Download Physical Chemistry 20130410 week 2 Wednesday April 10 2013

Physical Chemistry 20130410 week 2 Wednesday April 10, 2013 page 1
Competing first order reactions
A
k1
k2
k -1
k -2
C and A D and C
A and D
A k ' s are rate constants
→
→
→
→
D
K2 K1
A C
⇌ ⇌
K 1 ,K 2 are equilibrium constants
[C]
K 1 [A] [C]
=
=
K 2 [D] [D]
[A]
also
k 1 [C]
=
k 2 [D]
since dividing the formula for [C] by the formula for [D] for rate of reaction simplifies to k1/k2.
If k1/k2 >>1 then [C]>>[D] so C is favored kinetically.
If K1/K2<<1 then [D]>>[D] so D is favored thermodynamically.
See handout for 1,3 butadiene. HBr is a very strong acid. The intermediate of a reaction is always at the
bottom of the dip. 1,4 addition is favored thermodynamically because it has lower energy. 1,2 addition
is favored kinetically because the hump isn’t as high.
determination of rate laws
-has to be done experimentally
r=k[A]α[B]β[L]λ
How do we determine values for α, β, λ, and k?
1. half-life method
rate = k[A]n
a. If n=1 then t1/2 is independent of [A]0
b. If n≠1 then t 1/2 =
2n-1 -1 1
t 1/2 =
(n-1)k [A]n-1
0
logt 1/2 =log
2n-1 -1
(n-1)[A]n-1
0 k
take log
2n-1 -1
-(n-1)log[A]0
(n-1)k
ka=k A
Graph logt 1/2 against log[A]0 and the slope is -(n-1)=1-n and log
2n-1 -1
is the y intercept. ka=k A
(n-1)k
That’s only for n≠1. Start with concentration [A]0 and wait until [A] is [A]0/2. The amount of time
elapsed is the half life. This is where the data comes from for the graph to determine rate.
The fractional life method (tα)
Define: tα = time required for [A]0 to drop to α[A]0 so for t1/2 α=1/2.
α=1/2 for half life 0<α<1
If r=k[A]n
logt ∝ =
α=
[A]
[A]0
tα=
-lnα
k
log∝1-n -1
-(n-1)log[A]0
(n-1)k
n≠1
n≠1
t α =fractional life
α usually set to 0.75
2. Powell plot
r=k[A]n
α is dimensionless parameter
α=
For n=1 ϕ = kt
ϕ=fee=k[A]n-1
0 t
ϕ is dimensionless
[A]
(
)
[A]0
(α)1-n-1 = (n-1)ϕ
lnα = -kt = -ϕ
[A]
[A]0
1-n
=1+[A]n-1
0 (n-1)kt
n≠1
n≠1
n=1
ϕ = k[A]0n-1t
logϕ = logk[A]0n-1+logt
logϕ-logt = logk[A]0n-1 right side is constant so left side is constant
Slide plot of data left and right over master plot to get a match.
3. initial rate method
r0 = k[A]0α[B]0β[C]0λ
A+B+C-> product(s)
# is experiment number
Do two experiments, with [B]0 and [C]0 the same in both but different [A]0
#2
[B]#1
0 =[B]0
#2
[C]#1
0 = [C]0
#2
[A]#1
0 ≠[A]0
(r0 )#1 k[A]α0 [B]β0 [C]λ0 experiment 1
Divide the rate laws of the two experiments:
=
=
(r0 )#2 k[A]α [B]β [C]λ experiment 2
0
0
0
That gives:
(r0 )#1
=([A]0 )α
(r0 )#2
where [A]0 is a ratio of the two different [A]0