ert 108/3 physical chemistry sem 2 (2010/2011) ert 108/3 : physical chemistry chemical kinetics by;...
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ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
ERT 108/3 : PHYSICAL CHEMISTRY
Chemical KineticsBy; Mrs Hafiza Binti Shukor
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
TOPIC TOPIC COVERED…; 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).
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
CHEMICAL KINETICS??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
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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. 4
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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?
5
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Rate of Reaction: A variable quantity
• Rate of reaction is expressed as either:
or
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t
treacrateaction
tan
Re
t
productrateaction
Re
[ Negative value ]
[ Positive value ]
Disappearance of reactant
Formation of products
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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.7
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
• From the figure , determine the rate of decomposition of N2O5 at 1900s.
• What is the initial rate of reaction?
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Example 2:
Note: the rate of reaction can be expressed as the slope of a tangent line.
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Answer (Example 2)
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• Based on the graph of concentration of reactant vs time,
the slope of a tangent line at t=1900s,
114
52
106.2
800
/21.0
tan
Re
SLmolx
s
Lmolt
ON
gentofslope
rateaction
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
10
• The initial rate =
= 8.0 x 10-4 mol N2O5 L-1 s-1
s
LONmol
200
/40.124.1 52
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
The Rate Law for Chemical Reactions
• The rate law or rate equation – mathematical equation.
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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............ hHgGbBaA
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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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ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Example 3
• The data of three reactions involving S2O8
2- and I- were given in the below table.
(i) Use the data to establish the order of reaction with respect to S2O8
2-, the order with respect to I- & the overall order.
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aqIaqSOaqIaqOS 324
282 )(2)(3)(
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Cont…Example 3
(ii) Determine the value of k for the above reaction.
(iii) What is the initial rate of disappearance of S2O8
2- reaction in which the initial concentrations are [S2O8
2- ] =0.050M & [I-]=0.025M?
(iv) What is the rate of formation of SO4
2- in Experiment 1?
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ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Answer (Example 3)
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(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. 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
If 2m =2, then m =1. The reaction is first order in S2O8
2-.
2104.1
108.22
060.0038.0
060.0038.0)2(5
5
1
2
x
x
k
k
R
R mnm
nmm
nm IOSkR 82
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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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
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).
2104.1
108.22
030.0076.0
030.02076.05
5
3
2
x
x
k
k
R
R nnm
nnm
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
TRT401 Physical Chemistry BBLee@UniMAP 17
(ii) Use any one of the three experiments:
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 = 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.
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115
282
1
060.0038.0
104.1
LmolxLmol
sLmolx
IOS
Rk
IOSkR 82
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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• Based on the stoichiometry, 2 moles of SO42- are produced
for every mole of S2O82- consumed.
= 2.8 x 10-5 mol SO42-(L-1 s-1).
282
24112
825112
4 1
2104.1)(.
OmolS
SOmolxsLOSmolxsLSOmolNo
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Zero-order, First-order, Second-order Reactions
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ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Cont………
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ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Zero-order, First-order,
Second-order
Reactions
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Zero order
First order
Second order
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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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ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Answer (Example 4)
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(a) Rate of disappearance of N2O5 (R) = k [N2O5]
= 5.9 x 10-4 s-1
(b) For 1st order of reaction, to determine t, we can use:
log [A]0 = log [N2O5]0 = log 1.0 = 0.
log [A]t = log [N2O5]t = log 0.50 = -0.30.
use, k = 5.9 x 10-4 s-1.
Lmol
sLmolx
ON
Rk
/44.0
106.2 114
52
0log303.2
]log[ Atk
A t
0303.2
109.530.0
14
tsx
sxsx
xt 3
14101.1
109.5
30.0303.2
0ln]ln[ AktA
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Example 5
• 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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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
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Answer (Example 5)
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(a) Plot graph based on the data given in the Table.
(b) The slope of the 3rd graph:
Not Straight line – Not Zero order
Not Straight line – Not First order
Straight line – 2nd order
11 min12.0min25
/00.100.4
molLmolL
k
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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. 26
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Cont….• In 1889, Arrhenius noted that the k data for many
reactions fit the equation:
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:
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RTEaAek
ART
Ek a lnln A
RT
Ek a
1010 log303.2
log
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
• 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:
(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). 28
12
12
303.2log
1
2
TT
TT
R
Eakk
Cont….
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Example 6
(a) Use the figure given to find A and Ea for:
(b) Calculate Ea for a reaction where rate constant at room temperature is doubled by a 10Kelvin increase in T.
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2252 42 ONOON
Figure: Rate constant versus temperature for the gas-phase first order decomposition reaction
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Answer (Example 6a)
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• Tabulate the data as follows.
• Construct the Arrhenius plot of log10k versus 1/T for the reaction.
Intercept (log10A)=13.5
A = 3x1013s-1
Slope=-5500K,
Ea=25kcal/mol
=105 kJ/mol
Temp, 0C Temp, K 1/Temp, 1/K k, s-1 log10 k
25 298 0.0034 0.001 -3
R
Ea
303.25500
Figure: Arrhenius plot of log10 k versus 1/T for this reaction. Note: the long extrapolation needed to find A.
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Answer (Example 6b)
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• Based on the given info: k2 = 2k1 , T1 = room temperature (298K), T2=298+10 = 308K,• The Arrhenius equation:
• Substitute:
Ea = 53 kJ/mol
12
12
303.2log
1
2
TT
TT
R
Eakk
)298(308
298)308(
303.2log
1
12
R
Eakk
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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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ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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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ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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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ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
The Hydrogen-Iodine Reaction1st step: [Fast]
2nd step: [Slow]
Net:
• If the reversible step reaches a steady state condition: rate of disappearance of I2 = rate of formation of I2
35
)(22
1
2 gIgIk
k
)(2)(2 32 gHIgHgI k
gHIgHgI 2)()( 22
2221 ][][ IkIk
22
12 Ik
kI
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
The Hydrogen-Iodine Reaction• For the rate-determining
step: Rate of formation of HI = k3 [I]2[H2] where
= K[H2][I2] where (K=k1k3/k2)
36
2232
1 IHkk
k
)(2)(2 32 gHIgHgI k
2221 ][][ IkIk
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Example 7
• The thermal decomposition of ozone to oxygen: 2O3 (g) → 3O2 (g)
• The observed rate law: Rate of disappearance of O3 =
• Show that the following mechanism is consistent with this experiment rate law.
1st:
2nd:
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2
23
O
Ok
OOOk
k
23
2
1
2k
3 2O O O 3
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Answer (Example 7)
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• Assume the 1st step reaches the steady state condition: Rate of formation of O = Rate of disappearance of O k1 [O3] = k2 [O2] [O]
• Assume the 2nd step is the rate-determining step: Rate of disappearance of O3 = k3 [O][O3]
(where k = k1k3/k2)
22
31
Ok
OkO
2
23
22
3331
O
Ok
Ok
OOkk
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
Experimental methods for fast reactions• Many reactions are too fast to follow by the classical
methods.• Several ways to study fast reactions :
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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
ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
ASSIGNMENT 1
Write a short note for the following fast reaction:a) Rapid flow methodsb) Relaxation methodsc) Flash photolysisd) Shock tubee) Nuclear-magnetic-resonance (NMR)
spectroscopy
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ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)
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The End