Exercises — Rate law — order vs molecularity

Order ::: kisi concentration par experimentally measured rate law mein exponent; yeh batata hai ki concentration change karne par speed kaise respond karti hai. Molecularity ::: un molecules ki sankhya jo mechanism ke ek single (elementary) step mein ek saath collide karni chahiye; ek positive integer 1, 2, ya 3.

(a) Collide karne wale reactant molecules gino: ek $\text{NO}$ aur ek ${\text{O}}_3$ — yeh 2 molecules hain, isliye molecularity $=2$ (bimolecular).

(b) Sirf elementary step ke liye, exponents stoichiometric coefficients ke barabar hote hain (kyunki stoichiometry hi collision requirement hai). Har reactant ka coefficient 1 hai: $\text{Rate}=k[\text{NO}][{\text{O}}_3]$ Overall order $=1+1=2$. Yahan order aur molecularity match karte hain — yeh ek single elementary step ke liye guaranteed hai.

• (a) Order — yeh experimentally fitted exponent hai.
• (b) Order — orders fractional ho sakte hain (jaise chain reactions ke liye $\frac{1}{2}$), negative bhi. Molecularity nahi ho sakta.
• (c) Molecularity — yeh colliding particles ki count hai: 1, 2, ya 3.
• (d) Molecularity — "ek step mein kitne collide karte hain" ka matlab sirf ek single step ke liye hi banta hai.

Toh: Order $0$, fractional, ya negative ho sakta hai. Molecularity hamesha $\le 3$ ka ek positive integer hota hai. Ek worked negative-order case neeche L5.4 mein milega.

A mein Order — do aise trials chunno jahan sirf $[\text{A}]$ change ho: trials 1 aur 2 (B, 0.10 par fixed). Kyunki $[\text{B}]$ dono mein identical hai, $[\text{B}{]}^n$ factors cancel ho jaate hain aur sirf $[\text{A}{]}^m$ wala part bachta hai: $\frac{{\text{Rate}}_2}{{\text{Rate}}_1}={\left(\frac{[\text{A}{]}_2}{[\text{A}{]}_1}\right)}^m⟹\frac{0.0060}{0.0030}={\left(\frac{0.20}{0.10}\right)}^m⟹2=2^m⟹m=1.$ A ko double karne se rate double hui → A mein first order (figure mein yeh ×2 yellow arrow hai).

B mein Order — ab trials 2 aur 3 use karo jahan sirf $[\text{B}]$ change ho (A, 0.20 par fixed), toh $[\text{A}{]}^m$ factors cancel ho jaate hain: $\frac{0.0240}{0.0060}={\left(\frac{0.20}{0.10}\right)}^n⟹4=2^n⟹n=2.$ B ko double karne se rate chaar guna hui → B mein second order (yeh ×4 pink arrow hai).

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

Rate constant — trial 1 ko $\text{Rate}=k[\text{A}][\text{B}{]}^2$ mein plug karo: $0.0030=k(0.10)(0.10{)}^2=k(0.10)(0.01)=k(0.001)⟹k=3.0{\text{M}}^{-2}{\text{s}}^{-1}.$ Units check: overall-order-3 reaction ke liye, $k$ ke units ${\text{M}}^{1-3}{\text{s}}^{-1}={\text{M}}^{-2}{\text{s}}^{-1}$ hote hain. ✓

Log form kyun? First order ka matlab hai $\frac{d[\text{A}]}{dt}=-k[\text{A}]$; integrate karne par straight-line law milta hai $\ln [\text{A}]=\ln [\text{A}{]}_0-kt$. Hum ise isliye use karte hain kyunki yeh directly concentration ko time se connect karta hai. $\ln [\text{A}]=\ln (0.80)-(0.0231)(60)=-0.2231-1.386=-\mathrm{1.609.}$ $[\text{A}]=e^{-1.609}=0.20\text{M}.$ Intuition se check karo: $t_{1/2}=\ln 2/k=30$ s, toh 60 s exactly do half-lives hai: $0.80\to 0.40\to 0.20$ M. ✓ ($k$ first order ke liye units ${\text{s}}^{-1}$ hain, concentration se independent.)

Molecularity: Step 1 mein 2 colliding molecules hain → bimolecular (2). Step 2 mein bhi 2 hain → bimolecular (2).

Slow step se rate law. Sabse slow step bottleneck hai: overall kuch bhi us se faster nahi ja sakta (yeh figure s03 mein tall pehla barrier hai). Us elementary step ke liye exponents uske coefficients ke barabar hote hain: $\text{Rate}=k_1[\text{NO}][{\text{Cl}}_2].$ Overall order $=1+1=2$.

Note karo mismatch: balanced equation mein $2\text{NO}$ hai (coefficient 2), lekin slow step sirf ek NO use karta hai, isliye NO mein order 1 hai, 2 nahi. Order mechanism se decide hota hai, kabhi overall coefficients se nahi.

Agar yeh ek single termolecular step hota ($2{\text{NO}}_2+{\text{F}}_2$ ek saath collide karte), toh elementary-step rule force karta $\text{Rate}=k[{\text{NO}}_2{]}^2[{\text{F}}_2](\text{order 3}).$ Lekin experiment ${\text{NO}}_2$ mein order 1, total order 2 deta hai — nahi 3. Toh single-step claim data se disprove ho jaata hai.

Real mechanism mein ek slow bimolecular step hai ${\text{NO}}_2+{\text{F}}_2\to {\text{NO}}_2\text{F}+\text{F}$, jo $\text{Rate}=k[{\text{NO}}_2][{\text{F}}_2]$ deta hai, jo experiment se match karta hai. Rule: experiment referee hai; ek mechanism tab hi survive karta hai jab uska predicted rate law measured order se match kare.

Step 1 — slow step se rate. Slow step pace set karta hai: $\text{Rate}=k_2[{\text{N}}_2{\text{O}}_2][{\text{O}}_2].$ Lekin ${\text{N}}_2{\text{O}}_2$ ek intermediate hai — yeh kuch aisa nahi hai jise tum bottle mein rakh ke measure kar sako, isliye ise eliminate karna hoga.

Step 2 — fast equilibrium use karo. "Fast equilibrium" ka matlab hai step 1 ki forward aur reverse rates equal hain: $k_1[\text{NO}{]}^2=k_{-1}[{\text{N}}_2{\text{O}}_2]⟹[{\text{N}}_2{\text{O}}_2]=\frac{k_1}{k_{-1}}[\text{NO}{]}^2.$ Yeh allowed kyun hai: slow step ${\text{N}}_2{\text{O}}_2$ ko itna slowly drain karta hai ki step 1 essentially balanced rehta hai.

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

Har microscopic $k$ ke units kahan se aate hain? Rule simple hai: rate constant ke units woh hote hain jo apne elementary rate expression ke dono sides ko $\text{M}{\text{s}}^{-1}$ (concentration per time) mein laate hain.

• Forward step 1 NO mein bimolecular hai: $\text{rate}=k_1[\text{NO}{]}^2$. Yahan $[\text{NO}{]}^2$ ke units ${\text{M}}^2$ hain, aur humein $\text{M}{\text{s}}^{-1}$ chahiye, isliye $k_1$ ko $\frac{\text{M}{\text{s}}^{-1}}{{\text{M}}^2}={\text{M}}^{-1}{\text{s}}^{-1}$ supply karna hoga.
• Reverse step 1 ${\text{N}}_2{\text{O}}_2$ mein unimolecular hai: $\text{rate}=k_{-1}[{\text{N}}_2{\text{O}}_2]$. Yahan $[{\text{N}}_2{\text{O}}_2]$ ke units $\text{M}$ hain, isliye $k_{-1}=\frac{\text{M}{\text{s}}^{-1}}{\text{M}}={\text{s}}^{-1}$.
• Slow step 2 bimolecular hai (${\text{N}}_2{\text{O}}_2+{\text{O}}_2$): $\text{rate}=k_2[{\text{N}}_2{\text{O}}_2][{\text{O}}_2]$, do $\text{M}$'s ka product, isliye $k_2=\frac{\text{M}{\text{s}}^{-1}}{{\text{M}}^2}={\text{M}}^{-1}{\text{s}}^{-1}$.

$k_{\text{obs}}$ ke units: combine karne par, $\frac{k_1{k}_2}{k_{-1}}=\frac{({\text{M}}^{-1}{\text{s}}^{-1})({\text{M}}^{-1}{\text{s}}^{-1})}{{\text{s}}^{-1}}={\text{M}}^{-2}{\text{s}}^{-1}$ — exactly woh units jo overall-order-3 constant ke liye expected hain. ✓

Notice karo: order 3 aaya even though koi bhi step termolecular nahi tha — yeh ek equilibrium ko slow step mein fold karne se aaya.

Rate $\propto [\text{NO}{]}^2[{\text{O}}_2]$. NO ko teen guna karne se $3^2=9$ se multiply hota hai; ${\text{O}}_2$ ko aadha karne se $\frac{1}{2}$ se multiply hota hai. $\text{factor}=9\times \frac{1}{2}=\mathrm{4.5.}$ Rate 4.5 guna zyada ho jaati hai.

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

(b) Molecularity ek elementary step mein collide karne wale molecules count karta hai. Yeh reaction kai steps ki chain hai (Br₂ Br atoms mein dissociate hona, propagation, termination). "Poori reaction" ek single collision nahi hai, isliye molecularity ka iske liye koi matlab nahi — sirf individual steps ki molecularity hoti hai. $\frac{1}{2}$ ${\text{Br}}_2\rightleftharpoons 2\text{Br}$ dissociation equilibrium ka fingerprint hai jo chain mechanism mein square root ke andar aata hai.

(c) Rate $\propto [{\text{Br}}_2{]}^{1/2}$. Chaar guna karne par factor $4^{1/2}=2$ milta hai. Rate double ho jaati hai.

(d) Overall order $n=1.5$ ke liye, $k$ ke units ${\text{M}}^{1-n}{\text{s}}^{-1}={\text{M}}^{-0.5}{\text{s}}^{-1}$ hain (yaani ${\text{M}}^{-1/2}{\text{s}}^{-1}$).

$\frac{R_2}{R_1}=\frac{[\text{A}{]}_2}{[\text{A}{]}_1}\cdot {\left(\frac{[\text{B}{]}_2}{[\text{B}{]}_1}\right)}^{1/2}=\frac{0.080}{0.040}\cdot {\left(\frac{0.64}{0.16}\right)}^{1/2}.$ $=2\times (4{)}^{1/2}=2\times 2=4.$ Toh $R_2/R_1=4$: rate chaar guna ho jaati hai.

$k$ ke units: overall order $=1+\frac{1}{2}=\frac{3}{2}$, isliye $k$ ke units ${\text{M}}^{1-3/2}{\text{s}}^{-1}={\text{M}}^{-1/2}{\text{s}}^{-1}$ hain.

(a) Zero order: $\frac{d[\text{A}]}{dt}=-k$ (constant), isliye $[\text{A}]=[\text{A}{]}_0-kt$ — ek straight line. Yeh zero hoti hai jab $t=\frac{[\text{A}{]}_0}{k}=\frac{0.50}{2.0\times 10^{-3}}=250\text{s}.$ (Zero order ke liye $k$ ke units rate ke hote hain, $\text{M}{\text{s}}^{-1}$.)

(b) Zero order ka matlab hai rate concentration se independent hai — usually isliye ki ek catalyst surface (ya enzyme) saturated hai: har active site pehle se busy hai, isliye zyada $\text{A}$ add karne se speed up nahi hogi. Bottleneck fixed surface sites ki sankhya hai, solution mein collide karne wale molecules nahi. Isliye molecularity concept overall zero-order behaviour describe nahi karta — molecularity ek collision count hai ek single elementary step ke liye, aur yahan rate-limiting event "ek site ka turn over karna" hai, koi collision nahi. Mechanism ke andar step-by-step molecularity hoti hi hai, lekin kabhi bhi observed zero-order law par nahi. Yeh extreme reminder hai: order 0 ho sakta hai, molecularity kabhi 0 nahi ho sakti.

(a) $\text{Rate}=k[{\text{O}}_3{]}^2[{\text{O}}_2{]}^{-1}$ ki tarah rewrite karo. Toh ${\text{O}}_3$ mein order $=2$, ${\text{O}}_2$ mein order $=-1$, overall order $=2+(-1)=1$.

(b) Rate $\propto [{\text{O}}_2{]}^{-1}$. $[{\text{O}}_2]$ teen guna karne par rate $3^{-1}=\frac{1}{3}$ se multiply hoti hai: rate ek-tihaai reh jaati hai. Negative order ka matlab hai ki woh species add karne se reaction slow ho jaati hai.

(c) Mechanism mein ek fast pre-equilibrium hai ${\text{O}}_3\rightleftharpoons {\text{O}}_2+\text{O}$ jo ${\text{O}}_2$ produce karta hai. Extra ${\text{O}}_2$ add karna is equilibrium ko backward push karta hai (Le Chatelier), reactive $\text{O}$ atom ki concentration kam karta hai jo slow step drive karta hai. Kam $\text{O}$ atoms → slower reaction → ${\text{O}}_2$ negative exponent ke saath appear hota hai. Yeh L1.2 mein promised worked negative-order case hai.

$2\text{NO}+{\text{Cl}}_2\to 2\text{NOCl}$ mein NO ka order jab slow step $\text{NO}+{\text{Cl}}_2\to {\text{NOCl}}_2$ ho ::: 1 (slow step mein sirf ek NO hai, 2 nahi) L2.1 mein overall order ::: 3 Kya molecularity 0.5 ya $-1$ ho sakti hai? ::: Nahi — yeh ek whole-number collision count hai (1, 2, ya 3) $[{\text{Br}}_2{]}^{1/2}$ kahan se aata hai ::: chain mechanism mein Br₂ ⇌ 2Br dissociation equilibrium se $\text{Rate}=k[{\text{O}}_3{]}^2/[{\text{O}}_2]$ ka overall order ::: 1 (yaani $2+(-1)$)