Question 6.WP.3: Decomposition of di-2-methylpropan-2-yl peroxide produces pr......
Decomposition of di-2-methylpropan-2-yl peroxide produces propanone and ethane:
({\mathrm{CH}}_{3})_{3}{\mathrm{COOC}}({\mathrm{CH}}_{3})_{3}\to2{\mathrm{CH}}_{3}{\mathrm{COCH}}_{3}+{\mathrm{C}}_{2}{\mathrm{H}}_{6}
and the generally accepted mechanism is
({\mathrm{CH}}_{3})_{3}{\mathrm{COOC}}({\mathrm{CH}}_{3})_{3}\stackrel{k_{1}}{\longrightarrow}2({\mathrm{CH}}_{3})_{3}{\mathrm{CO}}^{•}
({\mathrm{CH}}_{3})_{3}C O^{•}\ {\stackrel{k_{2}}{\longrightarrow}}\ {\mathrm{CH}}_{3}C O C\mathrm{H}_{3}+{\mathrm{CH}}_{3}^{•}
\mathrm{CH}_{3}^{•}+\mathrm{CH}_{3}^{•}\stackrel{k_{3}}{\longrightarrow}\mathrm{C}_{2}\mathrm{H}_{6}
1. Explain why this is not a chain reaction.
2. Using the steady state assumption show that the reaction is first order throughout, even though it occurs in three consecutive steps. Formulate the rate of reaction in terms of production of \mathrm{C}_{2}\mathrm{H}_{6}.
3. Use of the steady state treatment requires that the radicals {\mathrm{CH}}_{3}^{•}{\mathrm{~and~}}({\mathrm{CH}}_{3})_{3}{\mathrm{CO}}^{•} are present in very, very low concentrations. Explain how this reaction can still be a standard source of {\mathrm{CH}}_{3}^{•}.
1. The intermediates are {\mathrm{CH}}_{3}^{•}{\mathrm{~and~}}({\mathrm{CH}}_{3})_{3}{\mathrm{CO}}^{•}. Although the ({ C H}_{3})_{3}{ C}O^{•} splits up to form {\mathrm{CH}}_{3}^{•} this does not regenerate ({ C H}_{3})_{3}{ C}O^{•} , and so this cannot be a chain reaction.
2. Steady states on the intermediates:
\frac{\mathrm{d}[({\mathrm{CH}}_{3})_{3}\,{\mathrm{CO}}^{•}]}{\mathrm{d}t}=2\,k_{1}[({\mathrm{CH}}_{3})_{3}{\mathrm{COOC}}({\mathrm{CH}}_{3})_{3}]-k_{2}[({\mathrm{CH}}_{3})_{3}{\mathrm{CO}}^{•}]=0 (6.5)
rate ~ of ~ production ∴ +\overset{\uparrow}{\mathrm{ve}} ~\overset{\uparrow}{\mathrm{rate ~ of ~ removal}} ∴ – ve
Note the factor of two: for each step 1, two ({ C H}_{3})_{3}{ C}O^{•} are formed, and the rate of step 1 is given in terms of production of ({ C H}_{3})_{3}{ C}O^{•}
∴ [({\bf C H}_{3})_{3}\mathrm{CO}^{•}]=\frac{2\,k_{1}}{k_{2}}[({\bf C H}_{3})_{3}\mathrm{COOC}(\mathrm{CH}_{3})_{3}] (6.6)
\frac{\mathrm{d}[\mathbf{CH}_{3}^{•}]}{\mathrm{d}t}=k_{2}[(\mathbf{CH}_{3})_{3}\mathbf{CO}^{•}]-2\,k_{3}[\mathbf{CH}_{3}^{•}]^{2}=0 (6.7)
rate ~of~ production ∴ +\overset{\uparrow}{\mathrm{ve}}~ \overset{\uparrow}{\mathrm{rate~ of~ removal}} ∴ – ve
Again a factor of two appears because two {\mathrm{CH}}_{3}^{•} radicals are removed for each act of recombination.
This is an equation in two unknowns and, unlike equation (6.5), cannot be solved. [{\mathrm{CH}}_{3}^{•}] can be found by either of two methods.
(a) Use the standard procedure of adding the steady state equations, giving
[\mathrm{CH}_{3}^{•}]=\left({\frac{k_{1}}{k_{3}}}\right)^{1/2}[(\mathrm{CH}_{3})_{3}\mathrm{COOC}(\mathrm{CH}_{3})_{3}]^{1/2} (6.8)
(b) Substitute for [({\bf C H}_{3})_{3}{\bf C}O^{•}] from (6.6) into (6.7), giving
[\mathrm{CH}_{3}^{•}]=\left(\frac{k_{1}}{k_{3}}\right)^{1/2}[(\mathrm{CH}_{3})_{3}\mathrm{COOC}(\mathrm{CH}_{3})_{3}]^{1/2} (6.9)
which is the same result, confirming the correctness of the algebra.
Rate of production of \mathrm{C}_{2}{\mathrm{H}}_{6}:
{\frac{\mathrm{d}[C_{2}{\mathrm{H}}_{6}]}{\mathrm{d}t}}=k_{3}[\mathrm{C{H}_{3}^{•}}]^{2} (6.10)
=\frac{k_{3}k_{1}}{k_{3}}[({\mathrm{CH}}_{3})_{3}{\mathrm{COOC}}({\mathrm{CH}}_{3})_{3}] (6.11)
=k_{1}[({\mathrm{CH}}_{3})_{3}{\mathrm{COOC}}({\mathrm{CH}}_{3})_{3}] (6.12)
which predicts the reaction to be first order in reactant, as is observed
experimentally, confirming that the mechanism fits the observed kinetics. Note: the rate of step 3 can be expressed in terms of
-\,\frac{\mathrm{d}[{\bf C H}_{3}^{•}]}{\mathrm{d}t}\quad\mathrm{or}\quad+\frac{\mathrm{d}[{\bf C}_{2}{\bf H}_{6}]}{\mathrm{d}t}
For each step of reaction, two {\mathrm{CH}}_{3}^{•} radicals are removed, and one \mathrm{C}_{2}\mathrm{H}_{6} is formed.
∴ \mathrm{rate~at~which~3~removes~CH_{3}^{•}=2~k_{3}[C H_{3}^{•}]^{2}} (6.13)
\mathrm{rate~at~which~C_{2}H_{6}~i s~f o r m e d}=+{\frac{\mathrm{d}[C_{2}H_{6}]}{\mathrm{d}t}}=k_{3}[C{\bf H}_{3}^{•}]^{2} (6.14)
so that
rate at which 3 removes \mathrm{CH}_{3}^{•}=+2\,{\frac{\mathrm{d}[C_{2}\mathbf{H}_{6}]}{\mathrm{d}t}} (6.15)
It is very important to understand these distinctions.
3. When the reaction is allowed to proceed without disturbance the {\mathrm{CH}}_{3}^{•} radicals are present in steady state concentrations. However, if they are removed continuously from the reaction vessel, the reaction never gets to the steady state.