Kinetics 5: Complex reactions: composite Arrhenius parameters; chain reactions

Question 1

Composite Arhenius parameters

Study the image. Assuming each elementary rate constant obeys an Arhenius law, with individual values of A and Ea, derive a composite expression for k_obs. Use a reaction profile to illustrate what the apparent Ea represents.

Answer 1

Question 2

What is a chain reaction?

Answer 2

A reaction comprised of non-linear elementary steps, which form chains in which the output of one step may be the input of an earlier step.

Question 3

What is a chain carrier?

Answer 3

An intermediate formed during a chain reaction, which is an input into earlier steps, and thus propogates the reaction.

Question 4

What are the general stages of a chain reaction?

Answer 4

• Initiation generates chain carriers
• Propogation maintains chain carriers
• Inhibition reduces reaction rate by destroying product (but not chain carriers)
• Termination destroys chain carriers
• Chain branching (sometimes): one chain carrier reacts to give 2+ carriers (results in explosion, eg combustion)

Question 5

The thermal chain reaction H₂ + Br₂ –> 2HBr is well-characterised. Illustrate the stages of this reaction.

Answer 5

Initiation

Br₂ + M –> 2Br + M where M is a molecule

Or photodissociation: Br₂ + hv –> 2Br

Propogation

Br + H₂ –> HBr + H

H + Br₂ –> HBr + Br

Inhibition

HBr + H –> H₂ + Br

Termination

2Br + M –> Br₂ + M (M absorbs excess energy from new Br2 molecule, such that the Br2 doesn’t immediately fall apart)

Question 6

The thermal chain reaction H₂ + Br₂ –> 2HBr is well-characterised. Which species would you expect to follow the steady state approximation?

Answer 6

Br and H, i.e. the chain carriers, since they are intermediates.

Propogation

• Br + H₂ –> HBr + H*
• H + Br₂ –> HBr + Br*

Question 7

The rate law for the thermal chain reaction H₂ + Br₂ –> 2HBr was found to be:

d[HBr]/dt = (k_a[H₂][Br₂]^3/2)/([Br₂] + k_b[HBr])

What does this indicate about the mechanism?

Answer 7

• Complex
• Fractional power indicates radicals are involved
• [HBr] on denominator indicates HBr inhibits the reaction, ie it goes slower as [HBr] increases — but when [HBr] is small in the initial phase, it can be omitted from the denominator and inhibition doesn’t occur

Question 8

Assuming the thermal chain reaction H₂ + Br₂ –> 2HBr has the following steps, use the ss approximation to derive an expression for the rate of formation of HBr.

Initiation

Br₂ + M –> 2Br + M (k₁) [1]

Propogation

Br + H₂ –> HBr + H (k₂) [2]

H + Br₂ –> HBr + Br (k₃) [3]

Termination

2Br + M –> Br₂ + M (k₄) [4]

note other reactions could be involved, but the question is whether they compete effectively enough to alter the rate law

Answer 8

Assume chain carriers Br and H are in the steady state

Expression for [H]

d[H]/dt = k₂[Br][H₂] - k₃[H][Br₂] = 0

[H] = (k₂[Br][H₂]) / (k₃[Br₂])

Expression for [Br]

d[Br]/dt = 2k₁[M][Br₂] - (k₂[Br][H₂] - k₃[H][Br₂​]) - 2k₄[M][Br]² = 0

but the quantity in brackets is just d[H]/dt = 0

= 2k₁[M][Br₂] - (d[H/dt) - 2k₄[M][Br]² = 0

= 2k₁[M][Br₂] - 2k₄[M][Br]² = 0

[Br] = (k₁[Br₂]/k₄)^1/2

Note the 2-coefficients are from the fact that there are 2Br in the reagents/products, so the rate of change of Br is doubled. Also the squared term for k4 is due to the stoichoimetric coefficient = 2

Plug [Br] into [H] expression

[H] = (k₂[Br][H₂]) / (k₃[Br₂])

= (k₂[H₂]/k₃[Br₂]) x (k₁[Br₂]/k₄)^1/2

= (k₂/k₃) x (k₁/k₄)^1/2 x ([H₂]/[Br₂]^1/2)

Expression for HBr, then plug in [H] and [Br]

d[HBr]_initial/dt = k₂[Br][H₂] + k₃[H][Br₂]

= k₂[H₂] x (k₁[Br₂]/k₄)^1/2

+ k₃[Br₂] x (k₂/k₃) x (k₁/k₄)^1/2 x ([H₂]/[Br₂]^1/2)

= 2k₂ x (k₁/k₄)^1/2 x [H₂] x [Br₂]^1/2

Question 9

Define chain length. Give its expression.

Answer 9

The average number of times that the closed cycle of steps producing products is repeated per chain carrier.

chain length, l = (overall reaction rate)/(rate of initiation)

Question 10

Expansion of the H₂ + Br₂ –> 2HBr reaction scheme to include [5], an inhibition step, gives the overall rate equation shown.

Derive an expression for the chain length in the limit of low [HBr].

Answer 10

Overall reaction rate, A

[HBr] low so can be removed from denominator, so:

d[HBr]/dt = 2k₂ x (k₁/k₄)^1/2 x [H₂] x [Br₂]^1/2

Rate of initiation, B

refer back to scheme

= 2k₁[M][Br₂]

Chain length = A/B

l = (2k₂ x (k₁/k₄)^1/2 x [H₂] x [Br₂]^1/2) / (2k₁[M][Br₂])

= (k₂ / (k₄k₁)^1/2) x ([H₂] / ([Br₂]^1/2[M]))

under typical conditions of pressure, chain length is order 10^13, indicating one Br produces many HBr before being terminated