Thermodynamics: Class 11 Chemistry Chapter 5 Hinglish mein

Chemistry Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

130 min
1 simulation
28 practice questions
30 MCQs

Is chapter mein aap kya seekhenge

System, surroundings aur boundary mein farq karo, aur systems ko open, closed ya isolated mein classify karo
State functions samjhao aur first law, ΔU = q + w, ko IUPAC sign conventions ke saath lagao
Ideal gas ke free, constant-pressure aur reversible isothermal expansions ke liye pressure–volume work calculate karo
Enthalpy define karo, Δn_gRT ke through ΔH ko ΔU se jodo, aur heat capacities use karo, Cp − Cv = R samet
Bomb aur constant-pressure calorimetry samjhao aur calorimetric data se ΔU aur ΔH calculate karo
Formation enthalpies, Hess's law, bond enthalpies aur Born–Haber cycles se reaction enthalpies calculate karo
Entropy aur second aur third laws se spontaneity samjhao, aur ΔG = ΔH − TΔS se use predict karo
Standard Gibbs energy change ko equilibrium constant se jodo aur ek se doosra calculate karo

NCERT Chemistry Class 11, Chapter 5, pp. 136–167 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Pressure cooker ka weight uthta hai aur seeti bajti hai, maachis ki teeli bhadak uthti hai, ek ice cube nimbu-paani ke glass mein gaayab ho jaata hai. Har ek aisa change hai jo apne aas-paas ke saath energy ka len-den karta hai, heat ke roop mein, work ke roop mein, ya dono. Thermodynamics (ऊष्मागतिकी) inhi len-den ka hisaab-kitaab hai.

Thermodynamics ko powerful woh cheez banati hai jise yeh ignore karti hai. Yeh kabhi individual molecules ko follow nahi karti, sirf un quantities ko jo lab mein lagbhag $10^{23}$ molecules ke poore sample par measure ho sakti hain, jaise pressure, volume aur temperature. Yeh kabhi nahi poochti ki change kitni tezi se hota hai ya kis mechanism se; yeh sirf compare karti hai ki change kahan se shuru hota hai aur kahan khatam. Iski keemat yeh hai ki yeh sirf equilibrium wale samples describe karti hai, jinki bulk properties badalna band ho chuki hain, aur aise states ke beech ke changes.

5.1 Thermodynamic Terms

Koi bhi calculation karne se pehle humein ek chhoti, precise vocabulary chahiye. Inmein se zyada tar roz ke words hain jinhe ek tez, saaf matlab de diya gaya hai.

5.1.1 System aur Surroundings

Test tube mein dilute hydrochloric acid ke andar zinc ka ek tukda daalo. Agar tum yahi reaction study karna chahte ho, to zinc, acid aur products milkar system (निकाय) banate hain: universe ka woh hissa jise hum dekhte aur measure karte hain. Baaki sab kuch surroundings (परिवेश) hai. Dono milkar universe banate hain.

\begin{array}{c} universe = system \\ + surroundings \end{array}

Practice mein sirf paas ki surroundings matter karti hain: tube ka glass, hawa aur bench, jo sach mein reaction ke saath energy ya matter exchange kar sakte hain. Door ki koi galaxy technically surroundings ka hissa hai, lekin hamara koi bhi measurement use kabhi notice nahi karega.

System kahan khatam hota hai, yeh tumhari choice hai, aur baantne wali surface ko boundary kehte hain. Yeh kisi asli wall ke saath chal sakti hai, jaise tube ka glass, ya ek imaginary surface ho sakti hai, maan lo sirf react kar rahe liquid ke chaaron taraf kheenchi gayi. Jo bhi isse paar hota hai use surroundings ke saath exchange hua heat, work ya matter gina jaata hai, isliye laaparwahi se chuni gayi boundary hisaab gadbad kar deti hai.

Definition

System

Universe ka woh hissa jo study ke liye chuna gaya hai; universe ka baaki hissa jo iske saath interact kar sakta hai, surroundings banata hai.

ExampleReaction flask ke andar ki cheezein.

5.1.2 System ke Types

Hum systems ko is hisaab se sort karte hain ki unki boundary ke paar kya ja sakta hai (Fig. 5.1).

Ek open system apni surroundings ke saath energy aur matter dono exchange karta hai. Stove par ubalta paani ka pan flame se heat leta hai aur steam ke roop mein paani khota hai.
Ek closed system energy exchange karta hai lekin matter nahi. Water bath mein garam kiya gaya food ka sealed can heat leta hai, lekin can mein na kuch andar jaata hai na bahar aata hai.
Ek isolated system dono mein se kuch bhi exchange nahi karta. Achhi tarah band kiya hua vacuum flask iske kareeb hai: kuch ghanton tak iski chai na leak hoti hai na zyada thandi hoti hai.

4/4Isolated: na energy paar jaati hai na matter.

Fig. 5.1Boundary ke paar kya ja sakta hai, yahi system ka type decide karta hai.NCERT Fig. 5.2 · p. 137

Flame par paani ka pan, ek sealed can aur ek vacuum flask saath-saath rakhe hain. Step one: pan ke chaaron taraf ek dashed boundary system ko surroundings se alag karti hai. Step two: ek dashed arrow pan se nikalti steam (matter) dikhata hai aur ek solid arrow energy ka paar jaana dikhata hai; ispar open likha hai. Step three: can ko ek energy arrow milta hai aur uske lid par matter ke liye cross; ispar closed likha hai. Step four: flask ko dono ke liye crosses milte hain aur ispar isolated likha hai.

Asli walls hamesha thoda leak karti hain, isliye 'isolated' ek approximation hai jo kuch samay tak hi sahi rehta hai. Is chapter mein hum jin reactions ko analyse karte hain, unmein se zyada tar ko closed systems maana jaata hai.

5.1.3 System ki State

Kuch bhi calculate karne ke liye humein theek-theek batana hoga ki change se pehle aur baad mein system kaisa hai. Mechanics mein tumne body ko uske har part ki position aur velocity se describe kiya tha, jo $10^{23}$ molecules ke liye namumkin hai. Thermodynamics kuch bulk properties se hi kaam chala leti hai jinhe thermometer, pressure gauge ya measuring cylinder padh sakta hai: pressure $p$ , volume $V$ , temperature $T$ , amount $n$ aur composition.

Inhe state variables kehte hain. Inki values system ki abhi ki condition batati hain aur uski history ke baare mein kuch nahi kehti. Yeh sab independent nahi hain. Ideal gas ki fixed amount ke liye, jaise hi tum $p$ aur $T$ chunte ho, volume $p V = n R T$ se tay ho jaata hai. Kaafi independent variables (pure gas ki fixed amount ke liye do) ki values chuno aur baaki sab apne aap aa jaate hain.

Aisi quantity jiski value sirf abhi ki state par depend karti hai, is par nahi ki woh state kaise pahunchi gayi, state function (अवस्था फलन) kehlaati hai. Iska change sirf initial aur final states par depend karta hai. Ek pahaadi is idea ko saaf kar deti hai (Fig. 5.2): do points ke beech tum jitni height chadhte ho woh fixed hai, chahe wahan tak pahunchne ke liye tum kitna bhi chalo.

4/4Dono routes theek +600 m chadhte hain. Height $U$ ki tarah behave karti hai; chali gayi distance $q$ ya $w$ ki tarah.

Fig. 5.2Chadhi gayi height state function hai; chali gayi distance path par depend karti hai.

Ek pahaadi jismein point A neeche ke paas 2000 m par aur point B chothi par, 2600 m par hai. Pehle step mein sirf do points aur unki heights dikhti hain. Doosre step mein 1.2 km ka seedha, khada route 1 unhe jodta hai. Teesre step mein 3.5 km ka dashed zig-zag route 2 judta hai. Chauthe step mein right side par ek double arrow height gain, +600 m, dikhata hai, jo dono routes ke liye same hai, jabki chali gayi distances alag hain.

Definition

State function

Aisi property jiski value sirf system ki abhi ki state par depend karti hai, isliye iska change sirf initial aur final states par depend karta hai.

Example $p$ , $V$ , $T$ , $U$ aur $H$ state functions hain; $q$ aur $w$ nahi hain.

Temperature ek jaana-pehchaana example hai. Doodh ke bartan ko seedhe 20 °C se 60 °C tak garam karo, ya pehle use 15 °C tak thanda hone do aur phir garam karo: dono tarah $Δ T = + 40$ °C. Mobile wallet bhi aise hi kaam karta hai. Agar uska balance ₹200 se ₹500 ho jaata hai, to usne ₹300 kamaaye, beech mein chahe kitne bhi top-ups aur payments hue hon. Balance state function jaisa hai; transactions ki list path jaisi hai.

5.1.4 State Function ke roop mein Internal Energy

Ek system kai roopon mein energy store karta hai: idhar-udhar udte aur ghoomte molecules ki kinetic energy, vibrate karte bonds ki energy, electrons aur nuclei ki chemical energy. Hum in sabko milakar internal energy $U$ (आंतरिक ऊर्जा) kehte hain. Yeh sirf teen tarah se badal sakti hai: heat andar ya bahar flow ho, system par ya system dwara work ho, ya matter boundary ke paar energy le jaaye. Matter sirf open systems mein paar jaata hai, isliye closed systems ke liye heat aur work kaafi hain.

Work

Jo wall heat ko poori tarah rok de use adiabatic (रुद्धोष्म) kehte hain; achhe vacuum flask ki double wall iske kareeb hai. Aisi walls mein band system mein koi bhi change adiabatic process hai. Jab heat paar hi nahi ja sakti, to system ki energy badalne ka sirf ek tareeka bachta hai: work.

Ab 25.0 °C par 200 g paani ki state do alag tareekon se badlo (Fig. 5.3). Pehle, use lakdi ke churner se mathon, jaise chhaachh ke liye mathani use hoti hai, jab tak 2.50 kJ mechanical work andar na chala jaaye. Doosre, phir se same state se shuru karke, usmein 50 W heating coil 50 s tak chalao, jo 2.50 kJ electrical work hai. Dono baar paani 28.0 °C par khatam hota hai. Same adiabatic work, kisi bhi tarah kiya jaaye, same change of state paida karta hai.

4/4Conducting walls aur koi work nahi: 2.50 kJ heat wahi 28.0 °C deti hai, isliye $Δ U = q .$

Fig. 5.3Barabar energy andar, kisi bhi route se, same final state deti hai.NCERT Fig. 5.3 · p. 138

25.0 °C par 200 g paani ka ek container, thermometer ke saath. Pehle step mein abhi kuch nahi daala gaya. Doosre step mein, insulated walls ke saath, ek churner 2.50 kJ work karta hai aur q zero hai; thermometer 25.0 se 28.0 °C tak badhta dikhata hai. Teesre step mein, phir se shuru karke, ek electric heater 2.50 kJ work karta hai aur rise same hai. Chauthe step mein walls conducting hain, koi work nahi hota, aur ek garam bath se 2.50 kJ heat andar aati hai, jo arrows se dikhayi gayi hai; rise phir se 25.0 se 28.0 °C hai.

Yeh experimental fact humein $U$ define karne deta hai. Agar barabar adiabatic work hamesha same change paida karta hai, to work zaroor ek aisi property ka change measure kar raha hai jo khud state ki hai:

Δ U = U_{2} - U_{1} = w_{ad}

Isliye $U$ ek state function hai. Hum IUPAC sign convention follow karte hain: system par kiya gaya work positive gina jaata hai, kyunki yeh system ki energy badhata hai, aur system jo work apni surroundings par karta hai woh negative gina jaata hai.

Heat

Starting state par wapas jao aur teesre tareeke se 28.0 °C tak pahuncho. Insulated walls ki jagah patli steel walls lagao, jo heat ko paar jaane deti hain, aur container ko garam paani ke bath mein rakho. Temperature difference ki wajah se energy andar flow hoti hai. Is tarah transfer hui energy heat, $q$ , hai. Koi work na hone aur volume na badalne par, yeh sab internal energy badhati hai, isliye $Δ U = q .$

Isi convention se, $q$ positive hota hai jab heat system ke andar flow kare aur negative jab system apni surroundings ko heat khoye.

General case

Aam taur par heat aur work saath kaam karte hain, aur internal energy ka change unka sum hota hai:

First law of thermodynamics

Δ U = q + w

(5.1)

Ab yeh baareek baat dekho. Ek system ko same do states ke beech do alag routes se le jao, aur heat aur work ka batwaara kaafi alag aa sakta hai, bilkul jaise pahaadi par chali gayi distance alag thi; phir bhi total hamesha match karta hai. Aur agar boundary ke paar kuch bhi nahi jaata, jaise isolated system mein, to $q = w = 0$ aur $U$ badal nahi sakti.

LawFirst law of thermodynamics

Isolated system ki total energy kabhi nahi badalti. Energy apna form badal sakti hai ya system aur surroundings ke beech move kar sakti hai, lekin yeh kabhi na banayi jaati hai na khatam hoti hai.

5/5500 J heat andar aur system dwara 120 J work: $Δ U = + 380 J .$

Fig. 5.4Har sign system ke point of view se gina jaata hai.

System likha ek box energy U rakhta hai. Heat ke arrows left par hain, heat in upar aur heat out neeche; dashed work arrows right par hain, work on upar aur work by neeche. Steps one se four tak har step ek arrow highlight karta hai aur uska sign deta hai: heat in se q positive, heat out se q negative, work on se w positive aur work by se w negative. Paanchve step mein heat in (500 J) aur work by (120 J) saath highlight hote hain, aur result Delta U = 500 − 120 = +380 J dikhaya jaata hai.

Example 5.1NCERT Problem 5.1 · p. 140

First law ko padhna

Har change ke liye $Δ U$ likho aur batao ki system ki boundary kis tarah ki honi chahiye. (a) Sealed, rigid steel bottle mein gas 300 J heat se garam hoti hai. (b) Insulated cylinder mein gas apne piston ko bahar dhakelti hai, 150 J work karti hai. (c) Cylinder mein gas 400 J work se compress hoti hai jabki woh cooling water ko 250 J heat khoti hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

5.2 Applications

Engine cylinder mein jalta fuel garam gas banata hai jo piston ko dhakelti hai; wahi fuel pan ke neeche jale to bas khaana garam karta hai. First law se dono par numbers lagane ke liye humein $w$ aur $q$ ke working formulas chahiye.

5.2.1 Work

Hum sirf pressure–volume work lete hain, jo tab hota hai jab gas kisi external pressure ke against expand ya compress hoti hai. Ek cylinder mein ideal gas ka ek mole socho, jo weightless, frictionless piston se band hai.

DerivationConstant external pressure ke against kiya gaya work

Maan lo piston ka area $A$ hai, aur us par constant external pressure $p_{ex}$ lag raha hai, to us par force hai $F = p_{ex} A .$ Agar piston $l$ distance andar move kare, to volume badalta hai $Δ V = V_{f} - V_{i} = - l A,$ jo negative hai kyunki gas dab rahi hai.

\begin{aligned} w & = F \times l \\ = p_{ex} A l \\ = - p_{ex} Δ V \end{aligned}

w = - p_{ex} (V_{f} - V_{i})

(5.2)

Minus sign se convention sahi aata hai. Compression par $V_{f} < V_{i}$ , isliye $w > 0$ : gas par work hota hai. Expansion par $V_{f} > V_{i}$ aur $w < 0$ : gas surroundings par work karti hai.

Ab process ke dauraan external pressure ko badalne do. Agar yeh kuch finite jumps mein badle, to total work stages par $- p_{ex} Δ V$ ka sum hai. Jumps ko chhota aur chhota karo. Limit mein external pressure har pal gas pressure se sirf infinitesimally alag hota hai, aur har stage volume ko infinitesimal $d V$ se badalti hai:

w = - \int_{V_{i}}^{V_{f}} p_{ex} d V

(5.3)

Aisa idealised process, jo anginat chhote steps mein hota hai aur jismein system kabhi bhi equilibrium se baal bhar se zyada door nahi hota, reversible process (उत्क्रमणीय प्रक्रम) hai: $p_{ex}$ ko baal bhar doosri taraf khisakao aur yeh ulta chalne lagta hai, apna path wapas trace karta hua. Har asli process, jo finite time leta hai, irreversible hai. $p V$ graph par work ek area hai, jisse comparison aasaani se dikhta hai (Fig. 5.5).

Definition

Reversible process

Ek idealised process jo equilibrium states ki ek lagaataar chain se guzarta hai, taaki conditions mein chhota sa change use same path par wapas bhej de.

ExampleEk gas jo expand hoti hai jab uske piston ka weight daana-daana karke hataya jaata hai.

4/4Reversible expansion sabse zyada work deta hai; reversible compression ko sabse kam chahiye.

Fig. 5.5Jitni zyada stages, area utna hi reversible maximum ke kareeb.NCERT Fig. 5.5 · p. 141

300 K par 1.00 mol ideal gas ke liye bar mein pressure vs litres mein volume, 5 L par 5.0 bar se 20 L par 1.25 bar tak ek dashed isotherm ke saath. Step one: 1.25 bar ke against ek stage; shaded rectangle, yaani work, 1.87 kJ hai. Step two: do stages, 10 L tak 2.49 bar ke against aur phir 1.25 bar; staircase 2.49 kJ hai. Step three: reversible expansion; isotherm ke neeche ka poora area, 3.46 kJ, shaded hai. Step four us par stage outlines overlay karta hai: reversible sabse zyada work deta hai.

DerivationIdeal gas ka reversible isothermal expansion

Reversible process mein $p_{ex} = p \pm d p,$ jahan $p$ gas pressure hai. Product $d p d V$ itna chhota hai ki ignore ho sakta hai, isliye $p_{ex}$ ki jagah $p$ rakh sakte hain:

w_{rev} = - \int_{V_{i}}^{V_{f}} p d V

(5.4)

$p_{ex}$ ke ulat, gas pressure pata hota hai: $n$ mol ideal gas ke liye $p = n R T / V$ . Constant temperature par (ek isothermal process) $n R T$ constant hai aur integral se bahar aa jaata hai, isliye reversible work hai

\begin{aligned} w & = - n R T \int_{V_{i}}^{V_{f}} \frac{d V}{V} \\ = - n R T \ln \frac{V_{f}}{V_{i}} \end{aligned}

\begin{aligned} w_{rev} & = - 2.303 n R T \\ \times \log \frac{V_{f}}{V_{i}} \end{aligned}

(5.5)

Eq. 5.2 ko first law mein rakhne se constant pressure ke against change ke liye $Δ U = q - p_{ex} Δ V$ milta hai. Isse kai special cases nikalte hain.

Free expansion. Vacuum mein expand hoti gas kisi cheez ke against nahi dhakelti: $p_{ex} = 0,$ isliye $w = 0$ . Joule ne paaya ki constant temperature par freely expand hoti ideal gas koi heat bhi absorb nahi karti, isliye $Δ U = 0.$
Constant volume. $Δ V = 0$ ke saath koi pressure–volume work nahi hota aur $Δ U = q_{V};$ subscript batata hai ki heat constant volume par di gayi thi.
Ideal gas ka isothermal change. Ideal gas ki internal energy sirf uske temperature par depend karti hai, isliye $Δ U = 0$ aur $q = - w$ . Constant pressure ke against, $q = p_{ex} (V_{f} - V_{i});$ reversibly, $q = n R T \ln (V_{f} / V_{i}) .$
Adiabatic change. $q = 0$ , isliye $Δ U = w_{ad} .$

Fig. 5.6Temperature, start aur end volumes aur stages ki sankhya set karo, phir piston chalane ke liye Play dabao aur work ko judte hue dekho.

Example 5.2NCERT Problem 5.2 · p. 142

Vacuum mein expansion

300 K par 1.00 mol ideal gas 5.0 L gherti hai. Yeh ek khaali (evacuated) space mein isothermally expand hoti hai jab tak iska volume 20.0 L na ho jaaye. $w$ , $q$ aur $Δ U$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5.3NCERT Problem 5.3 · p. 142

Steady pressure ke against expansion

Example 5.2 dobara karo, lekin gas ko 1.0 bar ke constant external pressure ke against expand hone do.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5.4NCERT Problem 5.4 · p. 142

Wahi expansion, reversibly

Ab Example 5.2 ki gas ko 300 K par 5.0 L se 20.0 L tak reversibly aur isothermally expand hone do.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

5.2.2 Enthalpy, H

Ek kaam ka naya state function

Zyada tar chemistry sealed, rigid vessel mein nahi hoti. School lab mein reactions khule beakers mein atmosphere ke diye pressure par bubble karte hain, aur jo bhi gas bante hain unhe jagah banane ke liye hawa ko hataana padta hai, jismein energy lagti hai (Fig. 5.7). Isliye khule bench par measure ki gayi heat $Δ U$ nahi hai, aur humein ek naya state function chahiye jise yeh heat measure kare.

DerivationConstant pressure par heat

Constant pressure par sirf work hai $w = - p Δ V,$ isliye first law banta hai $Δ U = q_{p} - p Δ V,$ jahan $q_{p}$ absorb hui heat hai. Initial state ko 1 aur final state ko 2 kaho:

\begin{aligned} q_{p} & = (U_{2} - U_{1}) \\ + p (V_{2} - V_{1}) \end{aligned}

\begin{aligned} q_{p} & = (U_{2} + p V_{2}) \\ - (U_{1} + p V_{1}) \end{aligned}

(5.6)

Combination $U + p V$ dono taraf aata hai, isliye hum ise ek naam dete hain, enthalpy $H$ (एन्थैल्पी):

H = U + p V

(5.7)

\begin{aligned} q_{p} & = H_{2} - H_{1} \\ = Δ H \end{aligned}

Yeh kamaal ka result hai. Heat aam taur par path par depend karti hai, phir bhi constant pressure par li gayi heat ek state function ke change ke barabar hai, kyunki $U$ , $p$ aur $V$ har ek sirf state par depend karte hain. Constant pressure par finite change ke liye,

Δ H = Δ U + p Δ V

(5.8)

To jab bhi tum constant pressure par heat measure karte ho, tum $Δ H$ measure kar rahe ho. Heat chhodne wala reaction exothermic hai aur uska $Δ H < 0;$ heat lene wala endothermic hai, jiska $Δ H > 0.$ Constant volume par $Δ V = 0,$ aur Eq. 5.8 deta hai $Δ H = Δ U = q_{V} .$

Definition

Enthalpy

$H = U + p V$ , ek state function jiska change constant pressure par system dwara absorb ki gayi heat ke barabar hai.

Example0 °C par 1 mol ice pighlana: $Δ H = + 6.01 k J .$

3/3Same $Δ U$ ke liye constant-pressure heat $p Δ V$ se zyada hai. Wahi heat $Δ H$ hai.

Fig. 5.7Constant pressure par heat ka kuch hissa hawa ko peeche dhakelne mein kharch hota hai.

Gas ka ek bolted rigid vessel, jis par fixed volume likha hai, aur ek cylinder jiske free piston ko hawa neeche dabaati hai, jis par fixed pressure likha hai. Step one: rigid vessel garam hota hai jabki cylinder faded hai. Step two: cylinder garam hota hai aur uska piston uthta hai. Step three: comparison ke liye dono garam hote hain, vessel ke neeche q V = Delta U aur cylinder ke neeche q p = Delta U + p Delta V.

$p Δ V$ term kitna bada hai? Solid ya liquid ka ek mole sirf kuch das millilitres gherta hai, isliye unke beech reactions ke liye $p Δ V$ kuch joules hota hai aur $Δ H \approx Δ U .$ Room temperature ke aas-paas gas ka ek mole lagbhag 25 L gherta hai, aur tab $p Δ V$ kilojoules tak pahunch jaata hai. Ideal gas law ise saaf-saaf deta hai.

DerivationGases ke reactions ke liye

Δ H

aur

Δ U

Constant $T$ aur $p$ par, maan lo gaseous reactants mein volume $V_{A}$ mein $n_{A}$ mol hain, aur gaseous products mein $V_{B}$ mein $n_{B}$ mol:

\begin{aligned} p V_{A} & = n_{A} R T \\ p V_{B} & = n_{B} R T \end{aligned}

Doosre mein se pehla ghatane par,

p Δ V = Δ n_{g} R T

(5.9)

jahan $Δ n_{g}$ gas ki amount mein change hai: product side par gas ke moles minus reactant side par gas ke moles. Eq. 5.8 mein rakhne par milta hai

Δ H = Δ U + Δ n_{g} R T

(5.10)

$Δ n_{g}$ mein sirf gases ginti hain. Units ka dhyaan rakho: $R = 8.314 J K^{- 1} m o l^{- 1},$ isliye $Δ n_{g} R T$ ko kJ wali value mein jodne se pehle kJ mein convert karo.

Example 5.5NCERT Problem 5.5 · p. 143

$Δ H$ se $Δ U$ tak

298 K par $N_{2} (g) + 3 H_{2} (g) \overset{}{\to} 2 N H_{3} (g)$ ke liye $Δ H = - 91.8 k J .$ Gases ko ideal maankar $Δ U$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Extensive aur intensive properties

Garam chai ke 1.0 L jug ko do cups mein daalo (Fig. 5.8). Har cup mein aadha volume, aadha mass aur aadhi enthalpy hai, phir bhi chai utni hi garam aur utni hi dense hai jitni pehle thi. Jo property matter ki amount ke saath badhti-ghatti hai, jaise mass, volume, $U$ , $H$ ya heat capacity, woh extensive property (विस्तीर्ण गुणधर्म) hai. Jo aisa nahi karti, jaise temperature, pressure ya density, woh intensive property (गहन गुणधर्म) hai.

2/2Har cup: 0.5 L, ab bhi 80 °C par.

Fig. 5.8Sample ko baantne se uski extensive properties aadhi ho jaati hain aur intensive properties waisi hi rehti hain.NCERT Fig. 5.6 · p. 144

Pehle step mein ek jug mein 80 °C par 1.0 L chai hai. Doosre step mein jug khaali hai aur do cups mein 80 °C par 0.5 L-0.5 L chai hai; neeche likha hai ki volume, mass aur enthalpy jaisi extensive properties aadhi ho jaati hain, jabki temperature aur density jaisi intensive properties same rehti hain.

Ek extensive property ko doosri se divide karne par intensive property milti hai. Density mass divided by volume hai, aur ek molar property $X_{m} = X / n$ kisi extensive property $X$ ki per mole value hai: molar volume $V_{m}$ , molar heat capacity $C_{m}$ , molar enthalpy $H_{m}$ . Inmein se koi bhi is par depend nahi karti ki tumhare paas kitna substance hai.

Heat capacity

Jab heat aise system mein flow hoti hai jo phase nahi badal raha ya react nahi kar raha, to uska temperature proportion mein badhta hai: $q = C Δ T .$ Proportionality ka constant, $C$ , heat capacity hai: temperature ke har kelvin rise ke liye zaroori heat. Yeh jitni badi ho, temperature utni hi dheere respond karta hai. Paani ki heat capacity bahut badi hai, yahi ek wajah hai ki samundar kinaare wale Mumbai mein temperature ke utaar-chadhaav andar ke Delhi se kaafi kam hote hain.

$C$ extensive hai, isliye hum ise per amount batate hain. Molar heat capacity $C_{m} = C / n$ woh heat hai jo 1 mol ko 1 K garam karne ke liye chahiye. Specific heat capacity $c$ woh heat hai jo 1 g ko 1 K garam karne ke liye chahiye; liquid paani ke liye $c = 4.18 J g^{- 1} K^{- 1} .$ Mass $m$ ke sample ke liye,

q = c m Δ T = C Δ T

(5.11)

Ideal gas ke liye Cp aur CV ka relation

Heat capacity garam karne ki conditions par depend karti hai. Constant volume par $q_{V} = C_{V} Δ T = Δ U;$ constant pressure par $q_{p} = C_{p} Δ T = Δ H .$

DerivationIdeal gas ke ek mole ke liye

C_{p} - C_{V}

1 mol ideal gas ke liye $p V = R T$ , isliye $Δ (p V) = R Δ T$ aur

Δ H = Δ U + R Δ T

(5.12)

Rakho $Δ H = C_{p} Δ T$ aur $Δ U = C_{V} Δ T,$ phir $Δ T$ se divide karo:

C_{p} - C_{V} = R

(5.13)

Constant pressure par gas ko expansion work bhi karna padta hai, isliye temperature mein same rise ke liye use zyada heat chahiye: $C_{p}$ hamesha $C_{V}$ se bada hota hai, aur dono ka farq hai $R = 8.314 J K^{- 1} m o l^{- 1} .$ Argon ke liye $C_{V} = 12.5$ aur $C_{p} = 20.8 J K^{- 1} m o l^{- 1} .$

Quick check

4 mein se 1

Kis reaction ke liye

Δ H

Δ U

ke barabar hai?

H_{2} (g) + C l_{2} (g) \overset{}{\to} 2 H C l (g)

2 C O (g) + O_{2} (g) \overset{}{\to} 2 C O_{2} (g)

C a C O_{3} (s) \overset{}{\to} C a O (s) + C O_{2} (g)

N_{2} O_{4} (g) \overset{}{\to} 2 N O_{2} (g)

Answer dekho

Answer (A) $H_{2} (g) + C l_{2} (g) \overset{}{\to} 2 H C l (g)$

Δ H = Δ U

jab

Δ n_{g} = 0 :

har side par gas ke do moles hon, jaise

H_{2} + C l_{2} \overset{}{\to} 2 H C l .

Baaki ke

Δ n_{g} = - 1,

+ 1

aur

+ 1

hain.

5.3 ΔU aur ΔH ka Measurement: Calorimetry

Heat measure karne ke liye use kisi aisi cheez ko garam (ya thanda) karne do jiski heat capacity tumhe pata ho, aur thermometer padho. Yahi calorimetry hai. Woh 'cheez' ek calorimeter hai, ek vessel aur uske chaaron taraf ka liquid, jo is tarah calibrate kiya gaya hai ki temperature change ka har kelvin joules ki ek jaani hui sankhya ke barabar ho: $q = C Δ T .$ Reaction band hai ya hawa ke liye khula, yahi decide karta hai ki woh heat humein kya batati hai (Fig. 5.9).

ΔU measurements

Bomb calorimeter ki khaas baat yeh hai ki iska reaction vessel expand nahi ho sakta. Sample ki ek pellet ek mote walls wale steel 'bomb' ke andar chhote cup mein rakhi hoti hai, jismein lagbhag 30 atm par oxygen bhari hoti hai. Bomb ko paani ki ek stirred, insulated balti mein utaara jaata hai aur sample ko ek patle wire se electrically jalaya jaata hai. Volume fixed hone se koi $p V$ work possible nahi, chahe gases banen, isliye nikli hui saari energy steel aur paani mein heat ban jaati hai: measure ki gayi heat $q_{V}$ hai, aur yeh reaction ke $- Δ U$ ke barabar hai. Thermometer $Δ T$ deta hai, aur calorimeter ki heat capacity, jo pehle se benzoic acid ka jaana hua mass jalakar nikali jaati hai, use joules mein badal deti hai.

ΔH measurements

$q_{p}$ measure karne ke liye reaction atmosphere ke liye khula chalta hai. Solution mein hone wale reactions ke liye, dheele lid wale do ek-ke-andar-ek foam cups school lab mein kamaal ka kaam karte hain. Is tarah constant pressure par measure ki gayi heat hai enthalpy of reaction, $Δ_{r} H .$ Iska sign system ko follow karta hai: agar solution garam hota hai, to reaction ne heat di aur $Δ_{r} H < 0;$ agar solution thanda hota hai, to reaction ne heat li aur $Δ_{r} H > 0.$

3/3Cup mein reaction atmospheric pressure par chalta hai, isliye heat $Δ H$ measure karti hai.

Fig. 5.9Constant volume

Δ U

deta hai; constant pressure

Δ H

deta hai.NCERT Fig. 5.7 · p. 145

Left: ek sealed steel bomb, sample cup aur ignition wires ke saath, ek stirred, insulated water bath mein jismein thermometer hai. Right: solution mein lid, stirrer aur thermometer wale do ek-ke-andar-ek foam cups. Step one dono ko kisi reaction se pehle dikhata hai. Step two: bomb mein flame dikhti hai, wavy arrows heat ko paani mein le jaate hain, uska thermometer chadhta hai, aur ek label kehta hai ki yeh Delta U measure karta hai. Step three: cup ka thermometer chadhta hai, ek note kehta hai ki uska dheela lid hawa ke liye khula hai, aur ek label kehta hai ki yeh Delta H measure karta hai.

Example 5.6NCERT Problem 5.6 · p. 146

Bomb mein naphthalene jalana

Naphthalene, $C_{10} H_{8}$ (molar mass 128.2 g mol⁻¹), ka 0.500 g sample ek bomb calorimeter mein poori tarah jalta hai jiski heat capacity 10.0 kJ K⁻¹ hai. Temperature 298 K ke aas-paas 2.01 K badhta hai. Naphthalene ke per mole $Δ_{c} U$ aur $Δ_{c} H$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Try karke dekho

Foam cups mein reaction enthalpy measure karo

Tumhe chahiye

lid wale do ek-ke-andar-ek foam cups
0.1 °C tak padhne wala thermometer
1.0 M hydrochloric acid ka 50 mL
1.0 M sodium hydroxide solution ka 50 mL
ek measuring cylinder

Pehle guess karo

Dono solutions ke barabar volumes milane se

mixture kuch degree thanda hogamixture kuch degree garam hogatemperature nahi badlega

Har solution ka temperature measure karo; dono 0.2 °C ke andar match hone chahiye.
Acid ko cups mein daalo, alkali jaldi se milao, lid lagao aur hilao.
Sabse zyada pahuncha hua temperature record karo.
$Δ T$ nikalo aur $q = (100 g) (4.18 J g^{- 1} K^{- 1}) Δ T,$ mixture ko paani maankar.

Table 5.1

Reading	Value
Shuruaati temperature (°C)
Sabse zyada temperature (°C)
Rise ΔT (K)
Heat q (J)

Tumhari entries apne aap save hoti hain.

Quick check

3 mein se 1

Bomb calorimeter fuel jalne par nikli heat measure karta hai. Yeh heat deti hai

Δ H

, kyunki reaction tez hai

Δ U

, kyunki volume fixed hai

Δ H

, kyunki oxygen excess mein hai

Δ U

, kyunki temperature fixed hai

Answer dekho

Answer (B) $Δ U$ , kyunki volume fixed hai

Sealed bomb expand nahi ho sakta, isliye koi work nahi hota aur

q_{V} = Δ U .

Temperature fixed nahi hai: uska rise measure karna hi to poora maqsad hai.

5.4 Reaction ka Enthalpy Change, ΔrH – Reaction Enthalpy

Reaction mein reactants products ban jaate hain. Iske saath hone wala enthalpy change hai reaction enthalpy, $Δ_{r} H :$ products ki total enthalpy minus reactants ki, har ek ko balanced equation mein uske coefficient se weight karke.

\begin{aligned} Δ_{r} H & = \sum_{i} a_{i} H_{prod} \\ - \sum_{i} b_{i} H_{react} \end{aligned}

(5.14)

jahan

$H_{prod}, H_{react}$: products aur reactants ki molar enthalpies

Yahan $a_{i}$ aur $b_{i}$ products aur reactants ke stoichiometric coefficients hain. Hydrogen ke jalne ke liye, jahan $H_{m}$ molar enthalpy ko darshata hai,

\begin{aligned} 2 H_{2} (g) + O_{2} (g) \\ \overset{}{\to} 2 H_{2} O (l) \end{aligned}

\begin{aligned} Δ_{r} H & = 2 H_{m} (H_{2} O) \\ - 2 H_{m} (H_{2}) \\ - H_{m} (O_{2}) \end{aligned}

Aise numbers practical sawaalon ka jawaab dete hain, jaise ammonia plant ko apne reactor ko zyada garam hone se rokne ke liye kitna bada cooling system chahiye, ya cement kiln ko har ghante kitna fuel jalana padega.

Reactions ki standard enthalpy

Methane ko aise jalao ki uska paani liquid ki jagah vapour ban kar nikle, aur $Δ H$ per mole 88 kJ badal jaata hai. Values tabhi compare ho sakti hain jab sab conditions par sehmat hon. Maani hui reference yeh hai ki har substance pure ho, theek 1 bar pressure par, ek bataye gaye physical form mein: yeh uski standard state hai. Jab har reactant aur product apni standard state mein ho, tab ka enthalpy change hai standard enthalpy of reaction, $Δ_{r} H^{\circ} .$ Temperature definition ka hissa nahi hai, isliye use batana padta hai; zyada tar tables 298 K use karti hain.

Superscript ° (aksar plimsoll sign, ⦵, ki tarah chhapta hai) ka matlab hai 'standard'. Jaise 400 K par 'standard' copper bas 400 K aur 1 bar par pure solid copper hai.

Phase transformations ke dauraan enthalpy changes

Melting, boiling aur sublimation bhi energy lete ya chhodte hain, jabki molecules ke andar koi bond nahi tootta. Constant pressure par har ek fixed temperature par hota hai: heat andar jaati hai, lekin thermometer tab tak ruka rehta hai jab tak change poora na ho jaaye (Fig. 5.10). Energy molecules ko unke aapsi attraction ke against alag karne mein lagti hai.

5/5Ab jaakar steam garam hoti hai.

Fig. 5.10Melting aur boiling ke dauraan heat phase badalti hai, temperature nahi.

1 mol paani ke liye degrees Celsius mein temperature vs kilojoules mein di gayi heat. Step one: ek chhoti, khadi line ice ko −20 se 0 °C tak garam hote dikhati hai. Step two: 0 °C par ek flat line, melting 6.0 kJ likhi hui. Step three: ek dhalaan wali line liquid paani ko 0 se 100 °C tak garam hote dikhati hai, lagbhag 7.5 kJ lekar. Step four: 100 °C par ek lambi flat line, boiling 40.7 kJ likhi hui, melting plateau se kaafi lambi. Step five: ek chhoti, khadi line steam ko 120 °C tak garam hote dikhati hai.

Standard enthalpy of fusion, $Δ_{fus} H^{\circ},$ woh enthalpy change hai jab 1 mol solid constant temperature aur 1 bar par pighalta hai; 273 K par ice ke liye yeh +6.01 kJ mol⁻¹ hai. Freezing yahi change ulta chalna hai, isliye freezer ko har mole paani ko ice banane ke liye woh 6.01 kJ bahar nikaalne padte hain. Standard enthalpy of vaporisation, $Δ_{vap} H^{\circ},$ woh enthalpy change hai jab 1 mol liquid constant temperature aur 1 bar par vapour banta hai; paani ke boiling point par yeh +40.7 kJ mol⁻¹ hai.

Sublimation solid ko seedhe vapour bana deta hai. Standard enthalpy of sublimation, $Δ_{sub} H^{\circ},$ woh enthalpy change hai jab 1 mol solid constant temperature aur 1 bar par sublime hota hai. Iodine crystals ko 298 K par lagbhag 62 kJ mol⁻¹ chahiye, aur puja mein use hone wali camphor (kapur) ki tablets sublime hokar dheere-dheere gaayab ho jaati hain.

Teeno enthalpies positive hain, kyunki melting, boiling aur subliming hamesha heat lete hain. Inka size particles ke beech attractions ki strength ke hisaab se hota hai (Table 5.2). Methane molecules ek doosre ko sirf halka attract karte hain aur hydrogen sulphide molecules thoda zyada. Ethanol aur paani hydrogen bonds banate hain, isliye inmein se kisi ka ek mole ubaalne mein methane ke ek mole se kai guna zyada heat lagti hai. Mercury atoms metallic bonding se jude hote hain, isliye is list mein mercury ki vaporisation enthalpy sabse zyada hai.

Table 5.21 bar par melting aur boiling (enthalpies kJ mol⁻¹ mein)NCERT Table 5.1 · p. 147

Substance	$T_{m}$ / K	$Δ_{fus} H^{\circ}$	$T_{b}$ / K	$Δ_{vap} H^{\circ}$
$C H_{4}$	90.7	0.94	111.7	8.19
$H_{2} S$	187.7	2.38	212.9	18.7
$C_{2} H_{5} O H$	159.0	4.93	351.4	38.6
$H_{2} O$	273.2	6.01	373.2	40.7
$H g$	234.3	2.29	629.9	59.1

Example 5.7NCERT Problem 5.7 · p. 148

Hand sanitiser thanda kyun lagta hai

Hand sanitiser ki ek squirt tumhari hatheli par 2.3 g ethanol (46.1 g mol⁻¹) chhodti hai, aur yeh 298 K par evaporate hota hai. Yeh tumhare haath se kitni heat leta hai, aur ethanol ki internal energy ka change kya hai? 298 K par $Δ_{vap} H (C_{2} H_{5} O H) = 42.3 k J m o l^{- 1};$ vapour ko ideal gas maano.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5.8NCERT Problem 5.8 · p. 148

Ice cubes banana

Ek ice tray mein 27 °C par 90 g paani hai. Ise 0 °C ki ice banane ke liye freezer ko kitni heat nikaalni padegi? Is change ke liye $Δ H$ aur $Δ U$ nikalo. Paani ke liye $c = 4.18 J g^{- 1} K^{- 1}$ lo aur $Δ_{fus} H = 6.01 k J m o l^{- 1} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Formation ki standard enthalpy

Har possible reaction ke liye $Δ_{r} H^{\circ}$ ki list banana namumkin hoga. Iski jagah hum har compound ke liye ek khaas reaction list karte hain: uske elements se uska banna. Standard molar enthalpy of formation, $Δ_{f} H^{\circ}$ (मानक संभवन एन्थैल्पी), woh enthalpy change hai jab apni standard state mein kisi compound ka 1 mol apne elements se banta hai jo apni reference states mein hon. Kisi element ki reference state 25 °C aur 1 bar par uska sabse stable form hai: hydrogen, nitrogen aur oxygen diatomic gases ke roop mein, bromine aur mercury liquids ke roop mein, sodium solid ke roop mein, carbon graphite ke roop mein aur sulphur rhombic crystals ke roop mein.

Definition

Standard enthalpy of formation

Woh enthalpy change jab apni standard state mein kisi compound ka 1 mol apne elements se banta hai jo apni reference states mein hon.

Example $N a (s) + \frac{1}{2} C l_{2} (g) \overset{}{\to} N a C l (s),$ $Δ_{f} H^{\circ} = - 411.2 k J m o l^{- 1} .$

Formation equation mein sirf elements se theek ek mole product banta hai. Ammonia, jiske liye $Δ_{f} H^{\circ} = - 45.9 k J m o l^{- 1},$ ek typical case hai:

\begin{aligned} \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \\ \overset{}{\to} N H_{3} (g) \end{aligned}

Definition se, apni reference state mein kisi element ka $Δ_{f} H^{\circ}$ zero hai: graphite se graphite banane mein koi change nahi hota. Kisi value ko formation enthalpy kehne se pehle shartein dhyaan se padho: har reactant apni reference state mein element hona chahiye, aur theek ek mole product banna chahiye. $S O_{2}$ ko $S O_{3}$ mein oxidise karna pehli shart mein fail hota hai; chlorine mein sodium ke do moles jalana (−822.4 kJ) doosri mein fail hota hai, aur NaCl ka $2 Δ_{f} H^{\circ}$ deta hai.

Table 5.3298 K par standard enthalpies of formation, kJ mol⁻¹ meinNCERT Table 5.2 · p. 149

Substance	$Δ_{f} H^{\circ}$	Substance	$Δ_{f} H^{\circ}$
$A l_{2} O_{3} (s)$	−1675.7	$H_{2} O (g)$	−241.8
$C$ (diamond)	+1.9	$H_{2} O_{2} (l)$	−187.8
$C H_{4} (g)$	−74.8	$K C l (s)$	−436.7
$C_{2} H_{4} (g)$	+52.3	$M g O (s)$	−601.7
$C_{2} H_{6} (g)$	−84.7	$N H_{3} (g)$	−45.9
$C O (g)$	−110.5	$N O (g)$	+90.3
$C O_{2} (g)$	−393.5	$N O_{2} (g)$	+33.2
$F e_{2} O_{3} (s)$	−824.2	$O_{3} (g)$	+142.7
$H_{2} O (l)$	−285.8	$S O_{2} (g)$	−296.8
$N a C l (s)$	−411.2	$S O_{3} (g)$	−395.7

Aisi table ke saath koi bhi reaction enthalpy is se nikalti hai

\begin{aligned} Δ_{r} H^{\circ} & = \sum a_{i} Δ_{f} H_{P}^{\circ} \\ - \sum b_{i} Δ_{f} H_{R}^{\circ} \end{aligned}

(5.15)

jahan

$Δ_{f} H_{P}^{\circ}, Δ_{f} H_{R}^{\circ}$: products aur reactants ki standard enthalpies of formation

Yeh isliye kaam karta hai kyunki enthalpy ek state function hai. Socho ki reactants ko unke elements mein tod diya aur phir unhi elements se products banaye: elements cancel ho jaate hain, aur sirf formation enthalpies bachti hain.

Example

Rails ki thermite welding

Railway track ki teams thermite reaction se steel rails jodti hain. Table 5.3 use karke is reaction ka $Δ_{r} H^{\circ}$ nikalo: $2 A l (s) + F e_{2} O_{3} (s)$ $\overset{}{\to} A l_{2} O_{3} (s) + 2 F e (s) .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Thermochemical equations

Jo balanced equation saath mein rakhe apna $Δ_{r} H,$ aur har species ki physical state (aur jahan zaroori ho, allotrope), use thermochemical equation kehte hain. Compressed natural gas aise jalti hai

\begin{aligned} C H_{4} (g) + 2 O_{2} (g) \overset{}{\to} \\ C O_{2} (g) + 2 H_{2} O (l) \end{aligned}

Iske liye $Δ_{r} H^{\circ} = - 890.3 k J m o l^{- 1} .$ Ise padhne ke teen conventions hain:

Har coefficient ko moles mein amount ki tarah padho. Isiliye ½ jaise fractions bilkul theek hain: aadha mole samajh aata hai, aadha molecule nahi.
$Δ_{r} H^{\circ}$ equation ka hai jaisi woh likhi gayi hai, aur uski unit kJ mol⁻¹ ka matlab hai 'per mole of reaction'. Har coefficient aadha karo aur $Δ_{r} H^{\circ}$ bhi aadha ho jaata hai, −445.2 kJ mol⁻¹, kyunki enthalpy extensive hai.
Equation ulti karo aur $Δ_{r} H^{\circ}$ ka sign badal jaata hai: carbon dioxide aur paani se methane banane mein per mole +890.3 kJ lagenge.

States bhi matter karti hain. Agar bana hua paani liquid ki jagah vapour hota, to methane ke per mole 2 × 44.0 = 88.0 kJ kam heat nikalti, kyunki itni heat paani ko vapour banane mein kharch hoti.

Hess's law of constant heat summation

Yeh phir se Fig. 5.2 wali pahaadi hai, ab reactions ke liye. Reactants aur products ki enthalpies fixed points hain, isliye unke beech steps ka koi bhi sequence, asli ho ya kalpanik, jud kar wahi total dega: $Δ_{r} H .$ Yahi Hess's law hai, aur iski practical taakat yeh hai ki yeh un reactions ki enthalpies deta hai jinhe saaf-saaf kiya hi nahi ja sakta.

LawHess's law

Jab ek overall reaction ko ek temperature par steps ki series mein toda ja sake, to steps ke enthalpy changes jodne se overall reaction ka enthalpy change milta hai.

\begin{aligned} Δ_{r} H & = Δ_{r} H_{1} + Δ_{r} H_{2} \\ + Δ_{r} H_{3} + \dots \end{aligned}

(5.16)

Sulphur ka jalna ise enthalpy diagram par dikhata hai (Fig. 5.11). Sulphur ko seedhe sulphur trioxide mein oxidise kiya ja sakta hai, ya pehle dioxide mein, jo phir aage oxidise hota hai, jaise sulphuric acid banane ke contact process mein. Dono routes same levels par shuru aur khatam hote hain.

3/3Steps jud kar direct value dete hain, jaisa Hess's law kehta hai.

Fig. 5.11Do routes, ek total: enthalpy change sirf end points par depend karta hai.

Ek enthalpy axis upar ki taraf hai. Upar wala level sulphur plus dedh O2 hai; neeche wala level SO3 gas hai. Step one: ek single arrow unke beech seedha neeche jaata hai, −395.7 kJ per mole likha hua. Step two: ek beech ka level, SO2 plus aadha O2, dikhta hai; −296.8 ka ek dashed arrow upar wale level se us tak jaata hai, aur −98.9 ka doosra dashed arrow usse SO3 tak. Step three: text dono steps jodta hai, −296.8 − 98.9, aur dikhata hai ki yeh −395.7 ke barabar hain.

Example

Diamond se graphite

298 K par $Δ_{c} H^{\circ}$ graphite ke liye −393.5 kJ mol⁻¹ aur diamond ke liye −395.4 kJ mol⁻¹ hai, dono carbon dioxide banakar jalte hain. Is change ke liye $Δ_{r} H^{\circ}$ nikalo: $C (d i a m o n d) \overset{}{\to} C (g r a p h i t e),$ jo seedhe measure karne ke liye bahut hi dheema hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

5.5 Alag-alag Types ke Reactions ki Enthalpies

Kuch tarah ke reactions itne common hain ki unki enthalpies ke apne naam aur symbols hain. Har ek is quantity ka special case hai: $Δ_{r} H .$

Combustion ki standard enthalpy

Har fuel, auto-rickshaw ki CNG se lekar rocket ke upper stage ki liquid hydrogen tak, is heat se rate kiya jaata hai jo uska ek mole oxygen mein poori tarah jalne par deta hai. Jab sab kuch apni standard state mein ho, tab measure ki gayi yeh heat hai standard enthalpy of combustion, $Δ_{c} H^{\circ};$ kyunki jalne mein hamesha heat nikalti hai, yeh hamesha negative hoti hai. Methane, jo CNG ki mukhya gas hai, uska $Δ_{c} H^{\circ} = - 890.3 k J m o l^{- 1}$ hai jab bana hua paani liquid ho.

Khaana bhi fuel hai. Tumhare cells sugars ko enzyme-controlled steps ki ek lambi chain se carbon dioxide aur paani mein oxidise karte hain, lekin Hess's law ke hisaab se overall enthalpy change unhe calorimeter mein jalane ke barabar hai. Sucrose ke liye $Δ_{c} H^{\circ} \approx - 5640 k J m o l^{- 1},$ isliye 4 g ki ek chammach cheeni (0.0117 mol) lagbhag 66 kJ deti hai. Yahi food label par likhi '16 Calories' hai, kyunki 1 Calorie = 1 kcal = 4.184 kJ.

Example 5.9NCERT Problem 5.9 · p. 152

Combustion se propane ki formation enthalpy

Propane, LPG ka ek component, ka 298 K par $Δ_{c} H^{\circ} = - 2220.0 k J m o l^{- 1}$ hai. $C O_{2} (g)$ (−393.5) aur $H_{2} O (l)$ (−285.8 kJ mol⁻¹) ke $Δ_{f} H^{\circ}$ use karke propane ka $Δ_{f} H^{\circ}$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Atomization ki enthalpy

Enthalpy of atomization, $Δ_{a} H^{\circ},$ woh enthalpy change hai jab kisi substance ka 1 mol gas phase mein poori tarah alag-alag atoms mein toot jaata hai. Oxygen ko todne, $O_{2} (g) \overset{}{\to} 2 O (g),$ mein 498 kJ mol⁻¹ lagte hain; ammonia ko poori tarah ek N aur teen H atoms mein todne mein 1173 kJ mol⁻¹ lagte hain. Potassium jaise metal ke liye, jiska solid pehle se atoms ka lattice hai, atomization aur sublimation ek hi change hain: $K (s) \overset{}{\to} K (g)$ mein 89.2 kJ mol⁻¹ lagte hain.

Bond enthalpy

Kisi bhi reaction ko molecules ke level par dekho to tumhe bonds tootte aur naye bante dikhenge. Bond todne mein hamesha energy lagti hai aur bond banne mein hamesha energy nikalti hai, isliye dono ka hisaab lagbhag reaction enthalpy deta hai. Chemists bond energies ko do thode alag tareekon se list karte hain.

Diatomic molecule mein sirf ek bond hota hai. Uski bond dissociation enthalpy us bond ke ek mole ko todne ka enthalpy change hai, jismein molecule aur tukde sab gas phase mein hon: 431 kJ mol⁻¹ is ke liye: $H C l (g) \overset{}{\to} H (g) + C l (g),$ aur $N_{2}$ ke triple bond ke liye 945 kJ mol⁻¹. Todne ke liye sirf ek bond hone se, diatomic molecule ko atoms mein todna atomization bhi hai, isliye dono numbers same aate hain.

Polyatomic molecule mein ek jaise bonds ko ek ke baad ek todne mein alag energies lagti hain, kyunki peeche bacha tukda badalta rehta hai. Paani yeh dikhata hai. Pehla hydrogen hatane, $H_{2} O (g) \overset{}{\to} H (g) + O H (g),$ mein lagbhag 497 kJ mol⁻¹ lagte hain, lekin doosra, OH se, hatane mein sirf lagbhag 430 kJ mol⁻¹. Calculations ke liye hum mean bond enthalpy (आबंध एन्थैल्पी) use karte hain, dono ka average: paani mein O–H ke liye 463.5 kJ mol⁻¹. Kisi bond ki mean values ek molecule se doosre mein thodi hi badalti hain, isliye har bond type ki ek listed value kaafi achha kaam karti hai.

Aise reaction ke liye jismein har species gas ho, bond enthalpies ek estimate deti hain:

\begin{aligned} Δ_{r} H^{\circ} & \approx \sum B E_{react} \\ - \sum B E_{prod} \end{aligned}

(5.17)

jahan

$B E_{react}, B E_{prod}$: reactants aur products ke bonds ki bond enthalpies

Do saavdhaaniyan. Mean values averages hain, isliye kai per cent ki errors ki ummeed rakho (ethene wala example ek typical error dikhata hai). Aur rule tabhi lagta hai jab har substance gas ho, kyunki yeh condense ya freeze hone ki energy ignore karta hai. Ise tab use karo jab formation enthalpies na milen. Kuch books bond formation enthalpies list karti hain, wahi numbers negative signs ke saath; unke saath products ka sum pehle aata hai.

Example

Bond enthalpies se enthalpy estimate karna

Ethene ke hydrogenation, $C_{2} H_{4} (g) + H_{2} (g) \overset{}{\to} C_{2} H_{6} (g),$ ka $Δ_{r} H^{\circ}$ mean bond enthalpies (kJ mol⁻¹) se estimate karo: C=C 612, H–H 436, C–C 348, C–H 413. Table 5.3 wali value se compare karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Lattice enthalpy

Kisi ionic solid ki lattice enthalpy (जालक एन्थैल्पी) woh enthalpy change hai jab uska 1 mol gaseous ions mein toot jaata hai, jaise $K C l (s) \overset{}{\to} K^{+} (g) + C l^{-} (g) .$ Yeh measure karti hai ki lattice kitni mazbooti se juda hai. Koi experiment yeh seedhe nahi karta, isliye hum aise steps ka ek closed loop banate hain jinki enthalpies measure ho sakti hain, ek Born–Haber cycle, aur Hess's law use karte hain: kisi bhi closed cycle mein enthalpy changes jod kar zero hote hain (Fig. 5.12).

Definition

Lattice enthalpy

Woh enthalpy change jab ionic solid ka 1 mol apne gaseous ions mein alag hota hai.

Example $K C l (s) \overset{}{\to} K^{+} (g) + C l^{-} (g),$ lagbhag +718 kJ mol⁻¹.

4/4Jo loop ko band kare wahi lattice enthalpy hai: +717.8 kJ mol⁻¹.

Fig. 5.12Closed cycle mein ghoom kar enthalpy changes jod kar zero hone chahiye.NCERT Fig. 5.9 · p. 155

kJ per mole mein enthalpy levels. Step one: solid potassium plus aadha mole chlorine zero par hain, aur −436.7 ka ek dashed arrow solid KCl tak girta hai. Step two: chhote arrows gaseous potassium ke liye +89.2 aur chlorine atoms ke liye +121.7 jodte hain. Step three: +418.8 ka ek arrow potassium ko ionise karta hai, upar K+ plus ek electron plus Cl tak pahunchta hai; −348.6 ka ek dashed arrow gaseous K+ aur Cl− tak girta hai. Step four: solid KCl se gaseous ions tak upar jaata ek mota arrow, +717.8, lattice enthalpy hai.

Example

Potassium chloride ki lattice enthalpy

kJ mol⁻¹ mein yeh values use karo: $Δ_{f} H^{\circ} (K C l) = - 436.7;$ K ka sublimation, +89.2; $C l_{2}$ ki aadhi bond enthalpy, +121.7; K ki ionization enthalpy, +418.8; Cl ki electron gain enthalpy, −348.6. KCl ki lattice enthalpy aur 298 K par uska matching internal energy change nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Solution ki enthalpy

Kisi solute ka ek mole solvent ki ek bataayi gayi amount mein ghol do, aur constant pressure par absorb hui heat hai uski enthalpy of solution, $Δ_{sol} H .$ Value solvent ki amount ke saath badalti rehti hai jab tak solution itna dilute na ho jaaye ki solute particles kabhi ek doosre se na milen; woh limiting value infinite dilution par enthalpy of solution hai.

Ionic solid ko gholna do kalpanik steps mein baanta ja sakta hai. Pehle lattice ko kheench kar gaseous ions mein alag kiya jaata hai, jismein lattice enthalpy lagti hai. Phir paani ke molecules har ion ke chaaron taraf jama ho jaate hain, aur nikalti hai hydration enthalpy, $Δ_{hyd} H :$

\begin{aligned} Δ_{sol} H & = Δ_{lattice} H \\ + Δ_{hyd} H \end{aligned}

KCl ke liye lattice mein +717.8 kJ mol⁻¹ lagte hain aur measure ki gayi enthalpy of solution +17.2 kJ mol⁻¹ hai, isliye ek $K^{+}$ aur ek $C l^{-}$ ko hydrate karne mein lagbhag 700.6 kJ nikalne chahiye. Result do bade numbers ke beech ka chhota sa farq hai: KCl thodi thandak ke saath ghulta hai. KCl typical hai, kyunki zyada tar salts ko gholne mein thodi heat lagti hai, aur aise salts aam taur par thande paani se zyada garam paani mein ghulte hain. Doosri extreme par, bahut kas kar juda lattice shayad hydration se kabhi wapas na chukaya ja sake, aur salt mushkil se ghulta hai.

Try karke dekho

Heat of solution mehsoos karo

Tumhe chahiye

teen chhote beakers ya steel ke glass
ek thermometer
paani
sodium chloride, ammonium chloride aur anhydrous calcium chloride, har ek 5 g

Pehle guess karo

Kaunsa salt apne paani ko saaf taur par zyada thanda karega?

calcium chlorideammonium chloridesodium chloride

Har container mein 50 mL paani daalo aur uska temperature note karo.
Har ek mein ek salt ka 5 g daalo, ghulne tak hilao, aur pahuncha hua sabse kam ya sabse zyada temperature note karo.

Table 5.4

Salt	Start (°C)	End (°C)
NaCl
NH₄Cl
CaCl₂

Tumhari entries apne aap save hoti hain.

Dilution ki enthalpy

Enthalpy of solution is par depend karti hai ki kitna solvent use hua. Table woh enthalpy change deti hai jab 298 K par 1 mol pure liquid sulphuric acid ko paani ki alag-alag amounts ke saath milaya jaata hai, NIST–JANAF thermochemical tables se calculate kiya hua.

Table 5.5298 K par 1 mol

H_{2} S O_{4} (l)

ko paani ke saath milana

Daala gaya paani (mol)	$Δ H$ / kJ mol⁻¹
1	−27.8
2	−41.5
4	−53.9
6.5	−61.4

Jaise-jaise zyada paani daalte hain, values size mein badhti hain, lekin kam se kam amounts se, infinite dilution par ek limit ki taraf. Do entries ke beech ka farq enthalpy of dilution hai: pehle se bane solution mein extra solvent daalne par exchange hui heat. Solution ko per mole acid 2 se 4 mol paani tak le jaane par aur 12.4 kJ mol⁻¹ nikalti hai:

\begin{aligned} Δ_{dil} H & = - 53.9 + 41.5 \\ = - 12.4 \end{aligned}

Dhyaan do ki number is par depend karta hai ki tum kahan se shuru karte ho (per mole acid 2 mol paani) aur kitna paani daalte ho (2 mol aur), isliye dono batana zaroori hai. Sulphuric acid ke liye yeh bada aur negative hai, isiliye har lab yeh rule follow karti hai: hamesha acid ko paani mein daalo, dheere-dheere aur hilaate hue. Concentrated acid mein paani daaloge to pehli boondein chhoote hi ubal kar acid ko bahar uchhaal sakti hain.

5.6 Spontaneity

First law humein batata hai ki energy conserve hoti hai, lekin yeh nahi ki cheezein kis taraf jaati hain. Phir bhi nature saaf taur par ek-taraf hai. Thande kamre mein garam chai thandi hoti hai; woh kabhi kamre se heat lekar khud garam nahi hoti. Perfume ki ek boond poore kamre mein phail jaati hai; khushboo kabhi wapas bottle mein jama nahi hoti. Nam hawa mein iron ko jang lagta hai; jang kabhi apne aap chamakdaar iron nahi banta. Direction kaun decide karta hai?

Jo process khud se ho sakta hai, bina bahar se lagaataar madad ke, woh spontaneous (स्वतः प्रवर्तित) hai. Yeh word speed ke baare mein kuch nahi kehta. Hawa mein mila petrol vapour saalon tak bina jale pada reh sakta hai, phir bhi reaction spontaneous hai: ek chingaari use shuru karti hai, aur phir woh khud chalta hai. Diamond ka graphite banna room temperature par spontaneous hai lekin itna dheema ki kisi ne kabhi hote nahi dekha. Aur apne haal par chhoda jaaye to spontaneous change kabhi ulta nahi chalta. Kisi cheez ko un-mix, un-burn ya dobara garam karne ke liye tumhe bahar ki madad se keemat chukaani padti hai, jaise pump, refrigerator ya battery.

Kya enthalpy mein kami spontaneity ka criterion hai?

Ek ball katori ki tali tak lud jaati hai; khincha hua rubber band wapas jhatak jaata hai. Mechanical systems wahan tik jaate hain jahan unki energy sabse kam ho, isliye yeh andaaza lagane ka mann karta hai ki reactions usi direction mein chalte hain jo enthalpy kam kare. Kai spontaneous reactions sach mein exothermic hain: methane ka jalna (−890.3 kJ mol⁻¹), magnesium ka apni chamakdaar safed flame ke saath jalna, nitrogen aur hydrogen se ammonia banna (ammonia ke per mole −45.9 kJ). Enthalpy diagram par inke products reactants se neeche hote hain.

Lekin yeh idea kuch bahut aam changes ke liye fail ho jaata hai. Garam din mein ice pighalti hai, 6.01 kJ mol⁻¹ absorb karke. Geele kapde se paani evaporate hota hai, heat absorb karke, isiliye kapda thanda lagta hai. Ammonium nitrate paani mein itna endothermic tareeke se ghulta hai ki sports injuries ke instant cold packs isi par tike hain. Yeh sab spontaneous hain, phir bhi enthalpy badhti hai. Enthalpy ka girna aksar change ki madad karta hai, lekin yeh poori kahani nahi ho sakti.

Entropy aur spontaneity

Ek aisa change dekho jismein koi enthalpy change hi nahi. Do alag ideal gases ek isolated box mein saath-saath hain, ek partition se alag (Fig. 5.13). Partition hatao aur woh mix ho jaati hain, spontaneously aur poori tarah, jabki $Δ H = 0.$ Woh kabhi alag nahi hotin. Kyun nahi?

Chances gino. Har molecule random ghoomta hai aur box ke kisi bhi half mein hone ka chance barabar hai. Ek khaas A molecule ke left half mein hone ka chance ½ hai. Figure ke saare 8 A molecules ke ek saath wahan hone ka chance $(1 / 2)^{8}$ hai, yaani 256 mein 1. Gas ke ek mole ke liye chance ½ ko khud se $6 \times 10^{23}$ baar guna karna hai, itna chhota ki unmixed state kabhi nahi dikhegi. Mixed state energy ki wajah se nahi jeetti, balki isliye ki mixed hone ke tareeke bahut, bahut zyada hain.

3/3Saare 8 circles wapas left mein: 256 mein 1 chance. $10^{23}$ molecules ke liye: kabhi nahi.

Fig. 5.13Mixing spontaneous hai kyunki mixed arrangements unmixed arrangements se kahin zyada hain.NCERT Fig. 5.11 · p. 158

Ek isolated box mein gas A ke 8 molecules circles ke roop mein aur gas B ke 8 molecules squares ke roop mein hain. Step one: ek partition box ko baantta hai, saara A left mein aur saara B right mein. Step two: partition hat jaata hai aur molecules tab tak move karte hain jab tak circles aur squares poore box mein phail na jaayen. Step three: neeche text batata hai ki saare 8 A molecules ke left half mein wapas aane ka chance 256 mein 1 hai, aur gas ke ek mole ke liye yeh kabhi nahi dikhta.

Is baat ko pakadne wali property entropy, $S$ (एन्ट्रॉपी) hai. Mote taur par, entropy system ki randomness ya disorder measure karti hai; zyada precisely, yeh measure karti hai ki uske particles aur energy kitne tareekon se arrange ho sakte hain bina yeh badle ki hum kya observe karte hain. Isolated system mein spontaneous change hamesha zyada entropy ki taraf jaata hai. Crystal, jismein har particle apni jagah par hai, kisi substance ki states mein sabse kam entropy rakhta hai, uska liquid zyada, aur uski gas sabse zyada. Jis reaction ke products reactants se kam ordered hon, jaise solids se gas banane wala, uska entropy change positive hota hai.

Definition

Entropy

Ek state function, $S$ , jo measure karta hai ki system ka matter aur energy kitna phaila hua hai, mote taur par uska disorder; har spontaneous change mein total entropy badhti hai.

SimulationTeen states mein order aur disorder

Ek band box mein paani ke particles, −20 se 120 degrees Celsius tak ke temperature slider ke saath. Kam temperature par particles ek regular lattice mein baithe rehte hain aur sirf vibrate karte hain, ice ki tarah. Garam karne par woh pighal kar box ki tali mein dhakka-mukki karta liquid ban jaate hain, aur 100 degrees se upar woh gas ki tarah poore box mein udte hain. Ek thermometer aur state label abhi ka temperature aur state dikhate hain.

Fig. 5.14Temperature ko −20 °C se 120 °C tak drag karo aur dekho particles kitni azaadi se move karte aur phailte hain.

Try karo

−20 °C se shuru karo aur dheere garam karo. Kin do temperatures par arrangement sabse achanak badalta hai?
Ice, liquid paani aur steam ko entropy ke hisaab se rank karo, yeh dekh kar ki particles kitne alag arrangements le sakte hain.
Ek hi state ke andar, kya garam karne se motion zyada random hota hai? 10 °C ko 90 °C se compare karo.

Entropy heat ke through bhi measure ki ja sakti hai. System mein di gayi heat uske particles ko zyada zor se move karati hai, isliye entropy badhti hai. Lekin wahi heat garam system se zyada thande system ke liye matter karti hai. Shaant library mein ek chheenk bheed bhare railway station mein wahi chheenk se kahin zyada khalal daalti hai. Isi tarah, heat ka ek joule thande, shaant system ko garam, pehle se hilte-dulte system se kahin zyada hilata hai, isliye isse bani entropy temperature badhne ke saath ghatni chahiye. Temperature $T$ par reversible process ke liye,

Δ S = \frac{q_{rev}}{T}

(5.18)

$U$ aur $H$ ki tarah entropy bhi state function hai, isliye do states ke beech $Δ S$ same hai chahe change kaise bhi ho. Pench yeh hai ki humein ise reversible path par calculate karna padta hai, use karke $q_{rev},$ tab bhi jab asli process irreversible ho. Spontaneous process ke liye system plus surroundings ka total entropy change positive hota hai:

\begin{aligned} Δ S_{total} & = Δ S_{sys} \\ + Δ S_{surr} \\ > 0 \end{aligned}

(5.19)

Isliye spontaneous change entropy ki pahaadi par chadhaai hai. Equilibrium choti hai: wahan pahunch kar koi bhi aage ka chhota change deta hai $Δ S_{total} = 0.$

Entropy un processes mein farq kar sakti hai jinhe energy nahi kar sakti. 1 mol ideal gas ko constant 298 K par apna volume double karne do. Expansion free ho ya reversible, $Δ U = 0,$ aur gas same final state tak pahunchti hai, isliye $Δ S_{sys}$ bhi same hai. Reversible path par $q_{rev} = R T \ln 2,$ isliye $Δ S_{sys} = R \ln 2 = + 5.76 J K^{- 1} .$ Reversible expansion mein surroundings wahi heat same temperature par deti hain: $Δ S_{surr} = - 5.76 J K^{- 1}$ aur $Δ S_{total} = 0.$ Free expansion mein surroundings tak kuch nahi pahunchta, isliye $Δ S_{total} = + 5.76 J K^{- 1},$ jo irreversible, spontaneous change ki pehchaan hai.

Example 5.10NCERT Problem 5.10 · p. 159

$Δ S$ ka sign predict karna

Kya system ki entropy badhti hai ya ghatti hai? (a) Dry ice, solid $C O_{2}$ , gas ban jaati hai. (b) Thande glass par water vapour condense hota hai. (c) $C a C O_{3} (s) \overset{}{\to} C a O (s) + C O_{2} (g)$ (d) $N_{2} (g) + 3 H_{2} (g) \overset{}{\to} 2 N H_{3} (g)$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5.11NCERT Problem 5.11 · p. 160

Magnesium jalana: universe ki entropy

298 K par $2 M g (s) + O_{2} (g) \overset{}{\to} 2 M g O (s)$ ke liye $Δ_{r} H^{\circ} = - 1203.4 k J m o l^{- 1}$ aur $Δ_{r} S^{\circ} = - 216.7 J K^{- 1} m o l^{- 1} .$ System ki entropy girti hai, kyunki ek gas use ho jaati hai. Phir reaction spontaneous kyun hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Gibbs energy aur spontaneity

$Δ S_{total}$ check karne ka matlab hai system ke saath surroundings ka bhi hisaab rakhna, jo mushkil hai. 1870s mein American scientist J. Willard Gibbs ne surroundings ko sirf system ki ek property mein samet lene ka tareeka dhoondha, ek naya state function jise ab Gibbs energy $G$ (गिब्ज़ ऊर्जा) kehte hain:

G = H - T S

(5.20)

$G$ ek extensive state function hai. Constant temperature par change ke liye,

Δ G = Δ H - T Δ S

(5.21)

Yeh Gibbs equation hai. Har term sirf system ka hai, aur har ek energy hai: K mein $T$ guna J K⁻¹ mein $Δ S$ joules deta hai.

Definition

Gibbs energy

$G = H - T S$ ; constant temperature aur pressure par process spontaneous hota hai jab $Δ G < 0.$

Derivation

Δ G < 0

ka matlab spontaneous kyun hai

Maan lo system apni surroundings ke same temperature $T$ par hai, constant pressure par. System jo bhi heat chhodta hai, surroundings use absorb karti hain, isliye $Δ H_{surr} = - Δ H_{sys}$ aur

\begin{aligned} Δ S_{surr} & = \frac{Δ H_{surr}}{T} \\ = - \frac{Δ H_{sys}}{T} \end{aligned}

Ise Eq. 5.19 mein rakho aur poore ko $T$ se guna karo:

\begin{aligned} T Δ S_{total} \\ = T Δ S_{sys} - Δ H_{sys} \\ = - Δ G_{sys} \end{aligned}

Spontaneous process ke liye $Δ S_{total} > 0,$ isliye uske liye

\begin{aligned} Δ G & = Δ H - T Δ S \\ < 0 \end{aligned}

(5.22)

To constant temperature aur pressure par: agar $Δ G < 0$ to process spontaneous hai; agar $Δ G > 0$ to non-spontaneous hai, aur ulta process spontaneous hai. Equation ka ek energy wala matlab bhi hai. Reaction jo enthalpy chhodta hai, usmein se $T Δ S$ wala hissa system ki entropy ke change mein bandh jaata hai aur kaam mein nahi laaya ja sakta. Jo bachta hai, $- Δ G$ , woh expansion work ke alawa sabse zyada useful work hai, jaise cell mein electrical work, jo reaction kabhi de sakta hai. Isiliye iska purana naam hai free energy.

Dono terms saath mil kar ya ek doosre ke khilaaf kheench sakte hain (Table 5.6, Fig. 5.15). Jab $Δ H$ aur $Δ S$ ka sign same ho, to temperature decide karta hai, aur changeover wahan aata hai jahan $Δ G = 0,$ yaani $T = Δ H / Δ S$ par (dono ko lagbhag constant maan kar). Ek reaction lo jo heat absorb karta hai lekin disorder banata hai, yaani $Δ H > 0$ aur $Δ S > 0.$ Bada entropy gain uska changeover temperature neeche le aata hai, shayad room temperature se bhi neeche; chhota gain use furnace temperatures tak upar dhakel deta hai, isiliye kilns aur blast furnaces itne garam chalte hain.

Table 5.6Kya reaction spontaneous hai? Temperature kaise decide karta haiNCERT Table 5.4 · p. 162

$Δ H$	$Δ S$	Kam $T$ par	Zyada $T$ par
−	+	haan	haan
−	−	haan	nahi
+	+	nahi	haan
+	−	nahi	nahi

5/5Chaaron cases ek saath.

Fig. 5.15

Δ G = Δ H - T Δ S

T

mein ek seedhi line hai jiska intercept

Δ H

aur slope

- Δ S

hai.

Temperature ke against Delta G; axis ke neeche ka hissa shaded hai aur spontaneous likha hai. Har line Delta H se shuru hoti hai aur uska slope minus Delta S hai. Step one: line 1 (Delta H negative, Delta S positive) zero se neeche shuru hoti hai aur girti hai: hamesha spontaneous. Step two: line 2 (dono negative) zero se neeche shuru hoti hai aur chadhti hai, T = Delta H over Delta S par cross karti hai. Step three: line 3 (dono positive) zero se upar shuru hoti hai aur axis ke paar girti hai. Step four: line 4 (Delta H positive, Delta S negative) zero se upar shuru hoti hai aur chadhti hai: kabhi spontaneous nahi. Step five chaaron ko numbered dikhata hai.

Fig. 5.16Ek reaction chuno ya apna

Δ H

aur

Δ S

set karo, phir temperature drag karke

Δ G

, changeover temperature aur

K

dekho.

Entropy aur second law of thermodynamics

Poore universe ke chaaron taraf ek boundary kheencho, system aur surroundings dono ek saath, aur tumhare paas sabse bada isolated system hai, jiski energy first law se fixed hai. Samay aage badhne ke saath jo badalta hai woh iski entropy hai, jo sirf badhti hi hai. Yahi second law of thermodynamics hai. Clausius ka version, ki heat apne aap thandi body se garam body mein nahi jaati, aur engineer ka version, ki cycle mein chalne wala koi engine heat ko poori tarah work mein nahi badal sakta, alag angles se dekha gaya wahi law hain.

Yeh ek pattern bhi samjhata hai jo tumne shayad notice kiya ho: zyada tar reactions jo 'bas chal padte hain' heat chhodte hain. Jo heat woh surroundings mein daalte hain woh entropy ka bada tohfa hai, $Δ S_{surr} = - Δ H / T,$ jo aam taur par system mein bane kisi bhi order par bhaari padta hai, jaise Example 5.11 mein magnesium jalane mein hua.

Absolute entropy aur third law of thermodynamics

Enthalpy ka koi absolute zero nahi hai, isiliye hum hamesha sirf uske changes list karte hain. Entropy alag hai. Kisi pure substance ke perfect crystal ko 0 K ki taraf thanda karo aur uske molecules ek ke baad ek apni azaadi kho dete hain, udne ki (translation), ghoomne ki (rotation) aur apne bonds kheenchne ki (vibration), jab tak sirf ek, perfectly ordered arrangement na bache. Third law of thermodynamics kehta hai ki tab uski entropy zero ho jaati hai. Solutions aur glassy, supercooled liquids is mein nahi aate, kyunki unmein kuch randomness jam jaati hai aur 0 K par bhi bachi rehti hai.

Shuru karne ke liye zero hone se absolute entropy sirf heat measurements se banayi ja sakti hai. Crystal ko 0 K ke paas se chhote reversible steps mein garam karo, har step ke liye aur raaste mein har melting ya boiling ke liye $q_{rev} / T$ jodo, aur 298 K par ruko. Total substance ki standard molar entropy $S^{\circ}$ hai. Reaction entropies phir usi products-minus-reactants sum se aati hain jo enthalpies ke liye use hota hai.

Table 5.7298 K par kuch standard molar entropies

Substance	$S^{\circ}$ / J K⁻¹ mol⁻¹
$C$ (diamond)	2.4
$H_{2} O (l)$	69.9
$H_{2} (g)$	130.7
$H_{2} O (g)$	188.8
$N_{2} (g)$	191.6
$N H_{3} (g)$	192.8
$O_{2} (g)$	205.1

Gases ki entropies solids aur liquids se kahin zyada hoti hain, aur liquid paani steam se kaafi neeche hai, bilkul jaise simulation ne dikhaya tha.

Example

Ammonia synthesis kab spontaneous hona band ho jaata hai?

Is reaction ke liye, $\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \overset{}{\to} N H_{3} (g),$ $Δ H^{\circ} = - 45.9 k J m o l^{- 1} .$ Table 5.7 use karke nikalo $Δ S^{\circ},$ phir 298 K par $Δ G^{\circ}$ , aur woh temperature jiske upar standard conditions mein reaction spontaneous nahi rehta.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

5.7 Gibbs Energy Change aur Equilibrium

Catalyst par nitrogen aur hydrogen milao aur ammonia banti hai; pure ammonia se shuru karo aur uska kuch hissa toot jaata hai. Dono tarah mixture same composition par khatam hota hai aur wahin rehta hai. Gibbs energy samjhati hai ki woh rukne ka point kahan hai.

Reaction aage badhne ke saath poore reacting mixture ki Gibbs energy ko socho (Fig. 5.17). Pure reactants ki ek value hai aur pure products ki doosri, lekin dono ka mixture dono se neeche hota hai, kyunki mixing entropy jodti hai. Isliye $G$ beech mein ek minimum se guzarta hai. Kisi bhi taraf se minimum ki taraf jaana $G$ ko kam karta hai aur spontaneous hai; door jaana use badhayega. Mixture minimum par tik jaata hai, ek dynamic equilibrium jismein forward aur backward reactions barabar rates par chalte rehte hain, aur

Equilibrium par

Δ_{r} G = 0

3/3Minimum par $Δ_{r} G = 0$ aur composition badalna band ho jaata hai.

Fig. 5.17Kisi bhi taraf se, reaction mixture

G

mein dhalaan se neeche equilibrium minimum tak khisakta hai.

Vertical axis G hai aur horizontal axis left mein all reactants se right mein all products tak jaata hai. Step one: do dots pure reactants ka G, zyada, aur pure products ka G, kam, mark karte hain; right par ek double arrow gap dikhata hai, standard Gibbs energy change. Step two: ek curve unhe jodta hai, dono se neeche jhukta hua kyunki mixing G ko kam karti hai. Step three: sabse neeche wale point par equilibrium likha hai, wahan ek flat dashed tangent hai, aur curve par arrows dikhate hain ki mixture dono taraf se minimum tak neeche khisakta hai.

Standard Gibbs energy change, $Δ_{r} G^{\circ},$ jismein har reactant aur product apni standard state mein ho, reaction ke equilibrium constant $K$ se juda hai:

\begin{array}{c} Δ_{r} G^{\circ} = - R T \ln K \\ = - 2.303 R T \log K \end{array}

(5.23)

Standard conditions mein Gibbs equation ke saath milkar, yeh heat measurements aur equilibrium ke beech pul banata hai:

\begin{aligned} Δ_{r} G^{\circ} & = Δ_{r} H^{\circ} \\ - T Δ_{r} S^{\circ} \\ = - R T \ln K \end{aligned}

(5.24)

Negative $Δ_{r} G^{\circ}$ ka matlab $K > 1$ , isliye products favoured hain; positive ka matlab $K < 1$ . $K$ exponentially depend karta hai is par: $Δ_{r} G^{\circ},$ isliye kuch das kilojoules hi 'mushkil se koi product' ko 'lagbhag poora' se alag karte hain (Table 5.8). Aam taur par enthalpy haavi rehti hai, isliye bahut exothermic reactions lagbhag poore ho jaate hain aur bahut endothermic reactions mushkil se shuru hote hain. Jab $Δ_{r} H^{\circ}$ chhota ho, to $T Δ_{r} S^{\circ}$ term balance ko kisi bhi taraf jhuka sakta hai, jaise Example 5.14 mein $N_{2} O_{4}$ ke liye hota hai.

Table 5.8298 K par standard Gibbs energy change aur K

$Δ_{r} G^{\circ}$ / kJ mol⁻¹	$K$
−50	$5.8 \times 10^{8}$
−10	57
0	1
+10	0.018
+50	$1.7 \times 10^{- 9}$

Practice mein Eq. 5.24 do-taraf ka pul hai. Ek chemical engineer tables se $Δ H^{\circ}$ aur $Δ S^{\circ}$ lekar planned plant ke temperature par $K$ calculate kar sakta hai aur dekh sakta hai ki yield faayde ki hogi ya nahi. Ek lab chemist $K$ measure karke use badal sakta hai is mein: $Δ G^{\circ} .$

Example 5.12NCERT Problem 5.12 · p. 163

Hawa nitric oxide kyun nahi ban jaati

298 K par $N_{2} (g) + O_{2} (g) \overset{}{⇌} 2 N O (g)$ ke liye $K = 4.5 \times 10^{- 31} .$ Nikalo $Δ_{r} G^{\circ} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5.13NCERT Problem 5.13 · p. 163

$Δ_{r} G^{\circ}$ se $K$ tak

298 K par $\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \overset{}{⇌} N H_{3} (g)$ ke liye $Δ_{r} G^{\circ} = - 16.4 k J m o l^{- 1},$ jaisa ammonia wale example mein mila tha. $K$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5.14NCERT Problem 5.14 · p. 163

Thoda dissociate hua dinitrogen tetroxide

25 °C aur 1.00 bar ke total pressure par $N_{2} O_{4} (g) \overset{}{⇌} 2 N O_{2} (g)$ equilibrium par pahunchta hai jab $N_{2} O_{4}$ ka 19 % dissociate ho chuka hota hai. Is temperature par $K$ aur $Δ_{r} G^{\circ}$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Agar kisi temperature par ek reaction ke liye

Δ_{r} G^{\circ} = 0

hai, to us temperature par uska equilibrium constant hai

01−1infinitely bada

Answer dekho

Answer (B) 1

\ln K = - Δ_{r} G^{\circ} / R T = 0,

isliye

K = 1

K

kabhi negative nahi ho sakta.

Recap

Summary aur key points

Thermodynamics matter ko bulk mein dekhti hai aur initial aur final states compare karti hai; yeh rates ke baare mein kuch nahi kehti (Section 5.1).
System open, closed ya isolated hota hai, is par depend karta hai ki matter aur energy uski boundary paar kar sakte hain ya nahi (Section 5.1.2).
$p$ , $V$ , $T$ , $U$ , $H$ , $S$ aur $G$ jaise state functions sirf state par depend karte hain; $q$ aur $w$ path par depend karte hain (Section 5.1.3).
First law: $Δ U = q + w,$ jismein $q$ aur $w$ positive hain jab energy system mein aati hai ((5.1)).
Pressure–volume work hai $w = - p_{ex} Δ V;$ ideal gas par reversible isothermal work hai $- n R T \ln (V_{f} / V_{i}),$ jo ek expansion de sakne wala sabse zyada work hai (Section 5.2.1).
Constant volume par heat $Δ U$ measure karti hai; constant pressure par heat $Δ H$ measure karti hai, jahan $H = U + p V$ aur $Δ H = Δ U + Δ n_{g} R T$ (Section 5.2.2, Section 5.3).
$q = c m Δ T = C Δ T,$ aur ideal gas ke liye $C_{p} - C_{V} = R$ (heat capacity).
Reaction enthalpies formation enthalpies se nikalti hain, products minus reactants coefficients se weight karke, ya Hess's law se steps ke kisi bhi route se (formation, Hess's law).
Phase changes, combustion, atomization, bond todna, lattice ka tootna, solution aur dilution, sabki apni naam wali enthalpy hai; bond enthalpies gas-phase reaction enthalpies estimate karti hain aur Born–Haber cycles lattice enthalpies deti hain (Section 5.5).
Spontaneous process khud chal sakta hai, chahe kitna bhi dheere; har spontaneous change mein total entropy badhti hai, aur $Δ S = q_{rev} / T$ (entropy).
Constant $T$ aur $p$ par change spontaneous hai jab $Δ G = Δ H - T Δ S < 0;$ jab $Δ H$ aur $Δ S$ ka sign same ho to temperature decide karta hai (Gibbs energy).
Third law: perfect crystal ki 0 K par entropy zero hoti hai, jisse absolute entropies possible hoti hain (third law).
Equilibrium par $Δ_{r} G = 0,$ aur $Δ_{r} G^{\circ} = - R T \ln K$ heat data ko equilibrium constants se jodta hai (Section 5.7).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 5.1	Practice Q1
NCERT Exercises 5.2	Practice Q2
NCERT Exercises 5.3	Practice Q3
NCERT Exercises 5.4	Practice Q4
NCERT Exercises 5.5	Practice Q5
NCERT Exercises 5.6	Practice Q6
NCERT Exercises 5.7	Practice Q7
NCERT Exercises 5.8	Practice Q8
NCERT Exercises 5.9	Practice Q9
NCERT Exercises 5.10	Practice Q10
NCERT Exercises 5.11	Practice Q11
NCERT Exercises 5.12	Practice Q12
NCERT Exercises 5.13	Practice Q13
NCERT Exercises 5.14	Practice Q14
NCERT Exercises 5.15	Practice Q15
NCERT Exercises 5.16	Practice Q16
NCERT Exercises 5.17	Practice Q17
NCERT Exercises 5.18	Practice Q18
NCERT Exercises 5.19	Practice Q19
NCERT Exercises 5.20	Practice Q20
NCERT Exercises 5.21	Practice Q21
NCERT Exercises 5.22	Practice Q22

Quick check: MCQs

General case4 questions
5.2.1 Work3 questions
Ideal gas ke liye Cp aur CV ka relation4 questions
ΔH measurements3 questions
Hess's law of constant heat summation4 questions
Dilution ki enthalpy4 questions
Absolute entropy aur third law of thermodynamics5 questions
5.7 Gibbs Energy Change aur Equilibrium3 questions

Glossary · 15

Key terms (English + Hindi)

adiabaticरुद्धोष्मruddhoshm: Aisi wall ya process jisse system aur surroundings ke beech koi heat paar nahi jaati.Chapter mein pehli baar yahan aaya →

bond enthalpyआबंध एन्थैल्पीaabandh enthalpy: Gaseous molecules mein kisi bond ke ek mole ko todne ke liye zaroori enthalpy; aam taur par milte-julte bonds ka mean.Chapter mein pehli baar yahan aaya →

enthalpyएन्थैल्पीenthalpy: $H = U + p V$ ; iska change constant pressure par absorb hui heat ke barabar hai.Chapter mein pehli baar yahan aaya →
entropyएन्ट्रॉपीentropy: Ek state function, $S$ , jo measure karta hai ki matter aur energy kitne phaile hue hain; har spontaneous change mein yeh badhta hai.Chapter mein pehli baar yahan aaya →
extensive propertyविस्तीर्ण गुणधर्मvistirn gundharm: Matter ki amount ke proportional property, jaise mass, volume ya enthalpy.Chapter mein pehli baar yahan aaya →

Gibbs energyगिब्ज़ ऊर्जाGibbs oorja: $G = H - T S$ ; constant $T$ aur $p$ par process spontaneous hai jab $Δ G < 0$ .Chapter mein pehli baar yahan aaya →

Hess's lawहेस का नियमHess ka niyam: Reaction enthalpy same rehti hai chahe reaction ek step mein ho ya kai steps mein, isliye steps ki enthalpies jud jaati hain.Chapter mein pehli baar yahan aaya →

intensive propertyगहन गुणधर्मgahan gundharm: Matter ki amount se independent property, jaise temperature, pressure ya density.Chapter mein pehli baar yahan aaya →
internal energyआंतरिक ऊर्जाaantarik oorja: System mein store hui total energy; sirf iska change, $Δ U$ , measure ho sakta hai.Chapter mein pehli baar yahan aaya →

lattice enthalpyजालक एन्थैल्पीjaalak enthalpy: Woh enthalpy change jab ionic solid ka 1 mol apne gaseous ions mein alag hota hai.Chapter mein pehli baar yahan aaya →

reversible processउत्क्रमणीय प्रक्रमutkramaneeya prakram: Equilibrium ke bilkul paas wali states se chalne wala process, taaki ek infinitesimal change use kisi bhi stage par ulta kar sake.Chapter mein pehli baar yahan aaya →

spontaneous processस्वतः प्रवर्तित प्रक्रमsvatah pravartit prakram: Aisa process jo bina lagaataar bahari madad ke khud aage badh sakta hai, chahe kitna bhi dheere.Chapter mein pehli baar yahan aaya →
standard enthalpy of formationमानक संभवन एन्थैल्पीmaanak sambhavan enthalpy: Woh enthalpy change jab 1 mol compound apne elements se banta hai, sab apni standard ya reference states mein.Chapter mein pehli baar yahan aaya →
state functionअवस्था फलनavastha phalan: Aisi property jo sirf abhi ki state par depend karti hai, isliye iska change sirf initial aur final states par depend karta hai.Chapter mein pehli baar yahan aaya →
systemनिकायnikaay: Universe ka woh hissa jo study ke liye chuna gaya hai; baaki sab jo iske saath interact kar sakta hai, surroundings hai.Chapter mein pehli baar yahan aaya →

5.1 Thermodynamic Terms

5.1.1 System aur Surroundings

5.1.2 System ke Types

5.1.3 System ki State

5.1.4 State Function ke roop mein Internal Energy

Work

Heat

General case

5.2 Applications

5.2.1 Work

5.2.2 Enthalpy, H

Ek kaam ka naya state function

Extensive aur intensive properties

Heat capacity

Ideal gas ke liye Cp aur CV ka relation

5.3 ΔU aur ΔH ka Measurement: Calorimetry

ΔU measurements

ΔH measurements

5.4 Reaction ka Enthalpy Change, ΔrH – Reaction Enthalpy

Reactions ki standard enthalpy

Phase transformations ke dauraan enthalpy changes

Formation ki standard enthalpy

Thermochemical equations

Hess's law of constant heat summation

5.5 Alag-alag Types ke Reactions ki Enthalpies

Combustion ki standard enthalpy

Atomization ki enthalpy

Bond enthalpy

Lattice enthalpy

Solution ki enthalpy

Dilution ki enthalpy

5.6 Spontaneity

Kya enthalpy mein kami spontaneity ka criterion hai?

Entropy aur spontaneity

Gibbs energy aur spontaneity

Entropy aur second law of thermodynamics

Absolute entropy aur third law of thermodynamics

5.7 Gibbs Energy Change aur Equilibrium

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye