Transcript PowerPoint 簡報

Energy Profile and
Reaction Mechanism
p.01
How does
the reaction
take place???
C. Y. Yeung (CHW, 2009)
p.02
How does a reaction occur?
A+BC
A
+
A
+
B
B
K.E.  Ea
collide in right
orientation
Intermediate
Single
Stage Rxn
C
C
Multi-Stage
Rxn
How to know that …. ?
p.03
From the
Chemical & Differential Rate Eqns !
e.g. 2A + B  3C
If after Kinetics Studies, it was found
that …
Rate = k[A][B]
** a “A” molecule and a “B” molecule are
involved in the “rate-determining step”.
i.e. The reaction is a 2-step reaction!
Step 1:
A + B  Intermediate (r.d.s.)
Step 2:
Intermediate + A  C
p.04
e.g. 2A + B  3C
If after Kinetics Studies, it was found
that …
Rate = k[A]2
** Two “A” molecules are involved in the
“rate-determining step”.
i.e. The reaction is a 2-step reaction!
Step 1:
A + A  Intermediate (r.d.s.)
Step 2:
Intermediate + B  C
p.05
e.g. 2A + B  3C
If after Kinetics Studies, it was found
that …
Rate = k[B]
** Only one “B” molecule is involved in the
“rate-determining step”.
i.e. The reaction is a 2-step reaction!
Step 1:
B  Intermediate (r.d.s.)
Step 2:
Intermediate + 2A  C
p.06
e.g. 2A + B  3C
If after Kinetics Studies, it was found
that …
Rate = k[A]2[B]
** ALL the molecules are involved in the
“rate-determining step”.
i.e. The reaction is a Single-step
reaction!
2A + B  3C
2A + B  3C
Energy Profile ??
Transition state
Rate = k[A][B]
higher Ea: r.d.s.
(slower step)
Transition state
Rate = k[A]2
intermediate
Rate = k[B]
Rate = k[A]2[B]
p.07
p.08
p. 79 Q.9 (1999 --- Differential Rate Eqn.
and Energy Profile)
(a) Rate = k[A]
(c) A + B  product
i.e.
Only a “A” is involved in r.d.s.
p.09
Catalyst changes the energy profile of rxn!
A+BC
If no catalyst … A + B  C, with high Ea.
With catalyst (X)…
A + X intermediate, with a lower Ea. (r.d.s.)
intermediate + B  C + X (becomes a 2-stage
rxn)
p.10
Energy Profiles
Catalyst …
with / without
(slow) [r.d.s.]
(fast)
Only 1 transition state
2 transition states
DH is not affected
by catalyst
p.11
p. 79 Q.11 (2001 --- Differential Rate Eqn.
and Energy Profile)
(a) Rate = k [I(g)]2 [Ar]
(c) 2 I  I2
i.e.
Two “I” atoms and one “Ar” atom
are involved in r.d.s.
3 possible energy profiles!!
p.12
3 Different
Stories are
possible!
p.13
I(g) + I(g) + Ar  I2(g) +
Ar*(g)
Role of Ar: acts as a third
body to absorb
energy from the colliding I(g)
atoms.
p.14
Step 1:
I(g) + Ar(g)  I-Ar(g)
Step 2:
I-Ar(g) + I  I2 + Ar(g) [slow]
Role of Ar:
homogeneous catalyst
[fast]
p.15
Step 1:
2I(g) + Ar(g)  I2Ar(g) [slow]
Step 2:
I2Ar(g)  I2 + Ar(g) [fast]
Role of Ar:
homogeneous catalyst
p.16
Expt. 9 Activation Energy
formed quickly!
5 Br- + BrO3- + 6 H+  3 Br2 + 3H2O
e.g. 3 moles are form
in 42 s
OH
OH
Br
Br
3 Br2 +
All the 3 moles
of Br2 reacts
with phenol.
 does not
bleach methyl
red indicator.
+
e.g. 1 mol
3 HBr
Br
At 42.1s, a new Br2 is formed, which will not
react with phenol, but bleach methyl red!
p.17
The time required for bleaching Methyl Red is
recorded. 
data treatment ...?
Temp./0C Time/s
25
30
41
52
rate/mol s-1
(if 3 mol of Br2 are formed)
92
56
30
9
i.e. time required ,
rate 
3/92
3/56
3/30
3/9
 rate  1/t
p.18
rate  1/t
 k  1/t
Therefore …
Ea 1
ln k = –
+ lnA
R T
Ea 1
ln (1/t) = –
+ lnA
R T
p.19
Assignment
p.74
Q.11, 12 [due date: 2/3(Mon)]
Lab Report: Expt. 9 Determination of Ea
[due date: 3/3(Tue)]
Quiz on Chemical Kinetics (Ch. 13-15)
[9/3(Mon)]
p.20
Next ….
Chemical Equilibria & Keq
(p. 88-100)