Exercises — Rate law — order vs molecularity

Order ::: the exponent on a concentration in the experimentally measured rate law; it says how the speed responds when you change that concentration. Molecularity ::: the number of molecules that must collide together in one single (elementary) step of the mechanism; a positive integer 1, 2, or 3.

(a) Count the reactant molecules that collide: one $\text{NO}$ and one ${\text{O}}_3$ — that is 2 molecules, so molecularity $=2$ (bimolecular).

(b) For an elementary step ONLY, the exponents equal the stoichiometric coefficients (because the stoichiometry is the collision requirement). Each reactant has coefficient 1: $\text{Rate}=k[\text{NO}][{\text{O}}_3]$ Overall order $=1+1=2$. Here order and molecularity happen to match — that is guaranteed for a single elementary step.

• (a) Order — it is the experimentally fitted exponent.
• (b) Order — orders can be fractional (e.g. $\frac{1}{2}$ for chain reactions), even negative. Molecularity cannot.
• (c) Molecularity — it is a count of colliding particles: 1, 2, or 3.
• (d) Molecularity — "how many collide in one step" only makes sense for a single step.

So: Order can be $0$, fractional, or negative. Molecularity is always a positive integer $\le 3$. A worked negative-order case appears in L5.4 below.

Order in A — pick two trials where only $[\text{A}]$ changes: trials 1 and 2 (B fixed at 0.10). Because $[\text{B}]$ is identical in both, the $[\text{B}{]}^n$ factors cancel and only the $[\text{A}{]}^m$ part survives: $\frac{{\text{Rate}}_2}{{\text{Rate}}_1}={\left(\frac{[\text{A}{]}_2}{[\text{A}{]}_1}\right)}^m⟹\frac{0.0060}{0.0030}={\left(\frac{0.20}{0.10}\right)}^m⟹2=2^m⟹m=1.$ Doubling A doubled the rate → first order in A (this is the ×2 yellow arrow in the figure).

Order in B — now use trials 2 and 3 where only $[\text{B}]$ changes (A fixed at 0.20), so the $[\text{A}{]}^m$ factors cancel: $\frac{0.0240}{0.0060}={\left(\frac{0.20}{0.10}\right)}^n⟹4=2^n⟹n=2.$ Doubling B quadrupled the rate → second order in B (this is the ×4 pink arrow).

Overall order $=m+n=1+2=3$.

Rate constant — plug trial 1 into $\text{Rate}=k[\text{A}][\text{B}{]}^2$: $0.0030=k(0.10)(0.10{)}^2=k(0.10)(0.01)=k(0.001)⟹k=3.0{\text{M}}^{-2}{\text{s}}^{-1}.$ Units check: for an overall-order-3 reaction, $k$ carries units ${\text{M}}^{1-3}{\text{s}}^{-1}={\text{M}}^{-2}{\text{s}}^{-1}$. ✓

Why the log form? First order means $\frac{d[\text{A}]}{dt}=-k[\text{A}]$; integrating gives the straight-line law $\ln [\text{A}]=\ln [\text{A}{]}_0-kt$. We use it because it directly connects concentration to time. $\ln [\text{A}]=\ln (0.80)-(0.0231)(60)=-0.2231-1.386=-\mathrm{1.609.}$ $[\text{A}]=e^{-1.609}=0.20\text{M}.$ Check with intuition: $t_{1/2}=\ln 2/k=30$ s, so 60 s is exactly two half-lives: $0.80\to 0.40\to 0.20$ M. ✓ ($k$ for first order has units ${\text{s}}^{-1}$, independent of concentration.)

Molecularity: Step 1 has 2 colliding molecules → bimolecular (2). Step 2 also has 2 → bimolecular (2).

Rate law from the slow step. The slowest step is the bottleneck: nothing overall can go faster than it (this is the tall first barrier in figure s03). For that elementary step the exponents equal its coefficients: $\text{Rate}=k_1[\text{NO}][{\text{Cl}}_2].$ Overall order $=1+1=2$.

Note the mismatch: the balanced equation has $2\text{NO}$ (coefficient 2), but the slow step uses only one NO, so the order in NO is 1, not 2. Order is decided by the mechanism, never by the overall coefficients.

If it were a single termolecular step ($2{\text{NO}}_2+{\text{F}}_2$ colliding at once), the elementary-step rule would force $\text{Rate}=k[{\text{NO}}_2{]}^2[{\text{F}}_2](\text{order 3}).$ But experiment gives order 1 in ${\text{NO}}_2$, total order 2 — not 3. So the single-step claim is disproved by the data.

The real mechanism has a slow bimolecular step ${\text{NO}}_2+{\text{F}}_2\to {\text{NO}}_2\text{F}+\text{F}$, giving $\text{Rate}=k[{\text{NO}}_2][{\text{F}}_2]$, matching experiment. Rule: experiment is the referee; a mechanism survives only if its predicted rate law matches measured order.

Step 1 — rate from the slow step. The slow step sets the pace: $\text{Rate}=k_2[{\text{N}}_2{\text{O}}_2][{\text{O}}_2].$ But ${\text{N}}_2{\text{O}}_2$ is an intermediate — it is not something you can put in a bottle and measure, so it must be eliminated.

Step 2 — use the fast equilibrium. "Fast equilibrium" means step 1's forward and reverse rates are equal: $k_1[\text{NO}{]}^2=k_{-1}[{\text{N}}_2{\text{O}}_2]⟹[{\text{N}}_2{\text{O}}_2]=\frac{k_1}{k_{-1}}[\text{NO}{]}^2.$ Why this is allowed: the slow step drains ${\text{N}}_2{\text{O}}_2$ so slowly that step 1 stays essentially balanced.

Step 3 — substitute. $\text{Rate}=k_2\cdot \frac{k_1}{k_{-1}}[\text{NO}{]}^2[{\text{O}}_2]=k_{\text{obs}}[\text{NO}{]}^2[{\text{O}}_2].$ Overall order $=2+1=3$, with $k_{\text{obs}}=\frac{k_1{k}_2}{k_{-1}}$.

Where do the units of each microscopic $k$ come from? The rule is simple: a rate constant's units are whatever makes both sides of its own elementary rate expression come out in $\text{M}{\text{s}}^{-1}$ (concentration per time).

• Forward step 1 is bimolecular in NO: $\text{rate}=k_1[\text{NO}{]}^2$. Here $[\text{NO}{]}^2$ has units ${\text{M}}^2$, and we need $\text{M}{\text{s}}^{-1}$, so $k_1$ must supply $\frac{\text{M}{\text{s}}^{-1}}{{\text{M}}^2}={\text{M}}^{-1}{\text{s}}^{-1}$.
• Reverse step 1 is unimolecular in ${\text{N}}_2{\text{O}}_2$: $\text{rate}=k_{-1}[{\text{N}}_2{\text{O}}_2]$. Here $[{\text{N}}_2{\text{O}}_2]$ has units $\text{M}$, so $k_{-1}=\frac{\text{M}{\text{s}}^{-1}}{\text{M}}={\text{s}}^{-1}$.
• Slow step 2 is bimolecular (${\text{N}}_2{\text{O}}_2+{\text{O}}_2$): $\text{rate}=k_2[{\text{N}}_2{\text{O}}_2][{\text{O}}_2]$, product of two $\text{M}$'s, so $k_2=\frac{\text{M}{\text{s}}^{-1}}{{\text{M}}^2}={\text{M}}^{-1}{\text{s}}^{-1}$.

Units of $k_{\text{obs}}$: combining, $\frac{k_1{k}_2}{k_{-1}}=\frac{({\text{M}}^{-1}{\text{s}}^{-1})({\text{M}}^{-1}{\text{s}}^{-1})}{{\text{s}}^{-1}}={\text{M}}^{-2}{\text{s}}^{-1}$ — exactly the units expected for an overall-order-3 constant. ✓

Notice: order 3 arose even though no step was termolecular — it came from folding an equilibrium into the slow step.

Rate $\propto [\text{NO}{]}^2[{\text{O}}_2]$. Tripling NO multiplies by $3^2=9$; halving ${\text{O}}_2$ multiplies by $\frac{1}{2}$. $\text{factor}=9\times \frac{1}{2}=\mathrm{4.5.}$ The rate becomes 4.5 times larger.

(a) Order in ${\text{H}}_2=1$; order in ${\text{Br}}_2=\frac{1}{2}=0.5$; overall $=1+0.5=1.5$.

(b) Molecularity counts molecules colliding in one elementary step. This reaction is a chain of many steps (Br₂ dissociating into Br atoms, propagation, termination). "The whole reaction" is not a single collision, so molecularity has no meaning for it — only individual steps have molecularity. The $\frac{1}{2}$ is the fingerprint of the ${\text{Br}}_2\rightleftharpoons 2\text{Br}$ dissociation equilibrium appearing under a square root.

(c) Rate $\propto [{\text{Br}}_2{]}^{1/2}$. Quadrupling gives factor $4^{1/2}=2$. The rate doubles.

(d) For overall order $n=1.5$, $k$ has units ${\text{M}}^{1-n}{\text{s}}^{-1}={\text{M}}^{-0.5}{\text{s}}^{-1}$ (i.e. ${\text{M}}^{-1/2}{\text{s}}^{-1}$).

$\frac{R_2}{R_1}=\frac{[\text{A}{]}_2}{[\text{A}{]}_1}\cdot {\left(\frac{[\text{B}{]}_2}{[\text{B}{]}_1}\right)}^{1/2}=\frac{0.080}{0.040}\cdot {\left(\frac{0.64}{0.16}\right)}^{1/2}.$ $=2\times (4{)}^{1/2}=2\times 2=4.$ So $R_2/R_1=4$: the rate quadruples.

Units of $k$: overall order $=1+\frac{1}{2}=\frac{3}{2}$, so $k$ has units ${\text{M}}^{1-3/2}{\text{s}}^{-1}={\text{M}}^{-1/2}{\text{s}}^{-1}$.

(a) Zero order: $\frac{d[\text{A}]}{dt}=-k$ (constant), so $[\text{A}]=[\text{A}{]}_0-kt$ — a straight line. It hits zero when $t=\frac{[\text{A}{]}_0}{k}=\frac{0.50}{2.0\times 10^{-3}}=250\text{s}.$ (For zero order $k$ has units of a rate, $\text{M}{\text{s}}^{-1}$.)

(b) Zero order means the rate is independent of concentration — usually because a catalyst surface (or an enzyme) is saturated: every active site is already busy, so adding more $\text{A}$ cannot speed things up. The bottleneck is the fixed number of surface sites, not molecules colliding in solution. So the molecularity concept does not describe the overall zero-order behaviour — molecularity is a collision count for a single elementary step, and here the rate-limiting event is "a site turning over," not a collision. Molecularity still applies step-by-step inside the mechanism, but never to the observed zero-order law. This is the extreme reminder: order can be 0, molecularity can never be 0.

(a) Rewrite as $\text{Rate}=k[{\text{O}}_3{]}^2[{\text{O}}_2{]}^{-1}$. So order in ${\text{O}}_3=2$, order in ${\text{O}}_2=-1$, overall order $=2+(-1)=1$.

(b) Rate $\propto [{\text{O}}_2{]}^{-1}$. Tripling $[{\text{O}}_2]$ multiplies the rate by $3^{-1}=\frac{1}{3}$: the rate drops to one-third. A negative order means adding that species slows the reaction down.

(c) The mechanism has a fast pre-equilibrium ${\text{O}}_3\rightleftharpoons {\text{O}}_2+\text{O}$ that produces ${\text{O}}_2$. Adding extra ${\text{O}}_2$ pushes this equilibrium backward (Le Chatelier), lowering the concentration of the reactive $\text{O}$ atom that drives the slow step. Fewer $\text{O}$ atoms → slower reaction → ${\text{O}}_2$ appears with a negative exponent. This is the worked negative-order case promised in L1.2.

The order in NO for $2\text{NO}+{\text{Cl}}_2\to 2\text{NOCl}$ with slow step $\text{NO}+{\text{Cl}}_2\to {\text{NOCl}}_2$ ::: 1 (only one NO in the slow step, not 2) Overall order in L2.1 ::: 3 Can molecularity be 0.5 or $-1$? ::: No — it is a whole-number collision count (1, 2, or 3) Where does the $[{\text{Br}}_2{]}^{1/2}$ come from ::: a Br₂ ⇌ 2Br dissociation equilibrium in the chain mechanism Overall order of $\text{Rate}=k[{\text{O}}_3{]}^2/[{\text{O}}_2]$ ::: 1 (that is $2+(-1)$)