Rate law — order vs molecularity — Hinglish notes

2.8.2 · Chemistry › Chemical Kinetics

1. Rate Law and Order

Derivation from first principles

Hum ise is tarah kyun likhte hain? Kyunki lab mein hum [A] aur [B] ko systematically change karte hain aur measure karte hain ki Rate kaise respond karti hai. Agar [A] double karne se rate double ho jaaye (jab [B] fixed ho), toh Rate ∝ [A]¹. Agar [B] double karne se rate chaar guna ho jaaye, toh Rate ∝ [B]². Exponents $m, n$ data ke saath pure empirical fits hain, balanced equation se guess nahi.

Order experimentally kaise determine karein:

Initial rates method: Alag-alag starting concentrations ke saath multiple trials chalao, har baar initial rate measure karo.
Unhe trials compare karo jahaan sirf ek concentration change hoti hai: $\frac{Rate _{2}}{Rate _{1}} = \frac{k [ A ] _{2}^{m} [ B ] _{2}^{n}}{k [ A ] _{1}^{m} [ B ] _{1}^{n}}$ Agar [B] constant hai: $\frac{Rate _{2}}{Rate _{1}} = (\frac{[ A ] _{2}}{[ A ] _{1}})^{m}$
Logarithms use karke $m$ solve karo: $m = \frac{l n ( Rate _{2} / Rate _{1} )}{l n ([ A ] _{2} / [ A ] _{1} )}$

Ye kyun kaam karta hai: Hum ek waqt mein ek reactant ka effect isolate kar rahe hain, baaki ko constant rakh ke. Jo exponent ratio ko match karta hai, wahi order hai.

Ye graphs kyun? Rate equation ko integrate karo taaki A mile. First order ke liye: $\frac{d [ A ]}{d t} = - k [A] ⟹ \int_{[A]_{0}}^{[A]} \frac{d [ A ]}{[ A ]} = - k \int_{0}^{t} d t ⟹ ln [A] - ln [A]_{0} = - k t$ Toh $ln [A] = ln [A]_{0} - k t$ (jo $t$ mein linear hai).

Figure — Rate law — order vs molecularity

Worked example Data se order determine karna

Diya gaya hai: Reaction A + B → products

Trial	[A]₀ (M)	[B]₀ (M)	Initial Rate (M/s)
1	0.10	0.10	0.020
2	0.20	0.10	0.080
3	0.10	0.20	0.020

Dhundho: A, B ke w.r.t. order, aur overall order.

Solution: Trials 1 aur 2 compare karo ([B] constant): $\frac{0.080}{0.020} = (\frac{0.20}{0.10})^{m} ⟹ 4 = 2^{m} ⟹ m = 2$ Ye step kyun? [A] double karne se rate chaar guna ho gayi, toh Rate ∝ [A]².

Trials 1 aur 3 compare karo ([A] constant): $\frac{0.020}{0.020} = (\frac{0.20}{0.10})^{n} ⟹ 1 = 2^{n} ⟹ n = 0$ Ye step kyun? [B] double karne ka koi asar nahi hua, toh rate [B] par depend nahi karti.

Rate law: Rate = $k [A]^{2} [B]^{0} = k [A]^{2}$ Overall order: 2 + 0 = 2

Trial 1 se $k$ calculate karo: $0.020 = k (0.10)^{2} ⟹ k = \frac{0.020}{0.01} = 2.0 M^{- 1} s^{- 1}$

2. Molecularity

3 se aage rare kyun? Simultaneous collision ki probability exponentially drop hoti hai. Termolecular steps bhi uncommon hote hain; reactions uni- aur bimolecular steps ke sequences ke through proceed karti hain.

Sirf elementary steps ke liye

CRITICAL: Ye overall multi-step reactions ke liye bilkul sach nahi hai. Stoichiometric equation $2 H_{2} + O_{2} \to 2 H_{2} O$ ka matlab ye nahi ki Rate = $k [H_{2}]^{2} [O_{2}]$ kyunki mechanism mein kaafi saare elementary steps hote hain.

Worked example Elementary vs overall reaction

Reaction: $2 NO_{2} + F_{2} \to 2 NO_{2} F$

Case 1: Elementary (one step) Agar ye ek single termolecular collision mein hota hai (molecularity = 3): Rate = $k [NO_{2}]^{2} [F_{2}]$ Order w.r.t. NO₂ = 2, F₂ = 1, overall = 3. Yahaan, order = molecularity (dono = 3).

Case 2: Multi-step mechanism (actual) Experiments dikhate hain ki real mechanism hai:

$NO_{2} + F_{2} \to NO_{2} F + F$ (slow, bimolecular)
$NO_{2} + F \to NO_{2} F$ (fast, bimolecular)

Slow step overall rate determine karta hai: Rate = $k_{1} [NO_{2}] [F_{2}]$ Order w.r.t. NO₂ = 1, F₂ = 1, overall = 2. Step 1 ki molecularity = 2, step 2 ki = 2. Overall order (2) ≠ stoichiometric coefficients ka sum (3). Individual steps ki molecularity (dono 2) overall order nahi hai.

Ye step kyun? Slow step bottleneck hai. Fast step 2 F ko jaise hi banta hai consume kar leta hai, toh [F] accumulate nahi hota aur rate law mein appear nahi karta.

3. Order vs Molecularity — Key Differences

Property	Order	Molecularity
Source	Experimental (rate law from lab data)	Theoretical (from proposed mechanism)
Applies to	Elementary steps AND overall reactions	Sirf elementary steps
Values	0, 1, 2, ya fractional/negative bhi (rare)	1, 2, ya 3 (sirf positive integers)
Depends on	Kaun se reactants rate law mein hain, unke exponents	Ek step mein kitne molecules collide karte hain
Relation to stoichiometry	Overall reactions ke liye koi relation nahi	Elementary steps ke liye stoichiometric coefficients ke barabar

Common mistake Overall reactions ko molecularity assign karna

Galat idea: "Reaction $H_{2} + I_{2} \to 2 HI$ ki molecularity 2 hai." Ye sahi kyun lagta hai: Do molecules reactants ke roop mein appear hote hain. Fix: Molecularity sirf elementary steps ke liye defined hai. Overall reaction ek multi-step mechanism ke through proceed karti hai (jaise I₂ dissociation, phir H₂ + I → HI + H, phir H + I₂ → HI + I). Har elementary step ki apni molecularity hoti hai (dissociation ke liye 1, har bimolecular step ke liye 2), lekin overall reaction ki koi molecularity nahi hoti, sirf ek overall order hota hai (jo is reaction ke liye experimentally 2 hai, Rate = $k [H_{2}] [I_{2}]$ se match karta hai, modern understanding mein slow bimolecular step se, lekin is mechanism ke liye ye coincidence hai, koi rule nahi).

4. Mechanism ke zariye order aur molecularity ko connect karna

Ye kaise relate hote hain:

Proposed mechanism likho (elementary steps ka sequence).
Har elementary step ki molecularity uski rate law bataati hai (exponents = coefficients).
Rate-determining step (RDS, sabse slow) identify karo.
RDS rate law (possibly fast-equilibrium pre-steps substitute karke) overall rate law ban jaati hai.
Overall rate law mein exponents hi orders hain.

Ye kyun kaam karta hai: Sabse slow step bottleneck hai. Usse pehle ke faster steps equilibrium reach kar lete hain (forward/reverse rates equal), aur uske baad ke faster steps intermediates ko instantly clear kar dete hain. Toh overall rate ≈ RDS ki rate.

Worked example Mechanism with intermediate

Reaction: $2 NO + O_{2} \to 2 NO_{2}$ Mechanism:

$2 NO ⇌ N_{2} O_{2}$ (fast equilibrium, $K_{eq} = k_{1} / k_{- 1}$ )
$N_{2} O_{2} + O_{2} \to 2 NO_{2}$ (slow, bimolecular)

Step-by-step derivation: Overall reaction ki rate = step 2 ki rate (RDS): $Rate = k_{2} [N_{2} O_{2}] [O_{2}]$ Kyun? Step 2 slow hai, toh iska rate overall rate control karta hai.

Lekin N₂O₂ ek intermediate hai (measurable nahi). Step 1 se equilibrium use karo: $K_{eq} = \frac{[ N _{2} O _{2} ]}{[ NO ] ^{2}} ⟹ [N_{2} O_{2}] = K_{eq} [NO]^{2}$ Kyun? Fast equilibrium ka matlab hai $k_{1} [NO]^{2} = k_{- 1} [N_{2} O_{2}]$ , toh $[N_{2} O_{2}] = \frac{k _{1}}{k _{- 1}} [NO]^{2} = K_{eq} [NO]^{2}$ .

Rate equation mein substitute karo: $Rate = k_{2} K_{eq} [NO]^{2} [O_{2}] = k_{obs} [NO]^{2} [O_{2}]$ jahaan $k_{obs} = k_{2} K_{eq}$ .

Result: NO ke w.r.t. order = 2, O₂ = 1, overall = 3. Step 1 (forward) ki molecularity = 2, step 2 ki = 2. Overall order (3) balanced equation mein coefficients ke sum ke barabar hai BY COINCIDENCE kyunki is specific mechanism ka RDS saare stoichiometric species ko involve karta hai. Ise generalize mat karo!

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