Question 6.FP.3: Using the steady state treatment deduce the rate expressions......

(a) Chain carriers are {I}^{•},\;{C}_{2}{H}_{4}{I}^{•}

\frac{\mathrm{d}[\mathrm{I^{•}}]}{\mathrm{dt}}= k_{1}[{\bf C}_{2}{\bf H_{4}}I_{2}]-k_{2}[I^{•}][{\bf C}_{2}{\bf H_{4}}I_{2}]+k_{3}[{\bf C}_{2}{\bf H_{4}}I_{}^{•}]-\,2\,k_{4}[I^{•}]^{2}[{\bf M}]=0         (I)

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

Add (I) and (II):

2\,k_{1}[C_{2}{\bf H}_{4}{\bf I}_{2}]=2\,k_{4}[{\bf I}^{•}]^{2}[{\bf M}]

[{ I}^{•}]=\left(\frac{k_{1}}{k_{4}}\right)^{1/2}\!\frac{[{\bf C}_{2}{\bf H}_{4}{\bf I}_{2}]^{1/2}}{[{\bf M}]^{1/2}}                 (III)

Choose to express the rate in terms of \left[I{}^{\bullet}\right]

-{\frac{\mathrm{d}[C_{2}{\mathrm{H}}_{4}{\mathrm{I}}_{2}]}{\mathrm{d}t}}=k_{1}[C_{2}{\mathrm{H}}_{4}{\mathrm{I}}_{2}]+k_{2}[C_{2}{\mathrm{H}}_{4}{\mathrm{I}}_{2}][{\mathrm{I}}^{•}]

\approx k_{2}[C_{2}{\mathrm{H}}_{4}{\mathrm{I}}_{2}][{\mathrm{I}}^{•}]

(i.e. by using the long chains approximation rate of initiation \ll\mathrm{rate}\;\;\mathrm{of} propagation)

=k_{2}\left(\frac{k_{1}}{k_{4}}\right)^{1/2}\!\!\frac{[C_{2}\mathbf{H_{4}}\mathbf{I}_{2}]^{3/2}}{[\mathbf{M}]^{1/2}}

or

+\frac{\mathrm{d}[\mathbf{I}_{2}]}{\mathrm{d}t}=k_{2}[\mathbf{C}_{2}\mathbf{H}_{4}\mathbf{I}_{2}][\mathbf{I}^{•}]+k_{4}[\mathbf{I}^{•}]^{2}[\mathbf{M}]

\approx k_{2}[C_{2}{\bf H}_{4}{\bf I}_{2}][{\bf I}^{•}]

(i.e. by using the long chains approximation rate of termination \ll\mathrm{rate}\;\;\mathrm{of} propagation)

=k_{2}\left(\frac{k_{1}}{k_{4}}\right)^{1/2}\!\!\frac{[C_{2}\mathbf{H}_{4}\mathbf{I}_{2}]^{3/2}}{[\mathbf{M}]^{1/2}}

It would not be sensible to express the overall rate in terms of

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

as this requires a knowledge of the as yet unknown [{C}_{2}{H}_{4}{I}^{•}],\ {\mathrm{which}} would have to be found by substituting into one or other of the steady state equations and solving for [{C}_{2}{H}_{4}{I}^{•}],\ {\mathrm{}}, or by using the long chains approximation in the form that the rates of the two propagation steps are equal.

k_{2}[I^{\bullet}][C_{2}\mathrm{H}_{4}\mathrm{I}_{2}]=k_{3}[C_{2}\mathrm{H}_{4}\mathrm{I^{\bullet}}]            ∴   [{{I^{\bullet}}}]={\frac{k_{3}[{\mathrm{C}}_{2}{\mathrm{H}}_{4}I^{\bullet}]}{k_{2}[{\mathrm{C}}_{2}{\mathrm{H}}_{4}{\mathrm{I}}_{2}]}}   (IV)

But

[{\bf l}^{•}]=\left(\frac{k_{1}}{k_{4}}\right)^{1/2}\!\frac{[{\bf C}_{2}{\bf H}_{4}{\bf I}_{2}]^{1/2}}{[{\bf M}]^{1/2}}       (III)

Solving the simultaneous equations (III) and (IV) gives

\left[C_{2}\mathrm{H}_{4}\mathrm{I}^{•}\right]=\frac{k_{2}}{k_{3}}\left(\frac{k_{1}}{k_{4}}\right)^{1/2}\!\frac{\left[C_{2}\mathrm{H}_{4}\mathrm{I}_{2}\right]^{3/2}}{\left[\mathrm{M}\right]^{1/2}}

+\frac{\mathrm{d}[C_{2}{\mathrm{H}}_{4}]}{\mathrm{d}t}=k_{3}[C_{2}{\mathrm{H}}_{4}{\mathrm{I}}^{•}]=k_{2}\biggl(\frac{k_{1}}{k_{4}}\biggr)^{1/2}{\frac{[C_{2}{\mathrm{H}}_{4}{\mathrm{I}}_{2}]^{3/2}}{[{\mathrm{M}}]^{1/2}}}

as previously.

(b) Chain carriers: C_{4}\mathrm{H}_{9}^{•},\;C_{4}\mathrm{H}_{8}\mathrm{I}^{•},\;\mathrm{I}^{•}

Note: this is a three-stage propagation reaction, but can still be handled in the standard manner.

Add (I), (II) and (III). There are three steady state equations because there are three chain carriers.

2\,k_{1}[{\bf C}_{4}{\bf H}_{9}{\bf I}]=2\,k_{5}[{\bf I}^{•}]^{2}[{\bf M}]                 (IV)

[{I}^{•}]=\left(\frac{k_{1}}{k_{5}}\right)^{1/2}\!\!\frac{[{\bf C}_{4}{\bf H}_{9}]^{1/2}}{[{\bf M}]^{1/2}}

The overall rate can be expressed in terms of

removal of \mathrm{C}_{4}{\mathrm{H}}_{9}{\mathrm{I}}; this requires [\mathrm{C}_{4}{\mathrm{H}}^{•}_{9}{\mathrm{}};] and \left[{I}^{•}\right]

\mathrm{formation~of~I_{2};t h i s~r e q u i r e s~}[I^{•}];

\mathrm{production~of~C_{4}H_{10};\ t h i s~r e q u i r e s~}\left[C_{4}H_{9}^{•}\right];

\mathrm{production~of~C_{4}H_{8};\ t h i s~r e q u i r e s~}\left[C_{4}H_{8}I^{•}\right].

The simplest procedure is to choose the rate which requires [{I}^{•}], , since it is already known

+\frac{\mathrm{d}[\mathbf{I}_{2}]}{d\,t}\!=\!k_{4}[I^{•}][\mathbf{C}_{4}\mathbf{H}_{9}\mathbf{I}]+k_{5}[I^{•}]^{2}[\mathbf{M}]

\approx k_{4}[I^{•}][C_{4}{\mathrm{H_{9}I}}]

(by the long chains approximation rate of termination \ll\mathrm{rate~of~} propagation)

=k_{4}\left(\frac{k_{1}}{k_{5}}\right)^{1/2}\!\!\frac{[C_{4}\mathrm{H}_{9}I]^{3/2}}{[{\mathrm M}]^{1/2}}

(i) could substitute for [I^{\bullet}] into (I), and find [C_{4}H_{9}^{•}],\mathrm{thence}+\mathrm{d}[C_{4}H_{10}]/\mathrm{d}t

(ii) could substitute for \left[C_{4}\mathrm{H}_{9}^{•}\right] into (III), and find \left[C_{4}\mathrm{H}_{8}I^{•}\right]  , thence +\mathrm{d}[\mathbf{C}_{4}{\mathrm{H}}_{8}]/\mathrm{d}t

(iii) having found both [I^{\bullet}] and \left[C_{4}\mathrm{H}_{9}^{•}\right] , thence -\mathrm{d}[\mathbf{C}_{4}{\mathrm{H}}_{9}I]/\mathrm{d}t This would be a three term rate expression involving k_{1},\;k_{2}\;\;\mathrm{and}\;\;k_{4}. The long chains approximation rate of initiation \ll\mathrm{rate~of~} propagation would eliminate the term in k_{1}.