Class 12 Chemistry · Chapter 3
Short answer:
Is chapter me total 18 exercise questions cover kiye gaye hain — rate of reaction ki expression, factors affecting rate, rate law aur rate constant, order vs molecularity, integrated rate equations (zero order aur first order), half-life numericals, aur Arrhenius equation ke activation energy wale sawaal. Har numerical me full step-by-step working diya gaya hai taaki calculation part exam me bina galti ke ho sake.
Chemical Kinetics wo chapter hai jo batata hai ki koi reaction "kitni jaldi" hoti hai — na ki kitna hoti hai (woh equilibrium ka kaam hai). Thermodynamics sirf ye batata hai ki reaction possible hai ya nahi, lekin practically usme kitna time lagega, ye kinetics decide karta hai. Is chapter me hum rate of reaction define karenge, uske factors dekhenge (concentration, temperature, catalyst, surface area), phir rate law aur order nikalna seekhenge, integrated rate equations se concentration-time relation banayenge, aur last me Arrhenius equation se dekhenge ki temperature rate ko kaise affect karti hai. Numericals is chapter ka backbone hain — formula yaad hone se zyada important hai unit aur log ka sahi use.
Chapter 3 Summary — 5 Minute Revision
1. Rate of Reaction
Rate of reaction = concentration of reactant/product ka change per unit time. Reaction R → P ke liye:
- Average rate = Δ[R]/Δt (reactant ghat raha hai isliye negative sign) = Δ[P]/Δt (product badh raha hai)
- Instantaneous rate = jab Δt → 0, yaani concentration-time graph pe ek point pe tangent ka slope
- General reaction
aA + bB → cC + dDke liye: rate = −(1/a)d[A]/dt = −(1/b)d[B]/dt = +(1/c)d[C]/dt = +(1/d)d[D]/dt - Units: mol L−1 s−1 (ya time unit change ho toh accordingly)
2. Factors Affecting Rate of Reaction
- Concentration — jyada concentration = jyada collisions = jyada rate
- Temperature — sty jyada temperature se molecules ki kinetic energy badhti hai, effective collisions badhte hain
- Catalyst — alternate path deta hai jiska activation energy kam hota hai, isliye rate badhti hai (catalyst khud consume nahi hota)
- Surface area — solid reactant ka surface area jyada = rate jyada (powder > lump)
- Nature of reactants — ionic reactions fast hoti hain, covalent bond-breaking wali slow
3. Rate Law aur Rate Constant
Reaction aA + bB → products ke liye rate law (experimentally determine hota hai, balanced equation se assume NAHI kar sakte):
Rate = k[A]x[B]y
Yahan k = rate constant (specific reaction rate) hai — jab [A] = [B] = 1 mol/L ho, tab Rate = k. k sirf temperature pe depend karta hai, concentration pe nahi.
4. Order of Reaction vs Molecularity — SABSE ZYADA CONFUSION YAHI HAI
Order of reaction experimentally determine hota hai, molecularity theoretically — dono ka matlab alag hai aur exam me yahi confusion sabse zyada hota hai.
| Order of Reaction | Molecularity |
|---|---|
| Rate law ke exponents ka sum (x+y) | Ek elementary step me part lene wale molecules/atoms/ions ki total sankhya |
| Experimentally nikalta hai (data se) | Theoretically/mechanism se decide hota hai |
| Zero, fractional, ya negative ho sakta hai | Hamesha whole number hota hai (1, 2, 3 — kabhi zero ya fraction nahi) |
| Poore reaction ke liye ek hi value | Multi-step reaction ke har elementary step ki alag molecularity ho sakti hai |
| Complex reaction me directly balanced equation se predict nahi hota | Sirf elementary reactions ke liye meaningful term hai; overall complex reaction ki molecularity nahi hoti |
↔ Table ko side me swipe karein
5. Units of Rate Constant (order ke saath change hoti hai)
| Order | Units of k |
|---|---|
| Zero order | mol L−1 s−1 |
| First order | s−1 (sirf time−1, concentration ka koi role nahi) |
| Second order | mol−1 L s−1 |
| nth order | mol1−n Ln−1 s−1 |
↔ Table ko side me swipe karein
6. Integrated Rate Equations
Zero Order Reaction
Rate concentration pe depend nahi karta (jaise catalyst ki surface pe hone wali reactions):
[A] = [A]0 − kt ya k = ([A]0 − [A]) / t
Graph: [A] vs t ek straight line hai jiska slope = −k
First Order Reaction
Rate ek reactant ki concentration ke first power pe depend karta hai (radioactive decay ka classic example):
k = (2.303/t) log([A]0/[A])
Graph: log[A] vs t ek straight line hai jiska slope = −k/2.303
7. Half-Life (t1/2)
Half-life = wo time jisme reactant concentration apne initial value ke aadhe reh jaaye.
- Zero order: t1/2 = [A]0/2k — initial concentration pe depend karta hai
- First order: t1/2 = 0.693/k — initial concentration se INDEPENDENT hota hai (ye ek exam-favourite point hai)
8. Pseudo First Order Reactions
Jab ek reactant bahut zyada excess me ho (jaise water hydrolysis reactions me), uska concentration practically constant maan liya jaata hai aur poori reaction first order jaisi behave karti hai, jabki actual order do ya zyada ho sakta hai. Example: ester ka acid-catalysed hydrolysis, cane sugar ka inversion.
9. Temperature Dependence — Arrhenius Equation
Jaise-jaise temperature badhta hai, rate constant k bhi exponentially badhta hai:
k = A e−Ea/RT
Yahan A = Arrhenius factor/pre-exponential factor (frequency of collisions), Ea = activation energy, R = gas constant, T = absolute temperature (Kelvin me).
Log form (do temperature/rate constant data se Ea nikalne ke liye use hota hai):
log k = log A − Ea/(2.303RT)
log(k2/k1) = (Ea/2.303R) × (1/T1 − 1/T2)
Rule of thumb: zyadatar reactions me temperature 10°C badhne se rate roughly double ho jaata hai — isse "temperature coefficient" bhi kehte hain (approx value ~2).
10. Activation Energy (Ea)
Reactants ko products me convert hone ke liye ek minimum extra energy chahiye hoti hai — usko activation energy kehte hain. Higher Ea = slower reaction (kyunki kam molecules is threshold ko cross kar paate hain).
11. Collision Theory (Brief)
- Reaction tabhi hoti hai jab molecules aapas me collide karein
- Sirf collision kaafi nahi — collision "effective" hona chahiye, matlab (a) sahi orientation me ho aur (b) threshold energy se zyada energy ho
- Effective collisions ki frequency hi rate decide karti hai — isliye catalyst effective collisions badha deta hai (Ea kam karke), aur temperature high-energy molecules ki fraction badha deta hai
Poore Class 12 Chemistry ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q18)
Q1. Reaction 2N₂O₅ → 4NO₂ + O₂ me N₂O₅, NO₂ aur O₂ ke concentration change ke terms me rate expression likhiye.
Reaction: 2N₂O₅ → 4NO₂ + O₂. General rule ke mutabik coefficient se divide karke rate likhte hain:
Rate = −(1/2) d[N₂O₅]/dt = +(1/4) d[NO₂]/dt = +d[O₂]/dt
Q2. Reaction A + B → C ke liye rate law Rate = k[A][B]² diya hai. Is reaction ka order kya hoga? Agar [A] constant rakh kar [B] double kar diya jaaye toh rate kitne guna badhega?
Overall order = exponents ka sum = 1 (A ka) + 2 (B ka) = 3 (third order).
Ratenew/Rateold = [Bnew]²/[Bold]² = (2)² = 4
Rate 4 guna badh jaayega, kyunki B ka order 2 hai.
Q3. Ek first order reaction ka half-life 3.00 hours hai. Iska rate constant nikaliye (per hour aur per second dono me).
First order half-life formula:
t1/2 = 0.693/k ⇒ k = 0.693/t1/2
k = 0.693 / 3.00 h = 0.231 h−1
Second me convert karne ke liye 3600 se divide karo:
k = 0.231 / 3600 = 6.42 × 10−5 s−1
Q4. Ek first order reaction me initial concentration 0.1 mol L−1 se 40 minutes me 0.025 mol L−1 reh jaata hai. Rate constant nikaliye.
First order integrated rate equation use karte hain:
k = (2.303/t) log([A]0/[A])
k = (2.303/40) log(0.1/0.025) = (2.303/40) log(4)
log 4 = 0.602 ⇒ k = (2.303 × 0.602)/40 = 1.387/40
k = 0.0347 min−1 (approx 3.47 × 10−2 min−1)
Q5. Ek zero order reaction ka rate constant 2 × 10−2 mol L−1 s−1 hai. Agar initial concentration 1 mol L−1 ho toh concentration 0.5 mol L−1 hone me kitna time lagega?
Zero order integrated rate equation:
[A]0 − [A] = kt ⇒ t = ([A]0 − [A]) / k
t = (1 − 0.5) / (2 × 10−2) = 0.5 / 0.02
t = 25 seconds
Q6. Ek first order reaction ka half-life 60 minutes hai. 240 minutes baad reactant ka kitna percent react ho chuka hoga?
240 minutes me half-lives ki sankhya = 240/60 = 4 half-lives.
Fraction remaining = (1/2)n = (1/2)4 = 1/16 = 0.0625
Yaani sirf 6.25% reactant bacha, matlab:
% reacted = 100 − 6.25 = 93.75%
Q7. First order reaction ka rate constant 60 s−1 hai. Reactant ka concentration apne initial value ke 1/10 tak kam hone me kitna samay lagega?
[A]0/[A] = 10 use karte hain:
t = (2.303/k) log([A]0/[A]) = (2.303/60) log(10)
log 10 = 1 ⇒ t = 2.303/60
t = 0.0384 seconds
Q8. Reaction A + B → C ke liye niche diye gaye initial rate data se rate law nikaliye:
Exp 1: [A]=0.1M, [B]=0.1M, Rate=0.02 mol L⁻¹ s⁻¹
Exp 2: [A]=0.2M, [B]=0.1M, Rate=0.08 mol L⁻¹ s⁻¹
Exp 3: [A]=0.1M, [B]=0.2M, Rate=0.02 mol L⁻¹ s⁻¹
Order w.r.t. A: Exp 1 → 2 me [A] double hua (0.1→0.2), [B] same raha, aur rate 4 guna badh gaya (0.02→0.08).
2x = 4 ⇒ x = 2 (order in A = 2)
Order w.r.t. B: Exp 1 → 3 me [B] double hua (0.1→0.2), [A] same raha, par rate same raha (0.02).
2y = 1 ⇒ y = 0 (order in B = 0)
Isliye rate law:
Rate = k[A]²[B]⁰ = k[A]²
Overall order = 2 + 0 = 2 (second order). k nikalne ke liye Exp 1 use karo:
k = Rate/[A]² = 0.02/(0.1)² = 0.02/0.01 = 2 mol−1 L s−1
Q9. Ek reaction ka rate constant, temperature 300 K se 320 K karne pe 4 guna ho jaata hai. Reaction ki activation energy nikaliye. (R = 8.314 J K−1 mol−1)
Two-temperature Arrhenius equation use karte hain:
log(k2/k1) = (Ea/2.303R) × (1/T1 − 1/T2)
k2/k1 = 4 ⇒ log 4 = 0.6021
1/T1 − 1/T2 = 1/300 − 1/320 = (320−300)/(300×320) = 20/96000 = 2.083 × 10−4 K−1
0.6021 = [Ea/(2.303 × 8.314)] × 2.083 × 10−4
0.6021 = [Ea/19.147] × 2.083 × 10−4
Ea = (0.6021 × 19.147) / (2.083 × 10−4) = 11.53 / 2.083 × 10−4
Ea ≈ 55,340 J mol−1 ≈ 55.3 kJ mol−1
Q10. Ek reaction ka rate constant 300 K par 2.5 × 10−4 s−1 hai aur activation energy 75 kJ mol−1 hai. 350 K par rate constant nikaliye. (R = 8.314 J K−1 mol−1)
log(k2/k1) = (Ea/2.303R) × (1/T1 − 1/T2)
1/300 − 1/350 = (350−300)/(300×350) = 50/105000 = 4.762 × 10−4 K−1
Ea/2.303R = 75000/19.147 = 3917.6
log(k2/k1) = 3917.6 × 4.762 × 10−4 = 1.866
k2/k1 = antilog(1.866) ≈ 73.4
k2 = 2.5 × 10−4 × 73.4 ≈ 1.83 × 10−2 s−1
Q11. Ek zero order reaction ka rate constant 298 K par 0.0030 mol L−1 s−1 hai. Agar initial concentration 0.10 mol L−1 ho toh half-life nikaliye.
t1/2 = [A]0 / 2k
t1/2 = 0.10 / (2 × 0.0030) = 0.10/0.006
t1/2 = 16.67 seconds
Q12. Molecularity aur order of reaction me teen antar likhiye.
1. Definition: Molecularity ek elementary step me part lene wale species ki sankhya hai (theoretical); order rate law ke exponents ka sum hai (experimental).
2. Values: Molecularity hamesha whole number hoti hai (kabhi zero/fraction nahi); order zero, fractional, ya negative bhi ho sakta hai.
3. Applicability: Molecularity sirf elementary reaction ke liye meaningful hai; complex multi-step reaction ki koi single molecularity nahi hoti, par order poori reaction ke liye define hota hai.
Q13. Pseudo first order reaction kya hoti hai? Ek example ke saath samjhaiye.
Jab ek reaction actually second (ya higher) order hoti hai, lekin ek reactant itni zyada excess me hota hai ki uska concentration reaction ke dauran practically constant reh jaata hai — toh woh reaction experimentally first order jaisi behave karti hai. Isse pseudo first order reaction kehte hain.
Example: Ester hydrolysis — CH₃COOC₂H₅ + H₂O (excess) → CH₃COOH + C₂H₅OH. Actual rate law Rate = k[ester][H₂O] hai (second order), lekin water itna excess hota hai ki [H₂O] constant maan lete hain, isliye Rate = k'[ester] ban jaata hai (pseudo first order, jahan k' = k[H₂O]).
Q14. Reaction rate temperature ke saath kaise badhta hai? Collision theory ke through samjhaiye ki catalyst rate ko kaise badhata hai.
Temperature badhne se molecules ki average kinetic energy badhti hai, jisse zyada molecules threshold/activation energy cross kar paate hain — effective collisions ki frequency badh jaati hai, isliye rate badhta hai (Arrhenius equation: k = Ae−Ea/RT, T badhne se k exponentially badhta hai).
Catalyst: catalyst reaction ka ek alternate pathway provide karta hai jiska activation energy (Ea) original pathway se kam hota hai. Kam Ea ka matlab hai zyada fraction of molecules us threshold ko cross kar paate hain, isliye effective collisions aur rate dono badh jaate hain — bina catalyst khud consume hue.
Q15. Zero order aur first order reaction ke concentration-time graph ka shape describe kijiye, aur bataiye slope kya represent karta hai.
Zero order: [A] vs t graph ek straight line hai jo neeche ki taraf jaati hai (negative slope). Slope = −k. Equation: [A] = [A]0 − kt.
First order: log[A] vs t graph ek straight line deti hai (seedha [A] vs t nahi — woh exponential curve hoga). Slope = −k/2.303. Equation: log[A] = log[A]0 − kt/2.303.
Dono cases me graph ka linear hona hi confirm karta hai ki reaction respective order ki hai — ye experimental order-determination ka ek common tarika hai.
Q16. Ek first order reaction 30% complete hone me 40 minutes leti hai. Rate constant nikaliye.
30% react ho gaya matlab 70% reactant bacha hai, yaani [A] = 0.70[A]0:
k = (2.303/t) log([A]0/[A]) = (2.303/40) log(1/0.70)
log(1/0.70) = log(1.4286) = 0.1549
k = (2.303 × 0.1549)/40 = 0.3568/40
k = 8.92 × 10−3 min−1
Q17. Rate constant ke units se order of reaction kaise identify karte hain? Agar ek reaction ke k ka unit mol−1 L s−1 hai, toh uska order kya hoga?
nth order reaction ke liye k ka general unit hota hai mol1−n Ln−1 s−1. Diye gaye unit mol−1 L s−1 ko compare karo:
mol1−n = mol−1 ⇒ 1 − n = −1 ⇒ n = 2
Ye second order reaction ka characteristic unit hai.
Q18. Sucrose ke inversion reaction (C₁₂H₂₂O₁₁ + H₂O → glucose + fructose) ko pehle order ki reaction kyu maana jaata hai, jabki iska rate law do concentrations pe depend karta hai?
Rate law hai Rate = k[sucrose][H₂O], jo technically second order hai. Lekin reaction dilute aqueous solution me hoti hai jahan water itni zyada excess me hoti hai ki reaction ke poore duration me [H₂O] practically constant reh jaata hai. Isliye [H₂O] ko k me merge karke k' = k[H₂O] define kar dete hain, aur reaction Rate = k'[sucrose] ban jaati hai — yaani experimentally ye pseudo first order reaction ki tarah behave karti hai.
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Rate of reaction (R → P) | Rate = −Δ[R]/Δt = +Δ[P]/Δt |
| Rate for aA + bB → cC + dD | Rate = −(1/a)d[A]/dt = −(1/b)d[B]/dt = (1/c)d[C]/dt = (1/d)d[D]/dt |
| Rate law | Rate = k[A]x[B]y (x, y experimentally milte hain) |
| Zero order integrated equation | [A] = [A]0 − kt |
| Zero order rate constant | k = ([A]0 − [A]) / t |
| Zero order half-life | t1/2 = [A]0 / 2k |
| First order integrated equation | k = (2.303/t) log([A]0/[A]) |
| First order half-life | t1/2 = 0.693/k (initial conc. se independent) |
| Units of k — zero order | mol L−1 s−1 |
| Units of k — first order | s−1 |
| Units of k — second order | mol−1 L s−1 |
| Units of k — nth order | mol1−n Ln−1 s−1 |
| Arrhenius equation | k = A e−Ea/RT |
| Arrhenius log form | log k = log A − Ea/(2.303RT) |
| Two-temperature Arrhenius form | log(k2/k1) = (Ea/2.303R)(1/T1 − 1/T2) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Order aur molecularity ko same maan lena. Order experimentally rate law se nikalta hai aur zero/fractional bhi ho sakta hai; molecularity sirf elementary step ke liye hota hai aur hamesha whole number hota hai. Complex reaction ki 'ek' molecularity nahi hoti, lekin order hoti hai.
- Balanced chemical equation se seedha rate law likh dena. Rate law sirf experimental data se milta hai, stoichiometric coefficients se order predict NAHI kar sakte (sirf elementary reactions me coefficients hi order ke barabar hote hain).
- Har order ke liye k ka ek hi unit use karna. Zero order ka k mol L−1 s−1 hai, first order ka sirf s−1, second order ka mol−1 L s−1 — order badalte hi unit badal jaata hai, isse question me hi order confirm ho sakta hai.
- First order equation me natural log (ln) aur common log (log) me confuse hona. NCERT formula k = (2.303/t) log([A]0/[A]) common log (base 10) use karta hai, isliye 2.303 factor (jo ln10 hai) hamesha saath likhna hai — sirf log lagakar 2.303 bhool jaana bahut common mistake hai.
- Half-life ko har order ke liye same treat karna. First order ka half-life concentration-independent hota hai (t1/2 = 0.693/k), lekin zero order ka half-life initial concentration pe depend karta hai (t1/2 = [A]0/2k) — ye difference bhool jaana bahut students karte hain.
- Arrhenius equation me temperature Celsius me use karna. Arrhenius equation aur two-temperature formula dono me T hamesha Kelvin me hona chahiye — Celsius use karne se puri calculation galat aa jaati hai. Convert karna mat bhoolo: T(K) = T(°C) + 273.
Board-Style Important Questions
- 1 mark: Molecularity of an elementary reaction kabhi zero ya fraction kyu nahi ho sakti?
- 1 mark: Rate constant ka unit s−1 ho toh reaction ka order kya hai?
- 2 marks: Rate law aur law of mass action me antar spashth kijiye.
- 2 marks: First order reaction ka half-life initial concentration se independent kyu hota hai, derive karke dikhaiye.
- 3 marks: Ek first order reaction ka data diya gaya hai (concentration aur time) — rate constant graphically kaise nikalenge, samjhaiye.
- 5 marks: Arrhenius equation likhiye aur samjhaiye ki activation energy rate constant ko kaise affect karti hai; do temperature par rate constants diye jaane par Ea nikalne ka method bhi dikhaiye.
Aksar Poochhe Jaane Wale Sawaal
Order of reaction aur molecularity me kya main difference hai?
Order experimentally rate law ke data se nikalta hai aur zero, fractional ya negative bhi ho sakta hai. Molecularity ek elementary reaction step me part lene wale species ki actual sankhya hai, jo hamesha whole number hoti hai aur sirf single-step (elementary) reactions ke liye meaningful hai.
First order reaction ka half-life independent of concentration kyu hota hai?
Kyunki first order ka formula t1/2 = 0.693/k hai — isme [A]0 (initial concentration) ka koi term hi nahi hai. Rate constant k sirf temperature pe depend karta hai, isliye chahe initial concentration jo bhi ho, half-life same rahega.
Rate constant k temperature ke alawa aur kis cheez pe depend karta hai?
Practically sirf temperature pe depend karta hai (aur nature of reactants/catalyst pe, jo reaction fix karte hi tay ho jaate hain). Concentration pe k depend NAHI karta — concentration change karne se sirf rate change hoti hai, k nahi.
Pseudo first order reaction kya hoti hai, ek line me?
Jab actual order do ya zyada ho, lekin ek reactant (usually water) itna excess me ho ki uska concentration practically constant reh jaaye, tab reaction experimentally first order jaisi behave karti hai — jaise ester hydrolysis ya cane sugar ka inversion.
Zero order reaction possible kaise hoti hai jabki concentration zero pe rate bhi zero honi chahiye?
Zero order sirf ek limited concentration range tak valid rehta hai — jaise catalytic surface reactions me jab tak surface fully saturate hai, rate concentration-independent rehti hai. Bahut low concentration pe ye assumption break ho jaata hai.
Catalyst activation energy kam karta hai ya reaction ki enthalpy change karta hai?
Sirf activation energy kam karta hai, ek alternate lower-energy pathway dekar. Reaction ka overall enthalpy change (ΔH) catalyst se bilkul unaffected rehta hai — catalyst sirf rate badhata hai, thermodynamics change nahi karta.
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