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12.9: Chain Reactions

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    A large number of reactions proceed through a series of steps that can collectively be classified as a chain reaction. The reactions contain steps that can be classified as

    • initiation step – a step that creates the intermediates from stable species
    • propagation step – a step that consumes an intermediate, but creates a new one
    • termination step – a step that consumes intermediates without creating new ones

    These types of reactions are very common when the intermediates involved are radicals. An example, is the reaction

    \[H_2 + Br_2 \rightarrow 2HBr \nonumber \]

    The observed rate law for this reaction is

    \[ \text{rate} = \dfrac{k [H_2][Br_2]^{3/2}}{[Br_2] + k'[HBr]} \label{exp} \]

    A proposed mechanism is

    \[Br_2 \ce{<=>[k_1][k_{-1}]} 2Br^\cdot \label{step1} \]

    \[ 2Br^\cdot + H_2 \ce{<=>[k_2][k_{-2}]} HBr + H^\cdot \label{step2} \]

    \[ H^\cdot + Br_2 \xrightarrow{k_3} HBr + Br^\cdot \label{step3} \]

    Based on this mechanism, the rate of change of concentrations for the intermediates (\(H^\cdot\) and \(Br^\cdot\)) can be written, and the steady state approximation applied.

    \[\dfrac{d[H^\cdot]}{dt} = k_2[Br^\cdot][H_2] - k_{-2}[HBr][H^\cdot] - k_3[H^\cdot][Br_2] =0 \nonumber \]

    \[\dfrac{d[Br^\cdot]}{dt} = 2k_1[Br_2] - 2k_{-1}[Br^\cdot]^2 - k_2[Br^\cdot][H_2] + k_{-2}[HBr][H^\cdot] + k_3[H^\cdot][Br_2] =0 \nonumber \]

    Adding these two expressions cancels the terms involving \(k_2\), \(k_{-2}\), and \(k_3\). The result is

    \[ 2 k_1 [Br_2] - 2k_{-1} [Br^\cdot]^2 = 0 \nonumber \]

    Solving for \(Br^\cdot\)

    \[ Br^\cdot = \sqrt{\dfrac{k_1[Br_2]}{k_{-1}}} \nonumber \]

    This can be substituted into an expression for the \(H^\cdot\) that is generated by solving the steady state expression for \(d[H^\cdot]/dt\).

    \[[H^\cdot] = \dfrac{k_2 [Br^\cdot] [H_2]}{k_{-2}[HBr] + k_3[Br_2]} \nonumber \]

    so

    \[[H^\cdot] = \dfrac{k_2 \sqrt{\dfrac{k_1[Br_2]}{k_{-1}}} [H_2]}{k_{-2}[HBr] + k_3[Br_2]} \nonumber \]

    Now, armed with expressions for \(H^\cdot\) and \(Br^\cdot\), we can substitute them into an expression for the rate of production of the product \(HBr\):

    \[ \dfrac{[HBr]}{dt} = k_2[Br^\cdot] [H_2] + k_3 [H^\cdot] [Br_2] - k_{-2}[H^\cdot] [HBr] \nonumber \]

    After substitution and simplification, the result is

    \[ \dfrac{[HBr]}{dt} = \dfrac{2 k_2 \left( \dfrac{k_1}{k_{-1}}\right)^{1/2} [H_2][Br_2]^{1/2}}{1+ \dfrac{k_{-1}}{k_3} \dfrac{[HBr]}{[Br_2]} } \nonumber \]

    Multiplying the top and bottom expressions on the right by \([Br_2]\) produces

    \[ \dfrac{[HBr]}{dt} = \dfrac{2 k_2 \left( \dfrac{k_1}{k_{-1}}\right)^{1/2} [H_2][Br_2]^{3/2}}{[Br_2] + \dfrac{k_{-1}}{k_3} [HBr] } \nonumber \]

    which matches the form of the rate law found experimentally (Equation \ref{exp})! In this case,

    \[ k = 2k_2 \sqrt{ \dfrac{k_1}{k_{-1}}} \nonumber \]

    and

    \[ k'= \dfrac{k_{-2}}{k_3} \nonumber \]

    This page titled 12.9: Chain Reactions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Patrick Fleming.