5.
62 Spring 2004
Lecture #33, Page 1
Macroscopic Reaction Rates from Microscopic Properties Bimolecular Reactions Collision Theory A+BC From kinetic theory, we found AB collision rate in a gas:
2 8k BT Z AB = d AB N ANB AB = gas kinetic v N A N B 1/ 2
(cm-3 sec-1)
[Note on notation: to distinguish Boltzmann constant k from rate coefficient k, we will write the former as kB!] Reaction rate
d [C ] = k[ A][ B] dt
Can we simply identify rate constant k with gas-kinetic collision rate vAB? NO typically k << vAB k has Arrhenius type behavior, exp(-Eact/RT), which does not appear anywhere in the expression for gas kinetic collision frequency Arrhenius behavior suggests reactive cross section, R, has energy dependence R(E) or equivalently, velocity dependence R (vAB).
dZ reactive N A N B R (v AB )v AB f (v AB )4 v AB 2 dv AB
or
Z reactive 2 = k = v AB R (v AB ) f (v AB )4 v AB dv AB N ANB 0
5.62 Spring 2004
Lecture #33, Page 2
k = AB 2 k BT
3/ 2
v AB v ( v ) e AB R AB 0
/ 2 k BT
4 v AB dv AB
2
In terms of relative kinetic energy, E=
1 v AB 2 2
1 8 k= k BT AB k BT
1/ 2
E
0
( E )e E / kBT dE
Now we just need an expression for Hard-Sphere limit:
R (E)
or
R (v AB ) .
Preact = 1, rmin d AB Preact = 0, rmin > d AB
2 R ( E ) = d AB
k (T ) = AB 2 k BT
3/ 2
2 AB
v vAB e AB
0
/ 2 k BT
4 v AB dv AB
2
8k T 2 k ( E ) = v R = B d AB AB
Estimate the magnitude of this expression:
1/ 2
d AB 4 108 cm v 5 104 cm / s, 300 K
k 5 1010 cm3molecule1 sec 1 1.5 1011 L mole 1 sec 1
5.62 Spring 2004
Lecture #33, Page 3
t1/2 would be on the order of 10-11 seconds at p = 1 bar (or c=1M)! Also predicted temperature dependence of k is T1/2
Whats missing??
Reactive Hard Sphere Model Both Arrehenius temperature dependence and energy profile for a reaction suggest an energy requirement for reaction to occur. One possibility,
R = 0, if E<E 0
2 R = d AB , if E>E 0
But look at three situations with same kinetic energy E:
(i) (iii)
Impact parameter b=0; maximum impact Impact parameter, b dAB, no collision occurs at all
Therefore we need a more detailed dynamical model. Energy along line of centers (Old Collision Theory)
Partition energy into along line of centers,
tangent to the line of centers, Et
b2 EC = E 1 , if , b dAB d AB
2 1 AB ( v|| ) 2
EC =
1 2 AB ( v ) ; v = v cos ; 2
Et =
5.62 Spring 2004
Lecture #33, Page 4
EC =
1 1 AB v 2 cos 2 = AB v 2 (1 sin 2 ) = E (1 sin 2 ) 2 2
b2 EC = E 1 2 , ( b d AB ) d AB EC = 0, ( b > d AB )
sin = b / d AB , thus
If E < E0 of course, EC < E0, Preact(E)=0, R(E)=0 For any collision energy E > E0, there is some critical impact parameter b0 for which
b2 EC = E 1 2 = E0 d AB E0 2 1 b02 = d AB E
Assume: all collisions with b < b0 (at energy E > E0) are reactive Thus
2 R ( E ) = b02 = d AB 1
R = 0
E0 E
for E E0 for E < E0
Now we only need to insert in previous expression to find k(T).
5.62 Spring 2004
Lecture #33, Page 5
1/ 2
1 8 k= k BT AB k BT 1 8 = k BT AB k BT
E / k BT ( ) E E e dE R
0
1/ 2
E0 E / kBT 2 1 E d dE AB e E E0
1 8 E / k BT 2 dE = d AB ( E E0 ) e k BT AB k BT E0
1/ 2
x=
( E E0 ) ; k BT
dx =
dE k BT
e E / kBT = e E0 / kBT e x
Thus
8k BT E0 / k BT x 2 k = xe dx d AB e AB 0 1/ 2 x 8k BT E0 / k BT x 2 k = d e xe dx ; xe dx = 1 AB AB 0 0 1/ 2
(mean relative speed)
8k T k = B AB
1/ 2
2 d AB
e E0 / kBT
(Arrhenius-type factor with E0 Eact)
(hard sphere gas kinetic speed)
RELATIONSHIP BETWEEN E0 (critical line of centers energy) and Eact, (empirical Arrhenius energy)
5.62 Spring 2004
Lecture #33, Page 6
k = Ae Eact / kBT d ln k d Eact Eact ln = A = dT dT k BT k B T 2 8k T 2 k CT = B d AB e E0 / kBT AB E0 1 E d ln k CT d 1 8k BT 2 = + + 02 ln ln d ( = AB ) dT dT 2 AB k BT 2T k BT
Now we require that:
1/ 2
d ln k d ln k CT = dT dT 1 Eact E0 = + k BT 2 2T k BT 2 1 Eact = k BT + E0 2
How well does this theory work? Predicted pre-exponential factor is:
2 8k BT d AB AB
1/ 2
( 3 1015 cm 2 )( 4 104 cm / s ) 10 1011 cm3 / sec
or about
6 1010 L / mol s
In most cases, A(T) is much smaller in magnitude! Some examples:
5.62 Spring 2004 Reaction CH3+CH3C2H6 Cl+H2HCL+H NO+O3NO2+O2 CH3+C2H6CH4+C2H5 Eact, kJ/mol 0.0 21 28 43.5 Measured Log10(A/T1/2) 9.32 9.69 8.60 6.99
Lecture #33, Page 7 Calculated Log10(ZAB/T1/2) 9.78 10.38 9.89 10.03 P 0.35 0.20 0.05 0.0009
Clearly the bimolecular rate constant < (sometimes <<) the gas-kinetic collision rate constant, even after accounting for activation energy. Sometimes this is represented by a steric factor p=A/ZAB
So
( corrected )
= p d
2 AB
8k T B AB
1/ 2
e E0 / kBT
But p is strictly a fudge factor! (In a little while, we will give a recipe for fudge.)
