Alright, let's dive into the fascinating world of chain reactions. This is a cornerstone of chemical kinetics, explaining everything from how we make plastics to how explosions work. We'll go through each of the topics from your questions, one by one, and build a solid understanding. Grab a coffee, and let's begin!
At its simplest, a chain reaction is a sequence of reactions where a reactive species (like a radical) is continuously regenerated, allowing the reaction to self-propagate. Think of it like a single domino (the initiator) toppling a whole line of dominoes (the propagation). These reactions are characterized by three distinct steps.
Because the chain carriers (R•) are so reactive, their concentration is always very low and (after a brief induction period) doesn't change much. This is the Steady-State Approximation (SSA). We assume the rate of their formation is equal to the rate of their destruction.
Let's apply this to a general chain reaction: A → P.
Our goal is the overall rate, d[P]/dt. From step 2, we see:
$$ \frac{d[P]}{dt} = k_p[\text{R}^\bullet][A] $$
We can't have [R•] in our final rate law, so we use SSA to find it.
Apply SSA to R•:
$$ \frac{d[\text{R}^\bullet]}{dt} = (\text{Rate of Formation}) - (\text{Rate of Destruction}) \approx 0 $$
$$ (2k_i[A]) - (2k_t[\text{R}^\bullet]^2) = 0 $$
Wait, why isn't kp in the SSA? Because in the propagation step, one R• is consumed and one R• is produced, so the net change in [R•] is zero.
Now, we solve for [R•]:
$$ 2k_i[A] = 2k_t[\text{R}^\bullet]^2 $$
$$ [\text{R}^\bullet]^2 = \frac{k_i[A]}{k_t} $$
$$ [\text{R}^\bullet] = \left( \frac{k_i}{k_t} \right)^{1/2} [A]^{1/2} $$
Finally, substitute this expression for [R•] back into our rate equation for P:
$$ \frac{d[P]}{dt} = k_p \left[ \left( \frac{k_i}{k_t} \right)^{1/2} [A]^{1/2} \right] [A] $$
$$ \frac{d[P]}{dt} = k_p \left( \frac{k_i}{k_t} \right)^{1/2} [A]^{3/2} $$
The overall rate constant is keff = kp(ki/kt)1/2. The key takeaway is the non-integer reaction order (1.5 in this case). This is a classic hallmark of a chain reaction mechanism!
This is a famous and beautiful example of the SSA applied to a real-world reaction: the thermal decomposition of acetaldehyde (CH3CHO) to methane (CH4) and carbon monoxide (CO).
The overall reaction is: CH3CHO(g) ⟶Δ CH4(g) + CO(g)
The mechanism (proposed by Rice and Herzfeld) involves two radical intermediates: the methyl radical (•CH3) and the acetyl radical (CH3CO•).
(Note: The •CHO radical from step 1 can also decompose, but this mechanism provides the best fit). The termination step is the recombination of the most abundant, long-lived radical, which is •CH3.
We want the rate of formation of methane, d[CH4]/dt. From step 2:
$$ \frac{d[\text{CH}_4]}{dt} = k_2 [\bullet\text{CH}_3] [\text{CH}_3\text{CHO}] $$
We need to find [•CH3] using SSA. This time we have two intermediates!
Apply SSA to CH3CO•:
$$ \frac{d[\text{CH}_3\text{CO}^\bullet]}{dt} = k_2[\bullet\text{CH}_3][\text{CH}_3\text{CHO}] - k_3[\text{CH}_3\text{CO}^\bullet] \approx 0 $$
$$ k_2[\bullet\text{CH}_3][\text{CH}_3\text{CHO}] = k_3[\text{CH}_3\text{CO}^\bullet] $$
Apply SSA to •CH3:
$$ \frac{d[\bullet\text{CH}_3]}{dt} = k_1[\text{CH}_3\text{CHO}] - k_2[\bullet\text{CH}_3][\text{CH}_3\text{CHO}] + k_3[\text{CH}_3\text{CO}^\bullet] - 2k_4[\bullet\text{CH}_3]^2 \approx 0 $$
Look! The two middle terms are $ -k_2[\bullet\text{CH}_3][\text{CH}_3\text{CHO}] $ and $ +k_3[\text{CH}_3\text{CO}^\bullet] $. From our first SSA, we know these are equal and opposite, so they cancel out!
This is a common feature in chain reactions. It means the propagation steps that consume one radical are balanced by the steps that produce it. This is called a "long chain" assumption.
Our •CH3 SSA simplifies beautifully to:
$$ k_1[\text{CH}_3\text{CHO}] - 2k_4[\bullet\text{CH}_3]^2 = 0 $$
$$ 2k_4[\bullet\text{CH}_3]^2 = k_1[\text{CH}_3\text{CHO}] $$
$$ [\bullet\text{CH}_3] = \left( \frac{k_1}{2k_4} \right)^{1/2} [\text{CH}_3\text{CHO}]^{1/2} $$
Now, substitute this back into our target rate law for methane:
$$ \frac{d[\text{CH}_4]}{dt} = k_2 \left[ \left( \frac{k_1}{2k_4} \right)^{1/2} [\text{CH}_3\text{CHO}]^{1/2} \right] [\text{CH}_3\text{CHO}] $$
$$ \frac{d[\text{CH}_4]}{dt} = k_2 \left( \frac{k_1}{2k_4} \right)^{1/2} [\text{CH}_3\text{CHO}]^{3/2} $$
And there it is! The reaction is 3/2-order in acetaldehyde, a non-integer order that was experimentally observed and brilliantly explained by this mechanism.
These two reactions look identical on paper, but their kinetics are dramatically different. This difference is a perfect lesson in how subtle changes in thermodynamics (i.e., bond energies) can have a massive impact on a mechanism.
This is a photochemical reaction, initiated by light (hν).
Both propagation steps are fast. Step kp1 is only slightly endothermic and has a low activation energy. This reaction has a very high quantum yield (Φ), meaning one photon can produce up to 106 molecules of HCl. It's a classic, efficient chain reaction.
The rate law (using SSA, similar to above) is found to be:
$$ \frac{d[\text{HCl}]}{dt} = k' [\text{H}_2] I_{abs}^{1/2} $$
(Where Iabs is the intensity of absorbed light). Notice the product, HCl, does not appear in the rate law.
This mechanism looks similar, but with one killer difference.
Because step k1 is so slow, its reverse reaction becomes significant:
This is the crucial difference. The product itself (HBr) acts as an inhibitor by scavenging the H• radical and turning it back into the less reactive Br• radical, thus slowing the chain. This is called autoinhibition.
This is the prime example of a chain reaction with product inhibition. The goal is to derive the complex rate law that shows [HBr] in the denominator.
The Full Mechanism:
Step 1: Write the target rate law for the product, HBr.
HBr is formed in steps 2 and 3, and consumed in step 4.
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] + k_2[\text{H}^\bullet][\text{Br}_2] - k_{-1}[\text{H}^\bullet][\text{HBr}] $$
Step 2: Apply the Steady-State Approximation (SSA) to the intermediates (H• and Br•).
SSA for H•:
$$ \frac{d[\text{H}^\bullet]}{dt} \approx 0 = (\text{Formation}) - (\text{Consumption}) $$
$$ 0 = k_1[\text{Br}^\bullet][\text{H}_2] - k_2[\text{H}^\bullet][\text{Br}_2] - k_{-1}[\text{H}^\bullet][\text{HBr}] $$
Let's solve this for [H•] since it looks simpler:
$$ k_1[\text{Br}^\bullet][\text{H}_2] = k_2[\text{H}^\bullet][\text{Br}_2] + k_{-1}[\text{H}^\bullet][\text{HBr}] $$
$$ k_1[\text{Br}^\bullet][\text{H}_2] = [\text{H}^\bullet] \left( k_2[\text{Br}_2] + k_{-1}[\text{HBr}] \right) $$
$$ [\text{H}^\bullet] = \frac{k_1[\text{Br}^\bullet][\text{H}_2]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} $$
This is our first key expression (Eq. A).
SSA for Br•:
$$ \frac{d[\text{Br}^\bullet]}{dt} \approx 0 = (\text{Formation}) - (\text{Consumption}) $$
$$ 0 = 2k_i[\text{Br}_2] - k_1[\text{Br}^\bullet][\text{H}_2] + k_2[\text{H}^\bullet][\text{Br}_2] + k_{-1}[\text{H}^\bullet][\text{HBr}] - 2k_t[\text{Br}^\bullet]^2 $$
This looks horrible! But wait... look at the middle three terms: $- k_1[\text{Br}^\bullet][\text{H}_2] + k_2[\text{H}^\bullet][\text{Br}_2] + k_{-1}[\text{H}^\bullet][\text{HBr}]$.
From our SSA for H•, we know that $k_1[\text{Br}^\bullet][\text{H}_2] = k_2[\text{H}^\bullet][\text{Br}_2] + k_{-1}[\text{H}^\bullet][\text{HBr}]$.
This means $- (k_1[\text{Br}^\bullet][\text{H}_2]) + (k_2[\text{H}^\bullet][\text{Br}_2] + k_{-1}[\text{H}^\bullet][\text{HBr}]) = 0$. The three middle terms cancel out perfectly!
The Br• SSA simplifies to:
$$ 0 = 2k_i[\text{Br}_2] - 2k_t[\text{Br}^\bullet]^2 $$
$$ 2k_i[\text{Br}_2] = 2k_t[\text{Br}^\bullet]^2 $$
$$ [\text{Br}^\bullet] = \left( \frac{k_i}{k_t} \right)^{1/2} [\text{Br}_2]^{1/2} $$
This is our second key expression (Eq. B). (For a photochemical reaction, $2k_i[\text{Br}_2]$ would be replaced by $k_i I_{abs}$, yielding $[\text{Br}^\bullet] = (\frac{k_i I_{abs}}{2k_t})^{1/2}$).
Step 3: Substitute the intermediate concentrations back into the target rate law.
Our target was: $\frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] + k_2[\text{H}^\bullet][\text{Br}_2] - k_{-1}[\text{H}^\bullet][\text{HBr}]$
Let's factor out [H•] from the last two terms:
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] + [\text{H}^\bullet] \left( k_2[\text{Br}_2] - k_{-1}[\text{HBr}] \right) $$
Now, substitute Eq. A (our expression for [H•]) into this:
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] + \left( \frac{k_1[\text{Br}^\bullet][\text{H}_2]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} \right) \left( k_2[\text{Br}_2] - k_{-1}[\text{HBr}] \right) $$
This still looks messy. Let's find a common denominator for the whole expression. Factor out $k_1[\text{Br}^\bullet][\text{H}_2]$:
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] \left[ 1 + \frac{k_2[\text{Br}_2] - k_{-1}[\text{HBr}]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} \right] $$
The "1" can be rewritten as $\frac{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]}$:
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] \left[ \frac{(k_2[\text{Br}_2] + k_{-1}[\text{HBr}]) + (k_2[\text{Br}_2] - k_{-1}[\text{HBr}])}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} \right] $$
The $k_{-1}[\text{HBr}]$ terms in the numerator cancel out!
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] \left[ \frac{2k_2[\text{Br}_2]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} \right] $$
This is much cleaner. Now, for the final step, substitute Eq. B (our expression for [Br•]) into this:
$$ \frac{d[\text{HBr}]}{dt} = k_1 \left[ \left( \frac{k_i}{k_t} \right)^{1/2} [\text{Br}_2]^{1/2} \right] [\text{H}_2] \left[ \frac{2k_2[\text{Br}_2]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} \right] $$
Combine the constants and [Br2] terms:
$$ \frac{d[\text{HBr}]}{dt} = \frac{ 2 k_1 k_2 (k_i/k_t)^{1/2} [\text{H}_2] [\text{Br}_2]^{3/2} }{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} $$
To get the "classic" textbook form, we divide the top and bottom of the fraction by $k_2[\text{Br}_2]$:
$$ \frac{d[\text{HBr}]}{dt} = \frac{ (2 k_1 k_2 (k_i/k_t)^{1/2} [\text{H}_2] [\text{Br}_2]^{3/2}) / (k_2[\text{Br}_2]) }{(k_2[\text{Br}_2] + k_{-1}[\text{HBr}]) / (k_2[\text{Br}_2])} $$
$$ \frac{d[\text{HBr}]}{dt} = \frac{ 2 k_1 (k_i/k_t)^{1/2} [\text{H}_2] [\text{Br}_2]^{1/2} }{1 + (k_{-1}/k_2) \frac{[\text{HBr}]}{[\text{Br}_2]} } $$
By defining new composite constants $k = 2 k_1 (k_i/k_t)^{1/2}$ and $k' = k_{-1}/k_2$, we get the famous equation:
$$ \frac{d[\text{HBr}]}{dt} = \frac{k [\text{H}_2] [\text{Br}_2]^{1/2}}{1 + k' \frac{[\text{HBr}]}{[\text{Br}_2]}} $$
And there it is! The rate is inhibited by the product [HBr] in the denominator.
This is how we make countless plastics (polystyrene, PVC, PMMA). We start with an Initiator (I) that decomposes (usually with heat) to form radicals, and a Monomer (M) that has a double bond.
The rate of polymerization, Rp, is the rate at which monomer is consumed. Monomer is consumed in steps 2 and 3. But since a typical polymer chain has thousands of units, the propagation step (3) consumes vastly more monomer than the initiation step (2). Thus, we can approximate:
$$ R_p \approx R_{propagation} = k_p [\text{M}] [\text{P}^\bullet] $$
Where [P•] is the total concentration of all growing polymer radicals (∑ [Pn•]). We assume kp is independent of chain length.
Now, we use SSA to find [P•]. The rate of initiation Ri must equal the rate of termination Rt.
Rate of Initiation (Ri): This is the rate at which *growing chains* are started. It's related to the initiator decomposition, but not all radicals R• successfully start a chain. We introduce an initiator efficiency factor, f (usually 0.3-0.8).
$$ R_i = 2 f k_d [\text{I}] $$
(The '2' is because one I gives two R•).
Rate of Termination (Rt):
$$ R_t = 2 k_t [\text{P}^\bullet]^2 $$
(The '2' is because two P• radicals are destroyed per event).
Apply SSA: Ri = Rt
$$ 2 f k_d [\text{I}] = 2 k_t [\text{P}^\bullet]^2 $$
$$ [\text{P}^\bullet]^2 = \frac{f k_d [\text{I}]}{k_t} $$
$$ [\text{P}^\bullet] = \left( \frac{f k_d}{k_t} \right)^{1/2} [\text{I}]^{1/2} $$
Now, substitute this back into our Rp equation:
$$ R_p = k_p [\text{M}] \left[ \left( \frac{f k_d}{k_t} \right)^{1/2} [\text{I}]^{1/2} \right] $$
$$ R_p = k_{eff} [\text{M}]^1 [\text{I}]^{1/2} $$
This is a critical result! The rate of polymerization is first-order in monomer concentration and half-order in initiator concentration. This tells you exactly how to control the speed of your reaction.
This is defined as the average number of monomer units added to a chain per active chain initiated. It's a ratio of rates:
$$ \nu = \frac{\text{Rate of propagation}}{\text{Rate of initiation}} = \frac{R_p}{R_i} $$
$$ \nu = \frac{k_p [\text{M}] [\text{P}^\bullet]}{2 f k_d [\text{I}]} $$
We can substitute our expression for [P•]:
$$ \nu = \frac{k_p [\text{M}] \left( \frac{f k_d}{k_t} \right)^{1/2} [\text{I}]^{1/2}}{2 f k_d [\text{I}]} $$
Simplifying (remember x1/2 / x = x-1/2):
$$ \nu = \left( \frac{k_p}{2 (k_t f k_d)^{1/2}} \right) \frac{[\text{M}]}{[\text{I}]^{1/2}} $$
This tells us how to control the molecular weight of the polymer. To get longer chains (higher ν), you should increase monomer concentration and decrease initiator concentration.
This is a chain reaction of a completely different nature, but the principles are the same. Instead of radicals, the chain carriers are neutrons.
The key event is the fission of a heavy nucleus like Uranium-235:
$$ ^{235}\text{U} + n^0 \rightarrow \text{Fission Fragments} + \text{Energy} + \sim 2.5 n^0 $$
The crucial part is that one neutron goes in, but (on average) 2.5 neutrons come out. This isn't just propagation; it's a branching chain reaction. One event *multiplies* the number of chain carriers.
This is the single most important concept in nuclear engineering. It's defined as:
$$ k = \frac{\text{Number of neutrons in one generation}}{\text{Number of neutrons in the preceding generation}} $$
$k$ represents the average number of neutrons from one fission that will successfully cause *another* fission. It determines the behavior of the system:
An explosion is just an extremely fast reaction, and this can happen for two main reasons:
The H2 + O2 mechanism is complex, but the key steps are:
Steps (Branching 1) and (Branching 2) are the problem. One H• becomes *three* radicals (OH, OH, H) in just two steps. This is an exponential cascade.
Termination steps are what fight this cascade:
The "explosion peninsula" for H2 + O2 is a plot of Pressure vs. Temperature that shows when the mixture will explode. It's explained by the competition between branching and termination.
This is a great puzzle. The formation of phosgene (COCl2) from CO and Cl2 involves light and radicals, but it's not a chain reaction. The Cl• radical is acting as a catalyst. Let's see why and derive the full rate law.
The reaction: CO + Cl2 ⟶hν COCl2
We'll use $R_i = k_i I_{abs}$ for the photochemical initiation rate, where $I_{abs}$ is the intensity of absorbed light.
The Mechanism:
Step 1: Write the target rate law for the product, COCl₂.
Phosgene is only formed in step 4.
$$ \frac{d[\text{COCl}_2]}{dt} = k_2 [\text{COCl}^\bullet] [\text{Cl}_2] $$
Step 2: Apply the Steady-State Approximation (SSA) to the intermediates (COCl• and Cl•).
SSA for COCl•:
$$ \frac{d[\text{COCl}^\bullet]}{dt} \approx 0 = (\text{Formation}) - (\text{Consumption}) $$
$$ 0 = k_1[\text{Cl}^\bullet][\text{CO}] - k_{-1}[\text{COCl}^\bullet] - k_2[\text{COCl}^\bullet][\text{Cl}_2] $$
Let's solve this for [COCl•]:
$$ k_1[\text{Cl}^\bullet][\text{CO}] = k_{-1}[\text{COCl}^\bullet] + k_2[\text{COCl}^\bullet][\text{Cl}_2] $$
$$ k_1[\text{Cl}^\bullet][\text{CO}] = [\text{COCl}^\bullet] \left( k_{-1} + k_2[\text{Cl}_2] \right) $$
$$ [\text{COCl}^\bullet] = \frac{k_1[\text{Cl}^\bullet][\text{CO}]}{k_{-1} + k_2[\text{Cl}_2]} $$
This is our first expression (Eq. C).
SSA for Cl•:
$$ \frac{d[\text{Cl}^\bullet]}{dt} \approx 0 = (\text{Formation}) - (\text{Consumption}) $$
$$ 0 = 2k_i I_{abs} - k_1[\text{Cl}^\bullet][\text{CO}] + k_{-1}[\text{COCl}^\bullet] + k_2[\text{COCl}^\bullet][\text{Cl}_2] - 2k_t[\text{Cl}^\bullet]^2 $$
Again, this looks complicated, but let's look at the middle terms: $- k_1[\text{Cl}^\bullet][\text{CO}] + k_{-1}[\text{COCl}^\bullet] + k_2[\text{COCl}^\bullet][\text{Cl}_2]$.
From our SSA for COCl•, we know that $k_1[\text{Cl}^\bullet][\text{CO}] = k_{-1}[\text{COCl}^\bullet] + k_2[\text{COCl}^\bullet][\text{Cl}_2]$.
This means $- (k_1[\text{Cl}^\bullet][\text{CO}]) + (k_{-1}[\text{COCl}^\bullet] + k_2[\text{COCl}^\bullet][\text{Cl}_2]) = 0$. Once again, these terms cancel out!
The Cl• SSA simplifies to:
$$ 0 = 2k_i I_{abs} - 2k_t[\text{Cl}^\bullet]^2 $$
$$ 2k_i I_{abs} = 2k_t[\text{Cl}^\bullet]^2 $$
$$ [\text{Cl}^\bullet] = \left( \frac{k_i I_{abs}}{k_t} \right)^{1/2} $$
This is our second expression (Eq. D).
Step 3: Substitute the intermediate concentrations back into the target rate law.
Our target was: $\frac{d[\text{COCl}_2]}{dt} = k_2 [\text{COCl}^\bullet] [\text{Cl}_2]$
First, substitute Eq. C (for [COCl•]) into the target:
$$ \frac{d[\text{COCl}_2]}{dt} = k_2 \left[ \frac{k_1[\text{Cl}^\bullet][\text{CO}]}{k_{-1} + k_2[\text{Cl}_2]} \right] [\text{Cl}_2] $$
Now, substitute Eq. D (for [Cl•]) into this new expression:
$$ \frac{d[\text{COCl}_2]}{dt} = k_2 \left[ \frac{k_1 [\text{CO}]}{k_{-1} + k_2[\text{Cl}_2]} \right] \left[ \left( \frac{k_i I_{abs}}{k_t} \right)^{1/2} \right] [\text{Cl}_2] $$
Rearranging this gives the full rate law:
$$ \frac{d[\text{COCl}_2]}{dt} = \frac{k_1 k_2 (k_i/k_t)^{1/2} [\text{CO}] [\text{Cl}_2] I_{abs}^{1/2}}{k_{-1} + k_2[\text{Cl}_2]} $$
In many experimental conditions, the COCl• intermediate is much more likely to decompose back to reactants (step -1) than it is to react with Cl2 (step 2). This means $k_{-1}$ is very large.
Assumption: $k_{-1} \gg k_2[\text{Cl}_2]$
If this is true, our denominator $(k_{-1} + k_2[\text{Cl}_2])$ simplifies to just $k_{-1}$.
The rate law then becomes:
$$ \frac{d[\text{COCl}_2]}{dt} \approx \frac{k_1 k_2 (k_i/k_t)^{1/2} [\text{CO}] [\text{Cl}_2] I_{abs}^{1/2}}{k_{-1}} $$
We can group all the rate constants into one new effective rate constant, $k_{eff} = \frac{k_1 k_2 (k_i/k_t)^{1/2}}{k_{-1}}$:
$$ \frac{d[\text{COCl}_2]}{dt} \approx k_{eff} [\text{CO}]^1 [\text{Cl}_2]^1 I_{abs}^{1/2} $$
This simpler form is often what's presented, and it shows the reaction is first-order in CO, first-order in Cl2, and half-order in light intensity. The half-order in light is the classic sign of a radical mechanism with a bimolecular termination step (2Cl• → Cl2).
Both are processes where molecules are broken down by high-energy inputs, but the *nature* of that energy is key.
This is nature's perfected photochemical process. It uses light to create chemical energy (ATP) and reducing power (NADPH), which are then used to fix CO2 into sugar. The "engine" for this is the light-dependent reactions, often called the Z-Scheme.
It involves two large protein complexes called Photosystem II (PSII) and Photosystem I (PSI).
(It's called the "Z-Scheme" because a diagram of the energy levels of the electrons looks like a letter 'Z' tipped on its side).
Summary: The Z-scheme uses two "boosts" of light energy to achieve two goals: 1. Make ATP (energy currency) 2. Make NADPH (reducing power) These two molecules then go to the Calvin Cycle (the light-independent reactions) to turn CO2 into sugar.
This is another incredible biological-photochemical process. It's shockingly sensitive; a single photon can trigger a nerve impulse. It all boils down to one molecule changing its shape.
In the rod cells of your retina (used for low-light vision), you have a protein called rhodopsin. This is a complex of two parts: 1. A large protein called opsin. 2. A small, light-absorbing molecule (a chromophore) called 11-cis-retinal, which is derived from Vitamin A.
In the dark, the 11-cis-retinal fits perfectly into a specific "pocket" in the opsin protein.
$$ \text{11-cis-retinal} \xrightarrow{h\nu} \text{all-trans-retinal} $$
So, in the end, "seeing" is just your brain processing the signal from a photochemical isomerization that happened in your eye. All thanks to a single photon changing a cis bond to a trans bond.
Later, enzymes in the retina will convert the all-trans-retinal back to 11-cis-retinal, which re-binds to opsin, resetting the trap for the next photon. And that's chemistry in action!
And there you have it! We've walked through every topic, from the basic definitions of chain reactions to the complex photochemistry of life itself. We have now included the full, step-by-step derivations for the most complex mechanisms, including the H2 + Br2 reaction (showing product inhibition) and the phosgene reaction (showing a non-chain radical mechanism).
The key takeaway is always the mechanism. By applying the Steady-State Approximation (SSA), we can break down highly complex reactions into manageable steps and derive rate laws that perfectly match experimental observations. Understanding these detailed derivations gives you the power to predict and control how chemical reactions behave. I hope this fully-explained breakdown makes everything crystal clear for your exam. Great work today!
Here are the 10 questions from your exam, each followed by a detailed, MSc-level answer. We will cover all the theory, mechanisms, and derivations in full.
A chain reaction is a multi-step reaction mechanism where a reactive intermediate, known as a chain carrier (often a free radical), is generated in an initial step and then continuously regenerated in a repeating cycle of subsequent steps. This cyclic process, or "chain," allows a single initiation event to cause a large number of reactant molecules to be converted to products, often resulting in high quantum yields (in photochemistry) or explosive rates.
Let's consider a general chain reaction where reactant A is converted to product P, and the chain carrier is the radical R•.
The Mechanism:
Assumptions:
Step 1: Write the rate law for the product, P.
From the propagation step (Step 2), the rate of formation of P is:
$$ \frac{d[P]}{dt} = k_p[\text{R}^\bullet][A] $$
This equation is not useful as it contains the concentration of an unmeasurable intermediate, $ [\text{R}^\bullet] $. We must find an expression for $ [\text{R}^\bullet] $.
Step 2: Apply the SSA to the intermediate $ \text{R}^\bullet $.
$$ \frac{d[\text{R}^\bullet]}{dt} \approx 0 $$
$$ (\text{Rate of Formation}) - (\text{Rate of Consumption}) = 0 $$
$ \text{R}^\bullet $ is formed in Step 1 (rate $ 2k_i[A] $). It is consumed in Step 3 (rate $ 2k_t[\text{R}^\bullet]^2 $). In the propagation step (Step 2), one $ \text{R}^\bullet $ is consumed, but one $ \text{R}^\bullet $ is also produced, so the net change is zero. Thus, the propagation step does not appear in the SSA for the total radical concentration.
$$ 2k_i[A] - 2k_t[\text{R}^\bullet]^2 = 0 $$
Step 3: Solve for $ [\text{R}^\bullet] $.
$$ 2k_t[\text{R}^\bullet]^2 = 2k_i[A] $$
$$ [\text{R}^\bullet]^2 = \frac{k_i[A]}{k_t} $$
$$ [\text{R}^\bullet] = \left( \frac{k_i}{k_t} \right)^{1/2} [A]^{1/2} $$
Step 4: Substitute the expression for $ [\text{R}^\bullet] $ back into the rate law for P.
$$ \frac{d[P]}{dt} = k_p[\text{R}^\bullet][A] = k_p \left[ \left( \frac{k_i}{k_t} \right)^{1/2} [A]^{1/2} \right] [A] $$
Combining the terms, we get the final overall rate law:
$$ \frac{d[P]}{dt} = k_p \left( \frac{k_i}{k_t} \right)^{1/2} [A]^{3/2} $$
This result is significant as it shows a non-integer reaction order (1.5), which is a strong experimental indicator of a chain reaction mechanism.
The Rice-Herzfeld mechanism is a general mechanism for the thermal decomposition of organic molecules, explaining how complex products and non-integer reaction orders can arise from a free-radical chain reaction. The decomposition of acetaldehyde (CH₃CHO) into methane (CH₄) and carbon monoxide (CO) is the classic example.
Overall Reaction: $ \text{CH}_3\text{CHO}(g) \xrightarrow{\Delta} \text{CH}_4(g) + \text{CO}(g) $
Experimentally, the reaction is found to be 3/2-order with respect to acetaldehyde. The Rice-Herzfeld mechanism perfectly explains this.
The chain carriers are the methyl radical ($ \bullet\text{CH}_3 $) and the acetyl radical ($ \text{CH}_3\text{CO}^\bullet $).
(Note: The $ \bullet\text{CHO} $ radical from Step 1 can also participate, but this simplified mechanism is sufficient to derive the observed rate law.)
We apply the Steady-State Approximation (SSA) to both radical intermediates, $ \bullet\text{CH}_3 $ and $ \text{CH}_3\text{CO}^\bullet $.
Step 1: Write the target rate law for the product, CH₄.
From the propagation step (ii), the rate of formation of methane is:
$$ \frac{d[\text{CH}_4]}{dt} = k_2 [\bullet\text{CH}_3] [\text{CH}_3\text{CHO}] $$
We must find an expression for the intermediate concentration $ [\bullet\text{CH}_3] $.
Step 2: Apply SSA to the intermediates.
SSA for $ \text{CH}_3\text{CO}^\bullet $ (the "simpler" intermediate):
$$ \frac{d[\text{CH}_3\text{CO}^\bullet]}{dt} \approx 0 = (\text{Formation}) - (\text{Consumption}) $$
$$ 0 = k_2[\bullet\text{CH}_3][\text{CH}_3\text{CHO}] - k_3[\text{CH}_3\text{CO}^\bullet] $$
Rearranging, we get a relationship between the two radicals:
$$ k_3[\text{CH}_3\text{CO}^\bullet] = k_2[\bullet\text{CH}_3][\text{CH}_3\text{CHO}] $$
SSA for $ \bullet\text{CH}_3 $:
$$ \frac{d[\bullet\text{CH}_3]}{dt} \approx 0 = (\text{Formation}) - (\text{Consumption}) $$
$$ 0 = (k_1[\text{CH}_3\text{CHO}] + k_3[\text{CH}_3\text{CO}^\bullet]) - (k_2[\bullet\text{CH}_3][\text{CH}_3\text{CHO}] + 2k_4[\bullet\text{CH}_3]^2) $$
This looks complex, but we can substitute our result from the first SSA. Notice that the terms $ k_3[\text{CH}_3\text{CO}^\bullet] $ and $ k_2[\bullet\text{CH}_3][\text{CH}_3\text{CHO}] $ are equal. Therefore, they cancel each other out in the SSA for $ \bullet\text{CH}_3 $:
$$ 0 = k_1[\text{CH}_3\text{CHO}] - 2k_4[\bullet\text{CH}_3]^2 $$
Step 3: Solve for $ [\bullet\text{CH}_3] $.
From the simplified SSA for $ \bullet\text{CH}_3 $:
$$ 2k_4[\bullet\text{CH}_3]^2 = k_1[\text{CH}_3\text{CHO}] $$
$$ [\bullet\text{CH}_3]^2 = \frac{k_1[\text{CH}_3\text{CHO}]}{2k_4} $$
$$ [\bullet\text{CH}_3] = \left( \frac{k_1}{2k_4} \right)^{1/2} [\text{CH}_3\text{CHO}]^{1/2} $$
Step 4: Substitute this expression back into the target rate law (from Step 1).
$$ \frac{d[\text{CH}_4]}{dt} = k_2 [\bullet\text{CH}_3] [\text{CH}_3\text{CHO}] $$
$$ \frac{d[\text{CH}_4]}{dt} = k_2 \left[ \left( \frac{k_1}{2k_4} \right)^{1/2} [\text{CH}_3\text{CHO}]^{1/2} \right] [\text{CH}_3\text{CHO}] $$
Combining the $ [\text{CH}_3\text{CHO}] $ terms ($ [A]^{1/2} \times [A]^1 = [A]^{3/2} $):
$$ \frac{d[\text{CH}_4]}{dt} = k_2 \left( \frac{k_1}{2k_4} \right)^{1/2} [\text{CH}_3\text{CHO}]^{3/2} $$
If we define an effective rate constant $ k_{eff} = k_2 \left( \frac{k_1}{2k_4} \right)^{1/2} $, the rate law is:
$$ \text{Rate} = k_{eff} [\text{CH}_3\text{CHO}]^{3/2} $$
This derivation perfectly matches the experimentally observed 3/2 reaction order, providing strong evidence for this radical chain mechanism.
On the surface, the reactions of H₂ with Cl₂ and Br₂ appear identical. However, their kinetics are drastically different due to a subtle difference in the thermodynamics (bond energies) of their propagation steps. This makes their comparison a classic study in chemical kinetics.
Overall Reaction: $ \text{H}_2 + \text{Cl}_2 \xrightarrow{h\nu} 2\text{HCl} $
Mechanism:
Characteristics:
Overall Reaction: $ \text{H}_2 + \text{Br}_2 \xrightarrow{h\nu \text{ or } \Delta} 2\text{HBr} $
Mechanism:
The Crucial Difference & Role of HBr as an Inhibitor:
The key difference lies in Propagation 1. The H-H bond is stronger than the H-Br bond, making this step highly endothermic ($ \Delta H \approx +67 $ kJ/mol) and therefore having a very high activation energy ($ E_a = 77 $ kJ/mol). This step is extremely SLOW.
Because Step 1 is slow and reversible, its reverse reaction becomes significant. This reverse reaction is:
This reaction, where the product $ \text{HBr} $ destroys a chain-propagating $ \text{H}^\bullet $ radical, is called autoinhibition. As the reaction proceeds and $ \text{HBr} $ builds up, it "poisons" its own formation by scavenging the $ \text{H}^\bullet $ radicals, dramatically slowing the overall rate.
We must use the full mechanism, including the inhibition step $ k_{-1} $.
Step 1: Write the target rate law for HBr.
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] + k_2[\text{H}^\bullet][\text{Br}_2] - k_{-1}[\text{H}^\bullet][\text{HBr}] $$
Step 2: Apply SSA to the intermediates $ \text{H}^\bullet $ and $ \text{Br}^\bullet $.
SSA for $ \text{H}^\bullet $:
$$ \frac{d[\text{H}^\bullet]}{dt} \approx 0 = k_1[\text{Br}^\bullet][\text{H}_2] - k_2[\text{H}^\bullet][\text{Br}_2] - k_{-1}[\text{H}^\bullet][\text{HBr}] $$
$$ [\text{H}^\bullet] \left( k_2[\text{Br}_2] + k_{-1}[\text{HBr}] \right) = k_1[\text{Br}^\bullet][\text{H}_2] $$
$$ [\text{H}^\bullet] = \frac{k_1[\text{Br}^\bullet][\text{H}_2]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} \quad \text{(Eq. A)} $$
SSA for $ \text{Br}^\bullet $:
(Assuming thermal initiation, Rate = $ 2k_i[\text{Br}_2] $. The logic is identical for photochemical, $ R_i = k_i I_{abs} $)
$$ \frac{d[\text{Br}^\bullet]}{dt} \approx 0 = 2k_i[\text{Br}_2] - k_1[\text{Br}^\bullet][\text{H}_2] + k_2[\text{H}^\bullet][\text{Br}_2] + k_{-1}[\text{H}^\bullet][\text{HBr}] - 2k_t[\text{Br}^\bullet]^2 $$
From the SSA for $ \text{H}^\bullet $, we know $ k_1[\text{Br}^\bullet][\text{H}_2] = k_2[\text{H}^\bullet][\text{Br}_2] + k_{-1}[\text{H}^\bullet][\text{HBr}] $.
This means $- (k_1[\text{Br}^\bullet][\text{H}_2]) + (k_2[\text{H}^\bullet][\text{Br}_2] + k_{-1}[\text{H}^\bullet][\text{HBr}]) = 0$. The three middle terms cancel out perfectly!
The Br• SSA simplifies to:
$$ 0 = 2k_i[\text{Br}_2] - 2k_t[\text{Br}^\bullet]^2 $$
$$ 2k_i[\text{Br}_2] = 2k_t[\text{Br}^\bullet]^2 $$
$$ [\text{Br}^\bullet] = \left( \frac{k_i}{k_t} \right)^{1/2} [\text{Br}_2]^{1/2} \quad \text{(Eq. B)} $$
Step 3: Substitute back into the target rate law.
Let's simplify the target rate law from Step 1 by factoring out $ [\text{H}^\bullet] $:
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] + [\text{H}^\bullet] \left( k_2[\text{Br}_2] - k_{-1}[\text{HBr}] \right) $$
Now substitute Eq. A into this:
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] + \left( \frac{k_1[\text{Br}^\bullet][\text{H}_2]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} \right) \left( k_2[\text{Br}_2] - k_{-1}[\text{HBr}] \right) $$
Factor out $ k_1[\text{Br}^\bullet][\text{H}_2] $ and find a common denominator:
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] \left[ 1 + \frac{k_2[\text{Br}_2] - k_{-1}[\text{HBr}]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} \right] $$
The "1" can be rewritten as $\frac{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]}$:
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] \left[ \frac{(k_2[\text{Br}_2] + k_{-1}[\text{HBr}]) + (k_2[\text{Br}_2] - k_{-1}[\text{HBr}])}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} \right] $$
The $k_{-1}[\text{HBr}]$ terms in the numerator cancel out!
$$ \frac{d[\text{HBr}]}{dt} = k_1[\text{Br}^\bullet][\text{H}_2] \left[ \frac{2k_2[\text{Br}_2]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} \right] $$
This is much cleaner. Now, for the final step, substitute Eq. B for $ [\text{Br}^\bullet] $ into this:
$$ \frac{d[\text{HBr}]}{dt} = k_1 \left[ \left( \frac{k_i}{k_t} \right)^{1/2} [\text{Br}_2]^{1/2} \right] [\text{H}_2] \left[ \frac{2k_2[\text{Br}_2]}{k_2[\text{Br}_2] + k_{-1}[\text{HBr}]} \right] $$
To get the famous textbook form, we define $ k = 2k_1(k_i/k_t)^{1/2} $ and $ k' = k_{-1}/k_2 $ and divide the numerator and denominator by $ k_2[\text{Br}_2] $:
$$ \frac{d[\text{HBr}]}{dt} = \frac{k [\text{H}_2] [\text{Br}_2]^{1/2}}{1 + k' \frac{[\text{HBr}]}{[\text{Br}_2]}} $$
This final, complex rate law shows precisely how the reaction is inhibited by its own product, $ \text{HBr} $, which appears in the denominator.
Free-radical chain polymerization is a major industrial process used to convert small unsaturated molecules (monomers, M) into long polymer chains. It follows a standard chain reaction mechanism, initiated by a molecule (Initiator, I) that easily decomposes to form radicals.
Let $ \text{I} $ be the initiator, $ \text{M} $ the monomer, $ \text{R}^\bullet $ the initiator radical, and $ \text{P}_n^\bullet $ a growing polymer chain radical of length $ n $.
The overall Rate of Initiation ($ R_i $), i.e., the rate of formation of *growing chains*, is: $ R_i = 2 f k_d [\text{I}] $
(We assume $ k_p $ is independent of the chain length $ n $).
The Rate of Termination ($ R_t $), which consumes two radicals, is: $ R_t = 2 k_t [\text{P}^\bullet]^2 $, where $ [\text{P}^\bullet] $ is the total concentration of all growing radicals.
Step 1: Define the Rate of Polymerization ($ R_p $).
$ R_p $ is the rate at which monomer is consumed. Monomer is consumed in step 2 (initiation) and step 3 (propagation). However, since thousands of monomer units are added in propagation for every one used in initiation (long chain assumption), we can approximate:
$$ R_p \approx R_{propagation} $$
$$ R_p = k_p [\text{M}] [\text{P}^\bullet] $$
Our goal is to find an expression for the intermediate $ [\text{P}^\bullet] $.
Step 2: Apply the SSA to the total radical concentration $ [\text{P}^\bullet] $.
The steady-state assumption states that the rate of radical formation (Initiation) equals the rate of radical destruction (Termination).
$$ R_i = R_t $$
$$ 2 f k_d [\text{I}] = 2 k_t [\text{P}^\bullet]^2 $$
Step 3: Solve for $ [\text{P}^\bullet] $.
$$ [\text{P}^\bullet]^2 = \frac{f k_d [\text{I}]}{k_t} $$
$$ [\text{P}^\bullet] = \left( \frac{f k_d}{k_t} \right)^{1/2} [\text{I}]^{1/2} $$
Step 4: Substitute this expression for $ [\text{P}^\bullet] $ back into the $ R_p $ equation.
$$ R_p = k_p [\text{M}] \left[ \left( \frac{f k_d}{k_t} \right)^{1/2} [\text{I}]^{1/2} \right] $$
Combining all constants into $ k_{eff} $, the final rate law is:
$$ R_p = k_{eff} [\text{M}]^1 [\text{I}]^{1/2} $$
This is a crucial result: the rate of polymerization is first-order in monomer concentration and half-order in initiator concentration.
Step 1: Define the Kinetic Chain Length ($ \nu $).
$ \nu $ (nu) is the average number of monomer units polymerized per chain-initiating radical. It is the ratio of the rate of propagation (how fast monomer is consumed) to the rate of initiation (how fast new chains are started).
$$ \nu = \frac{\text{Rate of Propagation}}{\text{Rate of Initiation}} = \frac{R_p}{R_i} $$
Step 2: Substitute the full expressions for $ R_p $ and $ R_i $.
From our work above:
$$ R_p = k_p [\text{M}] [\text{P}^\bullet] $$
$$ R_i = 2 f k_d [\text{I}] $$
$$ \nu = \frac{k_p [\text{M}] [\text{P}^\bullet]}{2 f k_d [\text{I}]} $$
Step 3: Substitute the expression for $ [\text{P}^\bullet] $ and simplify.
We use $ [\text{P}^\bullet] = \left( \frac{f k_d}{k_t} \right)^{1/2} [\text{I}]^{1/2} $:
$$ \nu = \frac{k_p [\text{M}] \left( \frac{f k_d}{k_t} \right)^{1/2} [\text{I}]^{1/2}}{2 f k_d [\text{I}]} $$
Now, we simplify by grouping terms (remember that $ \frac{x^{1/2}}{x} = \frac{1}{x^{1/2}} $):
$$ \nu = \left[ \frac{k_p}{2 k_t^{1/2}} \right] \left[ \frac{(f k_d)^{1/2}}{f k_d} \right] \frac{[\text{M}]}{[\text{I}]^{1/2}} $$
$$ \nu = \left[ \frac{k_p}{2 (k_t f k_d)^{1/2}} \right] \frac{[\text{M}]}{[\text{I}]^{1/2}} $$
The kinetic chain length $ \nu $ (which is directly proportional to the polymer's molecular weight) is first-order in monomer and inverse-half-order in initiator.
Practical Implication: To get a high molecular weight polymer (large $ \nu $), one must use a high monomer concentration and a low initiator concentration.
A branching chain reaction is a specific type of chain reaction where the number of chain carriers increases in at least one of the propagation steps. This leads to an exponential acceleration in the reaction rate, often culminating in an explosion.
In the nuclear fission of Uranium-235, the chain carrier is the neutron ($ n^0 $). The process works as follows:
This is the definitive "branching" step: one chain carrier ($ 1 n^0 $) goes in, but 2.5 chain carriers ($ 2.5 n^0 $) come out. Each of these new neutrons can, in principle, go on to initiate another fission event, leading to a rapidly multiplying cascade ($ 1 \to 2.5 \to 6.25 \to 15.6 \to ... $). This exponential growth in the number of fission events per unit time is what makes the reaction so powerful.
The neutron multiplication factor ($ k $) is the single most important parameter in nuclear engineering. It is defined as the average number of neutrons from one fission event that will successfully cause another fission event in the next "generation."
$$ k = \frac{\text{Number of neutrons in one generation}}{\text{Number of neutrons in the preceding generation}} $$
$ k $ is a real-world value that accounts for the fact that not all 2.5 neutrons produced will cause another fission. Some will be absorbed by non-fissionable material (like $ ^{238}\text{U} $ or control rods), and some will escape the material (leakage).
The value of $ k $ precisely determines the behavior of the nuclear system:
Explosive reactions are reactions characterized by an extremely rapid, self-accelerating rate that generates a large volume of gas and a shock wave. This acceleration can have two primary causes:
The H₂ + O₂ reaction (formation of water) is the classic example of a chain-branching explosion.
The reaction is deceptively simple on paper ($ 2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O} $), but its mechanism is highly complex. The key chain carriers are $ \text{H}^\bullet $, $ \text{O}^\bullet $, and $ \bullet\text{OH} $.
The two branching steps are the source of the explosion. One $ \text{H}^\bullet $ radical (from propagation) reacts with O₂ to create two new radicals. These two new radicals ($ \bullet\text{OH} $ and $ \text{O}^\bullet $) then react with H₂ to create three more radicals ($ \text{H}^\bullet $, $ \bullet\text{OH} $, and $ \text{H}^\bullet $). The total cycle is $ \text{H}^\bullet \to 3 $ radicals. This is an exponential cascade.
This cascade is only stopped by termination steps, which compete with branching:
The "explosion limits" are a plot of Pressure vs. Temperature that shows the boundary between slow reaction and explosion. The behavior is determined by the competition: Branching vs. Termination.
No Explosion. At very low pressure, radicals are few and far between. They can easily diffuse to the container walls (Termination 1) and be destroyed before they can find an O₂ molecule to react with.
(Termination > Branching)
Explosion! As pressure increases, the radicals (like $ \text{H}^\bullet $) collide with O₂ molecules (Branching 1) much more frequently than they hit the walls. The branching cascade is faster than termination.
(Branching > Termination)
No Explosion. This is counter-intuitive. At high pressure, the three-body collision (Termination 2) becomes very efficient. This step ($ \text{H}^\bullet + \text{O}_2 + \text{M} $) requires a high concentration of $ \text{M} $ (i.e., high pressure) to occur. It effectively "steals" the $ \text{H}^\bullet $ radicals and deactivates them into the less reactive $ \text{HO}_2^\bullet $, stopping the branching cascade.
(Termination > Branching)
Explosion! At very high temperatures, the reaction becomes so fast that it becomes a thermal explosion. The heat is generated faster than it can be dissipated, regardless of the chain-branching mechanism, and the system explodes.
The formation of phosgene ($ \text{COCl}_2 $) from carbon monoxide and chlorine gas is a classic example of a photocatalytic reaction. It is initiated by light and involves radicals, but it is not a true chain reaction because the chain carriers are not regenerated in the same way.
Overall Reaction: $ \text{CO} + \text{Cl}_2 \xrightarrow{h\nu} \text{COCl}_2 $
This reaction has a very high quantum yield, which is often a sign of a chain reaction. However, in this case, it is due to the $ \text{Cl}^\bullet $ radical acting as a catalyst, not a traditional chain carrier.
Why is this NOT a chain reaction?
In a true chain reaction (like H₂ + Cl₂), one carrier ($ \text{Cl}^\bullet $) generates a *different* carrier ($ \text{H}^\bullet $), which in turn regenerates the first ($ \text{Cl}^\bullet $). This is a "hand-off." In the phosgene mechanism, the $ \text{Cl}^\bullet $ radical is more like a catalyst: it is consumed in Step 2 to form an intermediate ($ \text{COCl}^\bullet $), and then it is regenerated in Step 3 when that intermediate reacts. The cycle is $ \text{Cl}^\bullet \to \text{COCl}^\bullet \to \text{Cl}^\bullet $. The $ \text{Cl}^\bullet $ facilitates the conversion of $ \text{CO} + \text{Cl}_2 \to \text{COCl}_2 $ many times before terminating.
We apply the SSA to both radical intermediates, $ \text{COCl}^\bullet $ and $ \text{Cl}^\bullet $. The rate of initiation is $ R_i = 2k_i I_{abs} $, where $ I_{abs} $ is the intensity of absorbed light.
Step 1: Write the target rate law for the product, $ \text{COCl}_2 $.
From Step 3, the rate of formation is:
$$ \frac{d[\text{COCl}_2]}{dt} = k_2 [\text{COCl}^\bullet] [\text{Cl}_2] $$
Step 2: Apply SSA to the intermediates.
SSA for $ \text{COCl}^\bullet $:
$$ \frac{d[\text{COCl}^\bullet]}{dt} \approx 0 = k_1[\text{Cl}^\bullet][\text{CO}] - k_{-1}[\text{COCl}^\bullet] - k_2[\text{COCl}^\bullet][\text{Cl}_2] $$
Solve for $ [\text{COCl}^\bullet] $:
$$ k_1[\text{Cl}^\bullet][\text{CO}] = [\text{COCl}^\bullet] \left( k_{-1} + k_2[\text{Cl}_2] \right) $$
$$ [\text{COCl}^\bullet] = \frac{k_1[\text{Cl}^\bullet][\text{CO}]}{k_{-1} + k_2[\text{Cl}_2]} \quad \text{(Eq. C)} $$
SSA for $ \text{Cl}^\bullet $:
$$ \frac{d[\text{Cl}^\bullet]}{dt} \approx 0 = 2k_i I_{abs} - k_1[\text{Cl}^\bullet][\text{CO}] + k_{-1}[\text{COCl}^\bullet] + k_2[\text{COCl}^\bullet][\text{Cl}_2] - 2k_t[\text{Cl}^\bullet]^2 $$
From the SSA for $ \text{COCl}^\bullet $, we know $ k_1[\text{Cl}^\bullet][\text{CO}] = k_{-1}[\text{COCl}^\bullet] + k_2[\text{COCl}^\bullet][\text{Cl}_2] $. The middle terms cancel out!
$$ 0 = 2k_i I_{abs} - 2k_t[\text{Cl}^\bullet]^2 $$
$$ 2k_i I_{abs} = 2k_t[\text{Cl}^\bullet]^2 $$
$$ [\text{Cl}^\bullet] = \left( \frac{k_i I_{abs}}{k_t} \right)^{1/2} \quad \text{(Eq. D)} $$
Step 3: Substitute back into the target rate law.
First, substitute Eq. C into the target rate law:
$$ \frac{d[\text{COCl}_2]}{dt} = k_2 \left[ \frac{k_1[\text{Cl}^\bullet][\text{CO}]}{k_{-1} + k_2[\text{Cl}_2]} \right] [\text{Cl}_2] $$
Now, substitute Eq. D for $ [\text{Cl}^\bullet] $ into this new expression:
$$ \frac{d[\text{COCl}_2]}{dt} = \frac{k_1 k_2 [\text{CO}] [\text{Cl}_2]}{k_{-1} + k_2[\text{Cl}_2]} \left[ \left( \frac{k_i I_{abs}}{k_t} \right)^{1/2} \right] $$
This is the full rate law. A common simplification occurs if we assume the equilibrium in Step 2 is fast and $ k_{-1} \gg k_2[\text{Cl}_2] $. The denominator simplifies to $ k_{-1} $, giving:
$$ \frac{d[\text{COCl}_2]}{dt} \approx k_{eff} [\text{CO}]^1 [\text{Cl}_2]^1 I_{abs}^{1/2} $$
The half-order in light intensity ($ I_{abs}^{1/2} $) is the classic signature of a radical mechanism with a bimolecular termination step ($ 2\text{Cl}^\bullet \to \text{Cl}_2 $).
Photolysis and Radiolysis both describe the decomposition of a substance using energy, but they differ fundamentally in the type of energy used and the mechanism of interaction.
Acetone absorbs a photon (~300 nm) at its $ \text{C=O} $ chromophore. From the excited state, it cleaves at the adjacent C-C bond, which is the weakest bond.
$$ \text{CH}_3\text{COCH}_3 + h\nu \rightarrow \bullet\text{CH}_3 + \bullet\text{COCH}_3 \text{ (methyl and acetyl radicals)} $$A gamma ray passing through water does not target a single molecule, but rather ionizes the bulk solvent. The primary event $ \text{H}_2\text{O} + \gamma \to \text{H}_2\text{O}^{\bullet+} + e^- $ is followed by a cascade, producing a mix of radicals:
$$ \text{H}_2\text{O} \xrightarrow{\text{radiolysis}} e^-_{aq} \text{ (solvated electron)}, \bullet\text{OH} \text{ (hydroxyl radical)}, \text{H}^\bullet \text{ (hydrogen atom)}, \text{H}_2, \text{H}_2\text{O}_2 $$| Feature | Photolysis | Radiolysis |
|---|---|---|
| Energy Source | UV-Vis Photons (low energy, eV) | Gamma rays, X-rays, e-beams (high energy, keV-MeV) |
| Mechanism | Specific absorption by chromophore $\to$ Excited State | Non-specific ionization of bulk medium |
| Selectivity | Highly Selective (targets specific bonds) | Non-selective (creates a "soup" of radicals) |
| Products | Predictable radicals | $ e^-_{aq} $, $ \bullet\text{OH} $, $ \text{H}^\bullet $, etc. (in water) |
The primary photochemical process in photosynthesis is the light-dependent reaction, which occurs in the thylakoid membranes of chloroplasts. Its sole purpose is to convert light energy into chemical energy in the form of two molecules: ATP (energy currency) and NADPH (reducing power). These two molecules are then used by the light-independent reactions (Calvin Cycle) to fix CO₂ into sugar.
This process is accomplished by two large protein complexes called Photosystem II (PSII) and Photosystem I (PSI), linked by an electron transport chain. The flow of electrons through this system is described by the "Z-Scheme".
The "Z-Scheme" gets its name because a diagram of the electron's energy level as it moves through the system looks like the letter 'Z' on its side.
Summary: The Z-Scheme uses two "kicks" of light energy. The first kick (at PSII) is strong enough to split water and use the electron's energy to make ATP. The second kick (at PSI) re-energizes that same electron to a high-enough level to make NADPH.
The photochemistry of vision is the remarkable biological process that converts a single photon of light into an electrical nerve impulse that the brain interprets as "sight." This entire process is initiated by a single, ultrafast photochemical reaction in the rod and cone cells of the retina.
In the rod cells of the retina (which are responsible for low-light, black-and-white vision), the key molecule is rhodopsin. Rhodopsin is a biological pigment composed of two parts:
In its "dark" or resting state, the retinal molecule is in a specific bent shape called the 11-cis-retinal isomer. This specific shape fits perfectly into a binding "pocket" within the opsin protein, holding the protein in an inactive conformation.
The entire process of vision begins when a single photon of light enters the eye and is absorbed by the 11-cis-retinal chromophore.
$$ \text{11-cis-retinal} \xrightarrow{h\nu \text{ (1 photon)}} \text{all-trans-retinal} $$
This single, tiny molecular shape-change is the only light-dependent step in the entire process of vision.
What follows is a brilliant example of biological amplification, where one molecular event is magnified into a large-scale cellular response.
In essence, "seeing" is the brain processing the electrical signal generated by the cis-to-trans isomerization of a single molecule in the retina.
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