Rate law — order vs molecularity: Question bank — Hinglish deep dive

2.8.2 · D5 · Chemistry › Chemical Kinetics › Rate law — order vs molecularity

Recall Do definitions jo baar baar chahiye hoti hain

Order ::: Rate law $Rate = k [A]^{m} [B]^{n}$ mein concentration par experimentally measure kiya gaya exponent; overall order $m + n$ hota hai. Molecularity ::: Reactant particles ki woh sankhya jo ek elementary step mein collide karti hain — mechanism se theoretical count, hamesha ek positive integer (1, 2, ya 3).

Toolbox refresher (taaki traps samajh aayein)

Trap questions se pehle, teen quick build-ups jin par traps depend karte hain: ln kyun aata hai first order mein, first-order plot kaisa dikhta hai, aur termolecular steps rare kyun hote hain. Agar yeh pehle se pata hai toh skip karo.

First order mein logarithm kyun aata hai

KYA karte hain — do variables alag karo. Har $[A]$ ko left side par aur har $t$ ko right side par laao, taaki har side par sirf ek hi variable ho: $\frac{d [ A ]}{[ A ]} = - k d t$

YEH STEP KYUN. Jab ek side par ek hi variable ho tab hum dono ko alag alag integrate kar sakte hain — "separate" karne ka poora reason yahi hai.

INTEGRATE KARNE PAR KYA MILTA HAI. Start ( $t = 0$ , concentration $[A]_{0}$ ) se time $t$ (concentration $[A]$ ) tak integrate karo: $\int_{[A]_{0}}^{[A]} \frac{d [ A ]}{[ A ]} = - k \int_{0}^{t} d t$ $1/ [A]$ ka integral hai $ln [A]$ — yahi woh jagah hai jahan logarithm paida hota hai. Isse milta hai: $ln [A] - ln [A]_{0} = - k t ⟹ ln [A] = ln [A]_{0} - k t$

KAISA DIKHTA HAI. Isko $y = c - k t$ se compare karo: yeh $t$ mein ek straight line hai jiska intercept $ln [A]_{0}$ hai aur slope $- k$ hai. Neeche figure mein plot hai.

Figure — Rate law — order vs molecularity

Figure padhein: raw $[A]$ curve (cyan) bend karti hai — ek curve, line nahi — lekin ise $ln [A]$ (amber) ke roop mein replot karne par ek straight line banti hai jiska slope $- k$ read hota hai. Woh straightness hi first-order test hai.

Initial-rates method ek exponent ko kaise isolate karta hai

Worked example Order ka ek worked isolation

$A + B \to P$ ke liye data:

Trial	$[A]_{0}$ (M)	$[B]_{0}$ (M)	Rate (M/s)
1	0.10	0.10	0.030
2	0.20	0.10	0.120
3	0.10	0.30	0.090

$m$ isolate karo (A mein order): trials 1→2 mein sirf $[A]$ change hota hai (×2), $[B]$ fixed. To $\frac{0.120}{0.030} = (\frac{0.20}{0.10})^{m} \Rightarrow 4 = 2^{m} \Rightarrow m = 2$ . $n$ isolate karo (B mein order): trials 1→3 mein sirf $[B]$ change hota hai (×3), $[A]$ fixed. To $\frac{0.090}{0.030} = (\frac{0.30}{0.10})^{n} \Rightarrow 3 = 3^{n} \Rightarrow n = 1$ . Rate law: $Rate = k [A]^{2} [B]$ , overall order $3$ . Ek waqt mein ek concentration badlana hi is baat ki wajah hai ki har ratio ek single unknown exponent tak reduce ho jaata hai.

Termolecular steps rare kyun hote hain

Figure padhein: do-body meeting (left) mein sirf do spheres ka overlap chahiye — yeh common hai. Teen-body meeting (right) mein teeno ka same tiny reaction volume mein same instant par hona zaroori hai — shaded overlap dramatically shrink ho jaata hai, isliye collision probability bahut kam ho jaati hai. Isi liye nature almost hamesha reactions ko uni- aur bimolecular steps mein todti hai, na ki ek chaar-ya-zyaada-body collision mein.

Dimensional analysis se $k$ ki units derive karna

True ya false — justify karo

True/False: Kisi reaction ka overall order hamesha stoichiometric coefficients ke sum ke barabar hota hai.

False. Yeh equality sirf ek single elementary step ke liye hold hoti hai; multi-step reaction ke liye order slow step se set hota hai aur aksar balanced coefficients se koi link nahi hota.

True/False: Molecularity zero ho sakti hai.

False. Molecularity ek step mein colliding particles count karti hai, isliye yeh kam se kam 1 hoti hai; sirf order zero ho sakta hai (kisi concentration par rate independent ho).

True/False: Order fractional ya negative ho sakta hai, lekin molecularity nahi.

True. Order ek empirical fit hai aur

\frac{1}{2}

(jaise H₂+Br₂ mein Br₂) ya negative (ek inhibitor) aa sakta hai; molecularity ek physical collision count hai, isliye yeh ek positive integer honi chahiye.

True/False: Elementary step

2 A + B \to P

ke liye rate law

Rate = k [A]^{2} [B]

hai.

True. Ek elementary step mein stoichiometry hi collision requirement hoti hai, isliye exponents coefficients ke barabar hote hain — lekin yeh tabhi allowed hai jab humein bataya gaya ho ki step elementary hai.

True/False: Zero-order reaction tab tak constant rate par proceed karta hai jab tak reactant khatam na ho jaaye.

True. Rate

= k

concentration se independent hai, isliye

[A]

time ke saath linearly girta hai (slope

- k

) depletion tak, jahan model break ho jaata hai.

True/False: Rate constant

k

ki units har reaction ke liye same hoti hain.

False. Units overall order

n

ke liye

M^{1 - n} s^{- 1}

follow karti hain:

M s^{- 1}

(zero),

s^{- 1}

(first),

M^{- 1} s^{- 1}

(second).

True/False: Molecularity overall reaction H₂+Br₂→2HBr ke liye defined hai.

False. Woh reaction ek single elementary step nahi hai, isliye uske liye molecularity undefined hai; sirf uske individual elementary steps ki molecularity hoti hai.

True/False: Second-order reaction mein har concentration double karne se rate double ho jaata hai.

False. Agar overall order 2 hai, to saari concentrations double karne par rate

2^{2} = 4

se multiply hota hai, 2 se nahi.

Error dhundho

"Reaction

2 NO + O_{2} \to 2 NO_{2}

hai, isliye yeh termolecular aur third order honi chahiye."

Error: molecularity aur order directly coefficients se read kiye ja rahe hain. Yeh truly termolecular hai ya nahi, yeh mechanism par depend karta hai, aur order measure karna padega — coefficients akele kuch prove nahi karte.

"B ke w.r.t. order zero aaya, isliye B reactant nahi hai."

Error: zero order ka matlab hai rate [B] par depend nahi karta, na ki B absent hai. B consume bhi ho sakta hai aur fast pre-equilibrium ke baad ya saturating species ke roop mein appear kar sakta hai.

"Kyunki molecularity hamesha ≥1 hoti hai, reaction kabhi zero order nahi ho sakti."

Error: yeh dono ideas ko confuse kar raha hai. Molecularity (mechanism count) aur order (measured exponent) independent hain — ek enzyme-saturated ya surface-catalysed step zero order measure ho sakti hai jabki har elementary step ki molecularity phir bhi ≥1 rehti hai.

"Slow step

NO_{2} + F_{2} \to NO_{2} F + F

hai, isliye Rate

= k [NO_{2}]^{2} [F_{2}]

Error: exponents slow step ke apne species se match karne chahiye, jo deta hai Rate

= k [NO_{2}] [F_{2}]

. Extra NO₂ sirf ek baad ke fast step mein appear hota hai aur rate law mein kabhi nahi aata.

\frac{1}{2}

jaisa fractional order matlab adha molecule collide karta hai."

Error: adha molecule meaningless hai. Fractional order ek effective exponent hota hai jo kai elementary steps ko combine karne se banta hai (jaise ek dissociation equilibrium), na ki ek collision ki picture.

2 A + B \to C

ke liye hum hamesha Rate

= k [A] [B]

paate hain kyunki slow step mein sirf ek A hota hai."

Error: yeh assume nahi kar sakte ki slow step mein ek A hai — yahi to experiment decide karta hai. Rate law measured slow step follow karta hai, jo bhi woh nikle.

Why questions

Molecularity 3 se zyaada practically kabhi observe nahi hoti, kyun?

Char ya zyaada particles ka same point par same instant mein sahi orientation ke saath aana practically impossible hota hai, isliye nature reactions ko uni- aur bimolecular steps ki sequences mein route karta hai.

Multi-step reaction ke liye balanced equation se order predict kyun nahi ho sakta?

Balanced equation sirf mass balance enforce karta hai; yeh kuch nahi batata ki kaun se collisions speed control karte hain. Order slowest elementary step se fix hota hai, jo overall equation mein chhupta hai.

Slow step overall rate law kyun determine karta hai?

Yeh bottleneck hai — poori reaction apne slowest stage se tez nahi ja sakti, isliye rate law un concentrations ko inherit karta hai jo us step mein appear hoti hain, jaise ek pipeline ka throughput uske sabse narrow section se set hota hai.

Suspected first-order reaction ke liye hum

ln [A]

vs time plot kyun karte hain?

d [A] / d t = - k [A]

integrate karne par

ln [A] = ln [A]_{0} - k t

milta hai, jo

t

mein straight line hai; ek straight

ln

plot isliye first order confirm karta hai aur uska slope

- k

deta hai.

Initial-rates method ek waqt mein sirf ek concentration kyun change karta hai?

Baaki ko fixed rakhne se ek single reactant ka influence isolate ho jaata hai, isliye rate ratio sirf ek exponent par depend karta hai aur order directly read ho sakta hai.

Order ek experimental quantity kyun hai jabki molecularity theoretical hai?

Order woh exponent hai jo measured rate data fit karta hai, isliye woh sirf lab se aa sakta hai; molecularity ek proposed mechanism ke collision count se deduce hoti hai, isliye woh theory mein rehti hai jab tak mechanism verify na ho jaaye.

Edge cases

Ek intermediate F jo fast step se consume hoti hai, uski [F] rate law mein kyun nahi aati?

Yeh utni hi tezi se khatam hoti hai jitni tezi se banti hai, isliye uski concentration kabhi build up nahi hoti; sirf rate-determining slow step mein maujood species observable rate law mein survive karti hain.

Jis reaction ka mechanism unknown ho, uski molecularity kya hai?

Undefined. Molecularity specific elementary steps se attach hoti hai; known mechanism ke bina koi elementary steps nahi hain jinhe count kiya ja sake.

Agar ek reaction truly elementary hai, to uske order aur molecularity mein kya relation hai?

Dono numerically coincide karte hain — exponents coefficients ke barabar hote hain, isliye overall order us single step ki molecularity ke barabar hota hai (jaise termolecular step ke liye dono 3 hote hain).

Zero-order rate law mein jab

[A] \to 0

tab kya hota hai?

Model zero concentration par bhi constant rate

k

predict karta hai, jo unphysical hai; reality mein zero-order behaviour depletion ke paas break down ho jaata hai (catalyst ya surface ab saturated nahi rahta) aur reaction slow ho jaati hai.

Kya same balanced equation wali do reactions ke alag alag orders ho sakte hain?

Haan. Identical stoichiometry ke peeche alag mechanisms ho sakte hain jinke alag slow steps hote hain, isliye unke measured rate laws — aur isliye orders — bilkul alag ho sakte hain.

Kya first-order rate law yeh proof hai ki reaction ek unimolecular step mein hoti hai?

Nahi. First order ek unimolecular step ke saath consistent hai lekin multi-step mechanism se bhi aa sakta hai; order akele mechanism ko uniquely fix nahi karta.