Exercises — Reaction mechanisms — elementary steps, rate-determining step

Count the molecules on the left of each arrow — molecularity is literally "how many things collide/break apart."

1. One molecule falls apart → unimolecular (molecularity 1).
2. Two molecules meet → bimolecular (molecularity 2).
3. Three particles must meet at once ($\text{H}$, $\text{H}$, and a third body $\text{M}$) → termolecular (molecularity 3). Termolecular steps are rare because a simultaneous three-body collision is unlikely.

Because it is stated to be elementary, Fact A applies — exponents = coefficients: $\text{Rate}=k[\text{NO}{]}^2[{\text{Cl}}_2]$ Molecularity here is $2+1=3$ (termolecular).

(a) Overall equation — add the two steps and cancel what appears on both sides: ${\text{H}}_2{\text{O}}_2+{\text{I}}^{-}+{\text{H}}_2{\text{O}}_2+{\text{IO}}^{-}\to {\text{H}}_2\text{O}+{\text{IO}}^{-}+{\text{H}}_2\text{O}+{\text{O}}_2+{\text{I}}^{-}$ ${\text{IO}}^{-}$ cancels (made in 1, eaten in 2); ${\text{I}}^{-}$ cancels (eaten in 1, remade in 2): $\boxed{2{\text{H}}_2{\text{O}}_2\to 2{\text{H}}_2\text{O}+{\text{O}}_2}$

(b) Rate law — Step 1 is slow, so it is the RDS. Apply Fact A to Step 1 (elementary): $\boxed{\text{Rate}=k_1[{\text{H}}_2{\text{O}}_2][{\text{I}}^{-}]}$ No intermediate appears in this expression, so we are done.

(c) ${\text{IO}}^{-}$ is an intermediate (made then consumed). ${\text{I}}^{-}$ is a catalyst (consumed in step 1 but regenerated in step 2, so it does not appear in the overall equation). Notice the rate law contains ${\text{I}}^{-}$: catalysts can appear in rate laws even though they cancel from the overall equation — see Activation Energy and Catalysts.

Both are first order. Multiply the effects: $\frac{{\text{Rate}}_{\text{new}}}{{\text{Rate}}_{\text{old}}}=(3{)}^1\times (2{)}^1=6$ The rate becomes 6× larger. (Note we used the experimental orders — first order in each — not the stoichiometric "2".)

RDS rate (Step 2 is slow): $\text{Rate}=k_2[{\text{NOBr}}_2][\text{NO}]$ ${\text{NOBr}}_2$ is an intermediate — we must remove it. Why an equilibrium expression? Step 1 is fast in both directions, so its forward and reverse rates equalise long before the slow step drains it. Equilibrium means forward rate = reverse rate: $k_1[\text{NO}][{\text{Br}}_2]=k_{-1}[{\text{NOBr}}_2]\Rightarrow [{\text{NOBr}}_2]=\frac{k_1}{k_{-1}}[\text{NO}][{\text{Br}}_2]=K[\text{NO}][{\text{Br}}_2]$ Substitute: $\text{Rate}=k_{2}K[\text{NO}][{\text{Br}}_2][\text{NO}]=k_{2}K[\text{NO}{]}^2[{\text{Br}}_2]$ With $k_{\text{obs}}=k_{2}K$: $\boxed{\text{Rate}=k_{\text{obs}}[\text{NO}{]}^2[{\text{Br}}_2]}$ Overall order $=2+1=3$. See Reaction Coordinate Diagrams for why the slow step sits at the tallest energy barrier.

Mechanism A (elementary, Fact A): $\text{Rate}=k[\text{NO}{]}^2[{\text{O}}_2]$. ✓ Mechanism B: RDS gives $\text{Rate}=k_2[{\text{N}}_2{\text{O}}_2][{\text{O}}_2]$; equilibrium gives $[{\text{N}}_2{\text{O}}_2]=K[\text{NO}{]}^2$; substitute → $\text{Rate}=k_{2}K[\text{NO}{]}^2[{\text{O}}_2]$. ✓

Both fit the same rate law. A rate law can rule out a wrong mechanism but can never prove one, because different mechanisms can share a rate law. To distinguish them you need extra evidence — detecting the intermediate ${\text{N}}_2{\text{O}}_2$ spectroscopically, or the temperature/pressure dependence. Mechanism A (a rare termolecular step) is physically less likely, favouring B.

Set up the steady state for $\text{O}$. List every step that makes or destroys $\text{O}$:

• made: Step 1 forward, rate $k_1[{\text{O}}_3]$
• destroyed: Step 1 reverse, rate $k_{-1}[{\text{O}}_2][\text{O}]$; and Step 2, rate $k_2[\text{O}][{\text{O}}_3]$

Steady state ⇒ making rate = destroying rate: $k_1[{\text{O}}_3]=k_{-1}[{\text{O}}_2][\text{O}]+k_2[\text{O}][{\text{O}}_3]$ Solve for the intermediate: $[\text{O}]=\frac{k_1[{\text{O}}_3]}{k_{-1}[{\text{O}}_2]+k_2[{\text{O}}_3]}$ Rate of consumption of ${\text{O}}_3$: ${\text{O}}_3$ is used in Step 1 forward and Step 2 (and remade in Step 1 reverse). The overall rate is governed by Step 2 forming products; using ${\text{O}}_3$ loss, $\text{Rate}=k_2[\text{O}][{\text{O}}_3]=\frac{k_1{k}_2[{\text{O}}_3{]}^2}{k_{-1}[{\text{O}}_2]+k_2[{\text{O}}_3]}$ $\boxed{\text{Rate}=\frac{k_1{k}_2[{\text{O}}_3{]}^2}{k_{-1}[{\text{O}}_2]+k_2[{\text{O}}_3]}}$

Check the limits (this is the payoff of steady state — it covers both regimes at once):

• If Step 2 is slow / ${\text{O}}_2$ abundant ($k_{-1}[{\text{O}}_2]\gg k_2[{\text{O}}_3]$): denominator $\approx k_{-1}[{\text{O}}_2]$, giving $\text{Rate}\approx \frac{k_1{k}_2}{k_{-1}}\frac{[{\text{O}}_3{]}^2}{[{\text{O}}_2]}$ — a rate inhibited by product ${\text{O}}_2$. This matches the fast-pre-equilibrium answer.
• If Step 2 is fast ($k_2[{\text{O}}_3]\gg k_{-1}[{\text{O}}_2]$): denominator $\approx k_2[{\text{O}}_3]$, giving $\text{Rate}\approx k_1[{\text{O}}_3]$ — first order, controlled by the initial dissociation.

See Steady-State Approximation for the full justification.

Inhibited-limit law: $\text{Rate}\propto \frac{[{\text{O}}_3{]}^2}{[{\text{O}}_2]}$. $\frac{{\text{Rate}}_{\text{new}}}{{\text{Rate}}_{\text{old}}}=\frac{(1/2{)}^2}{2}=\frac{1/4}{2}=\frac{1}{8}$ The rate drops to 1/8 of its original value.

Target: first order in each of ${\text{H}}_2$ and ${\text{I}}_2$. Idea: let ${\text{I}}_2$ pre-dissociate into atoms, then react.

• Step 1 (fast eq): ${\text{I}}_2\rightleftharpoons 2\text{I}$, $K=k_1/k_{-1}$
• Step 2 (slow): ${\text{H}}_2+2\text{I}\to 2\text{HI}$, rate $=k_2[{\text{H}}_2][\text{I}{]}^2$

Eliminate the intermediate $\text{I}$ via the fast equilibrium: $K=\frac{[\text{I}{]}^2}{[{\text{I}}_2]}\Rightarrow [\text{I}{]}^2=K[{\text{I}}_2]$ Substitute: $\text{Rate}=k_2[{\text{H}}_2]K[{\text{I}}_2]=k_{2}K[{\text{H}}_2][{\text{I}}_2]$ With $k=k_{2}K$: $\boxed{\text{Rate}=k[{\text{H}}_2][{\text{I}}_2]}$ ✓ — matches experiment.

Consistency check (Fact B, step 5): sum the steps. Step 1 makes $2\text{I}$; Step 2 consumes them: ${\text{I}}_2+{\text{H}}_2\to 2\text{HI}$. ✓ Overall equation recovered. Notice $[\text{I}{]}^2$ (order-2 in an intermediate) collapsed to order-1 in $[{\text{I}}_2]$ — this is why the observed order need not equal any single coefficient.

(a) The smallest rate constant means the slowest step (fewest events per second): $k_b=3.0\times 10^{-2}{\text{s}}^{-1}\text{ is smallest}\Rightarrow \text{Step b is the RDS.}$ It is $\frac{5.0\times 10^2}{3.0\times 10^{-2}}\approx 1.67\times 10^4$ times slower than step a — a genuine bottleneck.

(b) Rate constant shrinks as activation energy $E_a$ grows (Arrhenius: bigger barrier ⇒ fewer molecules make it over). Step b has the largest $E_a=95{\text{kJ mol}}^{-1}$, so yes — smallest $k$ ↔ tallest barrier. Consistent. ✓ (The RDS is not the first step here, defeating the "step 1 is always the RDS" reflex.)

Overall order comes from the RDS after intermediates are removed ::: the coefficients of the overall equation are almost never the exponents you observe.