Question 6.FP.5: The following is a possible mechanism for the decomposition ......

With this mechanism the main difficulty lies in figuring out what is going on.

Initiation produces \mathrm{CH}_{3}^{•} and C_{2}\mathrm{H}_{5}^{•}\cdot\mathrm{CH}_{3}^{•} is undoubtedly a chain carrier: it is used up in step 2, and in doing so produces {\mathrm{C}}_{3}{\mathrm{H}}_{7}^{•}, and it is regenerated in step 4, which uses up {\mathrm{C}}_{3}{\mathrm{H}}_{7}^{•},. Therefore, {\mathrm{C}}_{3}{\mathrm{H}}_{7}^{•},  is also a chain carrier.

{\mathrm{C}}_{2}{\mathrm{H}}_{5}^{•}, is first produced in initiation, and it then is used up in step 3, which produces {\mathrm{C}}_{3}{\mathrm{H}}_{7}^{•} . {\mathrm{C}}_{3}{\mathrm{H}}_{7}^{•},  then forms \mathrm{CH}_{3}^{•}, which participates in the \mathrm{CH}_{3}^{•} / {\mathrm{C}}_{3}{\mathrm{H}}_{7}^{•}, chain. But {\mathrm{C}}_{2}{\mathrm{H}}_{5}^{•}, is not regenerated and is, therefore, not a chain carrier. But, because it produces a chain carrier, it must be included in the steady state equations. Had it only been produced in initiation and never appeared again, it could have been ignored. It would then have eventually produced a very trace amount of a minor product by some reaction not associated with the chain. But because it produces a chain carrier it cannot be ignored, and a steady state equation must be produced for it. However, it must not be regarded as a chain carrier. Chain carriers are \mathrm{CH}_{3}^{•}   and {\mathrm{C}}_{3}{\mathrm{H}}_{7}^{•} .

+\frac{\mathrm{d}[C{\bf H}_{3}^{•}]}{\mathrm{d}t}= k_{1}[C_{3}\mathrm{H_{8}}]-k_{2}[C\mathrm{H}_{3}^{•}][C_{3}\mathrm{H_{8}}]+k_{4}[C_{3}\mathrm{H}_{7}^{•}]=0                   (I)

+\frac{\mathrm{d}[C_{2}{\mathrm{H}}_{5}^{•}]}{\mathrm{d}t}=k_{1}[C_{3}{\mathrm{H}}_{8}]-k_{3}[C_{2}{\mathrm{H}}_{5}^{•}][C_{3}{\mathrm{H}}_{8}]=0                         (II)

\frac{\mathrm{d}[{\bf C}_{3}{\bf H}_{7}^{•}]}{\mathrm{d}t}=k_{2}[{\bf C}{\bf H}_{3}^{•}][{\bf C}_{3}{\bf H}_{8}]+k_{3}[{\bf C}_{2}{\bf H}_{5}^{•}][{\bf C}_{3}{\bf H}_{8}]-k_{4}[{\bf C}_{3}{\bf H}_{7}^{•}]

-\;2\,k_{5}[{\bf C}_{3}{\bf H}_{7}^{•}]^{2}-2\,k_{6}[{\bf C}_{3}{\bf H}_{7}^{•}]^{2}=0             (III)

Add (I), (II) and (III):

2\,k_{1}[{\bf C}_{3}{\bf H}_{8}]=2(k_{5}+k_{6})[{\bf C}_{3}{\bf H}_{7}^{•}]^{2}=0\,

∴  [{{C}}_{3}{H}_{7}^{•}]=\left(\frac{k_{1}}{k_{5}+k_{6}}\right)^{1/2}[{{C}}_{3}{H}_{8}]^{1/2}

The overall rate of reaction can be expressed in terms of removal of reactant {\mathrm{C}}_{3}{\mathrm{H}}_{8}{\mathrm{:}} this requires [\,\mathrm{CH}_{3}^{\bullet}\,]\,\mathrm{~and~}[C_{2}\mathrm{H}_{5}^{\bullet}].

[\left.C_{2}H_{5}^{•}\right] appears in step 3 and this term will drop out from the overall rate because it is not a propagation step, and although it produces \mathrm{C}_{3}\mathrm{H}_{7}^{•} this step is not essential for the continuance of the chain. Rate of propagation \gg\mathrm{rate~of~}. step 3. This will leave a two-term equation, which can be approximated by the long chains approximation, since rate of initiation \ll\mathrm{rate~of~} propagation.

-{\frac{\mathrm{d}[C_{3}\mathrm{H}_{8}]}{\mathrm{d}t}}=k_{1}[C_{3}\mathrm{H}_{8}]+k_{2}[C\mathrm{H}_{3}^{•}][C_{3}\mathrm{H}_{8}]+k_{3}[C_{2}\mathrm{H}_{5}^{•}][C_{3}\mathrm{H}_{8}]

With the above approximations, this reduces to

-{\frac{\mathrm{d}[C_{3}\mathrm{H}_{8}]}{\mathrm{d}t}}=k_{2}[C\mathrm{H}_{3}^{•}][C_{3}\mathrm{H}_{8}]

[\mathrm{CH}_{3}^{•}] has to be found by substituting [{\bf C}_{3}{\bf H}_{7}^{•}]=(k_{1}/(k_{5}+k_{6}))^{1/2}[{\bf C}_{3}{\bf H}_{8}]^{1/2} into equation (I)

+\frac{\mathrm{d}[C{\mathrm{H}}_{3}^{•}]}{\mathrm{d}t}=k_{1}[C_{3}{\mathrm{H}}_{8}]-k_{2}[C{\mathrm{H}}_{3}^{•}][C_{3}{\mathrm{H}}_{8}] +\,k_{4}\biggl({\frac{k_{1}}{k_{5}+k_{6}}}\biggr)^{1/2}[C_{3}\mathrm{H}_{8}]^{1/2}=0

∴ k_{2}[\mathbf{CH}_{3}^{\bullet}|[\mathbf{C}_{3}\mathbf{H}_{8}]=k_{1}[\mathbf{C}_{3}\mathbf{H}_{8}]+k_{4}\biggl(\frac{k_{1}}{k_{5}+k_{6}}\biggr)^{1/2}[\mathbf{C}_{3}\mathbf{H}_{8}]^{1/2}

∴   [{\bf C H}_{3}^{\bullet}]={\frac{k_{1}}{k_{2}}}+{\frac{k_{4}}{k_{2}}}\left({\frac{k_{1}}{k_{5}+k_{6}}}\right)^{1/2}[{\bf C_{3}H _{8}}]^{-1/2}

-{\frac{\mathrm{d}[C_{3}\mathrm{H}_{8}]}{\mathrm{d}t}}=k_{2}[C\mathrm{H}_{3}^{•}][C_{3}\mathrm{H}_{8}]

=k_{1}[C_{3}{\mathrm{H}}_{8}]+k_{4}\biggl(\frac{k_{1}}{k_{5}+k_{6}}\biggr)^{1/2}[{\mathrm C_{3}{\mathrm{H}}_{8}}]^{1/2}

Because of long chains

k_{1}[\mathbf{C}_{3}\mathbf{H}_{8}]\ll k_{4}\biggl(\frac{k_{1}}{k_{5}+k_{6}}\biggr)^{1/2}[\mathbf{C}_{3}\mathbf{H}_{8}]^{1/2}

∴ \frac{\mathrm{d}[{\bf C}_{3}{\bf H}_{8}]}{\mathrm{d}t}=k_{4}\biggl(\frac{k_{1}}{k_{5}+k_{6}}\biggr)^{1/2}[{\bf C}_{3}{\bf H}_{8}]^{1/2}

If the simpler procedure of looking at the rate of production of the major product, [C_{2}\mathrm{H}_{4}], is used

+\frac{\mathrm{d}[C_{2}\mathbf{H}_{4}]}{\mathrm{d}t}=k_{4}[C_{3}\mathbf{H}_{7}^{•}]

and by substituting for \left[C_{3}H_{7}^{•}\right]

=k_{4}\left(\frac{k_{1}}{k_{5}+k_{6}}\right)^{1/2}[\mathbf{C}_{3}\mathbf{H}_{8}]^{1/2}

Note. The overall rate of reaction must not be expressed in terms of production of \mathrm{C}_{2}\mathrm{H}_{6} as this is only a minor side reaction and not a rate of propagation.

k_{\mathrm{obs}}=k_{4}\biggl(\frac{k_{1}}{k_{5}+k_{6}}\biggr)^{1/2}