Chemistry 415 Fall 2014
Assignment 5: I
1
I. Construct a Newton Diagram for K + HBr (v = 1) → KBr + H. Assume the K beam
comes from at oven at 350°C and the HBr from a 25°C oven and use the most probable
beam speeds in your diagram. The bond dissociation energies of HBr and KBr are 3.75
eV and 3.96 eV respectively, and the vibrational frequency of HBr is 2650cm-1. Make a
scale drawing of this diagram using the scale 1cm = 50 M/sec. Draw the circle
corresponding to the maximum speed with which the KBr can emerge from the collision,
and from your diagram predict the laboratory angle and speed the KBr would have if it
recoiled exactly backwards along the relative velocity vector.
If the units of v, E, m, T are 100 m/sec, kcal/mole, AMU, and °K, then
T
3
2
M ; (this assumes f(v)=Av exp(-mv /2kT). The extra factor of v
arises because the probability of a molecule entering the beam is proportional to v)
vmp = 1.572
vK =6.2830
vg =3.0152
MC = mK v K + mG v G
| C |=
CK =
( mK vK )2 + ( mG vG )2
M
= 2.88
m K vK
= 2.04
M
vr = vK 2 + vG 2 = 6.96 u K − u g = 6.96; m K u K + mg u g = 0; m K u K + mg ( u K − 6.96 ) = 0
(m
K
)
+ mg u K = 6.96mg so u K =
6.96mg
(m
K
+ mg
=
) (m
vr mg
K
+ mg
T = 1.53
⎛C ⎞
α CM = cos −1 ⎜ K ⎟ = 44.9 vK − ck =| uK | cos β ; β =25.5
⎝ C ⎠
(T+I) + (-∆E-I') = E + Q
-∆E=BDE(KBr)-BDE(HBr)=.21eV=4.84 Kcal/mol
I=2650cm-1=2650/8065 eV=.328eV=7.58KCal/mol
(E+Q)max = 4.84+7.58+1.53 = 13.95 kcal/mol
)
= 4.70
Chemistry 415 Fall 2014
Assignment 5: I
2
| u3 | =
2m4 (E + Q)
2mH (E + Q)
2
=
= 90.4m / s v3 = vK2 + ( uK − u3 ) − 2vK | uK − u3 | cos β
m3 M
mKBr M
v3 = 270.4 m/s
sin α
sin β
.43
=
; sinα =
560 = sin.8911
| uK − u3 |
v3
270
α =63.0
from graphical construction, KBr would be 63° from the K beam and the speed is about 270
m/sec
63.0°
3.01
R0.90
4.7
ß=25.6
using beam speeds
6.28
2.04
4.24
0
Chemistry 415 Fall 2014
Assignment 5: I
3
===========================================================
Another approach:
If the most probable speed in the gas is used,
2kT
2RT
=
.
m
M
vmp =
For K@350C (623K) vmp=515m/s
For HBr@25C (298K) vmp=247 m/s
(E+Q)max = 4.84+7.58+1.03 = 13.45 kcal/mol=9.35x10-20 J
vr = vk 2 + vHBr 2 = 572m / s
| C |=
T=
( mK vK )2 + ( mG vG )2
M
MC = mK v K + mG v G
= 236m / s CK =
m K vK
= 167m / s
M
⎛C ⎞
α = cos −1 ⎜ K ⎟ = 44.90
⎝ C ⎠
2
(mK mg )vr 2
µvr 2 (mK mg )vr
=
=
= 4.30kj / mole = 1.03kcal / mole = .045eV / molecule
2
2(mK + mg ) 2(mK + mg )
vr = vK 2 + vG 2 = 571m / s u K − u g = 571; m K u K + mg u g = 0; m K u K + mg ( u K − 571) = 0
(m
K
)
+ mg u K = 571mg so u K =
vK − ck =| uK | cos β ; cos β =
| u3 | =
571mg
(m
K
+ mg
=
) (m
vr mg
K
+ mg
)
= 384
346
= 25.70°
384
2mH ( E + Q )
2m4 (E + Q)
=
= 88.7
m3 M
mKBr M
v3 = vK2 + ( uK − u3 ) − 2vK | uK − u3 | cos β
v3 = 223.5m / s
sin α
sin β
.434
=
; sinα =
472 = .9179; α =66.6°
| uK − u3 |
v3
223.5
2
Chemistry 415 Fall 2014
Assignment 5: I
4
67.2°
2.46
R0.89
25.6°
44.9°
1.67
3.46
5.15