## Question

###### A simplified mechanism for this reaction is $$\begin{array}{c} \text { electric spark }+\mathrm{H}_{2}(\mathrm{g}) \Longrightarrow 2 \mathrm{H}(\mathrm{g}) \\ \mathrm{H}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \stackrel{k_{1}}{\longrightarrow} \mathrm{OH}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \\ \mathrm{O}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{OH}(\mathrm{g})+\mathrm{H}(\mathrm{g}) \\ \mathrm{H}_{2}(\mathrm{g})+\mathrm{OH}(\mathrm{g}) \stackrel{k_{3}}{\long

A simplified mechanism for this reaction is $$\begin{array}{c} \text { electric spark }+\mathrm{H}_{2}(\mathrm{g}) \Longrightarrow 2 \mathrm{H}(\mathrm{g}) \\ \mathrm{H}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \stackrel{k_{1}}{\longrightarrow} \mathrm{OH}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \\ \mathrm{O}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{OH}(\mathrm{g})+\mathrm{H}(\mathrm{g}) \\ \mathrm{H}_{2}(\mathrm{g})+\mathrm{OH}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}(\mathrm{g}) \\ \mathrm{H}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \stackrel{k_{4}}{\longrightarrow} \mathrm{HO}_{2}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \end{array}$$ A reaction that produces more molecules that can participate in chain-propagation steps than it consumes is called a branching chain reaction. Label the branching chain reaction(s), inititation reaction(s), propagation reaction(s), and termination reaction(s) for this mechanism. Use the following bond dissociation energies to evaluate the energy change for steps (2) and (3) $$\begin{array}{cc} \text { Molecule } & D_{0} / \mathrm{kJ} \cdot \mathrm{mol}^{-1} \\ \hline \mathrm{H}_{2} & 432 \\ \mathrm{O}_{2} & 493 \\ \mathrm{OH} & 424 \end{array}$$