Class 11 Chemistry · Chapter 6
Short answer: Class 11 Chemistry Chapter 6 Equilibrium NCERT solutions — Kc, Kp, reaction quotient, Le Chatelier's principle, aur ionic equilibrium (pH, Ka/Kb, buffer, hydrolysis, Ksp) ke saare intext aur exercise questions step-by-step board-exam-style working ke sath, taaki board exam ke numericals me koi step miss na ho.
Equilibrium Class 11 Chemistry ka sabse number-heavy chapter hai, aur rationalised 2026-27 session ke syllabus me ye ab Chapter 6 hai — Part II ki shuruaat, jahan chapters 5 se 9 tak rakhe gaye hain (Part I me 1-4). Pehle jab class 11 chemistry all chapters list ncert me 14 chapters the, tab Equilibrium chapter 7 pe tha. 2023 ki rationalisation ke baad States of Matter, Hydrogen, s-Block Elements, Some p-Block Elements aur Environmental Chemistry — ye 5 poore chapters hata diye gaye, isliye numbering shift ho gayi. Agar tum class 11 chemistry deleted syllabus topics dhoondh rahe ho — yahi 5 hain; Equilibrium ka koi hissa delete nahi hua, poora chapter jaisa tha waisa hi hai.
Chapter do bade hisso me bata hai. Pehla half Chemical Equilibrium ka hai — dynamic equilibrium, law of mass action, equilibrium constant Kc aur Kp, reaction quotient Qc, aur Le Chatelier's principle jo batata hai equilibrium concentration/pressure/temperature change se kaise shift hota hai. Dusra half Ionic Equilibrium ka hai — acids-bases (Arrhenius, Bronsted-Lowry, Lewis), pH scale, weak acid/base ionization (Ostwald dilution law), buffer solutions, salt hydrolysis, aur solubility product Ksp. Class 11 chemistry chapter 6 equilibrium ncert solutions dhoondhne wale zyadatar students yahi numericals ke liye aate hain, kyunki Kc/Kp aur pH calculation dono JEE-NEET aur board exam dono me heavy weightage rakhte hain — thermodynamics (chapter 5) ke free energy concepts yahin K se connect hote hain.
Chapter 6 Summary — 5 Minute Revision
Equilibrium wo state hai jahan forward aur reverse reaction ki rates barabar ho jaati hain — ye dynamic hai, static nahi, kyunki dono reactions chalti rehti hain, sirf net change zero dikhta hai. Physical equilibrium (liquid-vapour, solid-liquid, solute-solution) aur chemical equilibrium dono is chapter me cover hote hain.
Law of mass action se equilibrium constant Kc (concentration-based) aur Kp (partial pressure-based, sirf gases ke liye) define hote hain, jo Kp = Kc(RT)^Δn se related hain. Reaction quotient Qc ko Kc se compare karke pata chalta hai reaction kis direction me proceed karega. Homogeneous equilibria (sab same phase) aur heterogeneous equilibria (multiple phases, jahan pure solids/liquids ki concentration expression me nahi aati) dono discuss hote hain. Le Chatelier's principle batata hai ki concentration, pressure, volume, temperature change hone par equilibrium us direction shift karta hai jo change ko counteract kare — catalyst equilibrium shift nahi karta, sirf jaldi pahunchata hai.
Ionic equilibrium section me acid-base ki teen theories (Arrhenius, Bronsted-Lowry, Lewis) padhte hain, pH scale (pH = -log[H+]), weak acid/base ki ionization (Ostwald dilution law, Ka = Cα²/(1-α)), common ion effect, buffer solutions (Henderson-Hasselbalch equation), salt hydrolysis, aur solubility product Ksp jo sparingly soluble salts ki solubility predict karta hai.
In-Text Questions — Solutions
1. Ek sealed container me liquid apni vapour ke saath fixed temperature pe equilibrium me hai. Container ka volume achanak badha diya jaata hai. (a) Vapour pressure pe initial effect kya hoga? (b) Evaporation aur condensation ki rates initially kaise badlengi? (c) Jab equilibrium wapas aata hai, final vapour pressure kya hogi?
(a) Volume badhne se molecules zyada space me phail jaate hain, isliye vapour density kam ho jaati hai aur vapour pressure turant ghat jaati hai.
(b) Evaporation ki rate temporarily badh jaati hai (kyunki liquid surface area same par vapour space zyada hai), jabki condensation ki rate ghat jaati hai (vapour density kam hone se molecules kam frequency se surface se takrate hain).
(c) Jab tak evaporation rate = condensation rate nahi ho jaati, evaporation continue hoga. Equilibrium restore hone par vapour pressure sirf temperature pe depend karti hai, volume pe nahi — isliye final vapour pressure same rahegi jitni pehle thi (bashart temperature constant ho).
2. Reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ke liye equilibrium concentrations [SO₂] = 0.60 M, [O₂] = 0.82 M, [SO₃] = 1.90 M hai. Kc nikaalo.
Kc = [SO₃]² / ([SO₂]² × [O₂])
Kc = (1.90)² / ((0.60)² × 0.82) = 3.61 / (0.36 × 0.82) = 3.61 / 0.2952
Kc = 12.23 mol⁻¹ L
3. Ek certain temperature par total pressure 10⁵ Pa hai aur iodine vapour me 40% by volume I atoms hain, reaction I₂(g) ⇌ 2I(g) ke according. Kp nikaalo.
40% volume I atoms ka matlab hai mole fraction of I = 0.4, isliye I₂ ka mole fraction = 0.6.
P(I) = 0.4 × 10⁵ = 4 × 10⁴ Pa; P(I₂) = 0.6 × 10⁵ = 6 × 10⁴ Pa
Kp = [P(I)]² / P(I₂) = (4×10⁴)² / (6×10⁴) = 16×10⁸ / 6×10⁴
Kp = 2.67 × 10⁴ Pa
4. In equilibria ke liye Kc expressions likho: (i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) (ii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq) (iii) Fe³⁺(aq) + 3OH⁻(aq) ⇌ Fe(OH)₃(s)
(i) Kc = [NO]²[Cl₂] / [NOCl]²
(ii) Kc = [CH₃COOH][C₂H₅OH] / [CH₃COOC₂H₅] (H₂O solvent hai, pure liquid ki tarah treat hoke omit)
(iii) Kc = 1 / ([Fe³⁺][OH⁻]³) (Fe(OH)₃ solid hai, expression me nahi aata)
5. Kp se Kc nikaalo: (i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); Kp = 1.8×10⁻² at 500 K (ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 167 at 1073 K
Formula: Kp = Kc(RT)^Δn, isliye Kc = Kp / (RT)^Δn (R = 0.0831 L bar mol⁻¹ K⁻¹)
(i) Δn = 3 − 2 = 1
Kc = 1.8×10⁻² / (0.0831 × 500) = 1.8×10⁻² / 41.55 = 4.33 × 10⁻⁴
(ii) Δn = 1 (sirf CO₂ gas hai)
Kc = 167 / (0.0831 × 1073) = 167 / 89.16 = 1.87
6. Reaction NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g) ke liye 1000 K par Kc = 6.3×10¹⁴ hai. Reverse reaction ka Kc kya hoga?
Reverse reaction ka equilibrium constant = 1 / Kc(forward)
Kc(reverse) = 1 / (6.3×10¹⁴)
Kc(reverse) = 1.59 × 10⁻¹⁵
7. Equilibrium constant expression likhte waqt pure solids aur pure liquids ko kyun ignore kiya jaata hai?
Pure solid ya pure liquid ki molar concentration (density/molar mass) ek fixed temperature par constant hoti hai — quantity badle to bhi concentration same rehti hai, kyunki concentration = amount per unit volume aur density fix hai. Isliye is constant value ko equilibrium constant expression me merge kar diya jaata hai (activity = 1 le li jaati hai), aur ye separately nahi likha jaata.
8. 500 K par N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ke liye Kc = 0.061 hai. Ek certain time par [N₂] = 3.0×10⁻² M, [H₂] = 2.0×10⁻³ M, [NH₃] = 1.5×10⁻² M measure kiye gaye. Kya reaction equilibrium par hai? Agar nahi, to kis direction proceed karega?
Qc = [NH₃]² / ([N₂][H₂]³) = (1.5×10⁻²)² / (3.0×10⁻² × (2.0×10⁻³)³)
= 2.25×10⁻⁴ / (3.0×10⁻² × 8×10⁻⁹) = 2.25×10⁻⁴ / 2.4×10⁻¹⁰ = 9.4 × 10⁵
Qc (9.4×10⁵) >> Kc (0.061), isliye reaction equilibrium par nahi hai. Chunki Qc > Kc, reaction reverse direction (products se reactants ki taraf, yaani NH₃ decompose hoga) proceed karega taaki Qc, Kc ke barabar aa jaaye.
9. Bromine monochloride 2BrCl(g) ⇌ Br₂(g) + Cl₂(g) ke according decompose hota hai; 500 K par Kc = 32. Agar shuru me sirf pure BrCl 3.3×10⁻³ mol/L concentration par present hai, equilibrium par sab species ki concentration nikaalo.
Maan lo 2x mol/L BrCl decompose hua:
[BrCl] = 3.3×10⁻³ − 2x; [Br₂] = [Cl₂] = x
Kc = x² / (3.3×10⁻³ − 2x)² = 32 ⟹ x / (3.3×10⁻³ − 2x) = √32 = 5.657
x = 5.657(3.3×10⁻³ − 2x) = 1.867×10⁻² − 11.314x
12.314x = 1.867×10⁻² ⟹ x = 1.52×10⁻³
[Br₂] = [Cl₂] = 1.52 × 10⁻³ M; [BrCl] = 3.3×10⁻³ − 2(1.52×10⁻³) = 2.66 × 10⁻⁴ M
10. NO(g) + ½O₂(g) ⇌ NO₂(g) ke liye 298 K par ΔG° aur equilibrium constant nikaalo, given ΔfG°(NO₂) = 52.0 kJ/mol, ΔfG°(NO) = 87.0 kJ/mol, ΔfG°(O₂) = 0.
ΔG°(rxn) = ΔfG°(NO₂) − ΔfG°(NO) − ½ΔfG°(O₂) = 52.0 − 87.0 − 0 = −35.0 kJ/mol = −35000 J/mol
ΔG° = −RT ln K ⟹ ln K = 35000 / (8.314 × 298) = 35000 / 2477.6 = 14.13
K = e^14.13
ΔG° = −35.0 kJ/mol; K ≈ 1.37 × 10⁶ (large negative ΔG° ke saath large K expected hai)
11. In se kaunse Lewis acids hain — H₂O, BF₃, H⁺, NH₄⁺?
Lewis acid = electron-pair acceptor (empty orbital ya positive charge chahiye).
- BF₃ — Boron ke paas incomplete octet hai (6 electrons), isliye electron pair accept kar sakta hai → Lewis acid ✔
- H⁺ — bilkul empty 1s orbital, sabse strong electron-pair acceptor → Lewis acid ✔
- H₂O — oxygen ke paas lone pairs hain, ye electron-pair donate karta hai → Lewis base (acid nahi)
- NH₄⁺ — N ka octet complete hai aur koi empty orbital nahi, electron pair accept nahi kar sakta → Lewis acid nahi
Answer: BF₃ aur H⁺ Lewis acids hain.
12. Fluoroacetic acid ke 0.015 M solution ka pH = 2.34 hai. Degree of ionization aur ionization constant nikaalo.
[H⁺] = 10⁻²·³⁴ = 4.57 × 10⁻³ M
α = [H⁺]/C = 4.57×10⁻³ / 0.015 = 0.305 (30.5%)
Ka = Cα² / (1−α) = 0.015 × (0.305)² / (1−0.305) = 0.015 × 0.0930 / 0.695
α = 0.305 (30.5%); Ka = 2.01 × 10⁻³
13. pH nikaalo: (i) 0.003 M HCl (ii) 0.005 M NaOH (iii) 0.002 M HBr (iv) 0.002 M KOH
(i) pH = −log(0.003) = 2.52
(ii) pOH = −log(0.005) = 2.30 ⟹ pH = 14 − 2.30 = 11.70
(iii) pH = −log(0.002) = 2.70
(iv) pOH = −log(0.002) = 2.70 ⟹ pH = 14 − 2.70 = 11.30
14. Acetic acid ka Ka = 1.74×10⁻⁵ hai. 0.05 M solution me degree of dissociation, [CH₃COO⁻] aur pH nikaalo.
Ostwald dilution law: α = √(Ka/C) (kyunki α bahut chhota hai, 1−α ≈ 1)
α = √(1.74×10⁻⁵ / 0.05) = √(3.48×10⁻⁴) = 0.0187 (1.87%)
[CH₃COO⁻] = [H⁺] = Cα = 0.05 × 0.0187 = 9.33 × 10⁻⁴ M
pH = −log(9.33×10⁻⁴)
α = 1.87%; [CH₃COO⁻] = 9.33×10⁻⁴ M; pH = 3.03
15. Common ion effect kya hai? Ek weak acid ke ionization par common ion add karne se kya hota hai, briefly explain karo.
Common ion effect: jab kisi weak electrolyte ke solution me ek aisa ion add kiya jaata hai jo already us equilibrium ka part hai (common ion), to Le Chatelier's principle ke according equilibrium us direction shift karta hai jo added ion ki concentration kam kare — isliye weak electrolyte ki ionization/dissociation suppress ho jaati hai.
Example: CH₃COOH ⇌ CH₃COO⁻ + H⁺ me agar CH₃COONa (jo CH₃COO⁻ ion deta hai) add karo, to equilibrium reverse (left) shift karega, aur CH₃COOH ka degree of ionization ghat jaayega. Yehi principle buffer solutions ka base hai.
16. Solubility product Ksp aur molar solubility s me kya relation hota hai? AgCl jaise AB-type salt ke liye derive karo.
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq); agar molar solubility = s mol/L, to
[Ag⁺] = s, [Cl⁻] = s
Ksp = [Ag⁺][Cl⁻] = s × s = s²
AB-type salt ke liye Ksp = s², isliye s = √Ksp. (AB₂ ya A₂B type ke liye ye relation different hota hai, kyunki stoichiometric factors change ho jaate hain.)
Poore Class 11 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–Q38)
1. Equilibrium state ko define karo aur explain karo ki ise 'dynamic equilibrium' kyun kaha jaata hai, static kyun nahi.
Jab kisi reversible reaction (ya physical process) me forward aur backward, dono directions ki rates barabar ho jaati hain, to system equilibrium me kehlaata hai. Iss point par net observable change zero dikhta hai (concentrations, pressure, colour sab constant reh jaate hain).
Ye dynamic isliye kehlaata hai kyunki reaction rukta nahi — forward aur reverse dono reactions molecular level par continuously chalti rehti hain, sirf unki rates barabar hone ki wajah se bulk properties me koi change nazar nahi aata. Static equilibrium hota to macroscopic level ke saath-saath molecular level par bhi kuch move nahi karta — jo chemical systems me possible nahi hai.
2. Equilibrium state ki 4 main characteristics likho.
- Equilibrium sirf closed system me attain hota hai (constant temperature par).
- Ye dynamic hai — forward aur reverse reactions dono chalti rehti hain, equal rates par.
- Equilibrium ko reaction kisi bhi direction se attain kiya ja sakta hai — chahe pure reactants se shuru karo ya pure products se, final Kc same rahega (fixed T par).
- Kisi bhi catalyst ki presence equilibrium composition change nahi karti, sirf equilibrium jaldi attain hota hai.
3. Reaction H₂(g) + I₂(g) ⇌ 2HI(g) ke liye equilibrium concentrations [H₂] = 0.15 M, [I₂] = 0.25 M, [HI] = 0.85 M hain. Kc calculate karo.
Kc = [HI]² / ([H₂][I₂]) = (0.85)² / (0.15 × 0.25) = 0.7225 / 0.0375
Kc = 19.27
4. Certain temperature par PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) ke liye Kp = 1.8 hai. Agar PCl₅ ka initial pressure 2 atm hai (constant T, V par), to dissociation ka degree (α) aur equilibrium par sab species ka partial pressure nikaalo.
ICE table (partial pressure, initial PCl₅ = 2 atm):
P(PCl₅) = 2(1−α); P(PCl₃) = P(Cl₂) = 2α
Kp = (2α)(2α) / [2(1−α)] = 2α² / (1−α) = 1.8
2α² + 1.8α − 1.8 = 0 ⟹ α = [−1.8 + √(3.24+14.4)] / 4 = [−1.8+4.2]/4
α = 0.6 (60% dissociation)
P(PCl₅) = 2(0.4) = 0.8 atm; P(PCl₃) = P(Cl₂) = 2(0.6) = 1.2 atm each
Check: Kp = (1.2×1.2)/0.8 = 1.8 ✔
5. Kc se Kp nikaalne ke liye formula likho aur us reaction ka example do jahan Kp = Kc ho jaata hai.
Kp = Kc(RT)^Δn, jahan Δn = (moles of gaseous products) − (moles of gaseous reactants)
Jab Δn = 0 (yaani reactants aur products dono side gaseous moles equal hain), tab (RT)^0 = 1 aur Kp = Kc.
Example: H₂(g) + I₂(g) ⇌ 2HI(g) — 2 moles gas reactant side, 2 moles product side, Δn = 0, isliye Kp = Kc.
6. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) ke liye 250°C par Kp = 1.8 hai. Kc calculate karo.
T = 523 K, Δn = 1
Kc = Kp / (RT)^Δn = 1.8 / (0.0831 × 523) = 1.8 / 43.46
Kc = 0.0414
7. Pure PCl₅ ko evacuated vessel me 473 K par introduce kiya gaya. Equilibrium par [PCl₅] = 0.05 mol/L paaya gaya. Agar Kc = 8.3×10⁻³ hai, to equilibrium par [PCl₃] aur [Cl₂] nikaalo.
Pure PCl₅ se shuru hua hai, isliye [PCl₃] = [Cl₂] = x
Kc = x² / [PCl₅] = x² / 0.05 = 8.3×10⁻³
x² = 4.15×10⁻⁴ ⟹ x = √(4.15×10⁻⁴)
[PCl₃] = [Cl₂] = 2.04 × 10⁻² M
8. Equilibrium constant expression Kc = [NH₃]⁴[O₂]⁵ / ([NO]⁴[H₂O]⁶) diya hai. Balanced chemical equation likho jisse ye expression aaya ho.
Numerator me products (NH₃, O₂), denominator me reactants (NO, H₂O) aate hain. Coefficients power se match karke:
4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g)
Check — N: 4=4; H: 12=12; O: (4+6)=10 = (0+10)=10 ✔ balanced.
9. Ek gaseous reaction ke liye Kp = Kc(RT)^Δn diya hai. Δn ka sign positive, negative aur zero hone ka kya matlab hai?
- Δn > 0 (positive): products me gaseous moles zyada hain reactants se — Kp > Kc (jab RT > 1)
- Δn < 0 (negative): reactants me gaseous moles zyada hain — Kp < Kc
- Δn = 0: dono side gaseous moles barabar hain — Kp = Kc
10. NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g) reaction elementary bimolecular hai, forward aur reverse dono. 1000 K par Kc = 6.3×10¹⁴ hai. Iska use karke btao ki equilibrium par reactants ka fraction bahut kam kyun hoga.
Kc =
11. 0.482 mol N₂ aur 0.933 mol O₂ ko 10 L vessel me rakha gaya, reaction 2N₂(g) + O₂(g) ⇌ 2N₂O(g), jiska Kc = 2.0×10⁻³⁷ hai. Equilibrium composition determine karo.
[N₂]₀ = 0.0482 M, [O₂]₀ = 0.0933 M. Kc extremely chhota hai, isliye forward reaction negligible hoga aur equilibrium concentrations initial ke lagbhag barabar rahengi.
Kc = [N₂O]² / ([N₂]²[O₂]) = 2.0×10⁻³⁷
[N₂O]² ≈ 2.0×10⁻³⁷ × (0.0482)² × 0.0933 = 4.33×10⁻⁴¹ ⟹ [N₂O] ≈ 6.6×10⁻²¹ M
[N₂] ≈ 0.0482 M, [O₂] ≈ 0.0933 M, [N₂O] ≈ 6.6×10⁻²¹ M — yaani reaction practically forward direction me proceed hi nahi karta.
12. 2NO(g) + Br₂(g) ⇌ 2NOBr(g) reaction me 0.087 mol NO aur 0.0437 mol Br₂ mix kiye gaye ek closed container me. Equilibrium par 0.0518 mol NOBr bana. Equilibrium par NO aur Br₂ ki amount nikaalo.
Stoichiometry se: 2 mol NOBr banne ke liye 2 mol NO aur 1 mol Br₂ consume hota hai, isliye 0.0518 mol NOBr banne ke liye —
NO consumed = 0.0518 mol; Br₂ consumed = 0.0518/2 = 0.0259 mol
NO(eq) = 0.087 − 0.0518 = 0.0352 mol
Br₂(eq) = 0.0437 − 0.0259 = 0.0178 mol
NO = 0.0352 mol; Br₂ = 0.0178 mol
13. 1024 K par H₂(g) + Br₂(g) ⇌ 2HBr(g) ka equilibrium constant 1.6×10⁵ hai. Agar 10.0 bar pure HBr ek sealed container me introduce kiya jaaye, equilibrium par sabhi gases ki pressure nikaalo.
Δn = 0 isliye Kp = Kc = 1.6×10⁵ (formation direction). HBr ka decomposition consider karo: 2HBr ⇌ H₂ + Br₂, iska K = 1/(1.6×10⁵) = 6.25×10⁻⁶
Maan lo 2x bar HBr decompose hua:
K = x² / (10−2x)² = 6.25×10⁻⁶ ⟹ x/(10−2x) = 2.5×10⁻³
x = 0.025 − 0.005x ⟹ 1.005x = 0.025 ⟹ x = 0.0249
P(H₂) = P(Br₂) ≈ 0.025 bar; P(HBr) = 10 − 0.0498 ≈ 9.95 bar
14. H₃O⁺, OH⁻ aur H₂O involve karte hue jitne balanced equations likh sakte ho, likho (autoionization aur neutralization dikhate hue).
Autoionization of water: 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
Neutralization: H₃O⁺(aq) + OH⁻(aq) ⇌ 2H₂O(l)
Acid ionization (generic): HA + H₂O ⇌ H₃O⁺ + A⁻
Base ionization (generic): B + H₂O ⇌ BH⁺ + OH⁻
15. Reactions N₂(g)+O₂(g) ⇌ 2NO(g) [K₁] aur 2NO(g)+O₂(g) ⇌ 2NO₂(g) [K₂] diye hain. Reaction N₂(g)+2O₂(g) ⇌ 2NO₂(g) ka equilibrium constant K, K₁ aur K₂ ke terms me likho.
Reaction 3 = Reaction 1 + Reaction 2 (add karne par 2NO cancel ho jaata hai, dono side):
N₂ + O₂ ⇌ 2NO ... K₁
2NO + O₂ ⇌ 2NO₂ ... K₂
Sum: N₂ + 2O₂ ⇌ 2NO₂
Jab reactions add hoti hain, equilibrium constants multiply hote hain:
K = K₁ × K₂
16. Le Chatelier's principle ko state karo. Ye pressure aur temperature ke effect ko kaise predict karta hai?
Le Chatelier's principle: Agar equilibrium me kisi system par concentration, pressure, volume ya temperature me change laga di jaaye, to system us direction shift karta hai jo change ke effect ko counteract (kam) kare.
- Pressure badhana (volume ghatana): equilibrium us side shift karta hai jahan gaseous moles kam hain (system pressure kam karne ki koshish karta hai).
- Temperature badhana: exothermic reaction ke liye equilibrium reverse (endothermic direction) shift karta hai, aur endothermic reaction ke liye forward shift karta hai (heat absorb karke).
17. Constant volume par inert gas add karne ka aur constant pressure par inert gas add karne ka equilibrium par kya alag-alag effect hota hai?
Constant volume par: inert gas add karne se total pressure badhती hai, lekin har reacting species ka partial pressure/concentration same rehta hai (volume fix hai) — isliye equilibrium par koi effect nahi padta.
Constant pressure par: inert gas add karne ke liye volume badhana padta hai (taaki total pressure constant rahe), isse har reacting species ki concentration/partial pressure kam ho jaati hai — equilibrium us direction shift karta hai jahan gaseous moles zyada hain.
18. Bronsted-Lowry theory ke according classify karo — NH₃, HCOOH, HS⁻, HSO₄⁻ — ye acid hain, base hain ya dono (amphoteric)?
- NH₃ — proton accept kar sakta hai (NH₄⁺ banata hai) → Bronsted base
- HCOOH — proton donate karta hai (HCOO⁻ banata hai) → Bronsted acid
- HS⁻ — H⁺ donate karke S²⁻ ya H⁺ accept karke H₂S bana sakta hai → amphoteric (dono)
- HSO₄⁻ — H⁺ donate karke SO₄²⁻ ya accept karke H₂SO₄ bana sakta hai → amphoteric (dono)
19. Arrhenius, Bronsted-Lowry aur Lewis theory of acids-bases me main difference kya hai? Ek table me summarize karo.
| Theory | Acid ki definition | Base ki definition | Limitation |
|---|---|---|---|
| Arrhenius | H⁺ deta hai aqueous solution me | OH⁻ deta hai aqueous solution me | Sirf aqueous medium tak limited |
| Bronsted-Lowry | Proton (H⁺) donor | Proton (H⁺) acceptor | Non-protonic acids (jaise BF₃) explain nahi karta |
| Lewis | Electron-pair acceptor | Electron-pair donor | Sabse broad, sabhi cases cover karta hai |
↔ Table ko side me swipe karein
20. pH nikaalo: (i) 0.06 M HNO₃ (ii) 0.05 M Ba(OH)₂
(i) HNO₃ strong monoprotic acid hai: pH = −log(0.06) = 1.22
(ii) Ba(OH)₂ strong diprotic base hai: [OH⁻] = 2 × 0.05 = 0.10 M
pOH = −log(0.10) = 1.00 ⟹ pH = 14 − 1.00 = 13.00
21. In biological fluids ke pH diye hain — human blood 7.38, gastric juice 1.2, saliva 6.4. In sab ki [H⁺] concentration nikaalo.
Blood: [H⁺] = 10⁻⁷·³⁸ = 4.17 × 10⁻⁸ M
Gastric juice: [H⁺] = 10⁻¹·² = 6.31 × 10⁻² M
Saliva: [H⁺] = 10⁻⁶·⁴ = 3.98 × 10⁻⁷ M
22. Cyanic acid HCNO ke 0.1 M solution ka pH = 2.34 hai. Ionization constant aur degree of ionization nikaalo.
[H⁺] = 10⁻²·³⁴ = 4.57 × 10⁻³ M
α = [H⁺]/C = 4.57×10⁻³ / 0.1 = 0.0457 (4.57%)
Ka = Cα²/(1−α) = 0.1 × (0.0457)² / (1−0.0457) = 0.1 × 0.00209 / 0.9543
Ka = 2.19 × 10⁻⁴; α = 4.57%
23. 0.1 M acetic acid solution ka degree of dissociation aur pH nikaalo (Ka = 1.74×10⁻⁵).
α = √(Ka/C) = √(1.74×10⁻⁵ / 0.1) = √(1.74×10⁻⁴) = 0.0132 (1.32%)
[H⁺] = Cα = 0.1 × 0.0132 = 1.32 × 10⁻³ M
pH = −log(1.32×10⁻³)
α = 1.32%; pH = 2.88
24. NaNO₂ ke 0.04 M solution ka pH aur degree of hydrolysis nikaalo (Ka of HNO₂ = 4.5×10⁻⁴).
NaNO₂ ek strong base + weak acid ka salt hai, isliye hydrolysis se basic solution banta hai.
Kh = Kw/Ka = 10⁻¹⁴ / (4.5×10⁻⁴) = 2.22 × 10⁻¹¹
h = √(Kh/C) = √(2.22×10⁻¹¹ / 0.04) = √(5.56×10⁻¹⁰) = 2.36 × 10⁻⁵
[OH⁻] = Ch = 0.04 × 2.36×10⁻⁵ = 9.43×10⁻⁷ ⟹ pOH = 6.03
Degree of hydrolysis = 2.36×10⁻⁵; pH = 14 − 6.03 = 7.97
25. Phenol ka Ka = 1.0×10⁻¹⁰ hai. Uske 0.05 M solution me [C₆H₅O⁻] nikaalo. Agar solution me 0.01 M sodium phenolate bhi add kar diya jaaye, common ion effect se [H⁺] par kya asar padega?
Bina common ion:
[C₆H₅O⁻] = √(Ka × C) = √(1.0×10⁻¹⁰ × 0.05) = √(5×10⁻¹²)
[C₆H₅O⁻] = 2.24 × 10⁻⁶ M
0.01 M sodium phenolate add karne par: [C₆H₅O⁻] ab dominantly salt se aata hai (≈0.01 M), isliye
Ka = [H⁺][C₆H₅O⁻]/[C₆H₅OH] ⟹ [H⁺] = Ka × 0.05/0.01 = 1.0×10⁻¹⁰ × 5
[H⁺] = 5 × 10⁻¹⁰ M — common ion (phenolate) add hone se phenol ki ionization strongly suppress ho jaati hai (Le Chatelier's principle).
26. H₂S ka Ka₁ = 9.1×10⁻⁸ aur Ka₂ = 1.2×10⁻¹³ hai. 0.1 M H₂S solution me jisme [H⁺] = 0.1 M (HCl se) hai, [S²⁻] nikaalo.
Overall: Ka₁×Ka₂ = [H⁺]²[S²⁻] / [H₂S]
[S²⁻] = Ka₁Ka₂[H₂S] / [H⁺]² = (9.1×10⁻⁸)(1.2×10⁻¹³)(0.1) / (0.1)²
= 1.092×10⁻²¹ / 0.01
[S²⁻] = 1.09 × 10⁻¹⁹ M — extremely kam, isliye acidic medium me metal sulphides precipitate nahi hote jab tak Ksp bahut chhota na ho.
27. Ostwald's dilution law ko derive karo weak monobasic acid HA ke liye, aur batao ki dilution badhane se degree of ionization par kya asar padta hai.
HA ⇌ H⁺ + A⁻; initial concentration C, degree of ionization α:
| HA | H⁺ | A⁻ | |
|---|---|---|---|
| Initial | C | 0 | 0 |
| Equilibrium | C(1−α) | Cα | Cα |
↔ Table ko side me swipe karein
Ka = (Cα)(Cα) / [C(1−α)] = Cα² / (1−α)
Jab α bahut chhota ho (1−α ≈ 1): Ka ≈ Cα² ⟹ α = √(Ka/C)
Dilution (C ghatana) se α badhta hai — kyunki α ∝ 1/√C, jitna solution dilute hoga, weak acid utna zyada ionize hoga.
28. Buffer solution kya hota hai? Acetate buffer (0.1 M CH₃COOH + 0.1 M CH₃COONa, pKa = 4.74) ka pH nikaalo Henderson-Hasselbalch equation se.
Buffer solution wo hai jo thodi si acid ya base add karne par apna pH significantly change nahi hone deta — usually weak acid + uska conjugate base salt (ya weak base + conjugate acid salt) ka mixture.
Henderson-Hasselbalch: pH = pKa + log([salt]/[acid])
pH = 4.74 + log(0.1/0.1) = 4.74 + log(1) = 4.74 + 0
pH = 4.74 (jab salt aur acid concentrations equal hon, pH = pKa)
29. Salt hydrolysis ke 4 types (strong acid-strong base, weak acid-strong base, strong acid-weak base, weak acid-weak base) ka nature ek table me summarize karo.
| Salt type | Example | Solution nature |
|---|---|---|
| Strong acid + strong base | NaCl | Neutral (pH = 7), hydrolysis nahi hota |
| Weak acid + strong base | CH₃COONa | Basic (pH > 7) |
| Strong acid + weak base | NH₄Cl | Acidic (pH < 7) |
| Weak acid + weak base | CH₃COONH₄ | pH ≈ 7, Ka aur Kb ki relative strength par depend karta hai |
↔ Table ko side me swipe karein
30. AgCl ki molar solubility 298 K par 1.34×10⁻⁵ mol/L hai. Iska Ksp nikaalo.
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), isliye [Ag⁺] = [Cl⁻] = s = 1.34×10⁻⁵ M
Ksp = [Ag⁺][Cl⁻] = s² = (1.34×10⁻⁵)²
Ksp = 1.80 × 10⁻¹⁰
31. AgCl ka Ksp = 1.8×10⁻¹⁰ hai. Uski solubility 0.01 M NaCl solution me (common ion effect ke saath) nikaalo aur pure water ki solubility se compare karo.
Maan lo AgCl ki solubility 0.01 M NaCl solution me s' hai. Chunki NaCl se already 0.01 M Cl⁻ present hai aur s' << 0.01:
Ksp = s'(0.01 + s') ≈ s' × 0.01
s' = 1.8×10⁻¹⁰ / 0.01 = 1.8 × 10⁻⁸ M
Pure water me solubility s = √Ksp = √(1.8×10⁻¹⁰) = 1.34×10⁻⁵ M.
Common ion effect se solubility 1.34×10⁻⁵ M se ghatkar 1.8×10⁻⁸ M ho gayi — lagbhag 750 guna kam, jo Le Chatelier's principle ke according expected hai.
32. Catalyst equilibrium constant ya equilibrium position ko change kyun nahi karta, sirf equilibrium jaldi kyun attain karata hai?
Catalyst forward aur reverse — dono reactions ki activation energy equally kam karta hai, isliye forward aur reverse dono rates same factor se badhती hain. Chunki rates ka ratio (jo equilibrium constant define karta hai) same rehta hai, equilibrium position/Kc change nahi hoti — sirf equilibrium zyada jaldi attain ho jaata hai kyunki dono reactions fast ho jaati hain.
33. Homogeneous aur heterogeneous equilibrium me difference batao, ek-ek example ke saath.
Homogeneous equilibrium: sabhi reactants aur products same phase me hote hain.
Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (sab gas phase)
Heterogeneous equilibrium: reactants aur products alag-alag phases me hote hain.
Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g) (solid-solid-gas mix)
Heterogeneous equilibrium me pure solids/liquids ki concentration Kc/Kp expression me nahi aati.
34. pOH scale kya hai aur pH + pOH = 14 relation kaise derive hota hai (298 K par)?
pOH = −log[OH⁻], jaise pH = −log[H⁺] hai.
Water ka ionic product: Kw = [H⁺][OH⁻] = 10⁻¹⁴ (298 K par). Dono side −log lagane par:
−log[H⁺] + (−log[OH⁻]) = −log(10⁻¹⁴)
pH + pOH = 14 (298 K, aqueous solutions ke liye)
35. Bronsted acid-base concept me conjugate acid-base pair kya hote hain? HCl + H₂O ⇌ H₃O⁺ + Cl⁻ me identify karo.
Ek acid apna proton kho kar jo species banata hai wo uska conjugate base hota hai, aur base proton lekar jo species banata hai wo uska conjugate acid hota hai.
HCl (acid) → Cl⁻ (conjugate base)
H₂O (base) → H₃O⁺ (conjugate acid)
Conjugate pairs: HCl/Cl⁻ aur H₂O/H₃O⁺
36. Dimethylamine ka Ka(conjugate acid ka) diya hai as Kb = 5.4×10⁻⁴. 0.02 M solution me degree of ionization nikaalo.
α = √(Kb/C) = √(5.4×10⁻⁴ / 0.02) = √(0.027)
α = 0.164 (16.4%)
37. Le Chatelier's principle ka use karke explain karo ki Haber process (N₂+3H₂⇌2NH₃, exothermic) me NH₃ yield badhane ke liye high pressure kyun use hoti hai lekin bahut high temperature nahi.
High pressure: reactant side me 4 moles gas hain, product side me 2 moles — pressure badhane se equilibrium us side shift karta hai jahan gaseous moles kam hain, yaani NH₃ (product) ki taraf — yield badhती hai.
Temperature: reaction exothermic hai, isliye temperature badhane se equilibrium reactants ki taraf shift karega (Le Chatelier), yield ghatega. Isliye ek moderate temperature (~700 K) chuna jaata hai jo reasonable yield ke saath reasonable rate bhi de — bahut high temperature yield kam kar degi, bahut low temperature rate bahut slow kar degi.
38. Ksp aur ionic product (Qsp) me kya difference hai? In dono ko compare karke precipitation predict kaise karte hain?
Ksp equilibrium (saturated solution) par valid solubility product hai — ek fixed temperature par constant.
Ionic product (Qsp) kisi bhi given moment par ions ki actual concentrations ka product hai (equilibrium par ho ya na ho).
- Agar Qsp > Ksp → solution supersaturated hai, precipitation hoga jab tak Qsp = Ksp na ho jaaye.
- Agar Qsp = Ksp → solution saturated hai, equilibrium par hai (na precipitation, na further dissolution).
- Agar Qsp < Ksp → solution unsaturated hai, aur salt dissolve ho sakta hai (precipitation nahi hoga).
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| Law of mass action (equilibrium constant, Kc) | Kc = [Products]^coeff / [Reactants]^coeff |
| Kp vs Kc relation | Kp = Kc(RT)^Δn, jahan Δn = moles gaseous products − moles gaseous reactants |
| Reaction quotient rule | Qc > Kc → reverse direction; Qc < Kc → forward direction; Qc = Kc → equilibrium |
| Free energy — equilibrium relation | ΔG° = −RT ln K = −2.303 RT log K |
| Reverse reaction constant | K(reverse) = 1 / K(forward) |
| Combined reactions | Agar reaction 3 = reaction 1 + reaction 2, to K₃ = K₁ × K₂ |
| pH aur pOH | pH = −log[H⁺]; pOH = −log[OH⁻]; pH + pOH = 14 (298 K) |
| Ionic product of water | Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 298 K |
| Ka × Kb relation | Ka × Kb = Kw (conjugate acid-base pair ke liye) |
| Ostwald's dilution law | Ka = Cα²/(1−α); jab α chhota ho: α ≈ √(Ka/C) |
| pKa, pKb | pKa = −log Ka; pKb = −log Kb; pKa + pKb = 14 |
| Henderson-Hasselbalch equation (buffer) | pH = pKa + log([salt]/[acid]) |
| Degree of hydrolysis | h = √(Kh/C), jahan Kh = Kw/Ka (weak acid salt) ya Kw/Kb (weak base salt) |
| Solubility product (AB-type salt) | Ksp = [A⁺][B⁻] = s², s = molar solubility |
| Precipitation condition | Qsp > Ksp → precipitation; Qsp = Ksp → saturated; Qsp < Ksp → unsaturated |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Kc/Kp expression likhte waqt stoichiometric coefficients ko power (exponent) banana bhool jaana — jaise 2SO₂+O₂⇌2SO₃ me Kc = [SO₃]²/([SO₂]²[O₂]) hoga, [SO₃]/([SO₂][O₂]) nahi.
- Pure solids aur pure liquids (jaise CaCO₃(s), H₂O(l) jab solvent ho) ki concentration ko Kc expression me include kar dena — inhe hamesha omit karna hai kyunki inki concentration constant hoti hai.
- Kp aur Kc ko interchange kar dena, ya R ki wrong value use karna — atm-pressure ke saath R=0.0821 L atm mol⁻¹K⁻¹, aur bar-pressure ke saath R=0.0831 L bar mol⁻¹K⁻¹ use karo, mix mat karo.
- Kp = Kc(RT)^Δn me Δn ka sign galat lagana — Δn = (gaseous product moles) − (gaseous reactant moles), reverse nahi.
- pH calculation me [H⁺] = 10^(−pH) conversion ya log table/calculator me sign error kar dena, especially jab concentration 1 se chhoti ho (negative power ka log).
- Le Chatelier's principle ko galat apply karna — catalyst kabhi equilibrium shift nahi karta (sirf rate badhata hai), aur pure solid/liquid ki 'amount' badhane se bhi equilibrium shift nahi hota, sirf gaseous/aqueous species ki concentration change se hota hai.
Board-Style Important Questions
- 1 mark: Equilibrium constant Kc ko define karo. Iski units kis case me hoti hain aur kis case me nahi?
- 2 marks: Reaction A(g) + B(g) ⇌ C(g) + D(g) ke liye equilibrium concentrations [A]=0.2M, [B]=0.2M, [C]=0.6M, [D]=0.6M hain. Kc calculate karo.
- 3 marks: Le Chatelier's principle state karo aur explain karo ki N₂(g)+3H₂(g)⇌2NH₃(g) (exothermic, Δn negative) equilibrium ko pressure badhane aur temperature badhane se kaise alag-alag affect kiya jaata hai.
- 3 marks: 0.05 M acetic acid solution ka degree of dissociation aur pH nikaalo, given Ka = 1.8×10⁻⁵.
- 5 marks: (a) Ostwald's dilution law derive karo weak monobasic acid ke liye. (b) 0.1 M CH₃COONa solution ka pH nikaalo, given Ka(CH₃COOH) = 1.8×10⁻⁵. (c) Buffer solution ka ek example do aur uska working principle briefly explain karo.
Aksar Poochhe Jaane Wale Sawaal
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NCERT ki official site (ncert.nic.in) se Class 11 Chemistry Part II textbook free PDF download hoti hai, jisme Equilibrium chapter included hai. Agar aapko class 11 chemistry ncert solutions pdf chahiye step-by-step working ke saath (jaise is page pe diya gaya hai), to har intext aur exercise question ka detailed solution yahi ek hi jagah reference ke liye use kar sakte ho.
Class 11 Chemistry syllabus 2026-27 rationalised me kitne chapters hain aur Equilibrium kaunse number pe hai?
Rationalised 2026-27 session ke class 11 chemistry syllabus 2026-27 rationalised me total 9 chapters hain (pehle 14 the). Equilibrium ab Chapter 6 hai — Part II ki shuruaat (Chapters 5-9), jabki Part I me Chapters 1-4 hain.
Class 11 Chemistry deleted syllabus topics kaunse hain — kya Equilibrium chapter se kuch hataya gaya?
Nahi, Equilibrium chapter poora same hai. Jo 5 chapters class 11 chemistry deleted syllabus topics me aate hain woh hain: States of Matter (Gases and Liquids), Hydrogen, s-Block Elements, Some p-Block Elements, aur Environmental Chemistry — ye poore chapters hata diye gaye, individual topics nahi.
Equilibrium chapter ka Chapter 5 Thermodynamics se kya connection hai?
Haan, direct connection hai. Class 11 chemistry chapter 5 thermodynamics numericals me Gibbs free energy (ΔG) padha hota hai, aur Equilibrium chapter me formula ΔG° = −RT ln K use hota hai jo equilibrium constant K ko free energy change se relate karta hai — isliye thermodynamics ke concepts clear hone chahiye is chapter se pehle.
Class 11 Chemistry me Equilibrium ke baad kaunsa chapter aata hai aur exam ke liye priority kya rakhein?
Rationalised syllabus me Equilibrium (Chapter 6) ke baad Redox Reactions (Chapter 7) aata hai — agar redox reactions class 11 important questions bhi practice kar rahe ho, to dono chapters ka overlap (oxidation states, electron transfer) samajhna helpful hota hai. Chapter 1 (Some basic concepts of chemistry notes), Chapter 2 (Structure of Atom) aur Chapter 4 (Chemical bonding) ki foundation clear honi chahiye pehle.
Organic chemistry (Hydrocarbons, Basic Principles) shuru karne se pehle Equilibrium zaroori hai kya?
Directly zaroori nahi, lekin helpful hai — class 11 chemistry organic chemistry basic principles and techniques notes aur hydrocarbons class 11 ncert solutions me kai jagah reaction equilibrium/reversibility ka concept implicitly use hota hai (jaise esterification). Best sequence yahi hai ki inorganic/physical chapters (1-7) pehle complete karke phir organic chapters (8-9) pe jaayein.
Class 11 Chemistry — Saare Chapters
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Class 11 Chemistry ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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