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# 15.4: 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 and Attributions

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

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