Transcript Chemical Kinetics Mac 2011

ERT 108/3 :
PHYSICAL CHEMISTRY
Chemical Kinetics
By; Mrs Hafiza Binti Shukor
ERT 108/3 PHYSICAL CHEMISTRY
SEM 2 (2010/2011)
TOPIC COVERED…;
Experimental Chemical and Kinetics Reactions
First Order Reactions
Second Order Reactions
Reaction Rates and Reaction Mechanisms
Light Spectroscopy and Adsorption Chemistry
(Experimental methods for fast reactions).
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CHEMICAL KINETICS??
Also called reaction kinetics
Study of the rates & mechanisms of
chemical reactions
2 types of reaction;
a)homogeneous – reaction occurs
in 1 phase (gas @liquid phase)
b)heterogeneous – reaction occurs
in 2 @ > phase
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Experimental Chemical and
Kinetics Reactions
Rates of chemical Reactions:
 the rate of speed with which a
reactant disappears or a product
appears.
 the rate at which the concentration
of one of the reactants decreases or
of one of the products increases with
time.
 mol L-1 s-1.
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Example 1
• The decomposition of
dinitrogen pentoxide (N2O5) in
an inert solvent (carbon
tetrachloride) at 450C:
• The data of the formation of
O2(g) and the disappearance of
N2O5 is shown in Table 1.
• The initial concentration
[N2O5] = 1.40M.
What is the concentration, [N2O5]
at time, t=423s?
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Rate of Reaction: A variable quantity
• Rate of reaction is
expressed as either:
 reac tan t 
Re action rate 
t
[ Negative value ]
Disappearance of reactant
or
 product 
Re action rate 
t
[ Positive value ]
Formation of products
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Answer (Example 1)
• At t=0, Initial [N2O5] = 1.40M
• At t = ∞, Final [N2O5] = 0M [decomposes
completely]
• 5.93cm3 O2(g) is obtained at STP.
• After 423s, the volume of O2 (g) collected
is 1.32cm3 of a possible 5.93cm3.
• The fraction of the N2O5 decomposed is
1.32/5.93.
• The decrease in concentration of N2O5 at
this point
= (1.32/5.93) x 1.40M = 0.312 M.
• After 423s, [N2O5] remaining
undecomposed
= 1.40-0.31 = 1.09M.
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Example 2:
• From the figure , determine the rate of
decomposition of N2O5 at 1900s.
• What is the initial rate of reaction?
Graphical determination of the rate of reaction of dinitrogen pentoxide in
carbon tetrachloride
1.60
1.40
1.20
[N2O5], mol/L
Note: the rate of
reaction can be
expressed as the
slope of a tangent
line.
1.00
0.80
0.60
0.40
0.20
0.00
0
200
400
600
800
1000 1200 1400 1600 1800 2000 2200 2400
Time, s
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Answer (Example 2)
• Based on the graph of
concentration of reactant
vs time,
 the slope of a tangent
line at t=1900s,
Re action rate
  slope of tan gent
 N 2O5 

t
0.21mol / L

800 s
 2.6 x10  4 mol L1 S 1
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• The initial rate =
 1.24  1.40mol N 2O5 / L
200s
= 8.0 x 10-4 mol N2O5 L-1 s-1
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The Rate Law for Chemical
Reactions
• The rate law or rate equation – mathematical equation.
aA  bB  ......  gG  hH  ......
Reaction rate, r = k[A]m[B]n …..
 The rate, r at time t is experimentally found to be
related to the concentrations of species present at that
time, t .
 The exponents in the rate reaction are called the order
of the reaction.
 The term k in the equation is called the rate constant.
 it is a proportionality constant that is characteristic of
the particular reaction & is significantly dependent only
on temperature.
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Method of Initial Rates
• This simple method of establishing
the exponents in a rate equation
involves measuring the initial rate of
reaction for different sets of initial
concentration.
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Example 3
• The data of three reactions involving S2O82and I- were given in the below table.
(i) Use the data to establish the order of
reaction with respect to S2O82-, the order
with respect to I- & the overall order.
S 2O82 (aq)  3I  (aq)  2SO42 (aq)  I 3 aq 
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Cont…Example 3
(ii) Determine the value of k for the above
reaction.
(iii) What is the initial rate of disappearance of
S2O82- reaction in which the initial
concentrations are [S2O82- ] =0.050M &
[I-]=0.025M?
(iv) What is the rate of formation of SO42- in
Experiment 1?
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Answer (Example 3)
(i) In the experiments 1 & 2, [I-] is held constant & [S2O82-] is
increased by a factor of 2, from 0.038 to 0.076M.
 The reaction rate, R increased by a factor of 2 also.
 
R  k S 2O8  I
m
 n
R2 = k (0.076)m(0.060)n = k (2x0.038)m(0.060)n
= k (2)m (0.038)m (0.060)n = 2.8 x 10-5 mol L-1 s-1
R1 = k (0.038)m(0.060)n =1.4 x 10-5 mol L-1 s-1
5
R2 k (2) 0.038 0.060
2
.
8
x
10
m


2

2
m
n
5
R1
1.4 x10
k 0.038 0.060
m
m
n
If 2m =2, then m =1.
The reaction is first order in S2O82-.
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R2 = k (0.076)m(0.060)n = k (0.076)m(2x0.030)n
= k (0.076)m (2)n(0.030)n = 2.8 x 10-5 mol L-1 s-1
R3 = k (0.076)m(0.030)n =1.4 x 10-5 mol L-1 s-1
5
R2 k 0.076 2 0.030
2
.
8
x
10
n


2

2
m
n
5
R3
1.4 x10
k 0.076 0.030
m
n
n
If 2n =2, then n =1. The reaction is first order in I-.
The overall order of the reaction is
m + n = 1+1 = 2 (second order).
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(ii) Use any one of the three experiments:
R1
1.4 x105 mol L1s 1
k

2

1
1
S 2O8 I
0.038mol L x 0.060 mol L

 
k = 6.1 x 10-3 L mol-1 s-1.
(iii) Once the k value is determined, the rate law can be used to
predict the rate of reaction.
Reaction rate,
 
R  k S 2O8  I 
R = 6.1 x 10-3 L mol-1 s-1 x 0.050 mol L-1 x 0.025 mol L-1
= 7.6 x 10-6 mol L-1 s-1.
TRT401 Physical Chemistry
BBLee@UniMAP
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• Based on the stoichiometry, 2 moles of SO42- are produced
for every mole of S2O82- consumed.
2
2
mol
SO
4
No.mol SO42 ( L1s 1 )  1.4 x10 5 mol S 2O82 L1s 1 x
1 molS 2O82
= 2.8 x 10-5 mol SO42-(L-1 s-1).
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Zero-order, First-order, Second-order Reactions
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Cont………
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Zero order
First order
Second order
Zero-order,
First-order,
Secondorder
Reactions
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Example 4
(a) When [N2O5] =0.44M, the rate of
decomposition of N2O5 is 2.6 x 10-4 mol
L-1 s-1.
 what is the value of k for this first-order
reaction?
(b) N2O5 initially at a concentration of 1.0
mol/L in CCl4, is allowed to decompose
at 450C. At what time will [N2O5] be
reduced to 0.50M?
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Answer (Example 4)
(a) Rate of disappearance of N2O5 (R) = k [N2O5]
R
2.6 x10 4 mol L1s 1
= 5.9 x 10-4 s-1
k

N 2O5 
0.44 mol / L
(b) For 1st order of reaction, to determine t, we can use:
ln[ A]  kt  ln A0
 k 
log[ A]t  
t  log A0
 2.303 
log [A]0 = log [N2O5]0 = log 1.0 = 0.
 5.9 x104 s 1 
t  0
log [A]t = log [N2O5]t = log 0.50 = -0.30.  0.30  
2.303 

use, k = 5.9 x 10-4 s-1.
2.303x0.30
3
t

1
.
1
x
10
s
 4 1
5.9 x10 s
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Example 5
Time, min
[A], M
log [A]
1/[A]
0
1.00
0.00
1.00
5
0.63
-0.20
1.59
10
0.46
-0.34
2.17
15
0.36
-0.44
2.78
25
0.25
-0.60
4.00
• The data of the above table were
obtained for the decomposition
reaction: A → 2B + C.
(a) Establish the order of the reaction.
(b) What is the rate constant, k?
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Answer (Example 5)
(a) Plot graph based on the data given in the Table.
Not Straight line – Not Zero Not Straight line – Not First
order
order
Straight line – 2nd order
(b) The slope of the 3rd graph:

4.00  1.00 L / mol
k
 0.12 L mol 1 min 1
25 min
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Reaction rates:
Effect of temperature
• Chemical reactions tend to go faster at
higher temperature.
 slow down some reactions by lowering the
temperature.
• Increasing the temperature increases the
fraction of the molecules that have energies
in excess of the activation energy.
 this factor is so important that for many
chemical reactions it can lead to a doubling
or tripling of the reaction rate for a
temperature increase of only 100C.
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Cont….
• In 1889, Arrhenius noted that the k data for many reactions fit the
equation:
k  Ae
 Ea RT
where,
A & Ea are constants characteristics of the reaction
R = the gas constant.
Ea = the Arrhenius activation energy (kJ/mol or kcal/mol)
A = the pre-exponential factor (Arrhenius factor).
( the unit of A is the same as those of k.)
• Taking log of the above equation:
Ea
Ea
 log 10 A
ln k  
 ln A log 10 k  
2.303RT
RT
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Cont….
• If the Arrhenius equation is obeyed:
 a plot of log10 k versus 1/T is a straight line with
slope: -Ea/2.303 R and intercept log10 A.
 This enables Ea and A to be found.
• Another useful equation:
log




k2
k1

Ea  T2  T1 


2.303R  T2T1 
(eliminate the constant A).
T2 and T1 - two kelvin temperatures.
k2 and k1 - the rate constants at these temperatures.
Ea – the activation energy (J/mol)
R – the gas constant (8.314 Jmol-1 K-1).
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Example 6
(a) Use the figure given
to find A and Ea for:
2 N 2O5  4 NO2  O2
(b) Calculate Ea for a
reaction where rate
constant at room
temperature is
doubled by a
10Kelvin increase in
T.
Figure: Rate constant versus temperature for
the gas-phase first order decomposition
reaction
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Answer (Example 6a)
• Tabulate the data as follows.
Temp, 0C
Temp, K
1/Temp, 1/K
k, s-1
log10 k
25
298
0.0034
0.001
-3
• Construct the Arrhenius plot of log10k versus 1/T for the reaction.
 Intercept (log10A)=13.5
A = 3x1013s-1
 Slope=-5500K,
Ea
 5500 
2.303R
Ea=25kcal/mol
=105 kJ/mol
Figure: Arrhenius plot of log10 k versus 1/T for
this reaction.
Note: the long extrapolation needed to find A.
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Answer (Example 6b)
• Based on the given info:
 k2 = 2k1 ,
T1 = room temperature (298K),
T2=298+10 = 308K,
• The Arrhenius equation:
log
• Substitute:
log
2 k1
k1
k2
k1
Ea  T2  T1 



2.303R  T2T1 
Ea  (308)  298 



2.303R  308(298) 
Ea = 53 kJ/mol
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Reaction Mechanisms
• Each molecular event that significantly
alters a molecule’s energy or geometry
is called an elementary process
(reaction).
• The mechanism of a reaction:
 the sequence of elementary reactions
that add up to give the overall
reaction.
 A mechanism is a hypothesis about
the elementary steps through which
chemical change occurs.
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Reaction Mechanisms
• Elementary processes in which a single molecule
dissociates (unimolecular) or two molecules collide
(bimolecular) much more probable than a process
requiring the simultaneous collision of three bodies
(termolecular).
• All elementary processes are reversible and may
reach a steady-state condition. In the steady state
the rates of the forward & reverse processes
become equal.
• One elementary process may occur much more
slower than all the others. In this case, it
determines the rate at which the overall reaction
proceeds & is called the rate-determining/ limiting
step.
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The Hydrogen-Iodine Reaction
H2 (g) + I2 (g) → 2HI (g)
• Rate of formation of HI = k [H2][I2]
• The hydrogen-iodine reaction is
proposed to be a two-step mechanism
[Sullivan J. (1967). J.Chem.Phys.46:73].
 1st step: iodine molecules are believed
to dissociate into iodine atoms.
 2nd step: simultaneous collision of two
iodine atoms and a hydrogen
molecule.
(this termolecular step is expected to occur
much more slowly – the rate-determining step).
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The Hydrogen-Iodine Reaction
1st step:
2nd step:
Net:

I 2 g 
2 I ( g ) [Fast]

k2
k1
2I g   H 2 ( g )  2HI ( g )
k3
[Slow]
I 2 ( g )  H 2 ( g )  2HI g 
• If the reversible step reaches a steady state condition:
 rate of disappearance of I2 = rate of formation of I2
k1[ I 2 ]  k2 [ I ]
2
I 2

k1
I 2 
k2
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The Hydrogen-Iodine Reaction
• For the rate-determining
step:
Rate of formation of HI
k3
2I g   H 2 ( g ) 
2HI ( g )
= k3 [I]2[H2]
where
k1[ I 2 ]  k2 [ I ]
2
k1

k3 H 2 I 2 
k2
= K[H2][I2]
where
(K=k1k3/k2)
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Example 7
• The thermal decomposition of ozone to oxygen:
2O3 (g) → 3O2 (g)
• The observed rate law:
2
Rate of disappearance of O3 = k O3
 
O2 
• Show that the following mechanism is consistent with this
experiment rate law.
1st:
k
1

O3
O2  O

k2
2nd:
k3
O  O 3 
2O 2
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Answer (Example 7)
• Assume the 1st step reaches the steady state condition:
Rate of formation of O = Rate of disappearance of O
k1 [O3] = k2 [O2] [O]
k1 O3 
O  
k 2 O2 
• Assume the 2nd step is the rate-determining step:
Rate of disappearance of O3 = k3 [O][O3]
(where k = k1k3/k2)

k1k3 O3 O3 
O3 

k
O2 
k 2 O2 
2
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Experimental methods for fast reactions
• Many reactions are too fast to follow by the classical
methods.
• Several ways to study fast reactions :
1. Rapid flow methods:
(i) Continuous flow
(ii) Stopped flow
2. Relaxation methods:
(i) Temperature jump (T-jump)
method
(ii) Pressure jump method
(iii) Electric field jump method
3. Flash photolysis
4. Shock tube
5. Nuclear-magnetic-resonance (NMR) spectroscopy
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ASSIGNMENT 1
Write a short note for the following fast
reaction:
a)
b)
c)
d)
e)
Rapid flow methods
Relaxation methods
Flash photolysis
Shock tube
Nuclear-magnetic-resonance (NMR)
spectroscopy
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The End
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