2.8.11 · D3Reaction mechanisms — elementary steps, rate-determining ste
Yeh page ek workout hai. Hum the parent note ki machinery lete hain aur use har us tarah ke mechanism problem pe chalate hain jo ek exam de sakta hai . Shuru karne se pehle, ek vaada: har symbol kamaya gaya hai. Chalo un do ideas ko phir se anchor karte hain jo tumhe actually chahiye.
Intuition "Rate" ka matlab, picture meinSocho ek corridor mein log hain jo kai doors ki series se guzarne ki koshish kar rahe hain. Poori bheed ke bahar nikalne ki rate sabse chhote door se set hoti hai — yaani sabse slow step se. Woh chhota door hi rate-determining step (RDS) hai. Is page par sab kuch yahi hai: chhota door dhundho, aur us par intezaar kar rahi bheed ko un cheezein ke terms mein describe karo jo hum actually count kar sakte hain.
Do symbols jo hum baar baar use karte hain:
[ X ] k
[ X ] species X ", moles per litre mein measure kiya jata hai (mol L − 1 M [ X ] k rate constant hai — ek fixed number jo kisi given step ke liye given temperature par batata hai "yeh collision kitna eager hai succeed hone ke liye?" Bada k k
Ek elementary step aA + bB → products k [ A ] a [ B ] b ek saath collide karne wale har molecule ki sankhya hain — kyunki simultaneous meeting ki probability multiply hoti hai.
Har mechanism problem in cells mein se ek (ya inhi ka blend) hota hai. Neeche ke hamare 9 examples inhe sab cover karte hain — ek standalone Cell Z bhi hai jo formulas ko unki edges par stress-test karta hai.
Cell
Scenario class
Tricky kyun hai
Example
A Slow step pehle , usme koi intermediate nahi
"Easy" case — rate seedha padh lo
Ex 1
B Slow step baad mein , ek fast pre-equilibrium se fed
K Ex 2
C Intermediate ek reactant hai jo already use ho chuka hai (denominator appear hota hai)
Fast equilibrium ek reactant consume kar rahi hai, isliye woh denominator mein aata hai
Ex 3
D Unimolecular RDS (molecularity 1)Degenerate "collision" — ek single molecule toot jaati hai
Ex 4
E Termolecular / 3-body apparent orderOrder 3 tak pahunch sakta hai; kya ek true 3-body step realistic hai?
Ex 5
F Koi obvious slow step nahi to Steady-State Approximation Rate build-up/drain balance se set hoti hai, ek door se nahi
Ex 6
G Real-world word problem (atmospheric / catalysis)Prose ko steps mein, steps ko law mein translate karo
Ex 7
H Exam twist : given observed order, sahi valid mechanism chunoUlta direction; candidates test karo
Ex 8
Z Degenerate/limiting checks : zero concentration, huge/tiny k Kya formulas edges par sane rehte hain?
Ex 9
Worked example Example 1 — clean caseOverall: NO 2 + CO → NO + CO 2
Step 1 (slow ): NO 2 + NO 2 → NO 3 + NO = k 1 [ NO 2 ] 2
Step 2 (fast): NO 3 + CO → NO 2 + CO 2
Forecast: overall rate law guess karo. Stoichiometry se tempting answer: k [ NO 2 ] [ CO ]
Step 1 — Chhota door dhundho. Step 1 slow label hai, isliye yeh RDS hai.
Yeh step kyun? Poori bheed utni hi tezi se nikal sakti hai jitna chhota door allow kare. Step 2 fast hai, isliye woh kisi ko rokta nahi.
Step 2 — RDS ki apni stoichiometry se RDS rate likho. Do NO 2 Rate = k 1 [ NO 2 ] 2 . Yeh step kyun? Step 1 elementary hai, isliye exponents = colliding molecules ki sankhya = 2.
Step 3 — Law mein intermediates check karo. NO 3 k 1 [ NO 2 ] 2 nahi aata. Eliminate karne ko kuch nahi.
Yeh step kyun? Ek finished rate law mein sirf wahi species ho sakti hain jo hum measure kar sakte hain — reactants/products.
Result: Rate = k 1 [ NO 2 ] 2 NO 2 CO
Verify: Steps sum karo: 2 NO 2 + NO 3 + CO → NO 3 + NO + NO 2 + CO 2 NO 3 NO 2 NO 2 + CO → NO + CO 2 k 1 M s − 1 [ NO 2 ] 2 M 2 k 1 M − 1 s − 1
Yahan hum us trick se milte hain jo is page ka baaki hissa carry karti hai: ek intermediate RDS ke andar baitha hai aur hum use trade karna padte hain.
Figure s01 — Example 2 ka ek diagram. Left mein ek box mein reactants NO + Br2 k 1 k − 1 barabar length hain jo ise middle box se connect karti hain jisme intermediate NOBr2 k 2 2
Worked example Example 2 — pre-equilibrium eliminationOverall: 2 NO + Br 2 → 2 NOBr
Step 1 (fast equilibrium ): NO + Br 2 ⇌ NOBr 2 k 1 k − 1
Step 2 (slow ): NOBr 2 + NO → 2 NOBr = k 2 [ NOBr 2 ] [ NO ]
Forecast: NO
Step 1 — RDS rate. Step 2 slow hai:
Rate = k 2 [ NOBr 2 ] [ NO ] . Kyun: sabse slow door hi pace set karta hai.
Step 2 — Intermediate spot karo. NOBr 2 Kyun: rate laws mein sirf measurable reactants ho sakte hain.
Step 3 — Fast equilibrium use karo. "Fast forward aur reverse" ka matlab hai Step 1 khud ko balance karta hai: forward rate = reverse rate — yeh exactly figure s01 mein equal-length arrows hain.
k 1 [ NO ] [ Br 2 ] = k − 1 [ NOBr 2 ] . Yeh tool kyun aur koi nahi? Kyunki Step 1 dono directions mein fast hai, yeh equilibrium reach kar leta hai bahut pehle jab slow step use drain kare — isliye equilibrium condition (in-rate = out-rate) hold karti hai.
Step 4 — Intermediate ke liye solve karo.
[ NOBr 2 ] = k − 1 k 1 [ NO ] [ Br 2 ] . Yeh step kyun? Rate jo hum chahte hain (k 2 [ NOBr 2 ] [ NO ] [ NOBr 2 ] [ NOBr 2 ]
Step 5 — Substitute karo.
Rate = k 2 ⋅ k − 1 k 1 [ NO ] [ Br 2 ] ⋅ [ NO ] = k − 1 k 1 k 2 [ NO ] 2 [ Br 2 ] . Yeh step kyun? Hum Step-4 expression ko RDS rate mein [ NOBr 2 ] NO Br 2 k obs = k 1 k 2 / k − 1 Rate = k obs [ NO ] 2 [ Br 2 ] .
Result: forecast ka jawab — NO second order , pehla nahi, kyunki do NO molecules path ko feed karte hain (ek equilibrium mein jaata hai, ek slow step mein). Total order 2 + 1 = 3
Verify: Sum: NO + Br 2 + NOBr 2 + NO → NOBr 2 + 2 NOBr NOBr 2 2 NO + Br 2 → 2 NOBr k obs M s − 1 [ NO ] 2 [ Br 2 ] M 3 k obs M − 2 s − 1 k 1 k 2 / k − 1 M − 1 s − 1 ⋅ M − 1 s − 1 / s − 1 = M − 2 s − 1 k 1 = 2 , k − 1 = 4 , k 2 = 3 k obs = 2 ⋅ 3/4 = 1.5 [ NO ] = 0.1 , [ Br 2 ] = 0.2 = 1.5 ⋅ 0.01 ⋅ 0.2 = 0.003 M s − 1
Worked example Example 3 — ek reactant denominator mein chhup jaata haiOzone decomposition, 2 O 3 → 3 O 2
Step 1 (fast equilibrium ): O 3 ⇌ O 2 + O k 1 , k − 1
Step 2 (slow ): O + O 3 → 2 O 2 = k 2 [ O ] [ O 3 ]
Forecast: Step 1 ka reverse O 2 khata hai . Kya [ O 2 ]
Step 1 — RDS rate: Rate = k 2 [ O ] [ O 3 ]
Step 2 — Intermediate atomic O Use eliminate karo.
Step 3 — Equilibrium condition. Forward = k 1 [ O 3 ] = k − 1 [ O 2 ] [ O ] . Kyun: reverse reaction bimolecular hai (O 2 O [ O 2 ]
Step 4 — Solve karo.
[ O ] = k − 1 [ O 2 ] k 1 [ O 3 ] . Yeh step kyun? RDS rate abhi bhi intermediate [ O ] [ O ] [ O 2 ] divide out ho gaya — woh denominator mein slide kar gaya. Yahi precisely woh mechanism hai jisse ek product rate ko inhibit karta hai.
Step 5 — Substitute karo.
Rate = k 2 ⋅ k − 1 [ O 2 ] k 1 [ O 3 ] ⋅ [ O 3 ] = k − 1 k 1 k 2 [ O 2 ] [ O 3 ] 2 . Yeh step kyun? Step-4 expression ko RDS rate mein drop karne se [ O ] O 3 O 2 k obs = k 1 k 2 / k − 1 Rate = k obs [ O 3 ] 2 [ O 2 ] − 1 .
Result: O 3 + 2 O 2 − 1 negative order — product O 2 inhibit karta hai, exactly wahi jo ozone ke liye experiment dikhata hai.
Verify: Sum: O 3 + O + O 3 → O 2 + O + 2 O 2 O 2 O 3 → 3 O 2 k obs M s − 1 k obs ⋅ M 2 / M = k obs ⋅ M k obs s − 1 k 1 k 2 / k − 1 s − 1 ⋅ M − 1 s − 1 / M − 1 s − 1 = s − 1 k 1 = 6 , k − 1 = 3 , k 2 = 2 ⇒ k obs = 4 [ O 3 ] = 0.1 , [ O 2 ] = 0.5 = 4 ⋅ 0.01/0.5 = 0.08 M s − 1
Worked example Example 4 — ek single molecule toot jaati haiOverall: N 2 O 5 → 2 NO 2 + 2 1 O 2 unimolecular hai:
Step 1 (slow ): N 2 O 5 → NO 2 + NO 3 = k 1 [ N 2 O 5 ]
Step 2 (fast): NO 2 + NO 3 → NO 2 + O 2 + NO
Forecast: "molecularity 1" ka matlab hai kitne molecules milne chahiye? Order kya hai?
Step 1 — RDS rate. Molecularity = 1 Rate = k 1 [ N 2 O 5 ] . Yeh degenerate case kyun hai: "collision" zero partners ke saath matlab hai ek molecule itni vibrational energy reach karta hai ki toot sake — dekho Activation Energy and Catalysts . Exponent 1 0 [ N 2 O 5 ] sankhya double ho jaati hai.
Result: Rate = k 1 [ N 2 O 5 ] first order .
Verify: Units — rate M s − 1 k 1 ⋅ M k 1 s − 1 [ N 2 O 5 ] = 0 = 0
Worked example Example 5 — order kitna high ja sakta hai?Overall: 2 NO + O 2 → 2 NO 2 termolecular step:
Step 1 (slow , 3-body): NO + NO + O 2 → 2 NO 2 = k 1 [ NO ] 2 [ O 2 ]
Forecast: overall order (exponents ka sum) kya hai? Aur kya ek genuine 3-body collision physically likely hai?
Step 1 — Rate padho. Elementary, molecularity 3:
Rate = k 1 [ NO ] 2 [ O 2 ] , order = 2 + 1 = 3.
Step 2 — Physics par sanity. Teen molecules ka ek hi waqt milna rare hai — yaad karo Collision Theory se ki 2-body collisions bhi norm hain aur 3-body wale roughly hazaar guna kam frequent hote hain. Isliye yeh sahi law deta hai, lekin chemists Example 2 ka pre-equilibrium route prefer karte hain (same law, koi improbable triple collision nahi).
Yeh kyun mention karein? Ek exam dono mechanisms accept kar sakta hai agar dono observed order-3 law reproduce karte hain — lekin tumse flag karne ko kehta hai kaun sa zyada realistic hai.
Result: Rate = k 1 [ NO ] 2 [ O 2 ]
Verify: Yeh exactly parent note ke Example 2 ka k obs [ NO ] 2 [ O 2 ] same rate law , confirm karta hai ki ek rate law ek mechanism uniquely prove nahi kar sakta. k 1 M s − 1 = k 1 ⋅ M 3 k 1 M − 2 s − 1 k 1 = 5 , [ NO ] = 0.2 , [ O 2 ] = 0.3 = 5 ⋅ 0.04 ⋅ 0.3 = 0.06 M s − 1
Jab koi step "slow" flag nahi hoti, funnel mein do similar doors hote hain. Hum Steady-State Approximation use karte hain: ek reactive intermediate itni tezi se banta aur toot ta hai ki uski matra muskil se change hoti hai, isliye uski net rate of change approximately zero hoti hai .
Figure s02 — time ke against intermediate concentration [ O ] d [ O ] / d t
Worked example Example 6 — steady-state derivationOverall: 2 O 3 → 3 O 2
Step 1: O 3 k 1 O 2 + O
Step − 1 O 2 + O k − 1 O 3
Step 2: O + O 3 k 2 2 O 2
Forecast: kya hoga answer ke saath jab k 2 [ O 3 ] k − 1 [ O 2 ]
Step 1 — Intermediate O Yeh Step 1 se banta hai, Steps − 1 2 toot ta hai:
d t d [ O ] = k 1 [ O 3 ] − k − 1 [ O 2 ] [ O ] − k 2 [ O ] [ O 3 ] . Yeh tool kyun? Koi single door narrow nahi hai, isliye hum ek step seedha nahi padh sakte. Balki hum figure s02 mein [ O ]
Step 2 — Zero set karo (steady state).
k 1 [ O 3 ] = [ O ] ( k − 1 [ O 2 ] + k 2 [ O 3 ] ) ⇒ [ O ] = k − 1 [ O 2 ] + k 2 [ O 3 ] k 1 [ O 3 ] . Yeh step kyun? Plateau ka matlab hai "banana = toot na", yaani poori right-hand side sum karke zero hoti hai. Use zero set karna ek mushkil differential equation ko ek plain algebraic equation mein badal deta hai, jise hum phir rearrange karte hain un-measurable [ O ]
Step 3 — Product ki rate (Step 2 ke zariye).
Rate = k 2 [ O ] [ O 3 ] = k − 1 [ O 2 ] + k 2 [ O 3 ] k 1 k 2 [ O 3 ] 2 . Yeh step kyun? Product O 2 k 2 [ O ] [ O 3 ] [ O ]
Step 4 — Limiting check. Agar k − 1 [ O 2 ] k 2 [ O 3 ] k 2 [ O 3 ] Rate → k − 1 k 1 k 2 [ O 2 ] [ O 3 ] 2 , zyada general law hai aur pre-equilibrium uska special case hai.
Verify: Units: numerator k 1 k 2 [ O 3 ] 2 k − 1 [ O 2 ] M s − 1 k 1 = 6 , k 2 = 2 , k − 1 = 3 , [ O 3 ] = 0.1 , [ O 2 ] = 0.5 Rate = 3 ⋅ 0.5 + 2 ⋅ 0.1 6 ⋅ 2 ⋅ 0.01 = 1.7 0.12 ≈ 0.0706 M s − 1 0.08 k 2 [ O 3 ]
Worked example Example 7 — enzyme-style bottleneck words mein"Ek enzyme E S E S E S P E
Step 1 (fast eq): E + S ⇌ E S k 1 , k − 1
Step 2 (slow): E S k 2 E + P = k 2 [ E S ]
Forecast: bahut zyada enzyme aur thoda sa substrate ke saath, kya tum [ S ]
Step 1 — RDS rate. Rate = k 2 [ E S ]
Step 2 — E S Step 1 ka fast equilibrium:
k 1 [ E ] [ S ] = k − 1 [ E S ] ⇒ [ E S ] = k − 1 k 1 [ E ] [ S ] . Yeh step kyun? Hum equilibrium rearrange karte hain un-measurable complex [ E S ]
Step 3 — Substitute karo.
Rate = k − 1 k 1 k 2 [ E ] [ S ] . Yeh step kyun? Step-2 expression ko RDS rate mein drop karne se [ E S ] Scarce-substrate limit mein enzyme mostly free hoti hai, isliye [ E ] [ E ] 0 Rate ≈ k ′ [ S ] , k ′ = k − 1 k 1 k 2 [ E ] 0 .
Result: low [ S ] Enzyme Kinetics ka linear low-substrate regime (Michaelis–Menten). High [ S ]
Verify: Low-[ S ] = 1 [ S ] = 0 k 1 = 10 , k − 1 = 2 , k 2 = 5 , [ E ] 0 = 0.01 , [ S ] = 0.1 k ′ = ( 10 ⋅ 5/2 ) ⋅ 0.01 = 0.25 = 0.25 ⋅ 0.1 = 0.025 M s − 1
Worked example Example 8 — woh mechanism chuno jo data fit kareExperiment H 2 + 2 ICl → I 2 + 2 HCl Rate = k [ H 2 ] [ ICl ]
Candidate X: single step H 2 + 2 ICl → products Rate = k [ H 2 ] [ ICl ] 2 Candidate Y:
Step 1 (slow ): H 2 + ICl → HI + HCl = k 1 [ H 2 ] [ ICl ]
Step 2 (fast): HI + ICl → I 2 + HCl
Forecast: kaun sa predicted law observed k [ H 2 ] [ ICl ]
Step 1 — X ka law predict karo. Elementary termolecular k [ H 2 ] [ ICl ] 2 = 2 Match nahi karta (observed 1 hai).
Reject kyun: ICl par exponent galat hai, aur 3-body collisions waise bhi unlikely hain.
Step 2 — Y ka law predict karo. RDS Step 1 hai, elementary, deta hai k 1 [ H 2 ] [ ICl ] = 1 Match karta hai ✓.
Step 3 — Confirm karo ki Y sahi se sum hota hai. H 2 + ICl + HI + ICl → HI + HCl + I 2 + HCl HI H 2 + 2 ICl → I 2 + 2 HCl
Result: Mechanism Y accept hai (observed law aur stoichiometry fit karta hai; koi improbable termolecular step nahi).
Verify: Observed law order sum = 1 + 1 = 2 2 3 k 1 = 4 , [ H 2 ] = 0.5 , [ ICl ] = 0.2 = 4 ⋅ 0.5 ⋅ 0.2 = 0.4 M s − 1
Worked example Example 9 — kya formulas edges par sane rehte hain?Example 2 ka pre-equilibrium law lo, Rate = k obs [ NO ] 2 [ Br 2 ] k obs = k 1 k 2 / k − 1
Forecast: guess karo kaun sa edge rate ko vanish hone ki jagah blow up karta hai.
Step 1 — Zero reactant. [ NO ] = 0 Rate = k obs ⋅ 0 2 ⋅ [ Br 2 ] = 0. Yeh step kyun? Agar law mein aane wala ek reactant absent hai, to koi path nahi chal sakta — rate exactly zero honi chahiye. Hai ✓.
Step 2 — Ek species ka zero jo law mein nahi hai. Example 1 mein, [ CO ] = 0 = k 1 [ NO 2 ] 2 unchanged rahti hai.
Yeh step kyun? Zero-order species genuinely RDS ko gate nahi karti, isliye unhe remove karna Step 1 slow nahi kar sakta. Yeh woh counter-intuitive edge hai jis par students trip karte hain.
Step 3 — Bahut fast reverse (k − 1 → ∞ k obs = k 1 k 2 / k − 1 k obs → 0 ⇒ Rate → 0. Yeh step kyun? Ek zordaar reverse intermediate ko slow step ke use karne se pehle hi tod deta hai — equilibrium far left baithti hai, Step 2 ko starve karti hai. Vanishing rate physically sahi hai ✓.
Step 4 — Product-inhibited law near zero product. Example 3 lo, Rate = k obs [ O 3 ] 2 / [ O 2 ] [ O 2 ] → 0 Rate → + ∞. Yeh step kyun? Yeh forecast ka "blow-up" edge hai: denominator zero ki taraf jaate hue rate sky-high bhej deta hai. Physically yeh model yahan toot jaata hai (tumhare paas sach mein kabhi zero O 2
Verify: Edge 1 0 k 1 [ NO 2 ] 2 0 lim k − 1 → ∞ k 1 k 2 / k − 1 = 0 lim [ O 2 ] → 0 + k obs [ O 3 ] 2 / [ O 2 ] = + ∞
Recall Pre-equilibrium vs steady state kab use karein?Pre-equilibrium tab use karo jab ek fast reversible step clearly slow step ke baad aati hai (reverse forward-drain se bahut zyada hai). Steady-state tab use karo jab koi step slow flag nahi hai, ya safe rehne ke liye — steady-state ek general result hai jo pre-equilibrium ko ek limit ke roop mein contain karta hai.
Recall Negative reaction order kaun se cell se aata hai?Cell C ::: ek product (ya reactant) jo ek fast equilibrium ke reverse mein appear hota hai denominator mein land karta hai, negative order deta hai — woh species reaction ko inhibit karti hai.
"Slowest sets it, intermediates get evicted." RDS dhundho, uska law likho, phir kisi bhi intermediate ko equilibrium (ya steady state) use karke evict karo jab tak sirf measurable species na bach jaayein.
