Reaction mechanisms — elementary steps, rate-determining step: Question bank

2.8.11 · D5 · Chemistry › Chemical Kinetics › Reaction mechanisms — elementary steps, rate-determining ste

Yeh page ek workout hai, lecture nahi. Neeche har line ek claim ya question hai jiska answer ::: ke baad chhupa hua hai. Right side cover karo, pehle apna answer ek wajah ke saath zor se bolo, phir reveal karo. Sirf "haan/nahi" answers ka koi score nahi hoga — goal yeh hai ki is topic ke famous misconceptions ko pakda jaaye. Agar koi term shaky lagti hai, toh parent note the topic note ya kisi prerequisite jaise Rate Laws and Reaction Order par wapas jaao, phir yahan continue karo.

True or false — justify

Har ek ke liye: true ya false decide karo, phir wajah batao — wajah hi poora point hai.

True/False: Overall reaction ka rate law hamesha uske balanced coefficients se seedha padha ja sakta hai.

False — sirf elementary steps mein hi tum exponents stoichiometry se padh sakte ho; overall equation kai steps ki ek summary hai, isliye uska order mechanism (RDS ke zariye) ya experiment se aana chahiye.

True/False: Rate-determining step hamesha mechanism mein pehli likhee hui step hoti hai.

False — RDS sabse slow step hoti hai (sabse zyada activation-energy barrier), aur yeh sequence mein kahin bhi ho sakti hai; "pehli" aur "sabse slow" ka koi relation nahi hai.

True/False: Molecularity aur reaction order ek hi cheez hain.

False — molecularity ek elementary step mein collide hone wale molecules ki count hai (ek theoretical integer 1, 2, ya 3); order ek experimentally measured exponent hai overall reaction ke liye. Yeh sirf ek akele elementary step ke liye hi coincide karte hain.

True/False: Ek intermediate ko final rate law mein appear karne ki permission hai.

False — ek intermediate mechanism ke andar banta aur consume hota hai, isliye tum lab mein uski concentration measure nahi kar sakte; ise fast-equilibrium ya steady-state substitution use karke eliminate karna zaroori hai.

True/False: Ek catalyst overall balanced equation mein appear karta hai.

False — catalyst ek early step mein consume hota hai aur ek baad ki step mein regenerate hota hai, isliye yeh overall equation se cancel ho jaata hai (dekho Activation Energy and Catalysts); yeh rate law mein legitimately appear kar sakta hai.

True/False: Ek elementary step

2 A \to products

ke liye rate law

k [A]^{2}

hai.

True — kyunki yeh elementary hai, exponents coefficients ke barabar hote hain; do

A

molecules ko collide karna hoga, isliye collision probability

[A]^{2}

ke scale par hoti hai.

True/False: Ek valid mechanism mein saari elementary steps ka sum overall balanced equation ke barabar hona chahiye.

True — yeh primary consistency check hai; agar steps milakar observed overall reaction nahi banaate, toh mechanism galat hai chahe rate kitni bhi achhi kyun na predict kare.

True/False: Jo reaction overall third-order hai, usme zaroor ek single termolecular step hogi.

False — third-order kinetics usually ek fast pre-equilibrium se aati hai jo ek bimolecular slow step ko feed karti hai; sachchi three-body simultaneous collisions bahut rare hain (dekho Collision Theory).

True/False: Agar Step 2 slow step hai, toh reactants usse pehle pile up ho jaate hain aur products utni hi tezi se bante hain jitni tezi se Step 2 chalti hai.

True — slow step bottleneck hai; isse upstream ki har cheez equilibrate ya accumulate hoti hai, aur overall throughput sabse slow gate se cap ho jaata hai.

Spot the error

Har line mein ek (galat) student reasoning ka piece hai. Galti batao.

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

, isliye Rate

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

inspection se."

Error — overall equation ko elementary treat karna. Asli slow step hai

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

, jisse Rate

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

milta hai (har ek mein first order).

"RDS

N_{2} O_{2} + O_{2} \to 2 NO_{2}

hai, isliye answer hai Rate

= k_{2} [N_{2} O_{2}] [O_{2}]

Error — intermediate

[N_{2} O_{2}]

ko answer mein chhod dena. Fast equilibrium

[N_{2} O_{2}] = K [NO]^{2}

use karo taaki Rate

= k_{2} K [NO]^{2} [O_{2}]

mile.

"Do elementary rate laws ko add karne se overall rate law milti hai."

Error — tum rate laws ko add nahi karte; tum RDS ki rate lete ho aur intermediates ko substitute out karte ho. Alag steps ki rate laws kabhi bhi add nahi ki jaati.

"Molecularity of 4 theek hai kyunki reactant side par chaar molecules hain."

Error — 3 se zyada molecularity practically kabhi ek single step ke liye observe nahi hoti; ek "4-molecule reaction" ko zaroor molecularity 1, 2, ya (rarely) 3 wali elementary steps mein todna padega.

"Equilibrium constant

K = k_{1} / k_{- 1}

ki units ek rate constant jaisi hain."

Error —

K

forward aur reverse rate constants ka ratio hai, isliye unki units partly cancel ho jaati hain;

K

generally rate constant ki units nahi rakhta, aur

k_{obs} = k_{2} K

antar ko absorb kar leta hai.

"Chunki Step 1 fast hai, yeh rate law ko bilkul affect nahi kar sakti."

Error — ek fast pre-equilibrium step bilkul rate law ko shape karti hai, kyunki yeh intermediate concentration fix karti hai jo slow step ko feed karti hai (isi tarah

[NO]^{2}

andar aata hai).

"Ek intermediate aur ek catalyst ek hi tarah ki species hain."

Error — intermediate banta hai phir consume hota hai (pehle product ki tarah appear karta hai, baad mein reactant ki tarah); catalyst consume hota hai phir regenerate hota hai (pehle reactant ki tarah appear karta hai, baad mein product ki tarah). Dono overall equation mein nahi hote, lekin unke roles opposite hain.

Why questions

"Kyun?" — yahan samajh rehti hai. Reveal karne se pehle jawab do.

Kyun hum ek elementary step ka rate law seedha uske coefficients se likh sakte hain, lekin overall reaction ke liye nahi?

Kyunki ek elementary step actual physical collision event hai, isliye uski rate literally us hisaab se proportional hoti hai ki wo exact molecules kitni baar milte hain; overall equation ek bookkeeping sum hai jiske peeche koi single collision nahi hai.

Kyun ek fast reversible first step equilibrium reach kar leti hai slow step se pehle?

Uski forward aur reverse rates dono badi hoti hain, isliye yeh jaldi ek balance par settle ho jaati hai; slow step intermediate ko itne gently drain karti hai ki equilibrium practically disturb nahi hota aur poore time hold karta hai.

Kyun overall rate kabhi bhi sabse slow step ki rate se zyada nahi ho sakti?

Har ek product molecule ko us bottleneck se guzarna padta hai; upstream steps sirf us gate ko material supply kar sakti hain, bypass nahi kar sakti, isliye gate total throughput ko cap kar deta hai.

Kyun hume final rate law se intermediates eliminate karne padte hain?

Intermediates short-lived hote hain aur tiny, hard-to-measure amounts mein present hote hain; rate law tabhi useful hai jab yeh un species ke terms mein likha ho (reactants, kabhi kabhi products) jinki concentrations hum actually control aur measure kar sakein.

Kyun molecularity sirf 1, 2, ya rarely 3 tak restricted hai?

Kyunki yeh un particles ko count karta hai jo ek event mein simultaneously collide karte hain; teen particles ka ek saath sahi geometry ke saath milna already bahut kam probable hai, aur chaar-at-once practically negligible hai (yeh Collision Theory logic hai).

Kyun

2 NO + O_{2}

ka mechanism NO mein second order deta hai jabki slow step mein koi NO nahi hai?

Do NO molecules fast equilibrium mein combine hokar intermediate

N_{2} O_{2}

banate hain;

[N_{2} O_{2}] = K [NO]^{2}

substitute karne se woh squared dependence final rate law mein aa jaati hai.

Kyun chemists mechanisms derive karne ki takleef uthate hain jab experimental rate law unhe order pehle se bata deta hai?

Rate law ek mechanism ko constrain karta hai lekin prove nahi karta; mechanism molecular pathway explain karta hai, intermediates aur catalysts predict karta hai, aur activation energy aur Reaction Coordinate Diagrams se connect karta hai.

Edge cases

Woh scenarios jo tidy examples skip kar dete hain.

Edge case: Kya ho agar do steps ki almost identical (comparable) rates hon — kya tab bhi ek single RDS hogi?

Koi clean single RDS nahi hogi; jab barriers similar hote hain tum ek slow step ke approximation se kaam nahi chala sakte aur poora Steady-State Approximation treatment use karna padega.

Edge case: Kya ho agar slow step bilkul aakhri step ho aur pehle ki har step fast equilibrium ho?

Rate law phir bhi slow step ki rate hogi, lekin tumhe shayad kai equilibrium substitutions chain karni padein taaki har ek intermediate ko reactants ke terms mein express kar sako, jab tak measurable species nahi mil jaatein.

Edge case: Ek mechanism jiske steps correctly sum hote hain lekin wrong experimental order predict karta hai — kya yeh valid hai?

Nahi — summation check pass karna necessary hai lekin sufficient nahi; ek valid mechanism ko overall equation aur observed rate law dono reproduce karne chahiye. Rate law fail karna use rule out kar deta hai.

Edge case: Rate law ka kya hoga agar "slow" step actually fast ho jaaye (jaise ek catalyst uski barrier lower kar de)?

Bottleneck shift ho jaata hai jo bhi step ab sabse slow hai uski taraf, isliye RDS — aur isliye rate law ki poori form — badal sakti hai; RDS koi fixed label nahi hai balki current barriers ka comparison hai.

Edge case: Ek unimolecular elementary step

A \to products

— uska order aur molecularity kya hai?

Dono 1 hain: molecularity 1 hai (ek molecule decompose ho raha hai), aur kyunki yeh elementary hai order molecularity ke barabar hai, isliye Rate

= k [A]

Edge case: Agar ek intermediate ko ek doosre pathway se bhi feed kiya ja raha ho, kya tum phir bhi simple fast-equilibrium substitution use kar sakte ho?

Aksar nahi — ek intermediate ke liye multiple production/consumption routes usually Steady-State Approximation demand karte hain, jo ek single equilibrium assume karne ki jagah saare formation aur removal terms ko balance karta hai.

Edge case: Overall equation mein clearly appear karne wale reactant mein zero-order behaviour — contradiction hai?

Koi contradiction nahi — agar woh reactant sirf RDS ke baad enter karta ho (ya koi surface/enzyme site saturate kare jaise Enzyme Kinetics mein), toh uski concentration badlane se bottleneck ki rate nahi badlegi, jo zero order deta hai.

Recall Jaane se pehle ek one-line self-test

Sab kuch cover karo aur jawab do: "Woh do sawaal kaunse hain jo almost har mechanism trap ko defuse karte hain?" ::: (1) Kya yeh line ek elementary step hai ya ek summary? (2) Kya koi intermediate final rate law mein tab survive kar raha hai jab nahi karna chahiye?