Sir C.N. Hinshelwood shared the 1956 Nobel prize in Chemistry for his work on the mechanisms of high — Physical Chemistry — Kinetics Chemistry Question
Hinshelwood
Sir C.N. Hinshelwood shared the 1956 Nobel prize in Chemistry for his work on the mechanisms of high temperature reactions.
The pyrolysis of ethanal proceeds by the following simplified mechanism:
Reaction rate constant Ea / kJ mol –1
CH3CHO → CH3· + HCO· k1 358
CH3· + CH3CHO → CH4 + CH3CO· k2 8
CH3CO· → CH3· + CO k3 59
HCO· → H· + CO k4 65
H· + CH3CHO → H2 + CH3CO· k5 15
2CH3· → C2H6 k6 0
List each reaction as initiation, propagation or termination.
Model Answer
Reactions
initiation 1
propagation 2, 3, 4, 5
termination 6
Use the steady-state approximation on the radical intermediates to find expressions for the steady-state concentrations of the HCO, H, CH3 and CH3CO radicals.
Model Answer
(1) [HCO]' = k1[AcH] − k4[HCO] = 0 ⇒ [HCO] = (k1/k4)[AcH]
(2) [H]' = k4[HCO] – k5[H][AcH] = 0 ⇒ [H] = (k4/(k5[AcH]))[HCO] = k1/k5 (from 1)
(3) [Me]' = k1[AcH] − k2[Me][AcH] + k3[Ac] − 2 k6[Me]2 = 0
(4) [Ac]' = k2[Me][AcH] – k3[Ac] + k5[H][AcH] = 0
Add (3) + (4), and substitute from (2) and then from (1).
0 = 2k1[AcH] − 2k6[Me]2 ⇒ [Me] = (k1/k6)1/2 [AcH]1/2
Finally from (4).
[Ac] = (k2[Me] + k5[H]) / k3 [AcH] = (k2/k3)(k1/k6)1/2 [AcH]3/2 + (k1/k3)[AcH]
Find rate laws for the rate of loss of ethanal, and the rates of formation of methane, ethane, hydrogen and CO.
Model Answer
-[AcH]' = k1[AcH] + k2[Me][AcH] + k5[H][AcH] = 2k1[AcH] + k2(k1/k6)1/2[AcH]3/2
[CH4]' = k2[Me][AcH] = k2(k1/k6)1/2[AcH]3/2
[C2H6]' = k6[Me]2 = k1[AcH]
[H2]' = k5[H][AcH] = k1[AcH]
[CO]' = k3[Ac] + k4[HCO] = 2k1[AcH] + k2(k1/k6)1/2[AcH]3/2
There are two pathways for the dissociation of ethanal. Write a balanced equation for each reaction and for each find the order with respect to ethanal, and the activation energy.
Model Answer
Find this by analysing the rates of formation of the different products: the formation of ethane and hydrogen is first order in ethanal with equal rates, and the formation of methane is order 3/2. Both routes form CO. The first order route forms CO at twice the rate of ethane and hydrogen, and the order 3/2 rate forms it at the same rate as methane.
To get the activation energies use the Arrhenius equation. Because the activation energy is the exponent, when the effective rate constant is a product of elementary rate constants, their activation energies must be added (with related rules for division and powers).
(i) 2 CH3CHO → C2H6 + H2 + 2CO order: 1 Ea = 358 kJ mol–1
(ii) CH3CHO → CH4 + CO order: 3/2 Ea = 187 kJ mol–1