SLIDE 1
Eyring plot example: Internal rotation of N,N-dimethylnicotinamide
O H3 CH3 C N N
Foundations of Chemical Kinetics Lecture 12: Transition-state - - PowerPoint PPT Presentation
Foundations of Chemical Kinetics Lecture 12: Transition-state - - PowerPoint PPT Presentation
Foundations of Chemical Kinetics Lecture 12: Transition-state theory: Examples Marc R. Roussel Department of Chemistry and Biochemistry Eyring plot example: Internal rotation of N,N-dimethylnicotinamide O C H 3 N CH 3 N This elementary
SLIDE 2
SLIDE 3
Eyring plot example: Internal rotation of N,N-dimethylnicotinamide (continued)
For each point, we calculate ln(kh/kBT) and plot this vs T −1:
- 29
- 28.5
- 28
- 27.5
- 27
- 26.5
- 26
- 25.5
- 25
- 24.5
3.1 3.15 3.2 3.25 3.3 3.35 3.4 3.45 3.5 3.55 ln(kh/kBT) T -1/10-3K-1
Slope = −9185 K Intercept = 3.829
SLIDE 4
Example: Eyring plot (continued)
∆‡H◦
m = −R(slope) = −(8.314 472 J K−1mol−1)(−9185 K)
= 76 kJ mol−1 ∆‡S◦
m = R(intercept) = (8.314 472 J K−1mol−1)(3.829)
= 32 J K−1mol−1 ∆‡G ◦
m = ∆‡H◦ m − T ◦∆‡S◦ m
= 76 kJ mol−1 − (298.15 K)(32 × 10−3 kJ K−1mol−1) = 67 kJ mol−1 Interpret ∆‡S◦
- m. Do the sign and size of this quantity make sense?
What does the reaction profile look like?
SLIDE 5
Statistical factors
◮ The reaction AB + C → A + BC should, all other things being
equal, be slower than A2 + C → A + AC because in the latter case, C can attack either end of the molecule and a reaction can still occur.
◮ We deal with this in the statistical approach to transition state
theory by adding a statistical factor (L‡) to the rate constant: k = L‡ kBT c◦h Q‡ QX QY N exp
- −∆E0
RT
- (second-order reaction)
◮ The statistical factor is the number of distinct, equivalent
transition states that can be formed if we label equivalent atoms in the reactants.
SLIDE 6
Statistical factors: examples
- 1. H + D2 → HD + D
- 2. CH3 + C2H6 → CH4 + C2H5
SLIDE 7
Calculation of the rate constant for the reaction D + H2 → HD + H
Data: H2:
◮ R = 0.741 ˚
A
◮ ˜
ν = 4400 cm−1 Transition state:
◮ Linear with RDH = RHH = 0.930 ˚
A
◮ Classical barrier height: 40.2 kJ mol−1 ◮ Vibrational frequencies: ˜
νsym stretch = 1764, ˜ νbend = 870 cm−1 with gbend = 2 To do: Calculate the rate constant at 450 K.
SLIDE 8
D + H2 → HD + H (continued)
◮ L‡ = 2 ◮ The factor involving the partition functions can be factored
into terms for each type of motion: Q‡ QD QH2 =
- Q‡
QD QH2
- elec
- Q‡
QD QH2
- tr
Q‡ QH2
- rot
Q‡ QH2
- vib
◮ The first excited states of both D and H2 are very high in
energy, so their partition functions at moderate temperatures are just their respective ground-state degeneracies. The same turns out to be true for the transition state. gelec,H2 = 1, gelec,D = 2, gelec,TS = 2 (one unpaired electron) ∴
- Q‡
QD QH2
- elec
= 1
SLIDE 9
D + H2 → HD + H (continued)
◮ Recall
Qtr = V h3 (2πmkBT)3/2 ∴
- Q‡
QD QH2
- tr
= h3 V
- mTS
2πkBTmDmH2 3/2
◮ The masses are readily computed: mD = 3.344 × 10−27 kg,
mH2 = 3.347 × 10−27 kg, mTS = mD + mH2 = 6.692 × 10−27 kg.
◮ Putting in all the constants, we get
- Q‡
QD QH2
- tr
= 5.513 × 10−31 m3 V
SLIDE 10
D + H2 → HD + H (continued)
◮ The vibrational partition function is
Qvib = [1 − exp(−hc˜ ν/kBT)]−1, so Q‡ QH2
- vib
= 1 − exp(−hc˜ νH2/kBT) [1 − exp(−hc˜ νsym/kBT] [1 − exp(−hc˜ νbend/kBT)]2 .
◮ Putting in all the data, we get
- Q‡/QH2
- vib = 1.140.
SLIDE 11
D + H2 → HD + H (continued)
◮ For a linear triatomic molecule,
3 1
m m m R2
1 2
R
I = m1m3 m (R1 + R2)2 + m2 m (m1R2
1 + m3R2 2)
with m = m1 + m2 + m3.
◮ For the transition state of this reaction, m1 = mD,
m2 = m3 = mH and R1 = R2 = R‡: ITS = mH m (R‡)2(5mD + mH) = 3.979 × 10−47 kg m2
◮ For a diatomic molecule, I = µR2, which gives
IH2 = 4.595 × 10−48 kg m2.
SLIDE 12
D + H2 → HD + H (continued)
◮ The rotational partition function of a linear molecule is
Qrot = 8π2IkBT h2
◮ The rotational term in the transition-state equation is
therefore Q‡ QH2
- rot
= ITS IH2 = 8.661.
◮ Putting it all together now, we have
Q‡ QD QH2 = (1) 5.513 × 10−31 m3 V
- (8.661)(1.192)
= 5.693 × 10−30 m3 V
SLIDE 13
D + H2 → HD + H (continued)
◮ The zero-point energy of H2 is
1 2hc˜
νH2 = 4.370 × 10−20 J ≡ 26.32 kJ mol−1.
◮ The zero-point energy of the transition state is the sum of the
zero-point energies for each of its vibrational modes added to the height of the barrier: ZPETS = L 2hc (˜ νsym stretch + 2˜ νbend) + Eb = 61.16 kJ mol−1
◮ Therefore
∆E0 = 61.16 − 26.32 kJ mol−1 = 34.84 kJ mol−1
SLIDE 14
D + H2 → HD + H (continued)
◮ If we want the rate constant in L mol−1s−1, pick
V = 10−3 m3, N = 6.022 141 29 × 1023, and c◦ = 1 mol L−1.
◮ Plugging everything in, we get
k = 5.6 × 106 L mol−1s−1
◮ The experimental value is
k = 9 × 106 L mol−1s−1
◮ Main source of error: tunneling, which is very significant for