1. Lecture 15

2. Steady State Approximation
Assuming that after an initial induction period, the rates of change of concentrations of all reaction intermediates are negligibly small.
Substitute the above expression back to the rate law of B
0 ≈ [B] = (k1/ k2)[A]
Then
The integrated solution of the above equation is
[C] ≈ ( )[A]0

3. Example Derive the rate law for the decomposition of ozone in the reaction 2O3(g) → 3O2(g) on the basis of the following mechanism
O3 → O2 + O k1
O2 + O → O k1’
O + O3 → O2 + O k2
Solution:
First write the rate law for the reactant O3 and the intermediate product O
Applying the steady state approximation to [O]
Plug the above relationship back to the rate law of [O3]

4. Pre-equilibrium Consider the reaction: A + B ↔ I → P
when k1’ >> k2, the intermediate product, I, could reach an equilibrium with the reactants A and B.
Knowing that A, B, and I are in equilibrium, one gets:
and
When expressing the rate of formation of the product P in terms of the reactants, we get

5. Example: Show that the pre-equilibrium mechanism in which 2A ↔ I followed by I + B → P results in an overall third-order reaction.
Solution: write the rate law for the product P
Because I, and A are in pre-equilibrium
so [I] = K [A]2
Therefore, the overall reaction order is 3.

6. Kinetic isotope effect
Kinetic isotope effect: the decrease in the rate of a chemical reaction upon replacement of one atom in a reactant by a heavier isotope.
Primary kinetic isotope effect: the kinetic isotope effect observed when the rate-determining step requires the scission of a bond involving the isotope:
with
Secondary kinetic isotope effect: the variation in reaction rate even though the bond involving the isotope is not broken to form product:

7. Kinetic isotope effect

8. 25.8 Unimolecular reactions
The Lindemann-Hinshelwood mechanism
A reactant molecule A becomes energetically excited by collision with another A molecule:
A A → A* A
The energized molecule may lose its excess energy by collision with another molecule:
A A* → A A
The excited molecule might shake itself apart to form products P
A* → P
The net rate of the formation of A* is

9. If the reaction step, A + A → A
If the reaction step, A A → A* A, is slow enough to be the rate determining step, one can apply the steady-state approximation to A*, so [A*] can be calculated as
Then
The rate law for the formation of P could be reformulated as
Further simplification could be obtained if the deactivation of A* is must faster than A*  P, i.e., then
in case

10. The activation energy of combined reactions
Consider that each of the rate constants the following reaction
A A → A* A
A A* → A A
A* → P
has an Arrhenius-like temperature dependence, one can get
Thus the composite rate constant also has an Arrhenius-like form with activation energy,
E = E1 + E2 – E1’
Whether or not the composite rate constant will increase with temperature depends on the value of E,
if E > 0, k will increase with the increase of temperature

11. Combined activation energy

12. Chapter 26: Kinetics of Complex reactions

13. 26.1 The rate laws of chain reactions
Consider the thermal decomposition of acetaldehyde
CH3CHO(g) → CH4(g) + CO(g)
v = k[CH3CHO]3/2
it indeed goes through the following steps:
1. Initiation: CH3CHO → . CH CHO
v = ki[CH3CHO]
2. Propagation: CH3CHO + . CH3 → CH4 + CH3CO kp
Propagation: CH3CO. → .CH CO k’p
3. Termination: .CH CH3 → CH3CH3 kt
The net rates of change of the intermediates are:

14. Applying the steady state approximation:
Sum of the above two equations equals:
thus the steady state concentration of [.CH3] is:
The rate of formation of CH4 can now be expressed as
the above result is in agreement with the three-halves order observed experimentally.

15. Example: The hydrogen-bromine reaction has complicated rate law rather than the second order reaction as anticipated.
H2(g) + Br2(g) → 2HBr(g)
Yield
The following mechanism has been proposed to account for the above rate law.
1. Initiation: Br M → Br Br. + M ki
2. Propagation: Br H2 → HBr + H kp1
H Br2 → HBr Br kp2
3. Retardation: H HBr → H Br kr
4. Termination: Br Br. + M → Br M* kt
derive the rate law based on the above mechanism.

16. The net rates of formation of the two intermediates are
The steady-state concentrations of the above two intermediates can be obtained by solving the following two equations:
substitute the above results to the rate law of [HBr]

17. Effects of HBr, H2, and Br2 on the reaction rate based on the equation
continued
The above results has the same form as the empirical rate law, and the two empirical rate constants can be identified as
Effects of HBr, H2, and Br2 on the reaction rate based on the equation