Molecular Dynamics
Abbreviated Notes
Part I
A + BC ææÆ AB + C
Nonreactive Trajectory
RAB
RAC
B
RBC
C
C
b
A
B
R
RAB
RBC
A
t
!!!!
Reactive Trajectory
B
R
A
B
C
b
RBC
RAB
RBC
A
RAB
C
t1
!!!!
time
To trace the trajectories we simply solve for R as a function of time
F = ma = m
dV
dt
d (mV)
dt
1
=
dP
dt
1
K.E. = mV 2 =
P2
2
2m
˜ (x)
dV
F=dx
˜ (R AB ,R AC ,R BC )
V
†
†
[
˜ = D AB 1- e - _____
V
†
Solution Procedure
=
]
See page 10 of notes.
†
Equating
† Equations (1) and (2)
dPx
dt
=1
MolecularAbbreviated.doc
†
˜
dV
dx
(1)
(2)
Integrating
Px = Pxo + Ú -
˜
dV
˜ˆ
Ê dV
dt = Pxo + Á ˜Dt
dx
Ë dx ¯
˜ˆ
Ê dV
Py = Pyo + Á ˜Dt
Ë dy ¯
†
†
˜ˆ
Ê dV
Pz = Pzo + Á ˜Dt
Ë dz ¯
dx
dt
†
dy
dt
=
=
dz
dt
=
1
mA
1
mA
1
mA
Px fi x = x 0 +
Py fi y = y 0 +
Pz fi z = z 0 +
1
mA
1
mA
1
mA
Px Dt
Py Dt
Pz Dt
z
A
xo , yo , zo
†
†
We can trace out a trajectory
x
y
H = KE + P.E. =
∂H
†
∂Px
=
1
2m A
1
2m A
2Px =
dx
dt
[P
2
x
Px
mA
=
†
=
]
˜
+ Py2 + Pz2 + V
mV
m
=V =
dx
dt
∂H
∂Px
We really don’t care where we are in space, what we care about is location of
molecules with regard to one another. Define a new coordinate system – affine
transformation.
†
Carry Out Trajectory Calculations
2
MolecularAbbreviated.doc
The equations of motion used to calculate the trajectories in order to obtain the
internuclear distances RAB, RAC, and RBC are
B
∂Q j
∂t
∂Pj
∂t
=-
b
Pj
=
C Q1, Q2, Q3
A
m
˜ (R BC ,R AB )
∂V
Q4 , Q5 , Q6
!!!!!!!!!!
∂Q j
˜ (RAB, RBC, RAC) is the potential energy surface.
where P is the momentum and V
Q 1 , Q 2 , Q 3 {Location of C with B as the origin.
†
Ï
Q 4†
, Q 5 , Q 6 ÌLocation of A with the center of mass
Óof BC as the origin.
†
H=
†
1
m BC
†
1
1
2m A,BC
 Pi2 +
2m BC
1
3
=
1
mB
+
1
,
mC
∂Q i
i = 1, 2, 3
∂t
6
=
†
†
i = 1, 2, 3
∂t
†
m A,BC
∂H
∂Pi
∂H
∂Pi
∂Q i
=-
mA
1
m BC
=
∂V
∂Q i
=
1
m B + mC
Pi
Pi
t
Ú 0 Pidt
1
m A,BC
Pi
Pi (t) = Pio +
3
+
t
m BC
Q i (t) = Q i (0) +
†
MolecularAbbreviated.doc
=
1
1
∂t
†
∂Pi
=
∂Q j
†
i = 4, 5, 6
1
Ú 0 Pidt
=
R BC , R AC )
4
Q i (t) = Q io +
∂t
†
 Pi2 + V˜ (R AB ,
tÊ
∂V ˆ
Ú 0 Á ∂Q ˜dt
Ë
i¯
Let’s Begin
We Specify
Monte Carlo Chooses
b
RBC (R < R < R+)
VR
b
f presentation of BC molecule
u
q relative to A
J
h angular momentum of
molecule (i.e., which
derivation is it turning)
Part II
dPx
dt
dx
dt
=-
=
˜
dV
B
dx
1
mA
q
r
Px
f
C
A
†
i = 1, 2, 3
i = 4, 5, 6
†
dQ i
dt
dQ i
dt
∂Pi
∂t
†
†
=
=
1
m BC
1
m A,BC
=-
Q i = Q io +
Pi
Pi
†
˜
∂V
Pi = Pio +
∂Q i
t
˜
∂V
Ú 0 - ∂Q
dt
i
Sectional view along the bold
trajectory.
†
V˜
!!!
4
MolecularAbbreviated.doc
t
Ú 0 Pidt
EB=9.13 kcal
Reaction Coordinate
We Specify Monte Carlo Chooses
b
R – distance between B and C (R– < R < R+)
q¸
˝ orientation of the BC molecule to A (0 < q < p, 0 < f < 2p)
f˛
UR
J
h angular momentum of BC, the direction the BC pair is turning
u
When q rotates p degrees the H–H molecule, i.e., (BC) is the same as it was q!=!0.
†
However, f rotates the C molecule towards and away from T. A molecule, therefore it
can rotate 2p.
Initial conditions, A and the center of mass of BC lie x–y plane and A approaches along
z axis.
For the
with regard to
Q1 = RBC Sin q Cos f
Q2 = RBC Sin q Sin f
Q3 = RBC Cos q
P1!=!P (Sin f Cos h + Cos f Cos q Sin y)
P2!=!P (Cos f Cos h – Cos f Sin q)
P3!=!P Sin f, Sin h
P=
b Q5
r0
J(J +1) h/R+ (R+ is the out turning radius)
Q6
Q4
P4!=!0
P5!=!0
P6!=!mA,BC UR
†
For
with regard to the center of the mass of BC
Q4 = 0
Q5 = b
(
Q 6 = - ro2 - b 2
)
–(r0!–!b)
1/2
12
Let’s begin to calculate the trajectories to find to whether or not we will have a reaction.
†
Set b!=!0 VR = 1.17 106 cm/s J = 0, n = 0
B
b=0
A
C
5
MolecularAbbreviated.doc
No Reaction
Reaction
RAB
RAC
R
RBC
RAB
R
RAB
RBC
t
time
!
Now count up the number of reactions and the number of trials
Trials
Pr = Probability Reaction
Reactions
Pr =
Pr =
4
= 0.5
8
Number of trajectories that resulted in reaction
Total number of trajectories (trials) carried out
†
Pr
a = 0.4
†
N RXS
Pr = N Lim
!!!!!!!!
T Æ•N
Trials( T )
b=0
† Now set b = 0.5 and again count up the number of reactions and trials to find Pr at this
value of b.
a
!!!!!!
†
Pr =
Nr
NT
Again
fi
Vary
b
!!!!!
!!!!
Now let’s calculate the reaction cross section
6
MolecularAbbreviated.doc
Pr
!!!
0
bmax
db
b
!!!!!!
The differential reaction cross section is
dSr = Pr (b,U,J,u)2pbdb
integrating
Sr =
†
Ê
†
†
Pr (b,U,J,u)2pbdb
ˆ
˜
Ë 2 b max ¯
If we approximate Pr = a cos Á
then
b max
Ú0
Area = pb 2max
pb
Sr = (U r ,J,u) =1.45 a b 2max (a.u )
2
† U to U =1.95 ¥10 6 cm s . We find both “b ” and “a” both
Now increase
R
max
increase
†
1.17¥10 6 cm s
1.95¥10 6 cm s
a
†
Pr
†
†
bmax
Now plot the reaction cross sections as a function of velocity and energy
Sr
Sr
1
fi! E = 2m U !fi!
U
Threshold
Energy
ET = 5.69 k!cal
E
†
7
MolecularAbbreviated.doc
2
The threshold kinetic energy, ET, below which no reaction will occur for u!=!0 and J!=!0
is 5.69 kcal
J=2
J=0
u=1
u=0
Sr
Sr
ET (J=2) > ET (J=0) because
of steric effects
U
ER(U)
U(E)
!!!!!
Let’s compare Sr versus E from Molecular Dynamics (MD) with that obtained from
collision theory
2
Rigid sphere SR = pb max , E > E A
M. D. Trajector calculations E > EA
Line of centers E > EA
Sr
†
EA
E
r
r
r
VR = VA - VBC
U R = VA - VBC
From Collision Theory we had
†
•
-rA = ÈÍ Ú o Sr Uf (U)dU˘˙C A C BC
Î1442443˚
†
k
by analogy we have
0 •
-rA (u,J) = ÈÍFBC (J,u) Ú 0 Ú o U r Sr (u,J,U r )fA dVA fBC dVBC ˘˙C A C BC
† Î1444444442444444443˚
k ( u,J )
where FBC (u,J) is fraction of BC molecules in rotation state, J, and vibration state u.
-rA (u,J) = k (u,J)C A C BC
†
†
The total reaction rate is found by summing over all quantum states u and J.
†
-rA = Â k (u,J)C A C BC = kC A C BC
u
J
8
MolecularAbbreviated.doc
†
From first principles we evaluate all the parameters to calculate k at 300K and 1000K to
compare with experimental observation.
(
)
k cm 2 mol •s
Temperature
300
1000
Theory
0.00185
11.5
†
Exp
0.0017 Æ 0.006
11 Æ 22
From the Arrhenius equation we know
k = Ae -E RT
ÊE ˆ 1
ln k = ln A - Á ˜
ËR¯ T
Plotting (ln!k) as a function of (1/T) we find the activation energy to be T.4 kcal/mole.
†
E A = 7.4 kcal mole
ln k
†
1000
T
Comparison of Energies
E V + E T = 6.2 + 5.89 =11.89 kcal
E B = 9.13 kcal
E A = 7.4 kcal
E T = 5.69 kcal
E R = 5.35 kcal
J=5
E R = 0.35 kcal
J =1
†
9
MolecularAbbreviated.doc