Class 11 Chemistry · Chapter 5
Short answer:
Class 11 Chemistry Chapter 5 Chemical Thermodynamics — naye rationalised syllabus (2026-27 session) ke 9-chapter structure me energy, heat aur spontaneity ka chapter hai. Isme system-surroundings, first law (ΔU = q + w), enthalpy (ΔH), Hess's law, bond enthalpy, entropy (ΔS) aur Gibbs free energy (ΔG = ΔH − TΔS) cover hote hain. Agar aap class 11 chemistry chapter 5 thermodynamics numericals practice kar rahe ho, to sign convention (w = −pextΔV) aur ΔH = ΔU + ΔngRT — ye do formulas sabse zyada numericals decide karte hain.
Physics me thermodynamics energy ke transformations padhata hai — Chemistry me wahi laws lagakar hum samajhte hain ki koi reaction khud-ba-khud (spontaneously) hogi ya nahi, aur agar hogi to kitni heat release/absorb hogi. Yehi is chapter ka core idea hai.
Rationalised NCERT (2026-27 session) me Class 11 Chemistry ab sirf 9 chapters ki hai, do parts me split. Part I me Some Basic Concepts of Chemistry (jinke "some basic concepts of chemistry class 11 notes" alag se milte hain), Structure of Atom ("structure of atom class 11 formulas" aur "class 11 chemistry chapter 2 structure of atom important questions" jahan cover hote hain), Classification of Elements and Periodicity in Properties ("classification of elements and periodicity in properties important questions"), aur Chemical Bonding and Molecular Structure ("chemical bonding and molecular structure class 11 notes pdf" wale topics) aate hain. Part II chapter 5 se shuru hoti hai — yahi Chemical Thermodynamics hai, uske baad Equilibrium (chapter 6 — "class 11 chemistry chapter 6 equilibrium ncert solutions" is chapter ke concepts pe hi tikke hain), Redox Reactions (chapter 7 — "redox reactions class 11 important questions"), Organic Chemistry — Some Basic Principles and Techniques (chapter 8 — "class 11 chemistry organic chemistry basic principles and techniques notes"), aur Hydrocarbons (chapter 9 — "hydrocarbons class 11 ncert solutions"). Poori "class 11 chemistry all chapters list ncert" yahi 9 chapters hai.
Agar aap "class 11 chemistry syllabus 2026-27 rationalised" search kar rahe ho, to note karo — States of Matter: Gases and Liquids, Hydrogen, s-Block Elements, Some p-Block Elements, aur Environmental Chemistry — ye 5 poore units "class 11 chemistry deleted syllabus topics" me aate hain, ab is session me nahi padhaye jaate. Yani agar kahin purane 14-chapter wale notes ya "class 11 chemistry ncert solutions pdf" mile jisme States of Matter ya s-Block ho, wo outdated hai — sirf 9 chapter wale rationalised set pe hi bharosa karo.
Chapter 5 khud ek foundation chapter hai — Chapter 6 Equilibrium (Chemical aur Ionic Equilibrium) yahi ΔG aur spontaneity ke concepts aage extend karta hai, isliye Thermodynamics ko theek se samajhna Equilibrium ke liye bhi zaroori hai.
Chapter 5 Summary — 5 Minute Revision
Chapter 5 Chemical Thermodynamics system aur surroundings ki definition se shuru hota hai (open, closed, isolated system), phir state functions vs path functions ka farak samjhata hai. First Law of Thermodynamics (ΔU = q + w) energy conservation ka core statement hai — isse work (w = −pextΔV expansion ke liye, ya reversible isothermal case me w = −nRT ln(V2/V1)) aur heat calculate karna aata hai.
Enthalpy (H) constant pressure pe heat change measure karti hai — ΔH = ΔU + ΔngRT. Chapter alag-alag enthalpy types cover karta hai: enthalpy of reaction, formation, combustion, atomization, sublimation, ionization, solution aur dilution. Hess's Law batata hai ki total enthalpy change path-independent hai — jisse indirect (multi-step) reactions ka ΔH nikala ja sakta hai. Bond enthalpy se bhi ΔH estimate hota hai — bonds break karne me energy lagti hai, bonds banane me energy release hoti hai.
Second half me spontaneity ka concept hai — spontaneous process wo hai jo bina continuous external help ke khud ho sakta hai. Sirf ΔH negative hona spontaneity guarantee nahi karta, isliye entropy (S) introduce hoti hai — disorder ka measure. Second law kehta hai ΔSuniverse > 0 spontaneous process ke liye. Dono (ΔH aur ΔS) ko combine karke Gibbs free energy ΔG = ΔH − TΔS milta hai — ΔG < 0 spontaneous, ΔG > 0 non-spontaneous, ΔG = 0 equilibrium. Chapter ΔG° = −RT ln K se equilibrium constant K ka thermodynamic connection bhi dikhata hai, jo Chapter 6 Equilibrium ki neev banta hai.
In-Text Questions — Solutions
State function kya hota hai? Teen examples dijiye.
State function wo property hai jiski value sirf system ki current state (initial aur final) pe depend karti hai, path pe nahi — chahe reaction kisi bhi tareeke se ho, value same rahegi.
Examples: Internal energy (U), Enthalpy (H), Entropy (S), pressure (P), volume (V), temperature (T).
Intensive aur extensive properties me farak batao, examples ke saath.
Extensive property system ki quantity (matter ki amount) pe depend karti hai — jaise mass, volume, internal energy, enthalpy, entropy, heat capacity.
Intensive property matter ki amount pe depend nahi karti — jaise temperature, pressure, density, molar volume, refractive index.
Internal energy kya hai? Kya ye state function hai?
Internal energy (U) system ke andar mojood total energy hai — kinetic energy (translational, rotational, vibrational) + potential energy (bonds, intermolecular forces) ka sum. Ye ek state function hai kyunki iski value system ki state pe depend karti hai, us tak pahunchne ke path pe nahi.
Ek gas 2 L se 5 L tak constant external pressure 2 atm ke against expand hota hai. Kiya gaya work calculate karo (1 L atm = 101.3 J).
Formula: w = −pextΔV
ΔV = V₂ − V₁ = 5 − 2 = 3 L
w = −2 × 3 = −6 L atm
w = −6 × 101.3 = −607.8 J
Negative sign dikhata hai ki gas ne surroundings pe work kiya (energy system se bahar gayi).
q aur w path functions kyun kehlate hain?
Kyunki heat (q) aur work (w) ki value sirf initial aur final state pe depend nahi karti — ye is baat pe depend karti hai ki process reversible tarike se hua ya irreversible tarike se, kitne steps me hua. Sirf inka sum, ΔU = q + w, state function hai — q aur w individually nahi.
Enthalpy (H) ki definition do. Chemistry me ΔU ki jagah ΔH zyada use kyun hota hai?
Enthalpy H = U + pV, ek state function hai jo constant pressure pe heat change measure karti hai (ΔH = qp).
Zyada use hota hai kyunki lab reactions aksar open beakers/flasks me constant atmospheric pressure pe hoti hain, constant volume pe nahi — isliye ΔH directly measurable aur practically relevant hai.
First law of thermodynamics state karo.
Energy na to create ho sakti hai na destroy — sirf ek form se doosri form me convert hoti hai. Mathematically: ΔU = q + w, jahan q system ko di gayi heat hai aur w system pe kiya gaya work hai.
Standard enthalpy of reaction kya hoti hai? Standard state ki conditions kya hain?
Standard enthalpy of reaction wo enthalpy change hai jab reactants standard state me react karke products standard state me banate hain.
Standard state conditions: 1 bar pressure aur specified temperature (usually 298 K), pure substance ki most stable physical form me.
Spontaneous process kya hota hai? Ek spontaneous aur ek non-spontaneous process ka example do.
Spontaneous process wo hai jo bina kisi continuous external energy input ke apne aap ek particular direction me hota hai (thermodynamically favourable), chahe wo fast ho ya slow.
Spontaneous example: garam cheez ka thanda hona (heat flow hot se cold ki taraf).
Non-spontaneous example: cold body se hot body me heat flow apne aap nahi hota, external work chahiye (refrigerator jaisa).
Gibbs free energy change (ΔG) ke through spontaneity ki criteria batao.
ΔG = ΔH − TΔS
- ΔG < 0 → process spontaneous hai
- ΔG > 0 → process non-spontaneous hai
- ΔG = 0 → system equilibrium me hai
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–Q18)
Q1. State function aur path function me farak batao. In quantities ko classify karo: internal energy (U), heat (q), work (w), enthalpy (H).
State function: Jiski value sirf system ke initial aur final state par depend karti hai, path par nahi — jaise U (internal energy) aur H (enthalpy).
Path function: Jiski value process kis tarike se hua us par depend karti hai — jaise q (heat) aur w (work); inka sum (ΔU = q + w) hi state function hota hai, individually nahi.
Q2. First law of thermodynamics ka mathematical statement likho, aur IUPAC sign convention (kab positive, kab negative) explain karo.
Mathematical form:
ΔU = q + w
Sign convention (IUPAC):
- Agar system heat absorb kare — q positive.
- Agar system heat release kare — q negative.
- Agar system par work kiya jaaye (compression) — w positive.
- Agar system work kare surroundings par (expansion) — w negative.
Q3. Ek gas adiabatic condition me (q = 0) 5 L se 12 L tak constant external pressure 1 atm ke against expand hota hai. w aur ΔU calculate karo. (1 L atm = 101.3 J)
ΔV = 12 − 5 = 7 L
w = −pextΔV = −1 × 7 = −7 L atm = −7 × 101.3 = −709.1 J
Adiabatic hone ki wajah se q = 0, isliye:
ΔU = q + w = 0 + (−709.1) = −709.1 J
Gas ne surroundings par work kiya, isliye internal energy kam ho gayi.
Q4. 2 mol ideal gas 300 K par reversible isothermal expansion me 10 L se 20 L tak jaata hai. Kiya gaya work calculate karo (R = 8.314 J K-1 mol-1), aur ΔU, q bhi batao.
w = −nRT ln(V₂/V₁) = −2 × 8.314 × 300 × ln(20/10)
w = −4988.4 × 0.693 = −3457.5 J
Isothermal process me ideal gas ke liye ΔU = 0 (temperature constant), isliye:
q = ΔU − w = 0 − (−3457.5) = +3457.5 J
Gas ne surroundings par work kiya (w negative) aur utni hi heat surroundings se absorb ki (q positive), isliye ΔU zero raha.
Q5. Hess's Law use karke C(s) + ½O₂(g) → CO(g) ka ΔH nikaalo, diye gaye data se:
C(s) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ/mol
CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ/mol
C(s) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ/mol
CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ/mol
Target equation = Equation 1 − Equation 2 (taaki extra CO₂ aur ½O₂ cancel ho jaayein aur CO product side pe aaye).
ΔH₃ = ΔH₁ − ΔH₂ = (−393.5) − (−283.0) = −110.5 kJ/mol
Isliye C(s) + ½O₂(g) → CO(g) ka ΔH = −110.5 kJ/mol. Ye Hess's Law ka direct application hai — total enthalpy change path-independent hoti hai, isliye do known steps ko algebraically combine karke unknown step nikala ja sakta hai.
Q6. Bond enthalpies H−H = 436 kJ/mol, Cl−Cl = 242 kJ/mol, H−Cl = 431 kJ/mol diye gaye hain. H₂(g) + Cl₂(g) → 2HCl(g) ka ΔH bond enthalpy method se calculate karo.
ΔH = Σ BE(reactants) − Σ BE(products)
ΔH = (436 + 242) − 2(431) = 678 − 862 = −184 kJ/mol
Negative ΔH dikhata hai ki reaction exothermic hai — naye H−Cl bonds banne me jitni energy release hoti hai, wo H−H aur Cl−Cl bonds todne me lagi energy se zyada hai.
Q7. Methane ke combustion, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ka ΔU = −885.0 kJ/mol hai 298 K par. Iska ΔH calculate karo. (R = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹)
Δng = (gaseous moles products) − (gaseous moles reactants) = 1 − (1 + 2) = −2 (H₂O liquid hai, gas nahi count hota)
ΔH = ΔU + ΔngRT
ΔH = −885.0 + (−2)(8.314 × 10⁻³)(298)
ΔH = −885.0 − 4.96 = −889.96 kJ/mol ≈ −890.0 kJ/mol
Q8. Standard enthalpies of formation: C₃H₈(g) = −103.85 kJ/mol, CO₂(g) = −393.5 kJ/mol, H₂O(l) = −285.8 kJ/mol. Propane, C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), ke combustion ka ΔH°rxn calculate karo.
ΔH°rxn = Σ ΔHf°(products) − Σ ΔHf°(reactants)
= [3(−393.5) + 4(−285.8)] − [(−103.85) + 5(0)]
= [−1180.5 − 1143.2] − [−103.85] = −2323.7 + 103.85 = −2219.85 kJ/mol
O₂(g) ek element apni standard state me hai, isliye iska ΔHf° zero maana jaata hai.
Q9. Entropy (S) kya hai? 2 mol ideal gas 4 L se 8 L tak isothermal reversible expansion karta hai — ΔS calculate karo. (R = 8.314 J K⁻¹ mol⁻¹)
Entropy system ke disorder/randomness ka measure hai — ek state function.
ΔS = nR ln(V₂/V₁) = 2 × 8.314 × ln(8/4)
ΔS = 16.628 × 0.693 = 11.52 J K⁻¹
Positive ΔS dikhata hai ki expansion ke baad gas ki randomness/disorder badh gaya.
Q10. Ice ka latent heat of fusion 6.01 kJ/mol hai, 273 K par. Melting process ka ΔS calculate karo.
Melting reversible aur constant temperature (273 K, ice ka melting point) par hota hai, isliye:
ΔS = qrev/T = 6010 J mol⁻¹ ÷ 273 K = 22.01 J K⁻¹ mol⁻¹
Solid se liquid banne par disorder badhta hai, isliye ΔS positive hai — expected hi hai.
Q11. Ek reaction ka ΔH = +40 kJ/mol aur ΔS = +100 J K⁻¹mol⁻¹ hai. 300 K par ΔG calculate karo aur bataao spontaneous hai ya nahi. Wo temperature bhi nikaalo jispar reaction spontaneous ban jaata hai.
ΔG = ΔH − TΔS = 40000 − (300 × 100) = 40000 − 30000 = +10000 J = +10 kJ/mol
ΔG positive hai, isliye 300 K par reaction non-spontaneous hai.
ΔG = 0 hone wali temperature nikaalne ke liye:
T = ΔH/ΔS = 40000/100 = 400 K
Isliye 400 K se upar temperature par ye reaction spontaneous ho jaayega (kyunki ΔS positive hai, TΔS term ΔH se bada ho jaayega).
Q12. Ek reaction ka equilibrium constant K = 10 hai 298 K par. ΔG° calculate karo. (R = 8.314 J K⁻¹mol⁻¹)
ΔG° = −RT ln K = −8.314 × 298 × ln(10)
ΔG° = −2477.6 × 2.303 = −5705.8 J/mol ≈ −5.71 kJ/mol
Negative ΔG° confirm karta hai ki K > 1 wale reactions thermodynamically favourable (products-favoured) hote hain.
Q13. Ice ka melting endothermic (ΔH positive) hai, phir bhi 0°C se upar temperature par ye spontaneously hota hai. Explain karo Gibbs free energy ke through.
Ice melting: H₂O(s) → H₂O(l), ΔH = +6.01 kJ/mol (positive, heat absorb hoti hai). Lekin solid se liquid banne me disorder badhta hai, isliye ΔS bhi positive hai.
ΔG = ΔH − TΔS
Jab temperature high ho (jaise 0°C se upar), TΔS term ΔH se bada ho jaata hai, isliye ΔG negative ho jaata hai aur process spontaneous ban jaata hai — chahe ΔH khud positive ho. Isliye sirf ΔH dekhkar spontaneity decide nahi ki ja sakti, ΔG hi final criterion hai.
Q14. Extensive aur intensive properties me farak batao. In quantities ko classify karo: heat capacity, molar volume, mass, density.
Extensive property: System ki quantity (amount of matter) par depend karti hai — heat capacity aur mass extensive hain.
Intensive property: Matter ki amount par depend nahi karti — molar volume aur density intensive hain.
Q15. Hess's Law state karo aur explain karo ki ye kaam kyun karta hai (state function ke concept se).
Hess's Law: Kisi bhi reaction ka total enthalpy change us path pe depend nahi karta jispe reaction hoti hai — chahe reaction ek hi step me ho ya kai steps me, ΔHoverall hamesha same rahega, basharte initial aur final states same hon.
Kyun kaam karta hai: Enthalpy (H) ek state function hai — iski value sirf system ki initial aur final state par depend karti hai. Isliye ek reaction ko chahe kitne bhi intermediate steps me todo, saare steps ke ΔH ka sum hamesha overall reaction ke direct ΔH ke barabar hoga.
Q16. 100 mL 1 M HCl ko 100 mL 1 M NaOH ke saath neutralize karne par 5.73 kJ heat release hoti hai. Enthalpy of neutralization (per mole) calculate karo.
Moles HCl = 0.1 L × 1 mol/L = 0.1 mol (limiting/stoichiometric, NaOH bhi 0.1 mol)
ΔH = −5.73 kJ ÷ 0.1 mol = −57.3 kJ/mol
Ye standard strong acid–strong base neutralization enthalpy ke kareeb hai (lagbhag −57 kJ/mol), jo H⁺(aq) + OH⁻(aq) → H₂O(l) reaction ke liye characteristic hai.
Q17. Exothermic aur endothermic reactions me sign convention ke saath farak batao, do-do examples ke saath.
Exothermic: Reaction jisme heat release hoti hai, ΔH negative hota hai. Examples: combustion (jalaana), neutralization reactions.
Endothermic: Reaction jisme heat absorb hoti hai, ΔH positive hota hai. Examples: ice ka melting, photosynthesis.
Q18. 1 mol ideal gas 298 K par 10 L se 5 L tak compress hota hai. (a) Reversible isothermal compression me kiya gaya work nikaalo. (b) Single-step irreversible compression me, jahan external pressure final pressure ke barabar constant rakha jaaye, work nikaalo. (R = 0.0821 L atm K⁻¹mol⁻¹; 1 L atm = 101.3 J)
(a) Reversible:
wrev = −nRT ln(V₂/V₁) = −1 × 0.0821 × 298 × ln(5/10)
wrev = −24.47 × (−0.693) = +16.96 L atm = 16.96 × 101.3 = +1718.1 J
(b) Irreversible (constant Pext = final pressure):
P₂ (final pressure) = nRT/V₂ = (1 × 0.0821 × 298)/5 = 4.89 atm
wirrev = −Pext(V₂ − V₁) = −4.89 × (5 − 10) = +24.47 L atm = 24.47 × 101.3 = +2478.8 J
Conclusion: Irreversible compression (2478.8 J) me reversible compression (1718.1 J) se zyada work karna padta hai gas ko compress karne ke liye — reversible process hamesha compression ke liye minimum work maangta hai (aur expansion ke liye maximum work deta hai).
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| First Law of Thermodynamics | ΔU = q + w |
| Work (constant external pressure) | w = −pextΔV |
| Work (reversible isothermal, ideal gas) | w = −nRT ln(V2/V1) |
| Enthalpy definition | H = U + pV |
| Enthalpy vs internal energy change | ΔH = ΔU + ΔngRT |
| Heat capacity relation (ideal gas) | Cp − Cv = R |
| Constant volume heat | qv = ΔU = nCvΔT |
| Constant pressure heat | qp = ΔH = nCpΔT |
| Hess's Law | ΔHreaction = Σ ΔH (individual steps), path-independent |
| Enthalpy from formation data | ΔH°reaction = Σ ΔHf°(products) − Σ ΔHf°(reactants) |
| Enthalpy from bond enthalpies | ΔH = Σ BE(reactants) − Σ BE(products) |
| Entropy change (reversible process) | ΔS = qrev/T |
| Entropy change (isothermal ideal gas expansion) | ΔS = nR ln(V2/V1) |
| Second Law spontaneity criteria | ΔSuniverse = ΔSsystem + ΔSsurroundings > 0 |
| Gibbs free energy | ΔG = ΔH − TΔS |
| Spontaneity via ΔG | ΔG < 0 spontaneous · ΔG > 0 non-spontaneous · ΔG = 0 equilibrium |
| ΔG° and equilibrium constant | ΔG° = −RT ln K |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- w = −pextΔV formula me sign galat lagana — expansion (ΔV positive) me work negative hota hai (system surroundings par kaam karta hai), compression me positive; is sign ko ulta likhna common galti hai.
- ΔH aur ΔU ko same maan lena — jab reaction me gases involved hon aur Δng ≠ 0, tab ΔH = ΔU + ΔngRT term important ho jaata hai, sirf constant-volume reactions me hi ΔH ≈ ΔU hota hai.
- q aur w ko state function samajh lena — inki value path pe depend karti hai; sirf inka sum ΔU (= q + w) state function hai, q ya w individually nahi.
- Hess's Law apply karte waqt equation reverse karne par ΔH ka sign badalna bhool jaana — agar koi given equation reverse karni pade (products ko reactants bana ke), to uska ΔH bhi sign badal jaata hai.
- Sirf ΔH negative dekhkar spontaneity decide kar dena — spontaneity ka sahi criterion ΔG hai, na ki sirf ΔH; kai endothermic processes (jaise ice ka melting) bhi spontaneous hote hain jab TΔS, ΔH se bada ho jaaye.
- Bond enthalpy formula ulta laga dena — sahi formula hai ΔH = Σ BE(reactants) − Σ BE(products), isse ulta likhne par answer ka sign hi galat aa jaata hai.
Board-Style Important Questions
- 1 mark: First law of thermodynamics ka mathematical form likhiye.
- 2 marks: State function aur path function me antar spasht kijiye, ek-ek example ke saath.
- 3 marks: Hess's Law state kijiye aur ek numerical example se samjhaiye.
- 3 marks: Gibbs free energy (ΔG) ke through spontaneity ke teeno cases likhiye aur samjhaiye.
- 5 marks: Bond enthalpy method se kisi gas-phase reaction ka ΔH calculate karne ka tareeka numerical ke saath samjhaiye.
Aksar Poochhe Jaane Wale Sawaal
Chemical Thermodynamics chapter me sabse zyada numericals kis topic se aate hain?
Sabse zyada numericals work calculation (w = −pextΔV ya reversible isothermal w = −nRT ln(V₂/V₁)), ΔH aur ΔU ka conversion (ΔH = ΔU + ΔngRT), Hess's Law based multi-step problems, aur Gibbs free energy (ΔG = ΔH − TΔS, spontaneity check) se aate hain — inhi topics ki practice sabse zyada zaroori hai.
Class 11 Chemistry me is chapter ka number kya hai — Chapter 5 ya Chapter 6?
2026-27 ke rationalised (9-chapter) syllabus me ye Chapter 5 hai. Purani (pre-rationalisation, 14-chapter) numbering me thermodynamics Chapter 6 hota tha — kyunki States of Matter jaisa chapter beech me hota tha jo ab hata diya gaya hai. Agar kahin 'chapter 6 thermodynamics' likha mile to woh purani numbering hai.
ΔH negative hone ka matlab reaction spontaneous hai kya?
Nahi, zaroori nahi. Spontaneity ka sahi criterion ΔG hai (ΔG = ΔH − TΔS), sirf ΔH nahi. Bahut saare exothermic (ΔH negative) reactions spontaneous hote hain, lekin kuch endothermic (ΔH positive) reactions bhi spontaneous ho sakte hain agar ΔS itna positive ho ki TΔS term ΔH se bada ho jaaye — jaise ice ka melting 0°C se upar.
Hess's Law numericals solve karne ka sabse aasaan tareeka kya hai?
Sabse pehle target equation likho, phir diye gaye equations ko dekho ki unhe kaise add/subtract/multiply karna hai taaki extra species cancel ho jaayein aur target equation ban jaaye. Jo bhi algebraic operation equations par karo, wahi operation unke ΔH values par bhi karo (reverse karne par sign badlo, multiply karne par ΔH bhi multiply karo).
Class 11 Chemistry Chapter 5 Thermodynamics ke important formulas kaunse hain jo yaad rakhne chahiye?
Sabse important: ΔU = q + w (first law), w = −pextΔV, ΔH = ΔU + ΔngRT, ΔS = qrev/T, aur ΔG = ΔH − TΔS. Upar diye 'formulas' table me poori list step-by-step examples ke saath di gayi hai.
Ye chapter Chapter 6 Equilibrium se kaise connected hai?
ΔG° = −RT ln K formula hi Thermodynamics aur Equilibrium ke beech ka direct pul hai — is formula se pata chalta hai ki kisi reaction ka Gibbs free energy uske equilibrium constant K se seedha judaa hota hai. Isliye Chapter 6 Equilibrium shuru karne se pehle Chapter 5 ke ΔG aur spontaneity concepts clear hona zaroori hai.
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