termed an intermediate
Rate Laws of Elementary Processes
 rate law determined by the mechanism, not the balanced chemical equation, unless
mechanism consists of a single elementary step
 rate law of elementary step based on molecularity:
A  products ; Rate = k [A]
A + A  products ; Rate = k [A]2 ; etc.
 molecularity and order are the same for an elementary step
 NB: cannot tell from balanced chemical equation whether one or several elementary steps in
mechanism; experimentally determined
 do Example 13.6
13.4 Reaction Mechanisms and Rate Equations
 mechanism = sequence of elementary steps
 one slower? = rate-determining step
 toll-road analogy (or, bridge vs. Customs)
 slowest step in a multi-step chemical reaction determines overall rate; eg.:
NO 2 (g) + CO(g)  NO(g) + CO 2 (g)
Rate law (from expt.): Rate = k [NO 2 ]2
 propose mechanism consistent with this:
k1
NO 2 (g) + NO 2 (g) 
 NO3 (g) + NO(g) (slow)
k2
NO3 (g) + CO(g) 
 CO 2 (g) + NO 2 (g) (fast)
Chem 110 (’02), ch. 13, Chemical Kinetics
8
i.e.- k1 << k2
 Example 13.7
Mechanisms with an Initial Fast Step
 an intermediate involved in rate-determining step; eg.
2 NO(g) + Br2  2 NOBr(g)
Rate = k [NO]2 [Br2 ]
(expt'l)
 consistent mechanism?
 termolecular? unlikely, very rare
 2-step?
NO(g) + Br2 ( g)
k1

NOBr2 (g)

k -1
(fast)
k2
NOBr2 (g) + NO(g) 
2 NOBr (g)
 rate law governed by slow step:
(slow)
Rate = k 2 [NOBr2 ][NO]
 but [NOBr2] not known, probably low, not measurable; it has two fates:
 k2 step, which is slow
 k-1 step, assume fast; step 1 an equilibrium:
k1 [NO][Br2 ] = k -1[NOBr2 ]
forward rate
 [NOBr2 ] =
reverse rate
k1
[NO][Br2 ]
k -1
 substitute into rate law for rate-det. step:
Rate = k 2
k1
[NO][Br2 ][NO] = k [NO]2 [Br2 ]
k -1
i.e.- expt’l rate const., k = k2(k1/k-1)
 foregoing more probable, since only unimolecular and bimolecular processes
 generally, when a fast step precedes a slow one, solve for [intermediate] by assuming fast step
at equilibrium (pre-equilibrium assumption)
 (alternatively and more generally, can make steady-state assumption/approximation for
[intermediate])
Chem 110 (’02), ch. 13, Chemical Kinetics
9
 “soon” after reaction started the rates of production and consumption of an intermediate
become equal
 read for interest
13.5 Effect of Temperature on Reaction Rate
Arrhenius Equation
 most reaction rates increase with temperature, an effect on rate constant
 observed that most rate constant data obeyed the following equation:
k = A e- E a / RT





k - rate constant
Ea - activation energy
R - gas constant (8.314 J/mol-K)
T - temp on Kelvin scale
A - frequency factor (temp independent), related to the frequency of collisions and
probability of favorable orientations

exponential factor, e-Ea/RT, is the fraction (< 1) of molecules having the minimum
kinetic energy for reaction
 note: as Ea increases, k decreases
 deal with “e” factor by “linearizing”, take ln
Ea
+ ln A
RT
 plot ln k vs. 1/ T, slope = - Ea/ R, calc. Ea from graphical analysis, Fig. 13.9
 alternatively, determine rate constant at two temperatures and calculate Ea:
ln k = -
ln
k1
k2
=
Ea  1
1


R  T2
T1 
 if Ea is known, can use Arrhenius equation to calculate k at a different temp.
 Example 13.8
Collision Model
 microscopic view of reactions
 molecules must collide to react
 at higher temperature more collisions and they are more energetic, Fig. 13.10
 even so, only a small fraction of collisions lead to products (eg. 1 in 1013 !)
 collision frequency also enhanced by higher concentration
Activation Energy
 colliding molecules must possess some minimum kinetic energy to be converted to product(s)
- a barrier to be surmounted (see Fig. 13.11)
Chem 110 (’02), ch. 13, Chemical Kinetics
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 minimum energy termed activation energy, Ea , for a specific reaction
 state at the height of the barrier, where enough energy has been gained to “get” to product(s),
is activated complex or transition state
 in terms of present butene isomerization: “twisted” structure
 knowing Ea does not tell us anything about the structure of the transition state
 reactions can occur without all molecules having kinetic energies greater than minimum
through collisions
 recall: distribution of kinetic energies; see Fig. 13.10
 if reaction involves two reactants, must also collide with correct orientation (or, steric factor)
 read section 13. 6 for interest
13.7 Kinetics of Catalysis
 catalyst increases the rate of a chemical reaction without itself undergoing permanent change
 egs. virtually all reactions in living cells, many in the environment, many industrial processes
Homogeneous Catalysis
 i.e.- in same phase as the reactants
 eg. hydrogen peroxide decomposition:
2 H 2 O 2 (aq)  2 H 2 O(l) + O 2 (g)
 very slow, but can be catalyzed by bromine:
Br2 (aq) + H 2O 2 (aq)  2Br - (aq) + 2H + (aq) + O 2 (g)
2Br - (aq) + H 2O 2 (aq) + 2H + (aq)  Br2 (aq) + 2H 2O(l)
 sum of these two equations is the overall; no net change in Br2 (note: could also use Br - )
 Arrhenius equation: k = A e- Ea / RT
 increase k at constant T?
 by decreasing Ea or increasing A
 catalyst may do both, but especially Ea, see Figs. 13.17
Heterogeneous Catalysis
 catalyst exists in a phase different than the reactants
 eg. a gas phase or solution reaction might be catalyzed by a finely-divided solid (surface
area important)
 first step is binding of reactant(s) to surface of catalyst by adsorption, Fig. 13.15
 eg. oxidation of ammonia for nitric acid production
Chem 110 (’02), ch. 13, Chemical Kinetics
11
 eg. catalytic converter for auto exhaust Fig. 13.16 for:
CO, hydrocarbons, NO x  CO 2 , N 2
Enzyme Catalysis
 biological catalysts (homogeneous?)
 eg. for H2O2 decomposition - enzyme catalase in all cells
 enzyme active site accepts substrate by a lock-and-key complementarity (Fig. 13.18):
 binding due to set of weak interactions:
electrostatic, H-bonding, dipole - dipole, London dispersion
 substrate distorted upon binding (stretched on the “rack”) in the direction of the transition
state
 from “active site” concept, see how frequency factor, A, could be influenced as well as
rate (for all catalysts)
 kinetics: Michaelis-Menten model:
E + S  E.S  E + P


accounts for the phenomenon of saturation (Fig. 13.19)
initial rate (from steady-state approximation):
vi 
d[P]

dt

k 2 [ E] o [S]
k  k2
[S]  -1
k1
Vmax [S]
K m  [S]
Suggested Problems
1 – 11; 19 – 21; 25 – 29; 35, 37, 45, all odd
Chem 110 (’02), ch. 13, Chemical Kinetics