Chemical Kinetics Reaction Rates: Reaction Rate: The change in the concentration of a reactant or a product with time (M/s). Reactant → Products A → B
change in number of moles of B Average rate = change in time
∆()moles of B ∆[B] = = ∆t ∆t
∆[A] Since reactants go away with time: Rate=− ∆t
1 Consider the decomposition of N2O5 to give NO2 and O2:
2N2O5(g)→ 4NO2(g) + O2(g)
reactants products decrease with increase with time time
2 From the graph looking at t = 300 to 400 s
0.0009M −61− Rate O2 ==× 9 10 Ms Why do they differ? 100s 0.0037M Rate NO ==× 3.7 10−51 Ms− Recall: 2 100s
0.0019M −51− 2N O (g)→ 4NO (g) + O (g) Rate N O ==× 1.9 10 Ms 2 5 2 2 25 100s
To compare the rates one must account for the stoichiometry.
1 Rate O =×× 9 10−−61 Ms =× 9 10 −− 61 Ms 2 1
1 −51−−− 61 Rate NO2 =×× 3.7 10 Ms =× 9.2 10 Ms Now they 4 1 agree! Rate N O =×× 1.9 10−51 Ms−−− = 9.5 × 10 61Ms 25 2
Reaction Rate and Stoichiometry
In general for the reaction:
aA + bB → cC + dD
11∆∆∆∆[AB] [ ] 11[CD] [ ] Rate ====− − ab∆∆∆ttcdtt∆
3 Rate Law & Reaction Order The reaction rate law expression relates the rate of a reaction to the concentrations of the reactants. Each concentration is expressed with an order (exponent). The rate constant converts the concentration expression into the correct units of rate (Ms−1). (It also has deeper significance, which will be discussed later) For the general reaction: aA+ bB → cC+ dD
x and y are the reactant orders determined from experiment. Rate = k [A]xy [B] x and y are NOT the stoichiometric coefficients.
4 The Overall Order of a reaction is the sum of the individual orders: Rate (Ms−1) = k[A][B]1/2[C]2
Overall order: 1 + ½ + 2 = 3.5 = 7/2
or seven−halves order
note: when the order of a reaction is 1 (first order) no exponent is written.
Units for the rate constant:
The units of a rate constant will change depending upon the overall order.
The units of rate are always M/s or Ms–1
To find the units of a rate constant for a particular rate law, simply divide the units of rate by the units of molarity in the concentration term of the rate law.
Rate (Ms–1) = k[A] 1st order
Ms−1 k(units)= = s−1 M
5 Reaction Rates
Instantaneous rate (tangent line)
∆Y∆ [concentration] Average Rate = ∆∆Xtime concentration (M)
time (s)
Rules of logarithms
log (1) = 0 ln (1) = 0 log (10) = 1 ln (e) = 1 log (100) = 2 ln (ex) = x Log (10x) = x
log Ax = xlog A ln Ax = xln A
x AAx A logx == log x log BB B
A log(AB) = log A + log B log=− log A log B B
6 Determining Reaction Order: The Method of Initial Rates The reaction of nitric oxide with hydrogen at 1280°C is:
2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
From the following data, determine the rate law and rate constant.
Initial Rate Run [NO] (M) [H ] (M) o 2 o (M/min) 1 0.0100 0.0100 0.00600
2 0.0200 0.0300 0.144
3 0.0100 0.0200 0.0120
2NO(g) + 2H2(g) → N2(g) + 2H2O(g) The rate law for the reaction is given by:
x y Rate(M/min) = k [NO] [H2] Taking the ratio of the rates of runs 3 and 1 one finds:
xy Rate(3) k [NO](3) [H 2 ] (3) = xy Rate(1) k [NO](1) [H 2 ] (1)
0.0120M / min k [0.0100]xy [0.0200] [0.0200]y = 2 = = 0.00600M / min k [0.0100]xy [0.0100] [0.0100]y
y [0.0200]y 0.0200 y = [0.0100] 0.0100
7 xy y Rate(1) k [NO](1) [H 2 ] (1) 0.0200 = log = 2 Rate(2) k [NO]xy [H ] 0.0100 (2) 2 (2) x y 0.00600 k [0.0100] [0.0100] 0.0200 = x log= log 2 0.144 k [0.0200] [0.0300] 0.0100 x 1 11 = × ylog( 2) = log 2 24 23
x 11 y = 1 = 28 Now that “y” is known, we 11 xlog= log may solve for x in a similar 28 manner: x = 3
The Rate Law is:
3 Rate(M/min) = k [NO] [H2]
To find the rate constant, choose one set of data and solve:
M 3 0.0120= k()() 0.0100M 0.0200M min
M M 0.0120 0.0120 min min k = = 3 ()()0.0100M3 0.0200M ()()0.0100 0.0200 M4
M−3 k6.0010=×5 min
8 Integrated Rate Laws: time dependence of concentration For a first order process, the A → products rate law can be written:
∆[A] Rate(Ms−1 )=− = k[A] This is the “average rate” ∆t
If one considers the infinitesimal changes in concentration and time the rate law equation becomes: d[A] Rate(Ms−1 )=− = k[A] This is the “instantaneous rate” dt Integrating in terms of d[A] and dt:
[A]d[A] t where [A] = [A] at time t = 0 =−kdt o ∫∫[A]o [A] 0 and [A] = [A] at time t = t
[A]d[A] t [A] =−kdt ln= − kt ∫∫[A]o [A] 0 [A]o
Taking the exponent to each side of the equation:
[A] = e−kt or −kt [A]= [A]o e [A]o
Conclusion: The concentration of a reactant governed by first order kinetics falls off from an initial concentration exponentially with time.
9 Recognizing a first order process: AÆproducts
Whenever the conc. of a reactant falls off exponentially, the kinetics follow first order.
−kt [A]= [A]o e
Determining the Rate constant for a first order process
Taking the log of the integrated rate law for a first order process we find:
−kt ln() [A]= [A]o e ln[A]= ln[A]o −× k t
A plot of ln[A] versus time (t) is a straight line with slope -k and intercept ln[A]o
10 Example: The rate of decomposition of azomethane (C2H6N2) was studied by monitoring the partial pressure of the reactant as a function of time. Determine if the data below support a first order reaction. Calculate the rate constant for the reaction.
Time P ln (P) (s) (mmHg Plot of lnP vs. time 0 284 5.65 5.70 100 220 5.39 5.60 y = -0.0026x + 5.6485 150 193 5.26 5.50 R2 = 1 200 170 5.14 5.40 250 150 5.01 5.30 ln P 5.20 5.10 5.00 -3 -1 k = 2.6x10 s 4.90 0 50 100 150 200 250 300 time (s)
Reaction Half-Life
11 Reaction Half-life: Half-life is the time taken for the concentration of a reactant to drop to half its original value. [A] [A] = o 2
For a first order process the half life (t½ ) is found mathematically from: (1) ln A=− kt + ln A [ ] [ ]0 []A (5) ln0 = kt (2) ln[ A] −=− ln[ A] kt [A] 1 0 0 2 2 [A] (3) ln=− kt [A] 0 ln 2 0.693 t == [A] 1 kk (4) ln0 = kt 2 [A]
A certain reaction proceeds through t first order kinetics. The half-life of the reaction is 180 s. What percent of the initial concentration remains after 900s? Step 1: Determine the magnitude of the rate constant, k. ln 2 ln 2 ln 2 0.693 k== = 0.00385s−1 t 1 == t1 180s 2 kk 2 Using the integrated rate law, substituting in the value of k and 900s we find:
[A] −kt [A] −×0.00385 s−1 900 s = e = e = 0.0312 [A]o [A]o
Since the ratio of [A] to [A]0 represents the fraction of [A]o that remains, the % is given by: 100 × 0.0312 = 3.12%
12 For a Second Order Process: A → Products
Rate = k[A]2
d[A] Rate(Ms−12 )=− = k[A] dt Integrating as before we find: 11 =+kt []AA[]0
A plot of 1/[A] versus t is a straight line with slope k and intercept 1/[A]0 For a second order reaction, a plot of ln[A] vs. t is not linear.
11 =+kt []AA[]0
Non−linearity indicates that the reaction is not first order. Slope = k (rate constant)
13 Half-life for a second-order reaction Unlike a first order reaction, the rate constant for a second order process depends on and the initial concentration of a reactant.
11 =+kt AA []t0[]
1 at the half–life, []AA=×[] t02
1 Substituting and solving, t1/2 = kA []0
EXAMPLE: The reaction
2 NOBr (g) → 2 NO (g) + Br2 (g) is a second order reaction with respect to NOBr. k = 0.810 M-1⋅s-1 at 10o C.
-3 If [NOBr]o = 7.5 × 10 M, how much NOBr will be left after a reaction time of 10 minutes?
Determine the half-life of this reaction.
14 SOLUTION: One can solve for the amount of NOBr after 10 minutes by substituting the given data into the integrated rate law for a second-order reaction.
11 =+kt NOBr NOBr (Second Order) []t0[]
1 1 = (0.810 M-1⋅ s -1 )× (600 s) + NOBr −3 []t 7.5× 10 M
1 =×6.19 102-1 M NOBr []t
NOBr= 1.6× 10−3 M [ ]t
To determine the half-life for this reaction, we substitute the initial concentration of NOBr and the rate constant for the reaction into the equation for the half-life of a second-order reaction.
1 t = 1/2 kA []0
1 t = =160 s 1/2 0.810 M-1⋅× s−− 1 (7.5 10 3 M)
15 The Arrhenius Equation: Temperature dependence of the Rate Constant
Most reactions speed up as temperature increases. (example: food spoils when not refrigerated.)
kf(T)=
Where “f” is some function.
The magnitude of a first order rate constant is seen to increase exponentially with an increase in temperature.
Therefore one can conclude that:
k(T)∝ ef(T)
16 Why would k (along with the rate) increase with temperature? Let’s go back to Kinetic Molecular Theory to understand…
The Collision Model: In order for molecules to react they must collide. As temperature increases, the molecules move faster and the collision frequency increases. Therefore, the greater the number of collisions the faster the rate. Thus reaction rate should increase with an increase in temperature. Also, the more molecules present, the greater the probability of collision and the faster the rate. Thus reaction rate should increase with an increase in the concentration of reactant molecules.
Activation Energy
Arrhenius: Molecules must posses a minimum amount of energy to react. Why?
(1) In order to form products, bonds must be broken in the reactants. (2) Bond breakage requires energy. (3) Molecules moving too slowly, with too little kinetic energy, don’t react when they collide.
The Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.
Ea is specific to a particular reaction.
17 Consider the reaction:
N2O(g) + O2(g) → N2(g) + NO2(g)
The progress of a reaction can be described by a Reaction Coordinate
Reactants
Products
Recall for KMT that the temperature for a system of particles is described by a distribution.
E < Ea E > Ea
Since the probability of a molecule reacting increases, the rate increases.
At higher temps, more particles have enough energy to go over the barrier.
18 Orientation factors into the equation
The orientation of a molecule during collision can have a profound effect on whether or not a reaction occurs.
Some collisions do not lead to reaction even if there is sufficient energy.
The Arrhenius Equation
Arhenius discovered that most reaction-rate data obeyed an equation based on three factors:
(1) The number of collisions per unit time.
(2) The fraction of collisions that occur with the correct orientation.
(3) The fraction of the colliding molecules that have an energy
greater than or equal to Ea.
From these observations Arrhenius developed the aptly named Arrhenius equation.
19 Arrhenius equation
−Ea kAe= RT
k is the rate constant T is the temperature in K
E is the activation energy R is the ideal-gas a constant (8.314 J/Kmol)
A is known the frequency or pre–exponential factor
In addition to carrying the units of the rate constant, “A” relates to the frequency of collisions and the orientation of a favorable collision probability
Both A and Ea are specific to a given reaction.
Determining the Activation Energy
Ea may be determined experimentally. First take natural log of both sides of the Arrhenius equation:
−E a Ea RT ln k=− + ln A ln kAe= RT Golly–gee, what do we have once again… ln k A plot of ln k vs 1/T will
have a slope of –Ea/R and a y-intercept of ln A. 1/T
20 Determining the Activation Energy
One can determine the activation energy of a reaction by measuring the rate constant at two temperatures: Writing the Arrhenius equation for each temperature:
Ea Ea ln k1 =− + ln A ln k2 =− + ln A RT1 RT2
If one takes the natural log of the ratio of k2 over k1 we find that:
k2 ln = ln k21− ln k k1
Substituting in the values for Ea into the equation:
Ea Ea ln k21−= ln k −+ln A − −+ln A RT2 RT1
Lookie what happens… presto!
Knowing the rate at two temps yields the rate constant. k11E ln 2 =−a or kRTT112
Knowing the Ea and the rate constant at one temp allows one to find k(T2)
21 Example: The activation energy of a first order reaction is 50.2 kJ/mol at 25oC. At what temperature will the rate constant double?
k 2k E 11 (1) k2k= (2) ln 2 =ln 1 = ln(2) =−a 21 k k 1 1 RT12 T
103 J E 50.2 kJ/mol × (3) a = 1kJ = 6.04× 103 K J R 8.314 mol⋅ K o 11 A 10 C change of (4) ln(2)= 0.693 =×6.04 103 K ×− temperature doubles 298K T2 the rate!!
1 −−31 (5) =×3.24 10 K T = 308 K T2 2
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