22.5 The temperature dependence of reaction rates Arrhenius equation: A is the pre-exponential...
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Transcript of 22.5 The temperature dependence of reaction rates Arrhenius equation: A is the pre-exponential...
22.5 The temperature dependence of reaction rates
• Arrhenius equation:
A is the pre-exponential factor; Ea is the activation energy. The two quantities, A and Ea, are called Arrhenius parameters.
• In an alternative expression
lnk = lnA -
one can see that the plot of lnk against 1/T gives a straight line.
RTEaeAk /
RT
Ea
Example: Determining the Arrhenius parameters from the following data:
T/K 300 350 400 450 500
k(L mol-1s-1) 7.9x106 3.0x107 7.9x107 1.7x108 3.2x108
Solution:
1/T (K-1) 0.00333 0.00286 0.0025 0.00222 0.002
lnk (L mol-1s-1) 15.88 17.22 18.19 18.95 19.58
The slope of the above plotted straight line is –Ea/R, so Ea = 23 kJ mol-1.
The intersection of the straight line with y-axis is lnA, so A = 8x1010 L mol-1s-1
0
5
10
15
20
25
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035
Series1
The interpretation of the Arrhenius parameters
• Reaction coordinate: the collection of motions such as changes in interatomic distance, bond angles, etc.
• Activated complex
• Transition state
• For bimolecular reactions, the activation energy is the minimum kinetic energy that reactants must have in order to form products.
Applications of the Arrhenius principle
Temperature jump-relaxation method:
consider a simple first order reaction:
A ↔ B
at equilibrium:
After the temperature jump the system has a new equilibrium state. Assuming the distance between the current state and the new equilibrium state is x, one gets [A] = [A]eq + x; [B] = [B]eq - x;
][]['][ ' BkAk
dt
Adba
0dt
Ad ][
''' ][][' eqbeqa BkAk
22.6 Elementary reactions
• Elementary reactions: reactions which involves only a small number of molecules or ions.
A typical example:
H + Br2 → HBr + Br
• Molecularity: the number of molecules coming together to react in an elementary reaction.
• Molecularity and the reaction order are different !!! Reaction order is an empirical quantity, and obtained from the
experimental rate law; Molecularity refers to an elementary reaction proposed as an
individual step in a mechanism. It must be an integral.
• An elementary bimolecular reaction has a second-order rate law:
A + B → P
• If a reaction is an elementary bimolecular process, then it has second-order reaction kinetics; However, if the kinetics are second-order, then the reaction might be complex.
]][[][
BAkdt
Ad
22.7 Consecutive elementary reactions
• An example:
239U → 239Np →239Pu
• Consecutive unimolecular reaction
A → B → C
The rate of decomposition of A is:
• The intermediate B is formed from A, but also decays to C. The net formation rate of B is therefore:
• The reagent C is produced from the unimolecular decay of B:
][][
Akdt
Ad1
][][][
BkAkdt
Bd21
][][
Bkdt
Cd2
• Integrated solution for the first order reaction (A) is:
• Then one gets a new expression for the reactant B:
the integrated solution for the above equation is:
when assuming [B]0 = 0.
• Based on the conservation law [A] + [B] + [C] = [A]0
tkeAA 10
][][
][ e[A]][
1k-0 Bkk
dt
Bd t21
012
1 21 ])[(][ Aeekk
kB tktk
012
2112
1 ][][ Akk
ekekC
tktk
Example. In an industrial batch process a substance A produces the desired compound B that goes on to decay to a worthless product C, each step of the reaction being first-order. At what time will B be present in the greatest concentration?
Solution: At the maximum value of B
Using the equation 25.7.6 and taking derivatives with respect to t:
In order to satisfy
= 0
tmax =
The maximum concentration of B can be calculated by plugging the tmax into the equation.
0dt
Bd ][
12
210121
kk
ekekAk
dt
Bd tktk
)(][][
0dt
Bd ][
tktk ekek 2121
2
1
21
1
k
k
kkln
Steady State Approximation• Assuming that after an initial induction period, the rates of change of
concentrations of all reaction intermediates are negligibly small. • Substitute the above expression back to the rate law of B 0 ≈ [B] = (k1/ k2)[A]
• Then
• The integrated solution of the above equation is
[C] ≈ (1 - )[A]0
0dt
Bd ][
][][ BkAk 21
][][][
AkBkdt
Cd12
tkeAkdt
Cd1
01 ][
][
tke 1
Self-test 22.8 Derive the rate law for the decomposition of ozone in the reaction 2O3(g) → 3O2(g) on the basis of the following mechanism
O3 → O2 + O k1O2 + O → O3 k1’O + O3 → O2 + O2 k2
Solution: First write the rate law for the reactant O3 and the intermediate product O
Applying the steady state approximation to [O]
Plug the above relationship back to the rate law of [O3]
]][[]][[][][ '
3221313 OOkOOkOk
dt
Od
]][[]][[][][ '
322131 OOkOOkOkdt
Od
]][[]][[][ '3221310 OOkOOkOk
][][
][][
'321
31
OkOk
OkO
])[][(][][ '
3221313 OkOkOk
dt
Od
][][
]['
321
31
OkOk
Ok
Rate determining step
Simplifications with the rate determining step
• Suppose that k2 >> k1,
then k2 – k1 ≈ k2
therefore concentration C
can be reorganized as
[C] ≈ (1 - )[A]0
• The above result is the same as obtained with steady state approximation
tke 1
tktk ee 12
012
2112
1 ][][ Akk
ekekC
tktk
Kinetic and thermodynamic control of reactions
• Consider the following two reactionsA + B → P1 rate of formation of P1 = k1[A][B]
A + B → P2 rate of formation of P2 = k2[A][B]
• [P1]/[P2] = k1/k2 represents the kinetic control over the proportions of products.
• Problems 22.6 The gas phase decomposition of acetic acid at 1189 K proceeds by way of two parallel reactions:
(1) CH3COOH → CH4 + CO2, k1 = 3.74 s-1
(2) CH3COOH → H2C=C=O + H2O, k2 = 4.65 s-1
What is the maximum percentage yield of the ketene CH2CO obtainable at this temperature.
Pre-equilibrium
• Consider the reaction:
A + B ↔ I → P
when k1’ >> k2, the intermediate product, I, could reach an equilibrium with the reactants A and B.
• Knowing that A, B, and I are in equilibrium, one gets:
and
• When expressing the rate of formation of the product P in terms of the reactants, we get
]][[
][
BA
IK '
1
1
k
kK
]][[][][
BAKkIkdt
Pd22
Self-test 22.9: Show that the pre-equilibrium mechanism in which 2A ↔ I followed by I + B → P results in an overall third-order reaction.
Solution: write the rate law for the product P
Because I, and A are in pre-equilibrium
so [I] = K [A]2
Therefore, the overall reaction order is 3.
]][[][
BIkdt
Pd2
2][
][
A
IK
][][][
BAKkdt
Pd 22
Kinetic isotope effect
• Kinetic isotope effect: the decrease in the rate of a chemical reaction upon replacement of one atom in a reactant by a heavier isotope.
• Primary kinetic isotope effect: the kinetic isotope effect observed when the rate-determining step requires the scission of a bond involving the isotope:
with
• Secondary kinetic isotope effect: the variation in reaction rate even though the bond involving the isotope is not broken to form product:
with
eHCk
DCk
)(
)(
2112
/
~
)()(
CD
CH
kT
HCvhc
eHCk
DCk
)(
)(
2112
/
~~
)()}()({
CD
CH
kT
HCvHCvhc
Kinetic isotope effect
22.8 Unimolecular reactions• The Lindemann-Hinshelwood mechanism A reactant molecule A becomes energetically excited by collision
with another A molecule:A + A → A* + A
The energized molecule may lose its excess energy by collision with another molecule:
A + A* → A + A
The excited molecule might shake itself apart to form products PA* → P
The net rate of the formation of A* is
21 ][
*][Ak
dt
Ad
*]][[*][ ' AAk
dt
Ad1
*][*][
Akdt
Ad2
*][*]][[][*][ ' AkAAkAk
dt
Ad21
21
• If the reaction step, A + A → A* + A, is slow enough to be the rate determining step, one can apply the steady-state approximation to A*, so [A*] can be calculated as
Then
The rate law for the formation of P could be reformulated as
Further simplification could be obtained if the deactivation of A* is much faster than A* P, i.e., then
in case
0212
1 *][*]][[][*][ ' AkAAkAk
dt
Ad
21
21
kAk
AkA
][
][*][
'
2'1
221
2 ][
][*][
][
kAk
AkkAk
dt
Pd
*][*]][[' AkAAk 21 21 kAk ]['
][][
'A
k
kk
dt
Pd
1
21
21 kAk ]['
21 ][
][Ak
dt
Pd
• The equation can be reorganized into
• Using the effective rate constant k to represent
• Then one has
])[][
][(
][
2'1
21 AkAk
Akk
dt
Pd
2'1
221
2 ][
][*][
][
kAk
AkkAk
dt
Pd
2'1
21
][
][
kAk
Akkk
][
11
121
'1
Akkk
k
k
The Rice-Ramsperger-Kassel (RRK) model
• Reactions will occur only when enough of required energy has migrated into a particular location in the molecule.
• s is the number of modes of motion over which the energy may be dissipated, kb corresponds to k2
1*
1
s
E
EP
b
s
b kE
EEk
1*
1)(
The activation energy of combined reactions
• Consider that each of the rate constants of the following reactions A + A → A* + AA + A* → A + AA* → P
has an Arrhenius-like temperature dependence, one gets
Thus the composite rate constant also has an Arrhenius-like form with activation energy,
E = E1 + E2 – E1’
Whether or not the composite rate constant will increase with temperature depends on the value of E,
if E > 0, k will increase with the increase of temperature
RTE
RTERTE
eA
eAeA
k
kk/'
//
' '1
21
1
21
1
21
RTEEEeA
AA /)('
'121
1
21
Combined activation energy
• Theoretical problem 22.20
The reaction mechanism
A2 ↔ A + A (fast)
A + B → P (slow)
Involves an intermediate A. Deduce the rate law for the reaction.
• Solution: