Chain and photochemical kinetics

Chain reactions proceed through three kinds of steps, repeated many times over: Initiation generates a small number of reactive chain carriers (radicals or atoms), often thermally or photochemically, e.g. $\text{Br}_2 \xrightarrow{k_1} 2\text{Br}^{\bullet}$. Propagation lets a carrier react with a stable molecule to give product plus a fresh carrier, so the cycle regenerates itself: $\text{Br}^{\bullet}+\text{H}_2\xrightarrow{k_2}\text{HBr}+\text{H}^{\bullet}$, then $\text{H}^{\bullet}+\text{Br}_2\xrightarrow{k_3}\text{HBr}+\text{Br}^{\bullet}$. Termination removes carriers by recombination, e.g. $2\text{Br}^{\bullet}\xrightarrow{k_5}\text{Br}_2$, ending that particular chain. The efficiency of the whole process is captured by the chain length, the average number of propagation cycles per initiation event:

where: the numerator and denominator are both rates (mol L⁻¹ s⁻¹) of the propagation and initiation steps respectively. Chain lengths of $10^3$–$10^6$ are common, which is exactly why a handful of initiation events can consume a large amount of reactant — the figure below traces one such cycle.

The classic $\text{H}_2$–$\text{Br}_2$ mechanism adds one extra wrinkle, an inhibition step where product reacts back with a carrier:

Applying SSA to both radical intermediates and solving the simultaneous algebraic equations (worked out below) gives the famous rate law

where: $[\text{H}_2]$, $[\text{Br}_2]$ and $[\text{HBr}]$ are molar concentrations and $k_1$–$k_5$ are the five step rate constants above. Two features here are impossible for any single-step rate law: a half-order dependence on $[\text{Br}_2]$, and explicit inhibition by the product $\text{HBr}$ sitting in the denominator.

Explosion limits appear when branching steps (one radical producing two, as in $\text{H}^{\bullet}+\text{O}_2\to\text{OH}^{\bullet}+\text{O}^{\bullet\bullet}$ for $\text{H}_2$–$\text{O}_2$) outrun termination — carrier concentration then grows exponentially instead of holding steady, and the reaction accelerates uncontrollably. Whether branching wins depends on pressure and temperature, which is why $\text{H}_2$–$\text{O}_2$ mixtures show characteristic explosion-limit curves rather than a single ignition condition.

Photochemical kinetics governs reactions initiated by absorbed light and rests on two laws. The Grotthuss–Draper law says only light actually absorbed can cause photochemical change. The Stark–Einstein law (law of photochemical equivalence) says each absorbed photon activates exactly one molecule in the primary step, though that one activated molecule can go on to trigger many further thermal steps. Efficiency is measured by the quantum yield:

where: both numerator and denominator are counted in the same unit (molecules, or equivalently moles, since Avogadro's number cancels). A simple primary process with no chain gives $\Phi\approx1$; a secondary chain reaction (as in photochemical $\text{H}_2$–$\text{Cl}_2$) can push $\Phi$ up to $10^6$, since one photon starts a long thermal chain; and $\Phi<1$ signals a competing loss process such as fluorescence or recombination before reaction.

For the $\text{H}_2$–$\text{Br}_2$ chain reaction, $\dfrac{d[\text{HBr}]}{dt} = 1.2\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}$ is measured with $[\text{H}_2]=0.50\ \text{mol L}^{-1}$, $[\text{Br}_2]=0.20\ \text{mol L}^{-1}$, negligible $[\text{HBr}]$ (denominator $\approx1$), and composite constant $k_2(k_1/k_5)^{1/2}=2.68\times10^{-4}\ \text{L}^{1/2}\text{mol}^{-1/2}\text{s}^{-1}$. Verify the rate law fits, then use the same reaction context to check a photochemical alternative.

Step 1. Apply the early-time rate law: $\dfrac{d[\text{HBr}]}{dt}\approx2k_2(k_1/k_5)^{1/2}[\text{H}_2][\text{Br}_2]^{1/2}$.

Step 2. Substitute values: $(0.20)^{1/2}=0.4472$, so the rate is $2\times2.68\times10^{-4}\times0.50\times0.4472=1.198\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}$.

Step 3. Compare: the calculated $1.20\times10^{-4}$ matches the measured $1.2\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}$, confirming the half-order-in-bromine chain rate law. The predicted rate, $1.20\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}$, agrees with the measured value.

Competitive stretch

Now suppose instead $3.0\times10^{-3}$ mol of photons are absorbed by a $\text{Cl}_2$–$\text{H}_2$ mixture under UV, and $180\ \text{g}$ of $\text{HCl}$ ($M=36.5\ \text{g mol}^{-1}$) forms.

Step 1. Moles of $\text{HCl}$: $n=180/36.5=4.932\ \text{mol}$.

Step 2. Quantum yield: $\Phi=n_{\text{HCl}}/n_{\text{photons}}=4.932/(3.0\times10^{-3})\approx1.64\times10^{3}$.

Step 3. Interpret: $\Phi\gg1$ rules out a simple one-photon-one-molecule process, and points to the classic $\text{H}_2$–$\text{Cl}_2$ photochemical chain: photodissociation $\text{Cl}_2\xrightarrow{h\nu}2\text{Cl}^{\bullet}$ is the single primary act, but each $\text{Cl}^{\bullet}$ then propagates a long thermal chain before termination. $\Phi\approx1.64\times10^{3}$, directly reporting the chain length.

• Assuming the Stark–Einstein law caps $\Phi$ at 1. It only constrains the primary act; the overall measured $\Phi$ can be far above 1 (chain amplification) or below 1 (competing deactivation).
• Forgetting $\Phi$ is defined relative to photons absorbed, not photons incident — the fraction absorbed must be known separately.
• Misremembering the order in $[\text{Br}_2]$ as first order by analogy with simpler laws; the half order traces directly to the square root in Step 4 of the SSA derivation above.
• Confusing branching with ordinary propagation. Propagation conserves carrier number (one radical in, one out); branching produces more carriers than it consumes and drives explosive kinetics.

Ask the tutor: if you want the full simultaneous-SSA algebra carried out with explicit intermediate expressions for $[\text{H}^{\bullet}]$ substituted back in, or a look at how explosion-limit curves for $\text{H}_2$–$\text{O}_2$ are actually plotted against pressure and temperature.