I am trying to derive the overall rate equation for the photocatalysed halogenation (chlorination) of an alkane that proceed via this 4 step chain reaction:

$\ce{Cl2->2Cl^. \quad(1) initiation}$

$\ce{Cl^. +RH->R^. + HCl \quad(2) propagation}$

$\ce{R^. + Cl2-> RCl + Cl^. \quad(3) propagation}$

$\ce{R^. + Cl^. ->RCl \quad(4) termination}$

I have calculated the differential rate equations for $\ce{R^.}$ and $\ce{Cl^.}$ and set them to zero at steady state.

[The free radical dot symbols are omitted from this point onward for brevity]

This gives me $$\ce{For Cl ,2k1[Cl2]-k2[Cl][RH] +k3[R][Cl2] -k4[R][Cl]=0 \quad(5)}$$ $$\ce{For R, k2[Cl][RH] -k3[R][Cl2] -k4[R][Cl]=0 \quad(6)}$$ Adding these two equations, $$\ce{[Cl]=\frac{k1}{k4}\frac{[Cl2]}{[R]}}$$

Then I tried substituting this into equation (6), which gives me a quadratic equation. The solution of the quadratic equation is of the form $$\ce{[R]=constant +\sqrt{\mathrm{constant [RH]}}}$$

The overall rate law from this expression is a very complicated equation.

This looks wrong because the rate law for radical halogenation is usually of the form $\ce{k[RH]\sqrt{\mathrm{[Cl_2]}}}$ (Wikipedia)

Where am I going wrong?

  • 1
    $\begingroup$ The initiation rate may not depend on [Cl2]( for high enough [Cl2]), but on irradiation level only. i.e being a reaction of zero order . $\endgroup$ – Poutnik Jan 8 '20 at 18:56

In the book that Wikipedia cites (Advanced Organic Chemistry: Reaction Mechanisms By Reinhard Bruckner, ISBN 9780080498805), they have a different set of assumptions. They are saying the formation of chlorine radicals proceeds by a fast equilibrium. Then, they say that the organic radical is at steady state, ignoring reaction (4) given by the OP (the termination reaction).

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Where am I going wrong?

I'm not sure why your steady state approximation is less appropriate than that of the textbook. I'm sure that there is some experimental data (and some fitted rate constants) that would speak to that. Also, the appropriate approximation depends on the initial conditions.

[Comment from Poutnik] The initiation rate may not depend on [Cl2]( for high enough [Cl2]), but on irradiation level only. i.e being a reaction of zero order.

The derivation in the book is for the thermal reaction, but the OP asked about the photocatalyzed reaction, so there is a discrepancy there.

  • $\begingroup$ But I don't understand why using one termination reaction (R and Cl) instead of another termination reaction (Cl and Cl, which would give the equilibrium) would change the rate law. The text book's rate law is $\ce{\sqrt{[Cl2]}[RH]}$ whereas the rate law that I am getting is $\ce{\sqrt{const.+[RH]}[Cl2]}$. (Also this is a question from a book, so it must be a well known problem) $\endgroup$ – Shoubhik R Maiti Jan 11 '20 at 18:15
  • $\begingroup$ @ShoubhikRMaiti That is not the only difference. For your set of equations, [Cl] depends on [R]. For the assumptions in the textbook, it doesn't, so it becomes possible to get a simple expression for the steady state of [R], and for the rate of product formation. $\endgroup$ – Karsten Theis Jan 11 '20 at 20:29

The mechanism of the gas phase halogenation reactions is discussed in some detail in a paper by Benson and Bus[1]. In the following, I submarised section II ( mechanism and rate laws).

The mechanism of the reaction thermal reaction is

$\ce{X2 + M -> 2X^. + M \;\; k_1 \quad(1) }$

$\ce{2X^. + M -> X2 + M \;\; k_{-1} \quad(-1) }$

$\ce{X^. + RH -> R^. + HX \;\; k_2 \quad(2) }$

$\ce{R^. + HX -> X^. + RH \;\; k_{-2} \quad(-2)} $

$\ce{R^. + X2-> RX + X^. \;\; k_3 \quad(3) }$

$\ce{X^. + RX -> R^. + X2 \;\; k_{-3} \quad(-3) }$

$\ce{R^. + X^. (+ M) ->RCl (+ M) \;\; k_{4} \quad(4) }$

$\ce{R^. + R^. (+ M) -> R2 (+ M) \;\; k_{5} \quad(5)}$

Using the stationary state approximation for the intermediates $R^.$ and $X^.$, and neglecting the back reaction 4 which is generally unimportant[[1]] you find

$ \ce{\frac{|R|}{|X|} = \, \theta \; = \frac{k_2 |RH|}{\,k_3 |X_2|\,} \left(1 + \frac{k_2 |HX|}{k_3 |X_2|}\right)^{-1}}$

$ \ce{|X|_{ss} = \left(\frac{k_1 |M||X_2|}{k_{-2}|M| + k_4 \theta + k_5 \theta^2 } \right)^{1/2} }$

This equation results in

$ \ce{-\frac{d|X_2|}{dt} = k_1|RX||X|_{ss} \left( 1 + \frac{k_{-2}|HX|}{k_{3}|X_2|}\right)^{-1}} $

Depending on which of the postulated termination reactions predominates, the authors obtain three expressions for the velocity equation. Thus in case I,

$ \ce{-\frac{d|X_2|}{dt} = k'|RX||X_2|^{1/2} \left( 1 + \frac{k_{-2}|HX|}{k_{3}|X_2|}\right)^{-1}} $

that for low concentrations of HX reduces to

$ \ce{-\frac{d|X_2|}{dt} = k'|RX||X_2|^{1/2} } $

The equations for a photocatalized reaction are similar to the ones above but you must substitute $k_1$ by $I_a/|M||X_2|$ where $I_a$ is the specific rate of absorption of light.

[1]: S. W. Benson, and J. H. Buss, Kinetics of Gas Phase Halogenation Reactions, J. Chem. Phys. 28, 301 (1958); http://doi.org/10.1063/1.1744111


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