Question 6.FP.4: The pyrolysis of propanone is a chain reaction with possible......
The pyrolysis of propanone is a chain reaction with possible mechanism
\mathrm{CH}_{3}\mathrm{COCH}_{3}\stackrel{k_{1}}{\longrightarrow}\mathrm{CH}_{3}\mathrm{CO}^{•}+\mathrm{CH}_{3}^{•}
\mathrm{CH}_{3}C\mathrm{O}^{•}\stackrel{k_{2}}{\longrightarrow}C H_{3}^{•}+\mathrm{CO}
\mathrm{CH}_{3}^{•}+\mathrm{CH}_{3}\mathrm{COCH}_{3}\stackrel{k_{3}}{\longrightarrow}\mathrm{CH}_{4}+^{•}\mathrm{CH}_{2}\mathrm{COCH}_{3}
{}^{•}{\mathrm{CH}}_{2}{\mathrm{COCH}}_{3}\ {\stackrel{k_{4}}{\to}}\ {\mathrm{CH}}_{3}^{•}+{\mathrm{CH}}_{2}{\mathrm{CO}}
\mathrm{CH}_{3}^{•}+^{•}\mathrm{CH}_{2}\mathrm{COCH}_{3}\stackrel{k_{5}}{\rightarrow}\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COCH}_{3}
Using this mechanism explain what is meant by the terms initiation, propagation and termination, and identify the major and minor products. Find the order of the reaction and an expression for the observed rate constant.
If step 5 is replaced by
\mathrm{CH}_{3}^{•}+\mathrm{CH}_{3}^{•}\stackrel{k_{6}}{\longrightarrow}\mathrm{C}_{2}\mathrm{H}_{6}
predict what the order would become.
Likewise, predict the order if the major termination step became
^{•}{\mathrm{CH}}_{2}{\mathrm{COCH}}_{3}+^{•}{\mathrm{CH}}_{2}{\mathrm{COCH}}_{3}\stackrel{k_{7}}{\rightarrow}\mathrm{CH}_{3}{\mathrm{CO}}{\mathrm{}}_{}{\mathrm{CH}}_{2}{\mathrm{COCH}}_{3}
The essential first step here is to recognize the chain carriers and the propagation steps. There are three radicals, {\mathrm{CH}}_{3}^{•},\;{\mathrm{CH}}_{3}{\mathrm{CO}}^{•} and {}^{•}{\mathrm{CH}}_{2}{\mathrm{COCH}}_{3}. Two of these, \mathrm{CH}_{3}^{•} and {}^{•}C{\mathrm{H}}_{2}C{\mathrm{OCH}}_{3}, are recycled in steps 3 and 4, which are therefore the propagation steps and produce the major products, {\mathrm{CH}}_{4}\quad{\mathrm{and}}\quad{\mathrm{CH}}_{2}CO. These radicals are removed from the chain in step 5, which is thus a termination step, producing the minor product, \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COCH}_{3}\cdot\mathrm{CH}_{3}^{•}~\mathrm{and}~~\mathrm{CH}_{3}\mathrm{CO}^{•} are first produced in step 1, which is thus an initiation step. {\mathrm{CH}}_{3}C{\mathrm{O}}^{•} breaks down to give \mathrm{CH}_{3}^{•}, which is a chain carrier, and so this is also initiation. This mechanism thus involves a two-stage initiation. The breakdown of {\mathrm{CH}}_{3}C{\mathrm{O}}^{•} also gives CO. Since {\mathrm{CH}}_{3}C{\mathrm{O}}^{•} is not recycled, the CO will only be present in trace amounts.
Because {\mathrm{CH}}_{3}C{\mathrm{O}}^{•} generates a chain carrier it must be included in the steady state analysis.
+\frac{\mathrm{d}[\mathrm{CH}_{3}^{\bullet}]}{\mathrm{d}t}=k_{1}[\mathrm{CH}_{3}\mathrm{COCH}_{3}]+k_{2}[\mathrm{CH}_{3}\mathrm{CO}^{\bullet}]-k_{3}[\mathrm{CH}_{3}^{\bullet}][\mathrm{CH}_{3}\mathrm{COCH}_{3}]
+\,k_{4}[^{\bullet}{\bf C H}_{2}{\bf C O C H}_{3}]-k_{5}[{\bf C H}_{3}^{\bullet}][^{\bullet}{\bf C H}_{2}{\bf C O C H}_{3}] = 0 (I)
+\frac{{\mathrm{d}}[{\mathrm{CH}}_{3}{\mathrm{CO}}^{•}]}{{\mathrm{d}}t}=k_{1}[{\mathrm{CH}}_{3}{\mathrm{COCH}}_{3}]-k_{2}[{\mathrm{CH}}_{3}{\mathrm{CO}}^{•}]=0 (II)
+\frac{\mathrm{d}[\mathrm{^{•}CH}_{2}\mathrm{COCH}_{3}]}{\mathrm{d}t}=k_{3}[\mathrm{CH}_{3}^{•}][\mathrm{CH}_{3}\mathrm{COCH}_{3}]-k_{4}[^{•}\mathrm{CH}_{2}\mathrm{CH}_{3}]
-\,k_{5}[C{\bf H}_{3}^{•}][^{•}{\mathrm{CH}}_{2}{\mathrm{COCH}}_{3}]=0 (III)
Adding (I), (II) and (III):
2\,k_{1}[{\mathrm{CH}}_{3}{\mathrm{COCH}}_{3}]=2\,k_{5}[{\mathrm{CH}}_{3}^{\mathrm{•}}]^{\mathrm{}}[^{•}{\mathrm{CH}}_{2}{\mathrm{COCH}}_{3}] (IV)
This is the algebraic form of the steady state assumption, viz, the total rate of initiation is balanced by the total rate of termination.
Equation (IV) contains two unknowns, and so cannot be solved. A second independent equation in these two unknowns is required.
Since the chain carriers are not involved in any steps other than initiation,propagation and termination, the long chains assumption can be used in the form of ‘the rates of the two propagation steps are \mathrm{equal^{,}} .
k_{3}[\mathrm{CH}_{3}^{•}][\mathrm{CH}_{3}\mathrm{COCH}_{3}]=k_{4}[^{•}\mathrm{CH}_{2}\mathrm{COCH}_{3}] (V)
(\mathrm{IV})\,\,\mathrm{gives}\,\,[\mathrm{CH}_{3}^{•}]=\frac{k_{1}[\mathrm{CH}_{3}\mathrm{COCH}_{3}]}{k_{5}[^{•}\mathrm{CH}_{2}\mathrm{COCH}_{3}]}
(\mathrm{V})\ \mathrm{gives}\ \left[\mathrm{CH}_{3}^{\mathrm{•}}\right] = \frac{k_{4}[^{•}\mathrm{CH}_{2}\mathrm{COCH}_{3}]}{k_{3}[\mathrm{CH}_{3}\mathrm{COCH}_{3}]}
∴ [{}^{•}{\mathrm{CH}}_{2}{\mathrm{COCH}}_{3}]=\left({\frac{k_{1}k_{3}}{k_{4}k_{5}}}\right)^{1/2}[{\mathrm{CH}}_{3}{\mathrm{COCH}}_{3}] (VI)
Express the rate of reaction in terms of a rate expression for removal of reactant or formation of a major product. It is sensible to chose one that involves {}[^{\mathrm{•}}{\mathrm{CH}}_{2}{\mathrm{COCH}}_{3}].
+\frac{\mathrm{d}[C{\bf H}_{2}C{\bf O}]}{\mathrm{d}t} = k_{4}[^{•}\mathrm{CH}_{2}\mathrm{COCH}_{3}]
=\left({\frac{k_{1}\,k_{3}\,k_{4}}{k_{5}}}\right)^{1/2}[\mathrm{CH}_{3}\mathrm{COCH}_{3}]
Reaction is first order in reactant and k_{\mathrm{obs}}=(k_{1}\,k_{3}\,k_{4}/k_{5})^{1/2}.
Note: the rate of reaction in terms of removal of reactant or production of the other major product \mathrm{CH}_{4} requires [{\bf C H}_{3}^{•}].
The Rice–Herzfeld rules state the following.
Recombination of like radicals used up in the first propagation step gives overall 3/2 order in reactant.
Recombination of like radicals produced in the first propagation step gives overall 1/2 order in reactant.