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Chemistry LibreTexts

29.7: Some Reaction Mechanisms Involve Chain Reactions

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

\[H_2 + Br_2 \rightarrow 2HBr\]

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.

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

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

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

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

Solving for \(Br^\cdot\)

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

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

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

so

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

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 \(HBr\):

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

After substitution and simplification, the result is

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

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

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

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

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

and

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

Contributors

  • Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)