Question

1. derive Linde mann-Hinshelwood mechanism

2. show initiation, propagation, branching, termination steps for reaction

2 H₂ + O₂→ 2 H₂O

Answer 1

Ans- (1) The Lindemann-Hinshelwood Mechanism - First-order gas phase kinetics - Unimolecular Reactions-

The first successful explanation of unimolecular reactions was provided by Lindemann way back in 1921. This mechanism was later elaborated by Cyril Hinshelwood, and is hence now known as the Lindemann-Hinshelwood mechanisms,

representation of the Lindemann-Hinshelwood mechanism of unimolecular reactions. The species A is excited by collision with A, and the excited A molecule (A*) may either be deactivated by a collision with A or go on to decay by a unimolecular process to form products.

According to the Lindemann-Hinshelwood mechanism (see fig. 1), the reaction takes off by having a molecule A getting energetically through a collision with another A molecule:

The energized molecule A* might loose is excess energy either by:

(1) Collision with another A molecule (forming no products), i.e.:

(2) Shaking itself apart forming the product (the unimolecular step), i.e.:

If this last unimolecular step is slow enough such that it becomes the rate-determining step, then the overall reaction will have first-order kinetics. This can be demonstrated by applying the steady state approximation to the formation of A*:

which assuming equality solves to:

i.e. the rate of formation of P (which for now is not first-order) is given by:

Now if we assume that the rate of the reaction (2) is much greater than that of (3), i.e.:

(Note that this is particularly true at high concentrations (partial pressures) of A).

Then the rate of formation of the products will assume a first-order form, i.e.:

i.e.:

That this last equation is only valid when k`<sub>a</sub>[A] >> k<sub>b</sub>. Thus, if the concentration of [A] (i.e. the partial pressure) is reduced, then this assumption does not remain valid and instead at very low concentrations of A, we start getting k`_a[A] << k_b , i.e:

This means that if we let:

This means that is we take logs of both sides for the equation of k_obs we get:

If we plot log k_obs vs. log[A], we should get two straight line curves one at high concentrations with gradient 0 showing that k_obs is a constant, and one for low A concentrations, with gradient 1 (see fig. 2).

Note: We get two important advantages by taking logs of both sides: (1) the data for the lower concentrations gets more spaced out, and more importantly, (2) we get a big change in the gradient - from 0 at high [A] concentrations, to 1 at low concentrations.

()  Stationary or stable chain reactions. The concentration of reactive intermediates is constant in time or slowly decreasing. Example: CH3CHO → CH4 + CO Experimental observations: Small amounts of C2H6 and H2 are also produced, and the Rate of Reaction ∝ [CH3CHO]3/2. (These are signatures of a chain reaction mechanism)

Proposed mechanism for this reaction: k1 Initiation: CH3CHO CH3• + CHO• k2 Propagation: CH3• + CH3CHO •CH3CO + CH4 k3 •CH3CO CH3• + CO Termination: 2 CH3• k4 C2H6

“Side” Reactions: CHO• + M CO + H• + M k6 H• +CH3CHO H2 + •CH3CO

Kinetic Equations: d[CH4 ] = k2 [CH3 •][CH3CHO] dt d 3 [CH •] = k1 [CH3CHO] − k2 [CH3 •][CH3CHO] + k3 [• CH3CO] − 2k4 [CH3 •] 2 dt d[• CH3CO] dt = k2 [CH3 •][CH3CHO] − k3 [• CH3CO]

(3)