1. The activation energy of combined reactions
Consider that each of the rate constants of the following reactions
A A → A* A (E1)
A A* → A A (E1’)
A* → P (E2)
has an Arrhenius-like temperature dependence, one gets
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

2. Combined activation energy

3. Theoretical problem 22.20
The reaction mechanism
A2 ↔ A + A (fast)
A + B → P (slow)
Involves an intermediate A. Deduce the rate law for the reaction.
Solution:

4. Chain reactions
Chain reactions: a reaction intermediate produced in one step generates an intermediate in a subsequent step, then that intermediate generates another intermediate, and so on.
Chain carriers: the intermediates in a chain reaction. It could be radicals (species with unpaired electrons), ions, etc.
Initiation step:
Propagation steps:
Termination steps:

5. 23.1 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 ki
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:

6. 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.

7. Example: The hydrogen-bromine reaction has a 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.

8. 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]

9. 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

10. Self-test 23.1 Deduce the rate law for the production of HBr when the initiation step is the photolysis, or light-induced decomposition, of Br2 into two bromine atoms, Br.. Let the photolysis rate be v = Iabs, where Iabs is the intensity of absorbed radiation.
Hint: the initiation rate of Br. ?

11. Exercises 23.1b: On the basis of the following proposed mechanism, account for the experimental fact that the rate law for the decomposition
2N2O5(g) → 4NO2(g) + O2(g)
is v = k[N2O5].
N2O5 ↔ NO2 + NO k1, k1’
NO NO3 → NO2 + O2 + NO k2
NO + N2O5 → NO2 + NO2 + NO2 k3

12. Explosions
Thermal explosion: a very rapid reaction arising from a rapid increase of reaction rate with increasing temperature.
Chain-branching explosion: occurs when the number of chain centres grows exponentially.
An example of both types of explosion is the following reaction
2H2(g) O2(g) → 2H2O(g)
1. Initiation: H2 → H H.
2. Propagation H OH → H H2O kp
3. Branching: O H → .O OH kb1
.O H2 → .OH H Kb2
4. Termination H. + Wall → ½ H2 kt1
H. + O M → HO M* kt2

13. The explosion limits of the H2 + O2 reaction

14. Analyzing the reaction of hydrogen and oxygen (see preceding slide), show that an explosion occurs when the rate of chain branching exceeds that of chain termination.
Method: 1. Set up the corresponding rate laws for the reaction
intermediates and then apply the steady-state approximation.
2. Identify the rapid increase in the concentration of H.
atoms.
Applying the steady-state approximation to .OH and .O gives

15. Therefore,
we write kbranch = 2kb1[O2] and kterm = kt1 + kt2[O2][M], then
At low O2 concentrations, termination dominates branching, so kterm > kbranch. Then this solution corresponds
to steady combustion of hydrogen.
At high O2 concentrations, branching dominates termination, kbranch > kterm. Then
This is an explosive increase in the concentration of radicals!!!

16. Self-test 23.2 Calculate the variation in radical composition when rates of branching and termination are equal.
Solution:
kbranch = 2kb1[O2] and kterm = kt1 + kt2[O2][M],
The integrated solution is [H.] = vinit t