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

Har arrow ke left pe molecules ginlo — molecularity literally hai "kitni cheezein collide/break apart karti hain."

1. Ek molecule toot jaata hai → unimolecular (molecularity 1).
2. Do molecules milte hain → bimolecular (molecularity 2).
3. Teen particles ek saath milne chahiye ($\text{H}$, $\text{H}$, aur ek third body $\text{M}$) → termolecular (molecularity 3). Termolecular steps rare hote hain kyunki ek saath teen-body collision unlikely hai.

Kyunki yeh elementary bataya gaya hai, Fact A apply hoga — exponents = coefficients: $\text{Rate}=k[\text{NO}{]}^2[{\text{Cl}}_2]$ Molecularity yahan $2+1=3$ hai (termolecular).

(a) Overall equation — dono steps add karo aur jo dono sides pe aaye use cancel karo: ${\text{H}}_2{\text{O}}_2+{\text{I}}^{-}+{\text{H}}_2{\text{O}}_2+{\text{IO}}^{-}\to {\text{H}}_2\text{O}+{\text{IO}}^{-}+{\text{H}}_2\text{O}+{\text{O}}_2+{\text{I}}^{-}$ ${\text{IO}}^{-}$ cancel ho jaata hai (1 mein bana, 2 mein khatam); ${\text{I}}^{-}$ cancel ho jaata hai (1 mein khatam, 2 mein dobara bana): $\boxed{2{\text{H}}_2{\text{O}}_2\to 2{\text{H}}_2\text{O}+{\text{O}}_2}$

(b) Rate law — Step 1 slow hai, toh yeh RDS hai. Step 1 pe Fact A apply karo (elementary): $\boxed{\text{Rate}=k_1[{\text{H}}_2{\text{O}}_2][{\text{I}}^{-}]}$ Is expression mein koi intermediate nahi aaya, toh hum done hain.

(c) ${\text{IO}}^{-}$ ek intermediate hai (bana phir consume hua). ${\text{I}}^{-}$ ek catalyst hai (step 1 mein consume hua lekin step 2 mein regenerate ho gaya, toh overall equation mein appear nahi karta). Dhyan do ki rate law mein ${\text{I}}^{-}$ hai: catalysts rate laws mein aa sakte hain chahe overall equation se cancel ho jaayein — dekho Activation Energy and Catalysts.

Dono first order hain. Effects multiply karo: $\frac{{\text{Rate}}_{\text{new}}}{{\text{Rate}}_{\text{old}}}=(3{)}^1\times (2{)}^1=6$ Rate 6× zyada ho jaati hai. (Note karo ki humne experimental orders use kiye — har ek mein first order — ${\text{NO}}_2$ ke stoichiometric "2" ko nahi.)

RDS rate (Step 2 slow hai): $\text{Rate}=k_2[{\text{NOBr}}_2][\text{NO}]$ ${\text{NOBr}}_2$ ek intermediate hai — hume ise remove karna hoga. Equilibrium expression kyun? Step 1 dono directions mein fast hai, toh iska forward aur reverse rate slow step ke usse drain karne se pehle hi equalize ho jaata hai. Equilibrium ka matlab hai 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 karo: $\text{Rate}=k_{2}K[\text{NO}][{\text{Br}}_2][\text{NO}]=k_{2}K[\text{NO}{]}^2[{\text{Br}}_2]$ $k_{\text{obs}}=k_{2}K$ ke saath: $\boxed{\text{Rate}=k_{\text{obs}}[\text{NO}{]}^2[{\text{Br}}_2]}$ Overall order $=2+1=3$. Slow step sabse zyada energy barrier pe kyun baithta hai, yeh samajhne ke liye Reaction Coordinate Diagrams dekho.

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

Dono same rate law fit karte hain. Ek rate law galat mechanism ko rule out kar sakta hai lekin kisi ek ko prove nahi kar sakta, kyunki alag-alag mechanisms ek hi rate law share kar sakte hain. Inhe distinguish karne ke liye extra evidence chahiye — intermediate ${\text{N}}_2{\text{O}}_2$ ko spectroscopically detect karna, ya temperature/pressure dependence. Mechanism A (ek rare termolecular step) physically less likely hai, toh B zyada favorable hai.

$\text{O}$ ke liye steady state set up karo. Har woh step list karo jo $\text{O}$ banata ya destroy karta hai:

• banta hai: Step 1 forward, rate $k_1[{\text{O}}_3]$
• destroy hota hai: Step 1 reverse, rate $k_{-1}[{\text{O}}_2][\text{O}]$; aur 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: $[\text{O}]=\frac{k_1[{\text{O}}_3]}{k_{-1}[{\text{O}}_2]+k_2[{\text{O}}_3]}$ ${\text{O}}_3$ ki consumption ki rate: ${\text{O}}_3$ Step 1 forward aur Step 2 mein use hota hai (aur Step 1 reverse mein dobara banta hai). Overall rate Step 2 ke products banane se govern hoti hai; ${\text{O}}_3$ loss use karte hue, $\text{Rate}=k_2[\text{O}][{\text{O}}_3]=\frac{k_1{k}_2[{\text{O}}_3{]}^2}{k_{-1}[{\text{O}}_2]+k_2[{\text{O}}_3]}$ $\boxed{\text{Rate}=\frac{k_1{k}_2[{\text{O}}_3{]}^2}{k_{-1}[{\text{O}}_2]+k_2[{\text{O}}_3]}}$

Limits check karo (yahi steady state ka payoff hai — yeh dono regimes ek saath cover karta hai):

• Agar Step 2 slow hai / ${\text{O}}_2$ abundant hai ($k_{-1}[{\text{O}}_2]\gg k_2[{\text{O}}_3]$): denominator $\approx k_{-1}[{\text{O}}_2]$, deta hai $\text{Rate}\approx \frac{k_1{k}_2}{k_{-1}}\frac{[{\text{O}}_3{]}^2}{[{\text{O}}_2]}$ — ek rate jo product ${\text{O}}_2$ se inhibit hoti hai. Yeh fast-pre-equilibrium answer se match karta hai.
• Agar Step 2 fast hai ($k_2[{\text{O}}_3]\gg k_{-1}[{\text{O}}_2]$): denominator $\approx k_2[{\text{O}}_3]$, deta hai $\text{Rate}\approx k_1[{\text{O}}_3]$ — first order, initial dissociation se controlled.

Full justification ke liye Steady-State Approximation dekho.

Inhibited-limit law: $\text{Rate}\propto \frac{[{\text{O}}_3{]}^2}{[{\text{O}}_2]}$. $\frac{{\text{Rate}}_{\text{new}}}{{\text{Rate}}_{\text{old}}}=\frac{(1/2{)}^2}{2}=\frac{1/4}{2}=\frac{1}{8}$ Rate apni original value ki 1/8 reh jaati hai.

Target: ${\text{H}}_2$ aur ${\text{I}}_2$ dono mein first order. Idea: ${\text{I}}_2$ ko pehle atoms mein pre-dissociate karne do, phir react karo.

• 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 $=k_2[{\text{H}}_2][\text{I}{]}^2$

Intermediate $\text{I}$ ko eliminate karo fast equilibrium ke through: $K=\frac{[\text{I}{]}^2}{[{\text{I}}_2]}\Rightarrow [\text{I}{]}^2=K[{\text{I}}_2]$ Substitute karo: $\text{Rate}=k_2[{\text{H}}_2]K[{\text{I}}_2]=k_{2}K[{\text{H}}_2][{\text{I}}_2]$ $k=k_{2}K$ ke saath: $\boxed{\text{Rate}=k[{\text{H}}_2][{\text{I}}_2]}$ ✓ — experiment se match karta hai.

Consistency check (Fact B, step 5): steps sum karo. Step 1 $2\text{I}$ banata hai; Step 2 unhe consume karta hai: ${\text{I}}_2+{\text{H}}_2\to 2\text{HI}$. ✓ Overall equation recover ho gaya. Dhyan do ki $[\text{I}{]}^2$ (intermediate mein order-2) $[{\text{I}}_2]$ mein order-1 mein collapse ho gaya — isliye observed order kisi single coefficient se match nahi karta.

(a) Sabse chhota rate constant matlab sabse slow step (har second mein sabse kam events): $k_b=3.0\times 10^{-2}{\text{s}}^{-1}\text{ sabse chhota hai}\Rightarrow \text{Step b RDS hai.}$ Yeh step a se $\frac{5.0\times 10^2}{3.0\times 10^{-2}}\approx 1.67\times 10^4$ times slow hai — ek genuine bottleneck.

(b) Rate constant ghatta hai jaise activation energy $E_a$ badhti hai (Arrhenius: bada barrier ⇒ kam molecules usse cross kar paate hain). Step b ki sabse badi $E_a=95{\text{kJ mol}}^{-1}$ hai, toh haan — sabse chhota $k$ ↔ sabse lamba barrier. Consistent. ✓ (RDS yahan pehla step nahi hai, jo "step 1 hamesha RDS hota hai" wali reflex ko defeat karta hai.)

Overall order RDS se aata hai intermediates ko remove karne ke baad ::: overall equation ke coefficients almost kabhi woh exponents nahi hote jo tum observe karte ho.