Question 6.WP.8: The decomposition of methoxymethane, CH3OCH3, has already be......
The decomposition of methoxymethane,• \mathrm{CH}_{3}\mathrm{OCH}_{3}, has already been used to illustrate the characteristic features of a chain reaction:
\mathrm{CH}_{3}{\mathrm{OCH}}_{3}\stackrel{k_{1}}{\longrightarrow}\mathrm{CH}_{3}{\mathrm{O}}^{•}+\mathrm{CH}_{3}^{•}
\mathrm{CH}_{3}^{•}+\mathrm{CH}_{3}\mathrm{OCH}_{3}\stackrel{k_{2}}{\rightarrow}\mathrm{CH}_{4}+^{•}\mathrm{CH}_{2}\mathrm{OCH}_{3}
{}^{•}{\mathrm{CH}}_{2}\mathrm{OCH}_{3}\ {\stackrel{k_{3}}{\to}}\ \mathrm{HCHO}+{\mathrm{CH}}_{3}^{•}
\mathrm{CH}_{3}^{•}+\mathrm{CH}_{3}^{•}\stackrel{k_{4}}{\rightarrow}\mathrm{C}_{2}\mathrm{H}_{6}
^{•}{\mathrm{CH}}_{2}{\mathrm{OCH}}_{3}+^{•}{\mathrm{CH}}_{2}{\mathrm{OCH}}_{3}\stackrel{k_{5}}{\to}{\mathrm{CH}}_{3}{\mathrm{OCH}}_{2}{\mathrm{CH}}_{2}{\mathrm{OCH}}_{3}
\mathrm{CH}_{3}^{•}+~^{•}C H_{2}O C H_{3}\stackrel{k_{6}}{\rightarrow}C H_{3}C H_{2}O C H_{3}
Using this mechanism and considering step 4 to be the dominant termination step,show that this will lead to 3/2 order kinetics.
Can all the individual rate constants be determined from the reaction in its steady state?
+\frac{\mathrm{d}[\mathrm{CH}_{3}^{•}]}{\mathrm{d}t}=k_{1}[\mathrm{CH}_{3}\mathrm{OCH}_{3}]-k_{2}[\mathrm{CH}_{3}^{•}][\mathrm{CH}_{3}\mathrm{OCH}_{3}] +~k_{3}\mathrm{[}^{•}\mathrm{CH}_{2}O\mathrm{CH}_{3}|-2\,k_{4}[\mathrm{CH}_{3}^{•}]^{2}=0 (6.110)
Note the factor of two: for every act of termination two {\mathrm{CH}}_{3}^{•} are removed, and the rate is expressed in terms of production of {\mathrm{CH}}_{3}^{•}.
+\frac{\mathrm{d}^{•}[\mathrm{CH}_{2}\mathrm{OCH}_{3}]}{\mathrm{d}t}=k_{2}[\mathrm{CH}_{3}^{•}][\mathrm{CH}_{3}\mathrm{OCH}_{3}]-k_{3}[^{•}\mathrm{CH}_{2}\mathrm{OCH}_{3}]=0
Both are equations in two unknowns, and cannot be solved on their own. Adding (6.110) and (6.111) gives
k_{1}[\mathrm{CH}_{3}\mathrm{OCH}_{3}]-2\,k_{4}[\mathrm{CH}_{3}^{•}]^{2}=0 (6.112)
This is the algebraic statement of the steady state assumption that there is no buildup of chain carriers during reaction, and that termination balances initiation
∴ [{C}\mathrm{H}_{3}^{•}]=\left(\frac{k_{1}[\mathbf{C}\mathrm{H}_{3}{ O}\mathrm{C}\mathrm{H}_{3}]}{2\,k_{4}}\right)^{1/2} (6.113)
Rate of reaction can be expressed in three ways.
(a) +\frac{\mathrm{d}[\mathbf{CH}_{4}]}{\mathrm{d}t}=k_{2}[\mathbf{CH}_{3}^{•}][\mathrm{CH}_{3}\mathrm{OCH}_{3}] (6.114)
=k_{2}\left(\frac{k_{1}}{2\,k_{4}}\right)^{1/2}\!\left[\mathrm{CH}_{3}\mathrm{OCH}_{3}\right]^{3/2} (6.115)
(b) -\,{\frac{\mathrm{d}[\mathbf{CH}_{3}\mathbf{OCH}_{3}]}{\mathrm{d}t}}=k_{1}[\mathbf{CH}_{3}\mathbf{OCH}_{3}]+k_{2}[\mathbf{CH}_{3}^{•}][\mathbf{CH}_{3}\mathbf{OCH}_{3}] (6.116)
=k_{1}[\mathrm{CH}_{3}\mathrm{OCH}_{3}]+k_{2}{\left({\frac{k_{1}}{2\,k_{4}}}\right)}^{1/2}[\mathrm{CH}_{3}\mathrm{OCH}_{3}]^{3/2}\qquad{{(\mathrm{6}.1\,17})}
This can be simplified. The first term can be dropped, since rate of initiation \ll\mathrm{rate} of propagation—the long chains approximation—leaving the rate as
-{\frac{\mathrm{d}[\mathbf{CH}_{3}\mathbf{OCH}_{3}]}{\mathrm{d}t}}=k_{2}\left({\frac{k_{1}}{2\,k_{4}}}\right)^{1/2}[\mathbf{CH}_{3}\mathbf{OCH}_{3}]^{3/2} (6.115)
as in (a)
(c) +\frac{{\mathrm{d}}[{\mathrm{HCHO}}]}{{\mathrm{d}}t}=k_{3}{\mathrm{[}}^{•}{\mathrm{CH}}_{2}{\mathrm{OCH}}_{3}] (6.118)
{}[^{•}{\mathrm{CH}}_{2}{\mathrm{OCH}}_{3}] must be found. Substitute for \mathrm{[CH_{3}^{•}]}
{}[^{•}{\mathrm{CH}}_{2}{\mathrm{OCH}}_{3}]={\frac{k_{2}}{k_{3}}}\left({\frac{k_{1}}{2\,k_{4}}}\right)^{1/2}[{\mathrm{CH}}_{3}{\mathrm{OCH}}_{3}]^{3/2} (6.119)
giving
\frac{\mathrm{d}[\mathrm{HCHO}]}{\mathrm{d}t}=k_{3}[^{•}\mathrm{CH}_{2}\mathrm{OCH}_{3}] (6.120)
=k_{2}\left(\frac{k_{1}}{2\,k_{4}}\right)^{1/2}\!\left[\mathrm{CH}_{3}\mathrm{OCH}_{3}\right]^{3/2} (6.115)
as before.
Normally the easiest algebra is chosen: here \mathrm{[CH_{3}^{•}]} was found first from solution of the steady state equations, and so either (a) or (b) is chosen in preference. A very important point to notice. The steady state analysis shows that not all the individual rate constants can be found from the steady state reaction. In this case the steady state expression only involves three of the rate constants: k_{3} does not appear