Question 6.WP.12: The pyrolysis of ethanal CH3CHO(g) → CH4(g) + CO(g) has bee......

1. The Rice–Herzfeld rules indicate that 3/2 order occurs for termination via like radicals produced in initiation, which then undergo reaction of the type {R}^{•} + reactant → … Hence step 4 is the dominant termination.

\mathrm{CH}_{3}^{•}+\mathrm{CH}_{3}^{•}\stackrel{k_{4}}{\longrightarrow}\mathrm{C}_{2}\mathrm{H}_{6}

2. Applying the steady state treatment gives

+\frac{\mathrm{d}[\mathrm{CH}_{3}^{\bullet}]}{\mathrm{d}t}=k_{1}[\mathrm{CH}_{3}\mathrm{CH}\mathrm{O}]-k_{2}[\mathrm{CH}_{3}^{\bullet}][\mathrm{CH}_{3}\mathrm{CHO}]+k_{3}[\mathrm{CH}_{3}\mathrm{CO^{\bullet}}]-2k_{4}[\mathrm{CH}_{3}^{\bullet}]^{2}=0           (6.153)

Note the factor of two: for each termination step two {\mathrm{CH}}_{3}^{•} are removed and the rate is expressed in terms of production of {\mathrm{CH}}_{3}^{•}.

+\frac{\mathrm{d}[\mathrm{CH}_{3}C\mathrm{O}^{•}]}{\mathrm{d}t}=k_{2}[\mathrm{CH}_{3}^{•}][\mathrm{CH}_{3}\mathrm{CHO}]-k_{3}[\mathrm{CH}_{3}\mathrm{CO}^{•}]=0        (6.154)

Add (6.153) + (6.154)

k_{1}[\mathrm{CH}_{3}\mathrm{CHO}]=2\,k_{4}[\mathrm{CH}_{3}^{•}]^{2}                           (6.155)

Note this is the algebraic statement of the steady state assumption that there is no build-up of chain carriers during reaction, and so the total rate of production must be balanced by the total rate of removal.

∴       \mathrm{[CH_{3}^{•}]}=\left(\frac{k_{1}}{2\,k_{4}}\right)^{1/2}[\mathrm{CH_{3}C H O}]^{1/2}           (6.156)

The mechanistic rate must be expressed in terms of the overall removal of reactants or the total production of the major products. For removal of reactant, {\mathrm{CH}}_{3}{\mathrm{CHO}}\colon

-{\frac{\mathrm{d}[{\mathrm{CH}}_{3}{\mathrm{CHO}}]}{\mathrm{d}t}}=k_{1}[{\mathrm{CH}}_{3}{\mathrm{CHO}}]+k_{2}[{\mathrm{CH}}_{3}^{•}][{\mathrm{CH}}_{3}{\mathrm{CHO}}]         (6.157)

This approximates to

-{\frac{\mathrm{d}[{\mathrm{CH}_{3}{\mathrm{CHO}}]}}{\mathrm{d}t}}=k_{2}[{\mathrm{CH}_{3}^{•}}][{\mathrm{CH}_{3}{\mathrm{CHO}}]}                       (6.158)

This is a consequence of the long chains approximation, which states that initiation occurs only very infrequently compared to propagation.

Expressing the rate in terms of the major products:

+\frac{\mathrm{d}[C{\bf H}_{4}]}{\mathrm{d}t}=k_{2}[C{\bf H}_{3}^{•}][C{\bf H}_{3}C{\bf H}{\bf O}]    (6.159)

+\frac{\mathrm{d}[{\mathrm{CO}}]}{\mathrm{d}t}=k_{3}[{\mathrm{CH}}_{3}{\mathrm{CO}}^{•}]              (6.160)

Since [\mathrm{CH}_{3}^{\bullet}] has already been found, choose either (6.157) or (6.159).

-{\frac{\mathrm{d}[{\mathrm{CH}}_{3}{\mathrm{CHO}}]}{\mathrm{d}t}}\approx+{\frac{\mathrm{d}[{\mathrm{CH}}_{4}]}{\mathrm{d}t}}=k_{2}\biggl({\frac{k_{1}}{2\,k_{4}}}\biggr)^{1/2}[{\mathrm{CH}}_{3}{\mathrm{CHO}}]^{3/2}                 (6.161)

which agrees with experiment and confirms recombination of {\mathrm{CH}}_{3}^{•} as the dominant termination.

3. The initiation and second propagation steps are unimolecular.

\mathrm{CH}_{3}\mathrm{CHO}\stackrel{k_{1}}{\longrightarrow}\mathrm{CH}_{3}^{•}+\mathrm{CHO}^{•}

\mathrm{CH}_{3}\mathrm{CO}^{•}\stackrel{k_{3}}{\longrightarrow}\mathrm{CH}_{3}^{•}+\mathrm{CO}

Since these are complex species, with 15 and 12 vibrational degrees of freedom respectively, there will normally be no need for a third body. Reaction will be first order in both cases with the rate-determining step being reaction. It is only when activation is rate determining that there will be a need for a third body, in which case the rate of reaction would then be second order for both, and this would only occur at very low pressures and very high temperatures.

4. The recombination product is {\mathbf{C}}_{2}{\mathbf{H}}_{6}. This has 18 normal modes of vibration over which the exothermicity of the reaction can be spread, thereby stabilizing the C–C bond. A third body would not be required, unless at very low pressures and very high temperatures.

5. The alternative termination steps are

\mathrm{CH}_{3}^{•}+\mathrm{CH}_{3}\mathrm{CO}^{•}\stackrel{k_{5}}{\longrightarrow}\mathrm{CH}_{3}\mathrm{CO}\mathrm{CH}_{3}

\mathrm{CH}_{3}\mathrm{CO}^{•}+\mathrm{CH}_{3}\mathrm{CO}^{•}\stackrel{k_{6}}{\rightarrow}\mathrm{CH}_{3}\mathrm{COCOCH}_{3}

Step 5 is recombination of unlike radicals and would be predicted to show first order kinetics.

Step 6 is recombination of like radicals, where the radical is formed in the first propagation step and then undergoes an unimolecular reaction, and would be predicted to show 1/2 order kinetics.

6. The steady state equations for step 5 being the dominant termination are

+\frac{\mathrm{d}[\mathrm{CH}_{3}^{\bullet}]}{\mathrm{d}t}=k_{1}[\mathrm{CH}_{3}\mathrm{CHO}]-k_{2}[\mathrm{CH}_{3}^{\bullet}][\mathrm{CH}_{3}\mathrm{CHO}]+k_{3}[\mathrm{CH}_{3}\mathrm{CO}^{\bullet}]

-\,k_{5}[{\bf C H}_{3}{\bf C}\mathrm{O}^{•}][{\bf C H}_{3}^{•}]=0            (6.162)

+\frac{\mathrm{d}[\mathrm{CH}_{3}\mathrm{CO}^{•}]}{\mathrm{d}t}=k_{2}[\mathrm{CH}_{3}^{•}][\mathrm{CH}_{3}\mathrm{CH}\mathrm{O}]-k_{3}[\mathrm{CH}_{3}\mathrm{CO}^{•}]

-\,k_{5}[{\bf C H}_{3}{\bf C}\mathrm{O}^{•}][{\bf C H}_{3}^{•}]=0                   (6.163)

Add (6.162) + (6.163)

k_{1}[\mathrm{CH}_{3}\mathrm{CHO}]=2\,k_{5}[\mathrm{CH}_{3}\mathrm{CO}^{•}][\mathrm{CH}_{3}^{•}]             (6.164)

This is the steady state approximation.

\mathrm{[CH_{3}^{•}]}=\frac{k_{1}[\mathrm{CH}_{3}\mathrm{CHO}]}{2\,k_{5}[\mathrm{CH}_{3}\mathrm{CO}^{•}]}               (6.165)

This is an equation in two unknowns and cannot be solved. Another equation is needed.

As shown in Problems 6.3 and 6.8, there are two ways to tackle this:

(a) express either \mathrm{[CH_{3}^{•}]~o r~[C H_{3}C O^{•}]} in terms of the other, substitute in either steady state equation and solve the resulting quadratic equation; or

(b) use the long chains approximation, in the form that the rates of propagation are equal.

Method (b) is algebraically much simpler. However, the mechanism must be checked to see whether the approximation is legitimate. It is; the chain carriers are not involved in any steps other than initiation, propagation and termination, and so

k_{2}[{\mathrm{CH}}_{3}^{•}][{\mathrm{CH}}_{3}{\mathrm{CHO}}]=k_{3}[{\mathrm{CH}}_{3}{\mathrm{CO}}^{•}]         (6.166)

This is the long chains approximation.

∴   \mathrm{(CH_{3}^{•})}=\frac{k_{3}\mathrm{[CH_{3}C O^{•}]}}{k_{2}\mathrm{[CH_{3}C H O]}}         (6.167)

∴   \frac{k_{1}[\mathrm{CH}_{3}\mathrm{CHO}]}{2\,k_{5}[\mathrm{CH}_{3}\mathrm{CO}^{•}]}=\frac{k_{3}[\mathrm{CH}_{3}\mathrm{CO}^{•}]}{k_{2}[\mathrm{CH}_{3}\mathrm{CHO}]}    (6.168)

∴  \left[\mathrm{CH}_{3}\mathrm{CO}^{•}\right]=\left({\frac{k_{1}k_{2}}{2\,k_{3}k_{5}}}\right)^{1/2}\left[\mathrm{CH}_{3}\mathrm{CHO}\right]           (6.169)

Since [{\mathrm{CH}}_{3}{\mathrm{CO}}^{•}] has been found, it is simpler to express the overall rate in terms of production of CO, giving

\frac{\mathrm{d}[\mathbf{C}\mathbf{O}]}{\mathrm{d}t}=k_{3}[\mathbf{C}\mathbf{H}_{3}\mathbf{C}\mathbf{O}^{•}]       (6.160)

=k_{3}\left(\frac{k_{1}k_{2}}{2\,k_{3}k_{5}}\right)^{1/2}\!\left[\mathrm{CH}_{3}\mathrm{CHO}\right]                (6.170)

This is first order as predicted.

7. (a) The steady state expression for the 3/2 order mechanism is

-{\frac{\mathrm{d}[{\mathrm{CH}}_{3}{\mathrm{CHO}}]}{\mathrm{d}t}}\approx+{\frac{\mathrm{d}[{\mathrm{CH}}_{4}]}{\mathrm{d}t}}=k_{2}\left({\frac{k_{1}}{2\,k_{4}}}\right)^{1/2} \mathrm{(CH_{3}C H O)^{3/2}}                 (6.161)

This rate expression only contains three of the four individual rate constants of the mechanism. \,k_{3} does not appear and thus cannot be determined from the steady state analysis

• If the reaction can be photochemically initiated and occurs with the same mechanism, then the ratio of the photochemical and thermal rates should enable \,k_{1} to be found
• \,k_{2} is the rate constant for the H abstraction reaction, and \,k_{4} is the rate constant for the termination reaction. The termination reaction is a radical recombination reaction, which can be studied independently. It could also be calculated approximately as a collision number from collision theory or from transition state theory. Since it is a recombination reaction the activation energy can be taken as approximately zero.
• Since k_{\mathrm{obs}} is known, and if k_{1}{\mathrm{~and~}}k_{4}can be found as above, then \,k_{2} can be found. Alternatively \,k_{2} can be assumed to be equal to the rate constant given in tabulated data for this H abstraction reaction found from other reactions.
• As noted \,k_{3} cannot be found from the steady state data.

(b) The steady state rate for the first order mechanism is

\frac{\mathrm{d}[{\bf C}\mathrm{O}]}{\mathrm{d}t}=k_{3}\left(\frac{k_{1}k_{2}}{2\,k_{3}k_{5}}\right)^{1/2}[{\bf C H}_{3}{\bf C H O}]            (6.170)

All four individual rate constants for this mechanism can, in principle, be found from the steady state analysis.

• Again \,k_{1} can be found from the photochemical and thermal reactions,provided they have the same mechanism.
• \,k_{5} is a rate constant for the recombination reaction and can be found in a similar manner to \,k_{4} in the 3/2 order reaction.
• \,k_{2} is the rate constant for the H abstraction reaction, and \,k_{3} is a rate constant for a bond splitting process. The overall rate constant can be written as (k_{1}k_{2}k_{3}/2\,k_{5})^{1/2}. If k_{1}{\mathrm{~and~}}k_{5} can be found as above, this leaves only the product k_{2}k_{3} able to be found. Tabulated data for either of the two relevant reactions may well be necessary for the other to be found.

Indeed tabulated data may often be necessary to determine other ratem constants appearing in the steady state expression.