vs. Thermodynamics Kinetics
thermodynamics show that the reaction wants to proceed Kinetics describes how fast a reaction occurs and why
Reaction Rates and Mechanisms : The Central Focus of Chemical Kinetics
it is all about time
Fig. 16.1 from Silberberg
hydrogen balloon demo
Is the rate of a reaction important? think about airbags. Is the exact mechanism important? e.g. Ozone destruction (i) O3 + Cl O2 + ClO (ii) O + ClO O2 + Cl O3 + O 2 O2 Both rate of reaction and mechanism are vital to understanding this problem!
Instantaneous and Initial Rates
a more accurate determination of the rate is given by the instantaneous rate which is the slope of the tangent line (line a or d) initial rate is at time = 0 (line a) rates can change over time
Average vs. Instantaneous Rate
50 miles 1 hour 50 miles 1 hour 50 miles 1 hour 50 miles 1 hour
Rates of Reaction
the rate of the reaction can be expressed mathematically as: 1 [ NO 2 ] 1 [ NO] [O 2 ] Rate = = = 2 t 2 t t use this expression when you want to measure the change in the concentration of the species this gives the average rate of the reaction (if t is large)
2 NO2 2 NO + O2
avg rate = 200 miles/4 hours = 50 mph avg rate = 200 miles/4 hours = 50 mph
30 miles 1 hour 70 miles 1 hour 70 miles 1 hour 30 miles 1 hour
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Rates of Reaction
the calculus form of the rate of the reaction can be expressed mathematically as:
2 NO2 2 NO + O2
Rate = k [ NO 2 ] =
1 d [ NO 2 ] 1 d [ NO ] d [O 2 ] = = 2 dt 2 dt dt
A+BC This is used as a predictive tool Rate = k [A]x[B]y where: k is the rate constant
Rate Law of a Reaction
the is replaced by the d note sign for the product and the reactants this turns an average rate into an instantaneous rate (because dt is infinitely small)
A & B are reactants [A],[B] are the concentrations of A,B x,y exponents are the reaction order x,y can be positive, negative, zero, or fractional
A Rule of Kinetics:
The rate equation cannot be predicted by the overall reaction, it can only be measured empirically. concentrations of reactants must be measured experimentally the integrated form of the rate equation is used to determine if the proposed rate law is correct integrated form of the rate equation? Huh? More on this later
Reaction Order vs. Reactant Order
1st order: x = 1 or y = 1
2nd e.g. Rate = k [A] order: x = 2 or y = 2 or (x = 1, y = 1) e.g. Rate = k[A]2 or Rate = k[A][B]
2nd order in A, 2nd order over all 1st order in A, 2nd order over all
3rd order: x = 3 or (x = 2 , y = 1) etc. e.g. Rate = k[A]3 or Rate = k[A]2[B]
2 I- +1/2 H2O2 + H+ I2 + H2O Rate = k[I-]m[H2O2]n[H+]p
some times the rate law is complicated with lots of variables it can be simplified by taking it in parts
make [H2O2] and [H+] remain constant then Rate = k[I-]m[constant]n [constant]p or Rate = k[I-]m[constant]
Next Weeks Lab
2 I- +1/2 H2O2 + H+ I2 + H2O Rate = k [I-]m [H2O2]n[H+]p
Next Weeks Lab
linearize the rate law by taking the log( )
log(Rate) = log(k[I-]m[constant]) log(Rate) = log(k) + log([I-]m) + log([constant]) combine the constants log(k) + log([constant]) log(Rate) = log([I-]m) + log([constant]) log(Rate) = m log([I-]) + log([constant]) y = (slope)(x) + (y-intercept)
measure the Rate and [I-], plot y vs. x use Excel to get the trendline (slope and intercept) repeat by holding [I-] and [H+] constant
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Initial Rates
measured at the beginning of the experiment, before products start to complicate things one way to determine the Rate Law without running the full experiment
Method of Initial Rates
A +B C
[A] 0.200 0.100 0.200
Rate = k[A]n[B]m
[B] 0.050 0.050 0.025 Initial Rate 4.00 1.00 2.00
No change
m=1
Rate [A]n [B]m (1/2) = (1)n (1/2)m
Method of Initial Rates
A +B C Rate = k[A]n[B]1
[B] 0.050 0.050 0.025 Initial Rate 4.00 1.00 2.00
No change
Method of Initial Rates
A +B C in M
[A] 0.200 0.100 0.200
Rate = k[A]n[B]1
[B] 0.050 0.050 0.025 Initial Rate 4.00 1.00 2.00
[A] 0.200 0.100 0.200
in M/s
Rate [A]n [B]1 (1/4) = (1/2)n (1)1 n=2 Rate = k[A]2[B]1
Rate = k[A]2[B]1 just pick any line to determine k Rate = k = (4.00 M/s) = 2000 M-2 s -1 2[B]1 [A] [0.200 M]2[0.050 M]1
Method of Initial Rates
A +B C in M
[A] 0.200 0.100 0.200
Rate =
[B] 0.050 0.050 0.025
k[A]n[B]1 in M/s
Worksheet Initial rate Calculations
Initial Rate 4.00 1.00 2.00
Rate = k[A]2[B]1
taking experimental data to determine Rate Law the Rate Law tells you how the reaction rate depends on the concentration of the reactants changing [A] will have a larger effect than [B]
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Note something weird about k:
The definition of rate varies from reaction to reaction. e.g. Rate = k [A] [B] or Rate = k [C] Therefore the units of k vary from reaction to reaction. e.g. k is in M-1 sec-1 or k is in sec-1 Remember: Rate = conc. / time
Remember, the rate equation cannot be predicted, it can only be measured empirically. Rate = k [NO2]2
Find k from measurements Use the integrated form of the rate eqn. to determine the reaction order
There are two forms to know: First order: Second order: ln[A] = ln[A]o k t 1/[A] = 1/[A]o + k t
Determining The Reaction Order
the experimental concentration data forms a linear relationship with the equation that represents its reaction order You can use data to find k and reaction order
1st order eqn: 2nd order eqn: plot: ln[A] vs. t plot: 1/[A] vs. t ln[A] = ln[A]o k t 1/[A] = 1/[A]o + k t does this give a linear graph? does this give a linear graph?
what would you plot to form y = mx + b ?
How do you find the reaction order?
The one that gives the linear graph gives you the reaction order (assuming it is 1st or 2nd order)
Plot both.
How do you find k? the slope! 1/[A] = 1/[A]o + k t
400 350 300 250 200 150 100 50 0 0
2 NO2 2 NO + O2 ln[A] = ln[A]o k t
-3 -4 ln[NO2] -4 -5 -5 -6 -6 0 100 Time (s) 200 300
Plotting Kinetic Data
this works best in a spreadsheet
time (s) 0 10 20 30 [CH4] 0.05 0.0323 0.0238 0.0189 ln[CH4] -2.99573 -3.43269 -3.73807 -3.96859 1/[CH4] 20 30.95975 42.01681 52.91005
1st Order Plot R = 0.8389
2
2nd Order Plot
1/[NO2]
R = 0.9949
2
100 Time (s)
200
300
Only one will be close to linear. Rate = k [NO2]2
then plot ln[CH4] vs. t and 1/[CH4] vs. t
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Worksheet on Integrated Rate Laws
save part C for another day
Worksheet
SO2Cl2
Rate = k [SO2Cl2]
2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 0
3.5 3 2.5 2 1.5 1 0.5 0 0
2
1st order y = -0.1675x + 1.6013 R = 0.9999
2
ln(p)
5
2nd order y = 0.1695x - 0.0713 R = 0.9131
10 Time (hrs)
15
20
what is k? k= - slope = 0.168 hrs-1
15 20
1/p
5
10 Time(hr)
Worksheet
C8H12
Rate = k [C8H12]2
2nd order y = 0.0142x + 59.234 R = 0.9995
2
1st order
-4.1 -4.3 -4.5 -4.7 -4.9 -5.1 -5.3 -5.5 0 2000
y = -0.0001x - 4.1432 R = 0.9852
2
The integrated forms of the rate laws are important: You can predict [concentration] as a function of time!
8000
ln(p)
4000 Time (s)
6000
Example:
160 140 1/p 120 100 80 60 0 2000
what is k? k= slope = 0.0142 atm-1 sec-1
6000 8000
The decomposition reaction of NO2 is secondorder in [NO2], with a rate constant of 1.51 M1 min1. If the initial concentration is 0.041M, when will the concentration = 0.010 M?
2 NO2 2 NO + O2
4000 Time(s)
Rate Law of a reaction: Rate = k [NO2]2 The integrated solution: 1/[NO2] 1/[NO2]o = k t 1/0.010 - 1/0.041 = 1.51 t t= ? 50 min
1st Order parameter: Half-life
Radioactive decay = 1st order decay Integrated rate law: ln(At/Ao) = kt In the case that At = 0.5 Ao (half is left) t = t1/2 therefore: ln(0.5) = kt1/2 half-life is just a convenient time interval for 1st order reactions
5
Half-life example
Radiation (131I): Biology Time (days) 0 3 6 9
131I?
Things that affect the reaction rate
surface area (lycopodium powder demo)
12
molecules mix more efficiently there are more places to react
Mass 131I (g) 12 9.25 7.14 5.50 4.24 What is the half-life of ln(At/Ao) = kt ln(1/2) = kt1/2 ln(7.14/12)= k (6 days) 0.693 = .0865 (t1/2) k = 0.0865 days1 t1/2 = 8.0 days
temperature
molecules have more energy at higher temps they move faster and crash harder
a catalyst
reduces the barrier for the reaction
concentration/pressure
the more molecules present the higher probability of a reaction
Temperature Dependence
k depends on temperature and Arrhenius Activation energy.
What does Ea mean in terms of rate? k = A e Ea/RT e.g. e.g. big Ea e (big) = small k = small rate! small Ea e (small) = big k = big rate!
k = A e Ea/RT
A = constant (pre-exponential factor) Ea = Arrhenius Activation Energy R = 8.3145 J/mol K T = Temperature (in Kelvin)
Activation energy (Ea) is independent of concentration and temperature.
Definitions
transition state or activated complex
intermediate
a stable chemical that is formed then used up in a mechanism found in the valleys of a reaction diagram
Ea
activated complex (transition state) Erxn
an unstable chemical that is formed then used up in a mechanism found at the peaks of a reaction diagram
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An example:
activated complexes
The colorless gas N2O4 decomposes to the brown gas NO2 in a first-order reaction. The rate constant k = 4.5 103 s1 at 274 K and 1.0 x 104 s1 at 283 K. What is the k activation energy Ea? R ln k E= Use the correct equation: 11
2 a 1
intermediate
T1 T2
Ea = 8.3145 ln(4.5/10) / [1 /283 1/274 ] Ea = 57202 J/mol = 57 kJ/mol
Worksheet
Reaction at 823 K
-2.5 -2.7 -2.9 ln[CH4] -3.1 -3.3 -3.5 -3.7 -3.9 -4.1 0 y = -0.0322x - 3.0502 R2 = 0.9796 10 20 Time (s) 30 40 y = 1.0979x + 20.004 R2 = 1 1st order 2nd order 55 50 45 1/[CH4]
ln[CH4] -2.5 -3 -3.5 -4 -4.5 -5 0
Worksheet
Reaction at 873 K
y = 3.5015x + 19.987 R2 = 1 1st order 2nd order 135 115 1/[CH4] 95 75 55 y = -0.0599x - 3.184 R2 = 0.936 10 20 Time (s) 30 40 35 15
40 35 30 25 20 15
Worksheet
Mechanisms
Questions answered by mechanisms what steps does the reaction take to get from reactants to products? which steps are slow? which ones are fast? what is the rate law? how can the rate of the reaction be increased or decreased? it is all about molecular collisions!
k2 3.502 J 8.314 Kmol ln k1 kJ 1.098 = 139 Ea = = 11 1 1 mol 823 K 873 K T1 T2 R ln
(
)
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Determining the Reaction Rate: Two proposed mechanisms for 2 NO2 2 NO + O2 A) step 1: NO2 NO + O (slow) step 2: NO2 + O NO + O2 (fast) B) step 1: 2 NO2 NO3 + NO step 2: NO3 NO + O2
Which is correct???
rule of Kinetics: The number of molecules involved in a reaction mechanism greatly influences the rate of reaction.
(slow) (fast)
Definitions: Unimolecular rxn: N2O4 NO2 + NO2 Bimolecular rxn: NO2 + NO2 N2O4 Termolecular rxn: O2 + O2 + O2 2 O3
(very probable) (fairly probable) (improbable)
The textbook pictures:
Mechanisms: A Summary
all steps must add up to the overall chemical reaction
all intermediates must cancel out in the end
Determining the Reaction Rate: Two proposed mechanisms for 2 NO2 2 NO + O2
1 A) step 1: NO2 NO + O (slow) k2 step 2: NO2 + O NO + O2 (fast)
there can be many different proposed mechanisms for a given chemical reaction use your creativity to think of mechanistic variations
Try to use unimolecular and bimolecular reactions
k
Unimolecular, so Rate = k1[NO2]
k
the slowest step determines the reaction rate
1 B) step 1: 2 NO2 NO3 + NO k2 step 2: NO3 NO + O2
(slow) (fast)
Bimolecular, so Rate = k1[NO2]2
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Worksheet on Mechanisms
Rate = k [NO2]2
Lets do an example together...
Intermediates in Rate Law
Rate eqns which include fast equilibria H2 + Br2 2 HBr (overall)
Intermediates in Rate Law
Another rule: no intermediates are allowed in the rate law Br2 2 Br k
-1 2 H2 + Br HBr + H H + Br HBr
k1 k
What is the rate law? start with a proposed mechanism: Br2 2 Br k
-1
(fast) (slow) (fast)
not useful because of the intermediate
k1
(fast) (slow) (fast)
H2 + Br HBr + H H + Br HBr
k2
Rate = k2[H2][Br]
Intermediates in Rate Law
For an equilibrium, the forward rate is equal to the reverse rate Br2 2 Br k
-1
Determining Rate Laws
use the slow step to propose the rate law measure concentrations of reactants experimentally
k1
(fast)
Rate = k1[Br2] = k-1[Br]2 [Br] = (k1[Br2]/ k-1) Rate = k2[H2][Br] = kobs [H2][Br2] solve for [Br] substitute
use th...