Equilibrium: Class 11 Chemistry Chapter 6 Hinglish mein

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

Published 18 Oct 2026 · Updated 18 Oct 2026

125 min
1 simulation
43 practice questions
35 MCQs

Is chapter mein aap kya seekhenge

Samjhao ki physical aur chemical equilibria dynamic kyun hote hain, aur har tarah ke physical equilibrium mein kaun si quantity constant rehti hai, yeh pehchano
Kc aur Kp expressions likho, heterogeneous systems samet, Kp ko Kc se aur badli hui equations se jodo, aur K aur Q se reaction ka extent aur direction judge karo
ICE tables se initial data se equilibrium compositions calculate karo, aur K ko standard Gibbs energy change se jodo
Le Chatelier’s principle se predict karo ki concentration, pressure, inert gas, temperature aur catalyst equilibrium par kya asar daalte hain
Acids aur bases ko Arrhenius, Brønsted–Lowry aur Lewis definitions se classify karo aur conjugate acid–base pairs pehchano
pH, pOH, Ka, Kb aur degree of ionization calculate karo, aur salt aur buffer solutions ka pH bhi, common ion effect samet
Ksp ko molar solubility se jodo aur precipitation predict karo, common ion aur pH ke effects samet

NCERT Chemistry Class 11, Chapter 6, pp. 168–214 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Dukaan ki shelf par rakhi soda ki sealed bottle bilkul shaant dikhti hai: liquid ka level aur gas pressure kabhi nahi badalte. Andar carbon dioxide ke molecules lagataar liquid chhodte aur usmein waapas aate rehte hain, bilkul same rate par. Kuch hota hua dikhta nahi kyunki do ulti changes ek doosre ko cancel kar deti hain. Yeh chapter equilibrium (साम्यावस्था) ki yahi picture banata hai, pehle physical changes ke liye aur phir reactions ke liye.

Half-arrows ka joda aisi change dikhata hai jo ek saath dono taraf chalti hai, jaise $C O_{2} (g) \overset{}{⇌} C O_{2} (a q)$ . Jab composition badalna band ho jaata hai, to saath mein maujood reactants aur products milkar equilibrium mixture banate hain.

Reactions is baat mein bahut alag hote hain ki woh kitni door tak jaate hain. Oxygen mein jalne ke baad hydrogen lagbhag bachti hi nahi; paani ke saath hilaya silver chloride mushkil se dissolve hota hai; iodine vapour ke saath garam ki gayi hydrogen ke end mein teenon gases kaafi maatra mein hoti hain. Reaction kahan settle hoga, yeh temperature, pressure aur shuru ki amounts par depend karta hai, jinhe chemical plants apne product ke favour mein tune karte hain. Chapter ka doosra half yahi ideas paani mein ions par lagata hai: acids, bases, buffers aur sparingly soluble salts, yaani ionic equilibrium ka topic.

6.1 Physical Processes mein Equilibrium

Changes of state mein equilibrium dekhna sabse aasaan hai, kyunki kuch naya nahi banta: molecules sirf solid, liquid aur gas ke beech move karte hain. Yahan teen pairs important hain: $solid ⇌ liquid$ , $liquid ⇌ gas$ aur $solid ⇌ gas$ .

6.1.1 Solid–Liquid Equilibrium

Ice cubes aur paani ko ek achhe insulated flask mein 0 °C (273 K) aur 1 atm par saath rakho. Ghanton baad bhi ice aur paani ke masses, aur temperature, bilkul pehle jaise hote hain. Molecular scale par boundary bilkul bhi shaant nahi hoti: paani ke molecules ice ki surface se judte rehte hain jabki doosre usse alag hote rehte hain. Masses sirf isliye fixed rehte hain kyunki dono exchanges same rate par chalte hain. Yeh balance kisi given pressure par sirf ek hi temperature par hota hai. 1 atm par pure substance ke liye yeh temperature uska normal melting point hai, jo uska normal freezing point bhi hai. Do features abhi se saaf dikhte hain: ulte processes ek hi samay par chalte hain, aur barabar rates par chalte hain.

6.1.2 Liquid–Vapour Equilibrium

Socho ek sookha glass jar hai jiske lid mein pressure gauge laga hai. Andar paani ki ek dish rakho, jar seal karo aur use steady temperature par rakho. Gauge pehle tezi se badhta hai, phir dheere, aur aakhir mein ek fixed reading par ruk jaata hai, jabki paani ka level thoda gir jaata hai (Fig. 6.1).

Iski wajah do rates ke beech ki race hai. Evaporation ka rate temperature aur surface area se tay hota hai. Condensation zero se shuru hota hai aur badhta jaata hai jaise-jaise vapour jamta hai aur zyada molecules paani se takra kar chipakte hain. Jab woh barabar aa jaata hai, to vapour ki amount badalna band ho jaati hai: $H_{2} O (l) \overset{}{⇌} H_{2} O (g)$ , rate of evaporation = rate of condensation.

3/4Ab condensation evaporation ke barabar hai: vapour pressure steady hai.

Fig. 6.1Lid lagi ho to vapour pressure tab steady ho jaata hai jab condensation evaporation ke barabar aa jaata hai.NCERT Fig. 6.1 · p. 169

Paani ka sealed jar aur uske paas vapour pressure vs time ka graph; orange arrows evaporation dikhate hain, blue dashed arrows condensation. Step 1: sirf evaporation arrows, aur curve tezi se upar jaata hai. Step 2: zyada vapour aur ek condensation arrow; curve mudne lagta hai. Step 3: dono taraf teen-teen arrows aur ek flat curve jis par steady likha hai. Step 4: lid hata diya jaata hai, vapour bahar nikalta hai, paani ka level girta hai aur open jar label wala dashed curve neeche hi rehta hai.

Yeh steady pressure paani ka equilibrium vapour pressure hai, jise aksar sirf vapour pressure kehte hain, aur yeh temperature ke saath badhta hai. Same jar mein ethanol ya acetone same temperature par zyada reading deta hai. Jis liquid ka vapour pressure zyada hota hai woh zyada volatile hota hai aur kam temperature par boil karta hai. Agar dish ko khula chhod do to paani aakhir mein khatam ho jaata hai. Molecules pehle ki tarah evaporate hote hain, lekin room mein door chale jaate hain aur bahut kam waapas aate hain, isliye condensation kabhi barabar nahi aa paata. Dish ko sookhne mein kitna time lagega yeh liquid, uski amount aur temperature par depend karta hai, lekin balance kabhi nahi banta: open system equilibrium tak nahi pahunch sakta.

Boiling point par bhi liquid aur vapour saath rehte hain: 1 atm (1.013 bar) par 100 °C par paani aur steam equilibrium mein hote hain. Jis temperature par 1 atm par liquid aur uska vapour equilibrium mein hote hain, woh uska normal boiling point hai. Bahar ka pressure kam karo to boiling point gir jaata hai: Leh mein, lagbhag 3500 m ki oonchai par, paani 88 °C ke aas-paas boil karta hai, isiliye pahadon mein pressure cookers itne zaroori hain.

SimulationClosed aur open container mein evaporation

Ek container mein paani ke particles, temperature slider aur hatne wale lid ke saath. Lid lagi ho to molecules liquid chhodte hain, idhar-udhar takrate hain aur kuch waapas jud jaate hain, isliye vapour steady ho jaata hai. Lid hatne par escape counter badhta hi rehta hai.

Fig. 6.2Play dabao. Lid lagi ho to vapour tab tak jamta hai jab tak jitne molecules paani chhodte hain utne hi waapas na aane lagen.

Try karo

Lid lagi rehne do aur paani ke upar vapour ko dekho. Kya vapour particles ki sankhya hamesha badhti rehti hai?
Lid hata do. Escape counter ka kya hota hai, aur yeh container kabhi equilibrium tak kyun nahi pahunch sakta?
Lid waapas lagao aur temperature badhao. Sab settle hone ke baad vapour zyada hai ya kam?

6.1.3 Solid–Vapour Equilibrium

Kuch solids seedhe vapour ban jaate hain. Glass vessel mein seal kiye iodine crystals use violet vapour se bhar dete hain jiska rang gehra hota hai aur phir steady ho jaata hai: tab se $I_{2} (s) \overset{}{⇌} I_{2} (g)$ , jismein sublimation aur deposition barabar rates par hote hain. Camphor tablets aur naphthalene mothballs bhi aise hi behave karte hain. Ammonium chloride sirf sublime hota hua lagta hai: uska vapour ammonia aur hydrogen chloride hai, jo thanda hone par phir jud jaate hain.

6.1.4 Liquids mein Solids ya Gases ke Dissolution wala Equilibrium

Liquids mein solids. Ek glass paani mein cheeni ghol kar hilao to ek point aata hai jab us temperature par aur cheeni dissolve nahi hoti; extra crystals bas neeche pade rehte hain. Tab solution saturated hota hai. Garam paani mein bani gaadhi chashni, jaise gulab jamun ke liye, thandi hone par crystals banati hai kyunki kam temperature par cheeni kam soluble hoti hai.

Saturated solution mein $sugar(s) ⇌ sugar(aq)$ ek dynamic equilibrium hai: cheeni theek utni hi tezi se dissolve hoti hai jitni tezi se crystallise hoti hai. Radioactive tracers yeh two-way traffic dikhate hain. Radioactive carbon se bani cheeni ke kuch crystals ordinary cheeni ke saturated solution mein daalo. Jaldi hi detector solution mein aur bache hue crystals mein bhi radioactivity paata hai, jabki dissolve hui amount nahi badli, aur yeh tabhi mumkin hai jab dono processes chalte rahen.

Liquids mein gases. Soda carbon dioxide ke high pressure par bottle kiya jaata hai, isliye bahut gas dissolve hoti hai, $C O_{2} (g) \overset{}{⇌} C O_{2} (a q)$ . Ise Henry's law control karta hai: fixed temperature par, liquid ki di hui quantity jitni gas leti hai, woh us liquid ke upar us gas ke partial pressure ke proportional hoti hai. Cap kholne par $C O_{2}$ ka pressure hawa wali chhoti si value tak gir jaata hai, gas bubbles bankar nikalti hai jab tak solution us value se match na kare, aur drink flat ho jaati hai, garam ho to aur jaldi, kyunki garam paani mein gases kam soluble hoti hain.

Table 6.1Kuch physical equilibria mein kya fixed rehta haiNCERT Table 6.1 · p. 171

Process	Example	Given temperature par fixed
Solid ⇌ liquid	$H_{2} O (s) \overset{}{⇌} H_{2} O (l)$	given pressure par sirf ek temperature (melting point)
Liquid ⇌ vapour	$H_{2} O (l) \overset{}{⇌} H_{2} O (g)$	vapour pressure, $p_{H_{2} O}$
Solid ⇌ solution	sugar(s) ⇌ sugar(aq)	saturated solution ka concentration (solubility)
Gas ⇌ solution	$C O_{2} (g) \overset{}{⇌} C O_{2} (a q)$	ratio $[C O_{2} (a q)] / [C O_{2} (g)]$

6.1.5 Physical Processes wale Equilibria ki General Characteristics

Har measurable property, chahe mass, pressure, colour ya concentration ho, badalna band kar deti hai.
Phir bhi kuch ruka nahi hai: ulte processes barabar rates par chalte rehte hain, isliye state dynamic hai lekin stable hai.
Iske liye ek temperature par rakha closed system chahiye.
Given temperature par fixed ek quantity har equilibrium ko describe karti hai (Table 6.1); uska size batata hai ki balance hone se pehle process kitni door gaya.

6.2 Chemical Processes mein Equilibrium – Dynamic Equilibrium

Reactions bhi isi tarah settle hote hain. Closed vessel mein sirf A aur B se $A + B \overset{}{⇌} C + D$ shuru karo. Shuru mein sirf forward reaction chalta hai, sabse tez. Jaise-jaise C aur D jamte hain, reverse reaction shuru hokar tez hota jaata hai aur forward wala dheema padta hai, jab tak dono rates barabar na ho jaayen (Fig. 6.3). Uske baad har concentration constant rehta hai jabki dono reactions chalte rehte hain: yahi chemical equilibrium hai, aur yeh dynamic hai.

3/3Rates barabar hain, isliye concentrations ab nahi badalte, halaanki dono reactions chalte rehte hain.

Fig. 6.3Concentrations theek tab badalna band karte hain jab forward aur reverse rates barabar ho jaate hain.NCERT Fig. 6.2 · p. 172

Ek hi time axis par do graphs: reactants (solid, girta hua) aur products (dashed, badhta hua) ke concentrations, aur forward (solid) aur reverse (dashed) rates. Step 1 shuru ke pal dikhata hai: high forward rate, koi reverse rate nahi. Step 2 curves ko aage badhata hai jab forward rate girta hai aur reverse rate badhta hai. Step 3 ek shaded equilibrium region jodta hai jahan dono concentrations flat hain aur rate lines milti hain, jis par rates equal likha hai.

Har direction ko sirf apne reactants chahiye, isliye same equilibrium kisi bhi side se pahuncha ja sakta hai. Sirf C aur D se shuru karo to reverse reaction kaam karta hai. Given temperature par, same volume mein har tarah ke atom ki same total amount ke saath, dono raaste same composition par khatam hote hain (Fig. 6.4).

Fig. 6.4350 K par 0.10 M

N_{2} O_{4}

ya 0.20 M

N O_{2}

se: same equilibrium.NCERT Fig. 6.5 · p. 174

350 kelvin par dinitrogen tetroxide aur nitrogen dioxide ke do concentration–time graphs. Left: sirf 0.10 molar dinitrogen tetroxide se shuru. Right: sirf 0.20 molar nitrogen dioxide se shuru. Dashed guides dikhate hain ki dono same levels tak pahunchte hain, lagbhag 0.091 molar nitrogen dioxide aur 0.055 molar dinitrogen tetroxide.

Humein kaise pata ki reactions chalte rehte hain? Fritz Haber ke study kiye ammonia synthesis, $N_{2} (g) + 3 H_{2} (g) \overset{}{⇌} 2 N H_{3} (g)$ , ke saath isotopes ek saaf test dete hain. Do identical equilibrium mixtures banao, ek ordinary hydrogen se aur ek deuterium, $D_{2}$ , se, aur phir unhe mila do. Ammonia ki total amount nahi badalti, phir bhi mass spectrometer jaldi hi $N H_{2} D$ , $N H D_{2}$ aur $H D$ paata hai (Fig. 6.5). H aur D atoms molecules ke beech tabhi shuffle ho sakte hain jab ammonia banta aur tootta rahe: equilibrium par dono reactions abhi bhi chal rahe hain.

3/3Mass spectrometer NH₂D, NHD₂ aur HD paata hai: reactions kabhi ruke nahi.

Fig. 6.5Mixing ke baad H aur D ka scramble hona dikhata hai ki equilibrium par ammonia banta aur tootta rehta hai.

Ek box jo deewar se hydrogen mixture (NH3, H2, N2) aur deuterium mixture (ND3, D2, N2) mein banta hai. Step 1: do alag mixtures. Step 2: deewar hata di jaati hai aur ammonia ki total amount same rehti hai. Step 3: molecules ko NH2D, NHD2 aur HD jaise mixed forms ke roop mein relabel kiya jaata hai, jo dikhata hai ki H aur D scrambled hain.

Student activityNCERT Dynamic equilibrium – ek student activity · p. 173

Do straws se dynamic equilibrium ka model

Tumhe chahiye

do identical lambe, transparent glasses
do drinking straws, ek chauda aur ek patla
food colour ki ek boond wala paani
ek ruler

Pehle guess karo

Glass 1 aadha bhara hai aur glass 2 khaali. Tum chaude straw se glass 1 se glass 2 mein paani le jaate raho aur patle straw se waapas. Levels kahan settle honge?

Dono glasses mein barabarGlass 2 mein zyadaGlass 1 mein zyada

Glass 1 ko coloured paani se aadha bharo; glass 2 khaali rehne do. Chauda straw glass 1 mein aur patla straw glass 2 mein rakho.
Chauda straw glass 1 ki tali tak daalo, uska upar wala sira ungli se band karo aur use glass 2 mein khaali karo. Phir patle straw se glass 2 ka paani glass 1 mein le jao. Yeh ek round hai.
Har 5 rounds par dono levels naapo jab tak woh badalna band na kar dein, phir 10 rounds aur karo. Do barabar straws ke saath dohraao.

Tumhari readings (chauda straw glass 1 mein)

Rounds	Glass 1 mein level (cm)	Glass 2 mein level (cm)
5
10
20
30

Tumhari entries apne aap save hoti hain.

Levels pehle tezi se badalte hain aur phir settle ho jaate hain, halaanki paani dono taraf move karta rehta hai: yeh barabar aur ulte rates se bane constant concentrations ka model hai. Aadhi width wala straw har centimetre par ek chauthai paani uthata hai, isliye glass 2 glass 1 se lagbhag chaar guna gehra settle hota hai. Straw sizes, rate constants ki tarah, yeh tay karte hain ki balance kahan hai, yeh nahi ki balance banega ya nahi.

Quick check

3 mein se 1

Equilibrium par pahunchne ke baad

N_{2} + 3 H_{2} \overset{}{⇌} 2 N H_{3}

ko kaun sa statement describe karta hai?

Forward aur reverse rates barabar hain aur zero nahi hainDono reactions ruk gaye hain

[N_{2}] = [H_{2}] = [N H_{3}]

Ammonia abhi bhi dheere-dheere banta hai, isliye uska concentration dheere badhta hai

Answer dekho

Answer (A) Forward aur reverse rates barabar hain aur zero nahi hain

Rates barabar hain, isliye concentrations constant hain, lekin barabar nahi: K unka ratio fix karta hai.

6.3 Law of Chemical Equilibrium aur Equilibrium Constant

Teen sawaal apne aap aate hain. Equilibrium mixture mein concentrations aapas mein kaise jude hain? Shuru ki amounts se hum unhe kaise predict karein? Aur composition ko apni marzi ki taraf kaise push karein? Aakhri sawaal har baar important hai jab koi factory ammonia, hydrogen ya quicklime banati hai. 1864 mein Norwegian chemists Cato Guldberg aur Peter Waage ne kai reversible reactions study karke jawaab dhoondha: $A + B \overset{}{⇌} C + D$ ke liye equilibrium par ratio $[C] [D] / ([A] [B])$ ki given temperature par ek hi value hoti hai, shuru ki amounts chahe jo hon. Us samay concentration ko 'active mass' kaha jaata tha, isliye naam pada law of mass action.

Coefficients matter karte hain. Table 6.2 350 K par $N_{2} O_{4} (g) \overset{}{⇌} 2 N O_{2} (g)$ ke chaar runs dikhata hai, do $N_{2} O_{4}$ se shuru hue aur do $N O_{2}$ se. Seedha ratio $[N O_{2}] / [N_{2} O_{4}]$ lagbhag chaar guna tak badal jaata hai. $[N O_{2}]$ ko square karo, kyunki uska coefficient 2 hai, aur ratio har baar same aata hai (Fig. 6.6).

Table 6.2350 K par

N_{2} O_{4} \overset{}{⇌} 2 N O_{2}

ke chaar equilibrium runs (concentrations mol L⁻¹ mein)NCERT Tables 6.2, 6.3 · p. 175

Run	Start $[N_{2} O_{4}]$	Start $[N O_{2}]$	Equil. $[N_{2} O_{4}]$	Equil. $[N O_{2}]$	$\frac{[N O_{2}]}{[N_{2} O_{4}]}$	$\frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$
1	0.030	0	0.0103	0.0394	3.83	0.151
2	0.080	0	0.0409	0.0783	1.91	0.150
3	0	0.20	0.0547	0.0906	1.66	0.150
4	0	0.40	0.1301	0.1397	1.07	0.150

3/3Us line ka slope equilibrium constant hai, 350 K par 0.15.

Fig. 6.6Sirf

[N O_{2}]^{2} / [N_{2} O_{4}]

constant hai: power equation mein

N O_{2}

ka coefficient hai.

Step 1: chaar runs ke liye equilibrium nitrogen dioxide concentration ko dinitrogen tetroxide ke against plot karne par points ek curve par aate hain. Step 2: doosra graph nitrogen dioxide concentration ka square plot karta hai, aur points origin se guzarti seedhi line par aate hain. Step 3: us line par ek slope triangle uska gradient dikhata hai, jo equilibrium constant hai, lagbhag 0.15.

LawLaw of chemical equilibrium

Fixed temperature par equilibrium wale reaction ke liye, product concentrations ko aapas mein multiply karo, har ek ko balanced equation mein uske coefficient ki power dekar, aur reactant concentrations ko isi tarah treat karke unse divide karo. Result hamesha same number aata hai, equilibrium constant $K_{c}$ .

Equilibrium constant

K_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}

(6.4)

jahan

$a A + b B ⇌ c C + d D$: balanced equation
$[]$: equilibrium molar concentration, mol L⁻¹

Brackets mein equilibrium concentrations maani jaati hain, isliye 'eq' label ki zaroorat nahi, aur subscript c batata hai ki constant molar concentrations use karta hai. Expression mein state symbols aam taur par nahi likhe jaate. High temperature par methane ke jalne, $C H_{4} (g) + 2 O_{2} (g) ⇌ C O_{2} (g) + 2 H_{2} O (g)$ , ke liye expression $K_{c} = [C O_{2}] [H_{2} O]^{2} / ([C H_{4}] [O_{2}]^{2})$ hai. Kyunki $K_{c}$ ek particular equation ka hota hai, equation dobara likhne par yeh badal jaata hai. Equation ulti karne par expression ulta ho jaata hai; har coefficient double karne par yeh square ho jaata hai; aadha karne par iska square root aata hai (Table 6.3). K ko hamesha uski equation ke saath batao.

Table 6.3Equation dobara likhne par

K_{c}

kaise badalta haiNCERT Table 6.4 · p. 177

Equation mein change	Naya constant	Example: $H_{2} + I_{2} \overset{}{⇌} 2 H I$ , $K_{c} = 50$
Ulta karo	$1 / K_{c}$	$2 H I \overset{}{⇌} H_{2} + I_{2}$ : 0.020
Har coefficient ko $n$ se multiply karo	$K_{c}^{n}$	$2 H_{2} + 2 I_{2} \overset{}{⇌} 4 H I$ : 2500
Har coefficient aadha karo	$K_{c}^{1 / 2}$	$\frac{1}{2} H_{2} + \frac{1}{2} I_{2} ⇌ H I$ : 7.07
Do equations jodo	$K_{1} \times K_{2}$	dekho Section 6.11.5

Example 6.1NCERT Problems 6.1, 6.2 · p. 177

Equilibrium concentrations se Kc

Methanol 'synthesis gas' se banta hai, $C O (g) + 2 H_{2} (g) ⇌ C H_{3} O H (g)$ . 500 K par ek equilibrium mixture mein $[C O] = 0.050$ , $[H_{2}] = 0.10$ aur $[C H_{3} O H] = 9.0 \times 10^{- 3} M$ hai. $K_{c}$ nikalo, aur (a) reverse reaction aur (b) $\frac{1}{2} C O + H_{2} ⇌ \frac{1}{2} C H_{3} O H$ ke liye $K_{c}$ nikalo.

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

Quick check

3 mein se 1

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

ke liye

K_{c}

kaun sa hai?

\frac{[S O_{3}]}{[S O_{2}] [O_{2}]}

\frac{[S O_{2}]^{2} [O_{2}]}{[S O_{3}]^{2}}

\frac{2 [S O_{3}]}{2 [S O_{2}] [O_{2}]}

\frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}

Answer dekho

Answer (D) $\frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$

Products upar, reactants neeche, har ek uske coefficient ki power ke saath. Square karne ki jagah 2 se multiply karna ek common galti hai.

6.4 Homogeneous Equilibria

Homogeneous equilibrium mein har species same phase mein hoti hai: sab gases, jaise ammonia synthesis mein, ya sab ek hi solution mein. Paani mein dichromate aur chromate ions iska example hain: $C r_{2} O_{7}^{2 -} (a q) + H_{2} O (l) ⇌ 2 C r O_{4}^{2 -} (a q) + 2 H^{+} (a q)$ , left mein orange, right mein yellow.

6.4.1 Gaseous Systems mein Equilibrium Constant

Gases ke liye aksar concentrations se zyada aasaan pressures naapna hota hai. Ideal gas equation, $p V = n R T$ , se $p = (n / V) R T = c R T$ milta hai, isliye fixed temperature par har partial pressure us gas ke molar concentration ke proportional hota hai. $p$ bar mein aur $c$ mol L⁻¹ mein ho to $R = 0.0831 b a r L m o l^{- 1} K^{- 1}$ . Partial pressures se likha equilibrium law $K_{p}$ define karta hai.

DerivationKp ka Kc se kya relation hai

Maan lo gas A ke $a$ moles gas B ke $b$ moles se react karke C ke $c$ moles aur D ke $d$ moles dete hain. Partial pressures mein,

K_{p} = \frac{p_{C}^{c} p_{D}^{d}}{p_{A}^{a} p_{B}^{b}}

(6.12)

Ab har partial pressure ko concentration × $R T$ ke roop mein likho, jaise $p_{C} = [C] R T$ . Concentrations phir se $K_{c}$ bana dete hain, jabki numerator ko $(R T)^{c + d}$ aur denominator ko $(R T)^{a + b}$ milta hai, isliye $R T$ ke factors ek power mein jud jaate hain:

Kp aur Kc

K_{p} = K_{c} (R T)^{Δ n}

(6.15)

jahan

$Δ n$: balanced equation mein (gaseous products ke moles) − (gaseous reactants ke moles)

Isliye $K_{p} = K_{c}$ sirf tab hota hai jab $Δ n = 0$ ho (Fig. 6.7). Pressures bar mein hone chahiye, jo standard pressure hai, $1 b a r = 10^{5} P a$ . Solids aur liquids $Δ n$ mein nahi gine jaate.

Fig. 6.7Gas moles gino:

Δ n

R T

ki power tay karta hai.

Teen cards, har ek par ek equation, pehle aur baad ke gas molecules gine hue circles, aur result. Carbon monoxide plus steam: do se do, delta n zero, Kp equals Kc. Do sulphur dioxide plus oxygen: teen se do, delta n minus one, Kp equals Kc times RT to the minus one. Phosphorus pentachloride: ek se do, delta n plus one, Kp equals Kc times RT.

Example 6.4NCERT Problems 6.3, 6.4, 6.9 · p. 179

Quadratic solve karke equilibrium concentrations

$H_{2} (g) + I_{2} (g) \overset{}{⇌} 2 H I (g)$ ke liye 720 K par $K_{c} = 50$ hai. Ek flask $[H_{2}] = 0.20$ , $[I_{2}] = 0.10 M$ aur bina HI ke shuru hota hai. Equilibrium concentrations nikalo.

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

Quick check

2 mein se 1

Kis reaction ke liye

K_{p} = K_{c}

hai?

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

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

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)

Answer dekho

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

Sirf

H_{2} + C l_{2} \overset{}{⇌} 2 H C l

mein

Δ n = 2 - 2 = 0

hai. Limestone wale reaction mein

Δ n = 1

hai: solids nahi gine jaate, lekin

C O_{2}

gina jaata hai.

6.5 Heterogeneous Equilibria

Jab ek se zyada phase hissa lete hain, to equilibrium heterogeneous hota hai: closed flask mein paani aur uska vapour, ya apne saturated solution ke sampark mein slaked lime, $C a (O H)_{2} (s) ⇌ C a^{2 +} (a q) + 2 O H^{-} (a q)$ . Pure solids aur liquids expression ko simple bana dete hain. Pure solid ka 'concentration' fixed hota hai, jo uski density aur molar mass se tay hota hai: calcium carbonate ka bada tukda aur chhota tukda dono mein solid ke per litre same moles hote hain. Aise constant terms K mein mila diye jaate hain, isliye pure solids aur liquids chhod diye jaate hain; gases aur dissolved species, jinke concentrations badal sakte hain, rehte hain.

Closed kiln mein limestone ko garam karne se iska matlab samajh aata hai: $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ deta hai

K_{c}^{'} = [C O_{2}] ya K_{p} = p_{C O_{2}}

(6.18)

Given temperature par dono solids ke upar $C O_{2}$ ke pressure ki ek hi value hoti hai, chahe har solid kitna bhi ho (Fig. 6.8). Yeh 900 °C (lagbhag 1170 K) ke aas-paas lagbhag 1 bar tak pahunchta hai, isiliye lime kilns isse zyada garam chalaye jaate hain. Equilibrium ke liye kuch solid, chahe kitna bhi kam, maujood hona chahiye, halaanki woh K mein nahi aata.

3/3Kuch CO₂ pump karke nikalo: aur CaCO₃ decompose hota hai jab tak pressure waapas same value par na aa jaaye.

Fig. 6.8Fixed temperature par,

p_{C O_{2}}

solid ki amount par depend nahi karta.

Step 1: ek garam vessel jismein thoda calcium carbonate aur calcium oxide, kuch carbon dioxide aur ek pressure gauge hai. Step 2: same temperature par doosra vessel jismein bahut zyada solid hai; uska gauge same reading deta hai. Step 3: doosre vessel se kuch carbon dioxide pump karke nikali jaati hai; aur carbonate decompose hota hai, uska dher chhota hota hai aur gauge waapas same reading par aa jaata hai.

Do aur examples. Hydrogen se reduce hota iron oxide, $F e_{3} O_{4} (s) + 4 H_{2} (g) ⇌ 3 F e (s) + 4 H_{2} O (g)$ , ka $K_{c} = [H_{2} O]^{4} / [H_{2}]^{4}$ hai. Blue copper sulphate pentahydrate apne do waters kho sakta hai, jisse trihydrate aur water vapour bante hain, $K_{p} = p_{H_{2} O}^{2}$ ke saath: pentahydrate tabhi tak stable hai jab tak hawa mein kaafi water vapour ho, aur bahut sookhi hawa mein yeh dheere-dheere paani kho deta hai.

K ki units. Concentrations mol L⁻¹ mein ya pressures bar mein hon, to K ki unit hoti hai jab tak powers cancel na hon: $H_{2} + I_{2} \overset{}{⇌} 2 H I$ ke liye koi unit nahi, lekin $P C l_{5} \overset{}{⇌} P C l_{3} + C l_{2}$ ke $K_{c}$ ke liye mol L⁻¹ aur $K_{p}$ ke liye bar. Thermodynamics har concentration ko standard 1 mol L⁻¹ se aur har pressure ko 1 bar se divide karta hai, jisse K ek pure number ban jaata hai; kyunki standard states alag hain, $K_{c}$ aur $K_{p}$ phir bhi alag rehte hain jab tak $Δ n = 0$ na ho.

Example 6.6NCERT Problem 6.6 · p. 180

Solid ke saath ek heterogeneous equilibrium

$N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$ ka 298 K par $K_{p} = 0.11$ hai. Kuch solid ek flask mein daala jaata hai jismein pehle se 0.20 bar par ammonia hai. Equilibrium partial pressures nikalo.

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

Quick check

2 mein se 1

C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)

ke liye equilibrium constant kaun sa hai?

\frac{[C O] [H_{2}]}{[C] [H_{2} O]}

[C O] [H_{2}]

\frac{[C O] [H_{2}]}{[H_{2} O]}

\frac{[H_{2} O]}{[C O] [H_{2}]}

Answer dekho

Answer (C) $\frac{[C O] [H_{2}]}{[H_{2} O]}$

Solid carbon chhod diya jaata hai; steam gas hai, isliye woh denominator mein rehti hai.

6.6 Equilibrium Constants ke Applications

K use karne se pehle, iske baare mein jo pata hai woh ikattha karo:

Yeh tabhi apply hota hai jab concentrations badalna band kar dein.
Yeh shuru ke concentrations par depend nahi karta.
Har temperature par har balanced equation ke liye iski ek value hoti hai.
Equation ulti karne par yeh ulta (inverse) ho jaata hai; equation ko kisi number se multiply karne par yeh us power tak raise ho jaata hai.

K ke teen main uses hain: reaction kitni door jaata hai yeh judge karna, mixture kis taraf jaayega yeh predict karna, aur equilibrium composition calculate karna.

6.6.1 Reaction ka Extent Predict karna

K ke numerator mein products hote hain aur denominator mein reactants, isliye bada K matlab equilibrium mixture mein products zyada hain aur chhota K matlab reactants zyada hain. K speed ke baare mein kuch nahi batata: bahut bade K wale equilibrium tak pahunchne mein mixture ko saal lag sakte hain. Moti-moti bands (Fig. 6.9):

$K > 10^{3}$ : products zyada hote hain aur reaction lagbhag complete hota hai, jaise neutralisation mein, $H^{+} (a q) + O H^{-} (a q) ⇌ H_{2} O (l)$ , 298 K par $K = 10^{14}$ .
$K < 10^{- 3}$ : reactants zyada hote hain aur bahut kam product banta hai, jaise jab silver chloride paani mein dissolve hota hai, $K = 1.8 \times 10^{- 10}$ .
$10^{- 3}$ se $10^{3}$ : dono achhi maatra mein hote hain, jaise 720 K par $H_{2} + I_{2} \overset{}{⇌} 2 H I$ ( $K_{c} = 50$ ) ya 350 K par $N_{2} O_{4} \overset{}{⇌} 2 N O_{2}$ ( $K_{c} = 0.15$ ).

Fig. 6.9Logarithmic scale par K ka size dikhata hai ki reaction kitni door jaata hai.NCERT Fig. 6.6 · p. 182

Minus 16 se plus 16 tak ek log K axis jismein teen zones hain: minus 3 se neeche reactants, minus 3 aur plus 3 ke beech both, plus 3 se upar products. Pins silver chloride ka dissolve hona (lagbhag minus 9.7), acetic acid ka ionize hona (lagbhag minus 4.7), 720 kelvin par hydrogen plus iodine (lagbhag 1.7) aur hydrogen ion plus hydroxide se paani banna (14) dikhate hain.

6.6.2 Reaction ki Direction Predict karna

Jo mixture shayad equilibrium par na ho, uske liye reaction quotient $Q$ nikalo. Iska form K jaisa hi hai lekin yeh concentrations ( $Q_{c}$ ) ya partial pressures ( $Q_{p}$ ) us pal jaise hain waise use karta hai:

Reaction quotient

Q_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}} (kisi bhi pal)

(6.20)

Q ko K se compare karne par pata chalta hai ki mixture kis taraf jaayega (Fig. 6.10). Agar $Q < K$ hai to product bahut kam hai, isliye net reaction forward jaata hai. Agar $Q > K$ hai to product bahut zyada hai, isliye yeh backward jaata hai. Agar $Q = K$ hai to mixture pehle se equilibrium par hai.

1/3Product bahut kam: net reaction forward chalta hai jab tak Q = K na ho.

Fig. 6.10Mixture hamesha is tarah move karta hai ki Q, K ki taraf aaye.NCERT Fig. 6.7 · p. 182

Log-scale Q axis ke upar reactants aur products ke boxes, beech mein K marked. Step 1: Q pointer K ke left mein hai aur ek mota arrow reactants se products ki taraf hai, net forward. Step 2: pointer K par hai aur ulte arrows dikhate hain ki koi net change nahi. Step 3: pointer K ke right mein hai aur arrow waapas reactants ki taraf hai, net backward.

Example 6.7NCERT Problem 6.7 · p. 182

Mixture kis taraf jaayega?

$N_{2} O_{4} (g) \overset{}{⇌} 2 N O_{2} (g)$ ke liye 298 K par $K_{c} = 4.6 \times 10^{- 3}$ hai. 298 K par ek flask mein $[N_{2} O_{4}] = 0.020 M$ aur $[N O_{2}] = 0.015 M$ hai. Kya yeh equilibrium par hai? Agar nahi, to yeh kis taraf jaayega, aur tumhe kya dikhega?

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

6.6.3 Equilibrium Concentrations Calculate karna

Jab sirf shuru ki amounts pata hon, to ICE table (Initial, Change, Equilibrium) hisaab-kitaab saaf rakhti hai:

Balanced equation likho.
Uske neeche har species ka initial concentration, ek unknown $x$ ke terms mein uska change (coefficients use karke), aur uska equilibrium concentration likho.
K expression mein substitute karo aur $x$ solve karo, woh root rakho jisse har concentration positive aur possible rahe.
Har concentration nikalo, aur unhe waapas K mein daal kar check karo.

Example 6.8NCERT Problems 6.5, 6.8 · p. 183

Total pressure se Kp aur Kc

6.25 g $P C l_{5}$ (208.2 g mol⁻¹) ko ek khaali 1.00 L vessel mein 523 K tak garam kiya jaata hai. Equilibrium par, $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ , total pressure 2.18 bar hai. $K_{p}$ , $K_{c}$ aur dissociate hua fraction nikalo.

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

Quick check

3 mein se 1

Ek reaction ka

K_{c} = 4.0

hai. Reactants aur products ke ek mixture ka

Q_{c} = 9.0

hai. Aage kya hoga?

Net reaction forward jaata haiKuch nahi: yeh equilibrium par haiNet reaction backward jaata hai

K_{c}

badhkar 9.0 ho jaata hai

Answer dekho

Answer (C) Net reaction backward jaata hai

Q_{c} > K_{c}

: product bahut zyada, isliye net reaction backward chalta hai jab tak

Q_{c} = 4.0

na ho.

6.7 Equilibrium Constant K, Reaction Quotient Q aur Gibbs Energy G ke beech Relationship

Thermodynamics (Unit 5) batata hai ki K ki value kyun aisi hai. Constant temperature aur pressure par, $Δ G < 0$ wala mixture apne aap forward jaata hai, $Δ G > 0$ wala backward jaata hai, aur $Δ G = 0$ wala equilibrium par hai, change chalane ke liye koi Gibbs energy nahi bachi. $Δ G$ composition par Q ke through depend karta hai:

Δ G = Δ G^{\circ} + R T \ln Q

(6.21)

jahan

$Δ G^{\circ}$: standard Gibbs energy change (har species apni standard state mein)

DerivationK ko standard Gibbs energy se jodna

Equilibrium par $Δ G = 0$ aur $Q = K$ , isliye Eq. (6.21) ban jaata hai $0 = Δ G^{\circ} + R T \ln K$ :

Δ G^{\circ} = - R T \ln K

(6.22)

Antilogarithms lene par:

K = e^{- Δ G^{\circ} / R T}

(6.23)

To negative $Δ G^{\circ}$ se $K > 1$ milta hai, aur equilibrium par products favoured hote hain; positive $Δ G^{\circ}$ se $K < 1$ milta hai, sirf thode product ke saath. Fig. 6.11 dikhata hai ki har mixture same composition ki taraf kyun jaata hai: mixture ki Gibbs energy dono ends se ek minimum tak girti hai, aur woh minimum hi equilibrium hai.

3/3Minimum par ΔG = 0 aur Q = K: equilibrium mixture.

Fig. 6.11Kisi bhi side se shuru ho, mixture G mein downhill fisal kar same minimum par aata hai: equilibrium.

Pure reactants se pure products tak composition ke against mixture ki Gibbs energy, beech se thoda aage ek minimum ke saath. Step 1: reactant end ke paas ek ball aur forward likha downhill arrow. Step 2: product end ke paas ball aur backward likha downhill arrow. Step 3: ball minimum par ruki hai, jis par delta G equals zero, Q equals K likha hai.

Example 6.10NCERT Problems 6.10, 6.11 · p. 184

ΔG° se K aur waapas

(a) 310 K par ATP hydrolysis ka $Δ G^{\circ} = - 30.5 k J m o l^{- 1}$ hai. K nikalo. (b) 298 K par $N_{2} O_{4} \overset{}{⇌} 2 N O_{2}$ ka $K_{p} = 0.144$ hai. $Δ G^{\circ}$ nikalo.

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

Quick check

3 mein se 1

298 K par ek reaction ka

Δ G^{\circ} = + 10 k J m o l^{- 1}

hai. Kaun sa statement sahi hai?

Yeh bilkul nahi ho sakta

K > 1

: equilibrium par zyaadatar products

K = 0

K < 1

: equilibrium par thoda product

Answer dekho

Answer (D) $K < 1$ : equilibrium par thoda product

K = e^{- Δ G^{\circ} / R T} \approx 0.018 < 1

: kuch product phir bhi banta hai.

6.8 Equilibria ko Affect karne wale Factors

Industry kam se kam energy mein zyada se zyada product chahti hai. Ammonia plants, jo fertiliser industry ko supply karte hain (urea ammonia se banta hai), yahi choice face karte hain: duniya ka output saal mein 150 million tonnes se kaafi zyada hai. Agar equilibrium par kisi mixture ka concentration, pressure ya temperature badla jaaye, to net reaction tab tak chalta hai jab tak naya equilibrium na ban jaaye. Le Chatelier's principle direction predict karta hai:

PrincipleLe Chatelier’s principle

Agar equilibrium ko fix karne wali kisi bhi condition mein change kiya jaaye, to system us direction mein shift hota hai jo change ko kuch had tak counteract kare, jab tak naya equilibrium na ban jaaye. Yeh physical aur chemical dono equilibria par lagta hai.

Concentration aur pressure changes K ko nahi badalte lekin Q ko K se alag kar dete hain; temperature change khud K ko badal deta hai. Tum har factor Fig. 6.14 mein try kar sakte ho.

6.8.1 Concentration Change ka Effect

Reactant ya product jodo, to net reaction us direction mein chalta hai jo jode gaye substance ko use up kare.
Reactant ya product nikalo, to net reaction us direction mein chalta hai jo use replace kare.

Equilibrium par ammonia mixture mein nitrogen jodo. $Q_{c} = [N H_{3}]^{2} / ([N_{2}] [H_{2}]^{3})$ $K_{c}$ se neeche gir jaata hai, isliye forward reaction chalta hai, kuch $N_{2}$ aur $H_{2}$ use karke aur zyada $N H_{3}$ banakar (Fig. 6.12). Naye equilibrium par $[N_{2}]$ addition ke theek baad se kam hai lekin shuru se phir bhi zyada hai: change sirf kuch had tak undo hota hai.

4/4N₂ apne purane level se upar lekin peak se neeche settle hoti hai; ab NH₃ zyada hai.

Fig. 6.12

N_{2}

jodne se

Q < K

ho jaata hai; phir forward reaction jodi gayi nitrogen ka kuch hissa use kar leta hai.NCERT Fig. 6.8 · p. 185

Nitrogen (solid), hydrogen (dashed) aur ammonia (dotted) ke concentrations vs time. Step 1: teenon flat hain. Step 2: nitrogen jodne par achanak upar jaati hai. Step 3: nitrogen aur hydrogen neeche mudte hain aur ammonia upar mudta hai. Step 4: ek guide purana nitrogen level dikhata hai; naya level usse upar lekin peak se neeche hai, aur ammonia zyada hai.

Product nikalna bhi utna hi kaam karta hai aur bade scale par use hota hai. Ammonia ko liquefy karke reactor gases se nikal liya jaata hai, aur lime kiln mein $C O_{2}$ ko hata diya jaata hai taaki $C a C O_{3} \overset{}{\to} C a O + C O_{2}$ completion tak jaaye. Lagataar nikalne se Q, K se neeche rehta hai, isliye forward reaction kabhi nahi rukta.

ExperimentNCERT Concentration ka effect – ek experiment · p. 186

Iron(III)–thiocyanate equilibrium ko push karna

Tumhe chahiye

dilute iron(III) chloride ya nitrate
dilute potassium thiocyanate
dilute oxalic acid
chaar test tubes aur ek dropper
safed kaagaz

Pehle guess karo

Laal rang banne ke baad tum thiocyanate solution ki kuch aur boondein daalte ho. Laal rang

gehra hogahalka hogabilkul same rahega

20 mL paani mein 1 mL iron(III) solution mein thiocyanate ki 2 boondein daalo. Laal rang banta hai: $F e^{3 +} + S C N^{-} ⇌ [F e (S C N)]^{2 +}$ (yellow, colourless, gehra laal). Ise chaar tubes mein baant do; tube 1 reference hai.
Tube 2: thiocyanate ki 3 boondein daalo. Tube 3: iron(III) solution ki 3 boondein daalo. Tube 4: oxalic acid boond-boond karke daalo. Safed kaagaz ke saamne har ek ko tube 1 se compare karo.

Tumhare observations

Tube	Tumhara prediction	Tube 1 ke comparison mein rang
2: zyada SCN⁻
3: zyada Fe³⁺
4: oxalic acid

Tumhari entries apne aap save hoti hain.

6.8.2 Pressure Change ka Effect

Gas mixture ko compress karne se har partial pressure badhta hai. Agar reaction gas molecules ki sankhya badalta hai, to Le Chatelier's principle kam gas moles wali side ki taraf shift predict karta hai, jo pressure ko kuch had tak waapas kam karta hai (Fig. 6.13). Agar $Δ n = 0$ hai, to pressure composition ko affect nahi karta. Solids aur liquids ka volume mushkil se badalta hai, isliye sirf gases count hoti hain.

3/3Kuch NO₂ molecules jode bana lete hain: kam molecules, isliye pressure kuch had tak waapas girta hai.

Fig. 6.13

2 N O_{2} \overset{}{⇌} N_{2} O_{4}

ka volume aadha karne se yeh

N_{2} O_{4}

ki taraf shift hota hai, jo kam gas molecules wali side hai.

Piston wala ek cylinder jismein single nitrogen dioxide molecules aur jode wale dinitrogen tetroxide molecules hain. Step 1: aath singles aur do pairs cylinder bharte hain. Step 2: piston volume aadha kar deta hai aur wahi molecules paas-paas bhar jaate hain. Step 3: ab chaar singles aur chaar pairs, baarah ki jagah das molecules, kyunki kuch nitrogen dioxide jode bana leti hai.

Q bhi yahi kehta hai. $2 N O_{2} (g) \overset{}{⇌} N_{2} O_{4} (g)$ ke liye $Q_{c} = [N_{2} O_{4}] / [N O_{2}]^{2}$ . Volume aadha karne se har concentration double ho jaata hai: numerator double hota hai lekin denominator chaar guna, isliye $Q_{c} = K_{c} / 2$ aur forward reaction tab tak chalta hai jab tak $Q = K$ na ho. Isi tarah, high pressure water-gas reaction $C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)$ ko left ki taraf push karta hai.

6.8.3 Inert Gas Addition ka Effect

Constant volume par argon, ya koi bhi non-reacting gas, jodne se total pressure badhta hai lekin reacting gases ke partial pressures aur concentrations nahi badalte, isliye Q ab bhi K ke barabar hai aur kuch shift nahi hota. (Constant total pressure par vessel ko phailna padta hai, jisse har partial pressure kam hota hai, aur shift zyada gas moles ki taraf hota hai.)

6.8.4 Temperature Change ka Effect

Temperature alag hai: yeh khud K ko badalta hai. Heat ko reactant ya product ki tarah treat karo. Exothermic reaction ( $Δ H < 0$ ) mein heat product hai, isliye garam karne par equilibrium left shift hota hai aur K girta hai; endothermic reaction ke liye temperature ke saath K badhta hai. Ammonia synthesis ka $Δ H = - 92.4 k J m o l^{- 1}$ hai, isliye low temperatures ammonia ko favour karte hain, lekin wahan reaction bahut hi dheema hai.

DemonstrationNCERT Temperature ka effect – ek experiment · p. 187

Garam aur thande paani mein brown gas

Pehle guess karo

Brown

N O_{2}

ki do identical sealed tubes ko ice water aur garam paani mein rakha jaata hai.

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

ke liye

Δ H = - 57.2 k J m o l^{- 1}

. Ice water mein kya hota hai?

Brown rang gehra hota haiBrown rang halka hota haiKoi dikhne wala change nahi hota

Fume cupboard mein tumhare teacher brown $N O_{2}$ (copper aur concentrated nitric acid se) ko do identical tubes mein seal karte hain aur unhe room temperature par settle hone dete hain.
Ek tube namak wali ice mein jaati hai, doosri lagbhag 90 °C wale paani mein. Kuch minutes baad rang compare karo, phir tubes ki jagah badal do.

6.8.5 Catalyst ka Effect

Catalyst kam activation energy wala ek naya pathway deta hai. Yeh forward aur reverse dono reactions ke barrier ko bilkul same amount se kam karta hai, isliye dono same factor se tez hote hain. Equilibrium jaldi aata hai, lekin uska composition, aur K, nahi badalte. Catalyst na balanced equation mein aata hai na K expression mein.

Ammonia synthesis practice mein yeh balance dikhata hai. Low temperature achha yield deta hai lekin reaction dheema hota hai; high temperature tez hai lekin ammonia kam deta hai. Haber ka iron catalyst lagbhag 450–500 °C par useful rate deta hai, jahan yield ab bhi theek hai, aur high pressure, lagbhag 200 atm, madad karta hai kyunki gas ke 4 moles se 2 bante hain. Contact process mein, $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ ka K bahut bada hai lekin vanadium(V) oxide (ya platinum) catalyst ke bina yeh bahut hi dheema hai. Catalyst bahut chhote K wale reaction ko nahi bacha sakta: woh bas kharab equilibrium tak jaldi pahunchayega.

InteractiveEquilibrium lab

Fig. 6.14Q ko K ki taraf jaate dekhne ke liye Play dabao, phir substances jodo ya nikalo, volume ya temperature badlo, argon ya catalyst jodo, aur dekho mixture kaise respond karta hai.

Quick check

2 mein se 1

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

Δ H = + 57.2 k J m o l^{- 1}

ke liye, kaun sa change equilibrium par

N O_{2}

ki amount badhata hai?

Mixture ko compress karnaCatalyst jodnaTemperature badhanaConstant volume par argon jodna

Answer dekho

Answer (C) Temperature badhana

Forward reaction endothermic hai, isliye garam karne se K badhta hai. Compression

N_{2} O_{4}

ko favour karta hai; catalyst ya constant volume par argon kuch nahi badalte.

6.9 Solution mein Ionic Equilibrium

Iron(III)–thiocyanate equilibrium pehle se hi ions ke beech ka reaction tha, aur paani mein kai equilibria ions aur unse banne wale molecules se jude hote hain. Cheeni ka solution electricity conduct nahi karta, lekin namak ka solution karta hai, aur zyada namak behtar conduct karta hai.

Michael Faraday ne is test se substances ko sort kiya. Electrolytes paani mein ghulne par conduct karte hain; non-electrolytes nahi karte. Sodium chloride jaise strong electrolytes lagbhag poore ions ke roop mein hote hain, jabki acetic acid jaise weak electrolytes zyaadatar molecules hi rehte hain, 5 % se kam ionized (Fig. 6.15). Weak electrolyte mein ions aur molecules equilibrium mein hote hain, $C H_{3} C O O H (a q) ⇌ H^{+} (a q) + C H_{3} C O O^{-} (a q)$ : ek ionic equilibrium. Koi substance kitna ionize hoga, yeh tootne wale bond ki strength aur paani ions ko kitni mazbooti se solvate karta hai, is par depend karta hai. Sahi mein, dissociation solid (NaCl) mein pehle se maujood ions ko alag karta hai aur ionization ek neutral molecule (HCl) ko ions mein todta hai; yahan dono shabd ek doosre ki jagah use kiye gaye hain.

3/3Weak electrolyte mein molecules aur ions dynamic equilibrium mein hote hain.

Fig. 6.15Strong electrolyte poora ionized hota hai; weak electrolyte molecules aur kuch ions ke beech equilibrium banata hai.

Step 1: sodium chloride solution ka beaker jismein sirf positive aur negative ions hain, conducts well likha hai. Step 2: acetic acid solution ka beaker jismein aath neutral molecules aur bas ek pair ions hai, conducts weakly likha hai. Step 3: ulte arrows acetic acid mein molecules aur ions ko jodte hain, molecules in equilibrium with ions likha hai.

6.10 Acids, Bases aur Salts

Acids, bases aur salts har jagah hain. Stomach hydrochloric acid secrete karta hai, sirka dilute acetic acid hai, nimbu aur amla mein citric aur ascorbic acids hote hain, aur imli mein tartaric acid hota hai. Acid Latin shabd acidus se aaya hai, yaani khatta. Acids blue litmus ko laal karte hain aur active metals ke saath hydrogen chhodte hain; bases red litmus ko blue karte hain aur chhoone mein sabun jaise lagte hain, jaise washing soda. (Laboratory chemicals ko kabhi chakho mat.) Acids aur bases react karke sodium chloride, potassium nitrate aur barium sulphate jaise salts dete hain.

Namak itni aasaani se dissolve kyun hota hai? $N a^{+}$ aur $C l^{-}$ ko saath rakhne wala electrostatic force medium ke dielectric constant se divide hota hai, jo paani ke liye lagbhag 80 hai, isliye paani mein yeh lagbhag 80 guna kamzor hota hai. Har ion apne aas-paas oriented water molecules ka ek shell bhi ikattha karta hai, oxygen wale sire $N a^{+}$ ki taraf aur hydrogen wale sire $C l^{-}$ ki taraf. Yeh hydration energy chhodta hai aur ions ko alag aur move karne ke liye free rakhta hai.

6.10.1 Acids aur Bases ka Arrhenius Concept

Svante Arrhenius ne acid ko aisa substance define kiya jo paani mein hydrogen ions, $H^{+} (a q)$ , banata hai, aur base ko aisa jo hydroxide ions, $O H^{-} (a q)$ , banata hai: $H X (a q) \overset{}{\to} H^{+} (a q) + X^{-} (a q)$ aur $M O H (a q) ⟶ M^{+} (a q) + O H^{-} (a q)$ .

Ek akele proton (radius lagbhag $10^{- 15} m$ ) ka field itna tez hota hai ki woh paani mein kabhi free nahi ghoomta: woh water molecule ke lone pair se bond banata hai, jisse pyramidal hydronium ion banta hai, $H X + H_{2} O \overset{}{\to} H_{3} O^{+} + X^{-}$ . Hydronium aage aur hydrate hota hai ( $H_{5} O_{2}^{+}$ , $H_{9} O_{4}^{+}$ waghera ke roop mein), aur hydroxide bhi. Yahan $H^{+} (a q)$ aur $H_{3} O^{+} (a q)$ dono ka matlab hydrated proton hai. Arrhenius definitions sirf paani mein kaam karti hain, aur yeh nahi samjha sakti ki ammonia, jismein koi $O H$ group nahi hai, base kyun hai.

6.10.2 Brönsted–Lowry Acids aur Bases

1923 mein Johannes Brønsted (Denmark) aur Thomas Lowry (England) ne alag-alag is idea ko badaya: acid proton donor hai aur base proton acceptor hai. $N H_{3} (a q) + H_{2} O (l) ⇌ N H_{4}^{+} (a q) + O H^{-} (a q)$ (Fig. 6.16) mein paani ammonia ko proton deta hai, isliye paani acid hai aur ammonia base; reverse reaction mein $N H_{4}^{+}$ acid hai aur $O H^{-}$ base.

3/3NH₃/NH₄⁺ aur H₂O/OH⁻ conjugate pairs hain: har pair mein ek proton ka fark hai.

Fig. 6.16Brønsted acid–base reaction ek proton transfer hai; har acid aur jo base woh banta hai, milkar conjugate pair banate hain.

Equation ammonia plus water in equilibrium with ammonium plus hydroxide, species ke neeche base, acid, acid aur base likha hai. Step 1: water molecule par ek proton baitha hai. Step 2: ek curved arrow proton ko ammonia tak le jaata hai. Step 3: brackets ammonia ko ammonium se aur water ko hydroxide se jodte hain, har ek par conjugate pair likha hai.

Definition

Conjugate acid–base pair

Ek acid aur ek base jinmein theek ek proton ka fark ho. Acid mein uske conjugate base se ek $H^{+}$ zyada hota hai.

Example $N H_{4}^{+}$ aur $N H_{3}$ ; $H_{2} O$ aur $O H^{-}$ .

Hydrogen chloride ke saath paani doosra role nibhata hai: $H C l + H_{2} O \overset{}{\to} H_{3} O^{+} + C l^{-}$ mein woh proton leta hai, isliye yahan woh base ki tarah behave karta hai. $C l^{-}$ HCl ka conjugate base hai aur $H_{3} O^{+}$ paani ka conjugate acid. Jo species proton de bhi sake aur le bhi sake, jaise paani, woh amphiprotic hai. Acid jitna strong, uska conjugate base utna weak, aur ulta bhi.

Example 6.12NCERT Problems 6.12–6.14 · p. 191

Conjugate acids aur bases dhoondhna

Batao (a) $H C O O H$ , $H_{2} P O_{4}^{-}$ aur $N H_{4}^{+}$ ka conjugate base; (b) $C H_{3} N H_{2}$ aur $C O_{3}^{2 -}$ ka conjugate acid; (c) amphiprotic ion $H P O_{4}^{2 -}$ ke dono partners.

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

6.10.3 Lewis Acids aur Bases

G. N. Lewis (1923) ne focus electron pairs par kar diya: Lewis acid electron pair accept karta hai aur Lewis base ek donate karta hai. Brønsted bases bhi Lewis bases hain, kyunki woh lone pair dekar proton lete hain. Fayda yeh hai ki bina proton wali species bhi acids ho sakti hain: boron trifluoride, jismein boron ke aas-paas sirf chhah electrons hain, ammonia ka lone pair le leta hai, $B F_{3} + N H_{3} \overset{}{\to} F_{3} B - N H_{3}$ .

Doosre Lewis acids mein $A l C l_{3}$ jaise electron-deficient molecules, $F e^{3 +}$ , $C u^{2 +}$ aur $M g^{2 +}$ jaise metal ions, aur $C O_{2}$ shaamil hain, jo $O H^{-}$ se ek pair lekar $H C O_{3}^{-}$ banata hai. Lone pairs wale molecules aur ions, jaise $H_{2} O$ , $N H_{3}$ , $O H^{-}$ aur $C l^{-}$ , Lewis bases hain.

Example 6.15NCERT Problem 6.15 · p. 192

Lewis acids aur bases ko classify karna

Har species ko Lewis acid ya base mein classify karo aur batao ki woh kaise kaam karti hai: (a) $A g^{+}$ (b) $C N^{-}$ (c) $C O_{2}$ (d) $H_{2} O$ .

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

6.11 Acids aur Bases ka Ionization

Strong acids (perchloric, hydrochloric, hydrobromic, hydroiodic, nitric, aur pehle step mein sulphuric) aur strong bases (lithium, sodium, potassium, caesium aur barium ke hydroxides) paani mein lagbhag poore ionized hote hain. Brønsted terms mein, strong acid achha proton donor hai aur strong base achha proton acceptor.

Weak acid HA ke liye, reaction $H A + H_{2} O \overset{}{⇌} H_{3} O^{+} + A^{-}$ do acids, HA aur $H_{3} O^{+}$ , ke beech proton dene ke kaam ka muqabla hai. Equilibrium weaker acid aur weaker base wali side par hota hai: stronger acid apna proton stronger base ko de deta hai. HCl, HBr, HI, $H N O_{3}$ aur $H C l O_{4}$ ke liye yeh bahut right mein hota hai, isliye $C l^{-}$ , $B r^{-}$ , $I^{-}$ , $N O_{3}^{-}$ aur $C l O_{4}^{-}$ paani se bahut weaker bases hain. HF, $H N O_{2}$ aur acetic acid jaise weak acids zyaadatar molecules hi rehte hain, aur unke conjugate bases relatively strong hote hain. $N H_{2}^{-}$ , $O^{2 -}$ aur $H^{-}$ itne strong bases hain ki woh paani se protons kheench lete hain.

Indicators weak acids hain jinka acid form, HIn, aur conjugate base, $I n^{-}$ , alag rang ke hote hain: $H I n (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + I n^{-} (a q)$ . Phenolphthalein HIn ke roop mein colourless aur $I n^{-}$ ke roop mein pink hota hai; bromothymol blue acid mein yellow aur alkali mein blue hota hai. Kaun sa form zyada hoga yeh $[H^{+}]$ par depend karta hai, isiliye indicators titration ka end point bata sakte hain.

6.11.1 Water ka Ionization Constant aur uska Ionic Product

Kyunki paani amphiprotic hai, ek water molecule doosre ko proton de sakta hai: $H_{2} O (l) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + O H^{-} (a q)$ . Iske equilibrium constant, $K = [H_{3} O^{+}] [O H^{-}] / [H_{2} O]$ , mein pure paani ka concentration hai. Woh constant hai, lagbhag 1000 g L⁻¹ ÷ 18.0 g mol⁻¹ = 55.6 mol L⁻¹, isliye use ek naye constant mein mila diya jaata hai, ionic product of water:

Ionic product of water

K_{w} = [H^{+}] [O H^{-}]

(6.27)

298 K par pure paani mein measurement se $[H^{+}] = [O H^{-}] = 1.0 \times 10^{- 7} M$ hai, isliye

298 K par

K_{w} = (1.0 \times 10^{- 7})^{2} = 1.0 \times 10^{- 14}

(6.28)

Kisi bhi pal har $10^{9}$ mein se sirf lagbhag 2 molecules ionized hote hain ( $10^{- 7} / 55.6 = 1.8 \times 10^{- 9}$ ), isliye equilibrium un-ionized paani ki taraf bahut zyada hota hai. Kisi bhi equilibrium constant ki tarah, $K_{w}$ temperature ke saath badalta hai; paani garam karne par yeh badhta hai. Koi bhi aqueous solution acidic hai agar $[H_{3} O^{+}] > [O H^{-}]$ , neutral hai agar dono barabar hain aur basic hai agar $[H_{3} O^{+}] < [O H^{-}]$ .

6.11.2 pH Scale

Hydrogen ion concentrations das ki kai powers tak phaile hote hain, isliye unhe logarithmic scale par batate hain. Sahi mein, pH hydrogen ions ki activity, $a_{H^{+}}$ , ka negative base-10 logarithm hai, jo ek pure number hai. Dilute solutions mein (lagbhag 0.01 M se neeche) activity molarity ko 1 mol L⁻¹ se divide karne ke barabar hoti hai, isliye

pH ki definition

p H = - \log a_{H^{+}} = - \log [H^{+}]

To 0.001 M nitric acid ka pH 3 hai, aur $[O H^{-}] = 10^{- 3} M$ wale solution ka $[H^{+}] = 10^{- 11} M$ aur pH 11 hai. 298 K par pure paani ka pH 7 hai. 298 K par acidic solutions ka pH < 7, neutral ka pH = 7 aur basic ka pH > 7 hota hai. $K_{w}$ ka negative logarithm lene se ek aur useful relation milta hai:

298 K par

p H + p O H = p K_{w} = 14

(6.29)

Kyunki scale logarithmic hai, ek pH unit $[H^{+}]$ mein 10 ka factor hai, aur do units 100 ka factor (Fig. 6.17). $K_{w}$ temperature ke saath badalta hai, lekin aam working range mein pH par iska effect chhota hota hai aur aam taur par ignore kiya jaata hai.

3/3Ek pH unit matlab [H⁺] mein 10 ka factor; do units matlab 100.

Fig. 6.17Typical pH values; ek unit ka har step

[H^{+}]

mein das guna change hai.NCERT Table 6.5 · p. 195

0 se 14 tak ek pH bar, vermillion se green hote hue blue tak rangi hui. Step 1: pins nimbu ka ras 2.3 ke paas, dahi 4.5 ke paas aur doodh 6.7 ke paas dikhate hain. Step 2: neeche ke pins pure paani 7.0 par, blood 7.4 par, baking soda solution 8.3 ke paas aur lime water 12.4 ke paas jodte hain. Step 3: pH 9 se 10 aur 10 se 11 ke upar brackets har ek hydrogen ion concentration mein das guna change dikhate hain.

Universal indicator paper jaldi se estimate deta hai. Alag dyes ki chaar strips wale papers, jinmein har ek alag range mein rang badalti hai, pH 1–14 ko lagbhag ±0.5 tak padh sakte hain. pH meter ek electrical potential naapta hai jo pH par depend karta hai aur 0.01 ya usse behtar tak padhta hai; pen jitne meters ab itne saste hain ki kisaan mitti test karne ke liye use kar sakte hain.

Example 6.16NCERT Problems 6.16, 6.17 · p. 194

[H⁺] se pH, aur ek bahut dilute strong acid

(a) Imli ke ek drink ka $[H^{+}] = 2.5 \times 10^{- 3} M$ hai. Iska pH nikalo. (b) $2.0 \times 10^{- 8} M$ nitric acid ka pH nikalo.

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

Quick check

2 mein se 1

298 K par 0.0050 M barium hydroxide,

B a (O H)_{2}

, ka pH kya hai?

Answer dekho

Answer 12

Har formula unit 2

O H^{-}

deti hai:

[O H^{-}] = 0.010 M

, pOH = 2.00, pH = 12.00.

6.11.3 Weak Acids ke Ionization Constants

Maan lo ek weak acid HX ka initial concentration $c$ hai, aur uske molecules ka ek fraction $α$ , yaani uska degree of ionization, ionize hota hai:

Table 6.4

H X (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + X^{-} (a q)

ke liye ICE tableNCERT p. 195

	$[H X]$	$[H_{3} O^{+}]$	$[X^{-}]$
Initial	$c$	0	0
Change	$- c α$	$+ c α$	$+ c α$
Equilibrium	$c (1 - α)$	$c α$	$c α$

Acid ionization constant

K_{a} = \frac{[H^{+}] [X^{-}]}{[H X]} = \frac{c α^{2}}{1 - α}

(6.30)

$K_{a}$ acid ionization constant hai. Given temperature par $K_{a}$ jitna bada, acid utna strong. 1 M ko standard concentration lene se $K_{a}$ ek pure number ban jaata hai. pH ki tarah, $p K_{a} = - \log K_{a}$ (6.31), isliye chhota $p K_{a}$ matlab stronger acid. Table 6.5 mein kuch values di gayi hain.

Jab $α$ chhota ho, $1 - α \approx 1$ , isliye $K_{a} \approx c α^{2}$ , $α \approx \sqrt{K_{a} / c}$ aur $[H^{+}] \approx \sqrt{K_{a} c}$ . Isliye weak acid ko dilute karne se uska degree of ionization badhta hai, halaanki $[H^{+}]$ girta hai. Agar $α < 0.05$ hai to yeh approximation safe hai. Percent ionization (6.32) $100 α$ hai: ionized ÷ initial concentration × 100.

Table 6.5298 K par kuch weak acids aur bases ke ionization constantsNCERT Tables 6.6, 6.7 · p. 195

Substance	Kahan milta hai	$K_{a}$ ya $K_{b}$	$p K$
Formic acid	cheenti ka dank	$1.8 \times 10^{- 4}$	3.74
Lactic acid	dahi, thaki hui muscles	$1.4 \times 10^{- 4}$	3.86
Benzoic acid	food preservative	$6.3 \times 10^{- 5}$	4.20
Acetic acid	sirka	$1.8 \times 10^{- 5}$	4.74
Carbonic acid (pehla step)	soda water	$4.3 \times 10^{- 7}$	6.37
Hypochlorous acid	bleach, pool ka paani	$3.0 \times 10^{- 8}$	7.52
Hydrocyanic acid	kadve badam ki giri	$6.2 \times 10^{- 10}$	9.21
Methylamine (base)	machhli jaisi smell	$4.4 \times 10^{- 4}$	3.36
Ammonia (base)	ghar ke cleaners	$1.8 \times 10^{- 5}$	4.74
Aniline (base)	dye industry	$4.3 \times 10^{- 10}$	9.37

Weak electrolyte ka pH nikalne ka general raasta: maujood acids aur bases pehchano; har proton-transfer equilibrium likho; sabse bade K wale ko main reaction lo (paani ka apna ionization aam taur par bahut kam jodta hai); uski ICE table $α$ ya $x$ ke terms mein banao; solve karo, har approximation check karte hue; phir concentrations aur pH nikalo.

Example 6.18NCERT Problems 6.18–6.20 · p. 196

Weak acid ka pH aur degree of ionization

(a) 0.050 M formic acid ( $K_{a} = 1.8 \times 10^{- 4}$ ) ka pH aur percent ionization nikalo. (b) Ek unknown monobasic acid ke 0.10 M ka pH 3.20 hai. $K_{a}$ nikalo.

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

6.11.4 Weak Bases ka Ionization

Weak base B paani se proton lekar kuch had tak ionize hota hai, $B (a q) + H_{2} O (l) ⇌ B H^{+} (a q) + O H^{-} (a q)$ . (Hydroxide MOH ke liye, $M O H (a q) ⇌ M^{+} (a q) + O H^{-} (a q)$ .) Paani ka constant concentration base ionization constant mein mila diya jaata hai:

Base ionization constant

K_{b} = \frac{[B H^{+}] [O H^{-}]}{[B]} = \frac{c α^{2}}{1 - α}

(6.33)

Values Table 6.5 mein di gayi hain, aur $p K_{b} = - \log K_{b}$ (6.34). Kai organic bases amines hain, yaani ammonia jismein ek ya zyada hydrogens carbon groups se replace kiye gaye hain. Methylamine, nicotine, codeine aur quinine sab chhote $K_{b}$ wale weak bases hain.

Example 6.21NCERT Problems 6.21, 6.22 · p. 198

Solution ke pH se base ka Kb

0.010 M methylamine, $C H_{3} N H_{2}$ , ka pH 11.28 hai. $K_{b}$ nikalo.

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

6.11.5 Ka aur Kb ke beech Relation

DerivationConjugate pair ke liye Ka × Kb = Kw

Pair $N H_{4}^{+}$ / $N H_{3}$ lo. Acid ki tarah, $N H_{4}^{+} + H_{2} O \overset{}{⇌} H_{3} O^{+} + N H_{3}$ ka $K_{a}$ hai; base ki tarah, $N H_{3} + H_{2} O \overset{}{⇌} N H_{4}^{+} + O H^{-}$ ka $K_{b}$ hai. Dono equations jodne par $2 H_{2} O \overset{}{⇌} H_{3} O^{+} + O H^{-}$ milta hai, jiska constant $K_{w}$ hai. Jab equations jodi jaati hain, to unke constants multiply hote hain:

K_{net} = K_{1} \times K_{2}

(6.35)

K_{a} \times K_{b} = K_{w}

(6.36)

Algebra se bhi yahi nikalta hai: $K_{b}$ expression ke upar aur neeche $[H^{+}]$ se multiply karo, aur yeh $[H^{+}] [O H^{-}]$ divided by $[H^{+}] [B] / [B H^{+}]$ mein toot jaata hai, yaani $K_{w} / K_{a}$ . Negative logarithms lene par:

p K_{a} + p K_{b} = p K_{w} = 14

(6.37)

Ammonia ke liye $K_{b} = 1.8 \times 10^{- 5}$ hai, isliye ammonium ion ka $K_{a} = 1.0 \times 10^{- 14} / 1.8 \times 10^{- 5} = 5.6 \times 10^{- 10}$ hai. Strong acid ka conjugate base bahut weak hota hai, aur ulta bhi.

Example 6.23NCERT Problem 6.23 · p. 199

Weak base aur uska conjugate acid

0.10 M pyridine ( $K_{b} = 1.7 \times 10^{- 9}$ ) ka degree of ionization aur pH nikalo, aur pyridinium ion ka $K_{a}$ bhi.

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

6.11.6 Di- aur Polybasic Acids aur Di- aur Polyacidic Bases

Kuch acid molecules do ya teen protons de sakte hain, jaise oxalic, sulphuric, carbonic aur phosphoric acid: yeh polybasic (polyprotic) hain. Yeh steps mein ionize hote hain, har step ka apna constant hota hai. $H_{2} X$ ke liye, $H_{2} X \overset{}{⇌} H^{+} + H X^{-}$ ka $K_{a 1} = [H^{+}] [H X^{-}] / [H_{2} X]$ hai, aur $H X^{-} \overset{}{⇌} H^{+} + X^{2 -}$ ka $K_{a 2} = [H^{+}] [X^{2 -}] / [H X^{-}]$ hai. $H_{3} P O_{4}$ jaise tribasic acid ke teen hote hain.

Aage ke constants tezi se girte hain, kyunki pehle se negative ion se positive proton ko kheenchna zyada mushkil hai. Phosphoric acid ke $K_{a 1} = 7.5 \times 10^{- 3}$ , $K_{a 2} = 6.2 \times 10^{- 8}$ aur $K_{a 3} = 4.8 \times 10^{- 13}$ hain; carbonic acid ke $K_{a 1} = 4.3 \times 10^{- 7}$ aur $K_{a 2} = 4.7 \times 10^{- 11}$ hain. Isliye lagbhag saara $H_{3} O^{+}$ pehle step se aata hai. Do ya zyada protons accept karne wale bases, jaise ethane-1,2-diamine ya carbonate, ke bhi isi tarah har step ka ek $K_{b}$ hota hai.

6.11.7 Acid Strength ko Affect karne wale Factors

Kuch acids doosron se strong kyun hote hain? Poora jawaab complicated hai, lekin moti baat yeh hai ki H–A bond $H^{+}$ aasaani se tab deta hai jab woh weaker ho aur more polar ho. Kaun sa factor jeetega yeh is par depend karta hai ki A periodic table mein kahan hai:

Compare

Acid strength ke do trends

Group mein neeche: bond strength jeetti hai

A bada hota jaata hai, H–A bond lamba aur weak hota hai.
$H F ≪ H C l < H B r < H I$
$H_{2} S$ $H_{2} O$ se stronger acid hai.

Period mein aage: polarity jeetti hai

Atoms ka size milta-julta hai; A ki electronegativity badhti hai.
$C H_{4} < N H_{3} < H_{2} O < H F$
Zyada polar H–A bond $H^{+}$ aasaani se chhodta hai.

6.11.8 Acids aur Bases ke Ionization mein Common Ion Effect

Acetic acid mein sodium acetate jodo. Extra acetate $C H_{3} C O O H \overset{}{⇌} H^{+} + C H_{3} C O O^{-}$ ko left ki taraf push karta hai, isliye $[H^{+}]$ girta hai aur acid aur bhi kam ionize hota hai. Strong acid se $H^{+}$ jodne ka bhi same effect hai. Yeh Le Chatelier's principle ka hi ek case hai.

Definition

Common ion effect

Kisi equilibrium ka woh shift jo aisa substance jodne se hota hai jo pehle se usmein hissa le rahe ion ko supply kare. Yeh weak electrolyte ke ionization ko dabata hai aur sparingly soluble salt ki solubility kam karta hai.

ExampleSodium acetate acetic acid ka ionization kam karta hai.

Example 6.24NCERT Problem 6.24 · p. 201

Common ion aur partial neutralisation

$K_{a} = 1.8 \times 10^{- 5}$ ke saath, pH nikalo (a) 0.10 M acetic acid ka; (b) 0.10 M acetic acid + 0.10 M sodium acetate ka; (c) 40 mL 0.10 M acetic acid + 20 mL 0.10 M NaOH ka.

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

6.11.9 Salts ka Hydrolysis aur unke Solutions ka pH

Salts paani mein poore ionized hote hain, lekin unke kuch ions phir paani se react karte hain. Strong bases ke cations ( $N a^{+}$ , $K^{+}$ , $C a^{2 +}$ , $B a^{2 +}$ ) aur strong acids ke anions ( $C l^{-}$ , $B r^{-}$ , $N O_{3}^{-}$ , $C l O_{4}^{-}$ ) sirf hydrate hote hain, isliye NaCl neutral solution deta hai. Weak acid ya base se aaye ions paani se react karte hain, jise hydrolysis kehte hain, aur pH shift karte hain:

Weak acid + strong base, jaise sodium acetate: $C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-}$ . Hydroxide jamta hai, isliye solution basic hai (pH > 7).
Strong acid + weak base, jaise ammonium chloride: $N H_{4}^{+} + H_{2} O \overset{}{⇌} N H_{3} + H_{3} O^{+}$ . Solution acidic hai (pH < 7). (Purani books base ko $N H_{4} O H$ likhti hain.)
Weak acid + weak base, jaise ammonium acetate: dono ions hydrolyse hote hain, aur pH is par depend karta hai ki kaun stronger hai.

Pehle do types ke liye, hydrolyse hone wala ion bas ek weak base ya acid hai jiska constant $K_{w} / K_{a}$ ya $K_{w} / K_{b}$ hai, isliye uska pH Section 6.11.4 ki tarah nikalta hai. Weak acid aur weak base ke salt ke liye, poori treatment dikhati hai ki degree of hydrolysis concentration par depend nahi karta aur

p H = 7 + \frac{1}{2} (p K_{a} - p K_{b})

(6.38)

jahan

$p K_{a}$: weak acid ka
$p K_{b}$: weak base ka

Solution basic hai agar $p K_{a} > p K_{b}$ , yaani agar acid weaker partner hai, aur acidic hai agar $p K_{a} < p K_{b}$ .

Example 6.25NCERT Problem 6.25 · p. 202

Weak acid aur weak base ke salt ka pH

Ammonium formate solution ka pH nikalo. Formic acid ka $p K_{a} = 3.74$ aur ammonia ka $p K_{b} = 4.75$ hai.

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

Quick check

3 mein se 1

Kaun sa salt paani mein basic solution deta hai?

K N O_{3}

N H_{4} C l

N a_{2} C O_{3}

N a C l

Answer dekho

Answer (C) $N a_{2} C O_{3}$

Carbonate, jo weak acid se aata hai, paani se protons leta hai aur

O H^{-}

chhodta hai.

N H_{4} C l

acidic hai; baaki neutral hain.

6.12 Buffer Solutions

Blood pH 7.35 aur 7.45 ke beech rehta hai; kuch dashamlav ka bhi badlaav medical emergency hai. Kai reactions, medicines, eye drops aur shampoos ko bhi fixed pH chahiye. Jo solutions dilute karne par ya thoda acid ya alkali jodne par pH change ka virodh karte hain, woh buffer solutions hain. Acetic acid aur sodium acetate ka mixture pH 4.75 ke paas buffer karta hai, aur ammonium chloride aur ammonia ka mixture pH 9.25 ke paas. Trick yeh hai ki weak acid aur uske conjugate base dono ka reserve rakho. Jode gaye $H^{+}$ ko base sokh leta hai aur jode gaye $O H^{-}$ ko acid (Fig. 6.18), isliye $[H^{+}]$ mushkil se badalta hai; Fig. 6.19 mein tum buffer ko plain water se compare kar sakte ho. Buffer capacity tum aage ki classes mein padhoge.

2/3Acetate ions jode gaye H⁺ ko le lete hain: pH sirf 4.65 tak girta hai.

Fig. 6.180.010 mol acid ya alkali 0.10 M acetate buffer ke ek litre ko 0.1 pH se bhi kam badalta hai; plain water 5 units badal jaata.

pH meter ke paas acetic acid aur acetate ke bars. Step 1: barabar bars, pH 4.74. Step 2: hydrogen ions jode jaate hain; acetate acetic acid mein badalta hai, bars badalte hain aur pH 4.65 padhta hai, jabki plain water 7 se 2 par gir jaata. Step 3: iske bajaye hydroxide joda jaata hai; acetic acid acetate mein badalta hai aur pH 4.83 padhta hai, jabki paani 12 tak badh jaata.

6.12.1 Buffer Solution Design karna

DerivationHenderson–Hasselbalch equation

Acidic buffer ke liye weak acid ke equilibrium, $H A + H_{2} O \overset{}{⇌} H_{3} O^{+} + A^{-}$ , se shuru karo, aur $K_{a}$ ko $[H_{3} O^{+}]$ ke liye rearrange karo:

[H_{3} O^{+}] = K_{a} \frac{[H A]}{[A^{-}]}

Dono sides ka $- \log$ lo:

Henderson–Hasselbalch

p H = p K_{a} + \log \frac{[A^{-}]}{[H A]}

(6.39)

Weak acid mushkil se ionize hota hai, isliye $[H A]$ liye gaye acid ka concentration hai; lagbhag saara $A^{-}$ poore ionized salt se aata hai. Isliye

p H = p K_{a} + \log \frac{[salt]}{[acid]}

(6.40)

Acid aur salt ke barabar concentrations par $\log 1 = 0$ aur $p H = p K_{a}$ . Isliye buffer banane ke liye aisa acid chuno jiska $p K_{a}$ chahiye wale pH ke paas ho, phir ratio ko fine-tune karo: pH 4.8 ke aas-paas ke liye acetic acid ( $p K_{a}$ 4.76) sodium acetate ke saath.

Basic buffer weak base aur uske conjugate acid ko use karta hai, jaise $N H_{3}$ ke saath $N H_{4} C l$ : $p O H = p K_{b} + \log ([B H^{+}] / [B])$ (6.41). $p H = 14 - p O H$ aur $p K_{a} + p K_{b} = 14$ use karne se $p H = p K_{a} + \log ([B] / [B H^{+}])$ (6.42) milta hai, jismein $B H^{+}$ ka $p K_{a}$ hai: ammonium ke liye 9.25, isliye barabar amounts pH 9.25 deti hain. Buffer ko dilute karne se dono concentrations same factor se badalte hain, isliye ratio, aur pH, same rehte hain.

ExampleNCERT p. 203

Acetate buffer design karna

pH 5.00 ka buffer banane ke liye 500 mL 0.10 M acetic acid ( $p K_{a} = 4.76$ ) mein kitne grams sodium acetate, $C H_{3} C O O N a$ (82.0 g mol⁻¹), gholna hoga?

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

Fig. 6.19Paani, strong acid, weak acid, weak base ya buffer chuno, concentration set karo, phir acid ya alkali jodo aur pH dekho.

Quick check

3 mein se 1

Aise buffer ka pH nikalo jo

N H_{3}

mein 0.20 M aur

N H_{4} C l

mein 0.10 M hai (

p K_{b} = 4.75

Answer dekho

Answer 9.55

p O H = 4.75 + \log (0.10 / 0.20) = 4.45

, isliye pH = 9.55. Ya:

p H = 9.25 + \log (0.20 / 0.10) = 9.55

6.13 Sparingly Soluble Salts ke Solubility Equilibria

Ionic solids ki solubility mein bahut zyada fark hota hai. Anhydrous calcium chloride hawa se water vapour kheenchta hai aur drying agent ki tarah use hota hai, jabki silver chloride aur barium sulphate ko aksar insoluble kaha jaata hai. Solubility do cheezon ka muqabla hai: lattice enthalpy, jo ions ko saath rakhti hai, aur solvation enthalpy, jo solvent molecules ke ions ko gherne par release hoti hai (hamesha negative). Non-polar solvent mein bahut kam energy release hoti hai, isliye salts dissolve nahi hote. Moti baat yeh hai ki salt achhe se tabhi dissolve hota hai jab solvation lattice todne ki keemat chuka de. Solubility temperature par bhi depend karti hai. Room temperature par solubility ke hisaab se salts ko aksar soluble (0.1 M se upar), slightly soluble (0.01–0.1 M) aur sparingly soluble (0.01 M se neeche) groups mein rakha jaata hai. Yahan hum sparingly soluble salt aur uske saturated solution ke beech ka equilibrium dekhte hain.

6.13.1 Solubility Product Constant

Silver chloride ko paani ke saath hilao to bahut thoda dissolve hota hai, jiske baad $A g C l (s) ⇌ A g^{+} (a q) + C l^{-} (a q)$ equilibrium par hota hai. Solid ka concentration constant hai, isliye use K mein mila diya jaata hai:

Solubility product

K_{s p} = [A g^{+}] [C l^{-}]

(6.43)

$K_{s p}$ solubility product constant hai; AgCl ke liye 298 K par yeh $1.8 \times 10^{- 10}$ hai. Agar per litre $S$ mol dissolve hote hain (molar solubility), to $S^{2} = 1.8 \times 10^{- 10}$ aur $S = 1.34 \times 10^{- 5} m o l L^{- 1}$ . Barium sulphate, jo stomach X-rays ke liye 'barium meal' hai, ka $K_{s p} = 1.1 \times 10^{- 10}$ hai: itna kam dissolve hota hai ki toxic $B a^{2 +}$ nuksaan nahi karta.

General roop mein, salt $M_{x} X_{y}$ jo $x$ cations $M^{p +}$ aur $y$ anions $X^{q -}$ deta hai ( $x p = y q$ ke saath), uske liye $[M^{p +}] = x S$ aur $[X^{q -}] = y S$ :

K_{s p} = (x S)^{x} (y S)^{y} = x^{x} y^{y} S^{x + y}

(6.44)

S = {(\frac{K_{s p}}{x^{x} y^{y}})}^{1 / (x + y)}

(6.45)

Table 6.6298 K par kuch salts ke solubility productsNCERT Table 6.9 · p. 205

Salt	$K_{s p}$	$S$ ke terms mein $K_{s p}$
$A g C l$	$1.8 \times 10^{- 10}$	$S^{2}$
$A g B r$	$5.0 \times 10^{- 13}$	$S^{2}$
$B a S O_{4}$	$1.1 \times 10^{- 10}$	$S^{2}$
$C a F_{2}$	$3.9 \times 10^{- 11}$	$4 S^{3}$
$M g (O H)_{2}$	$1.8 \times 10^{- 11}$	$4 S^{3}$
$A g_{2} C r O_{4}$	$1.1 \times 10^{- 12}$	$4 S^{3}$
$P b I_{2}$	$7.1 \times 10^{- 9}$	$4 S^{3}$
$F e (O H)_{3}$	$2.8 \times 10^{- 39}$	$27 S^{4}$
$C a_{3} (P O_{4})_{2}$	$2.0 \times 10^{- 33}$	$108 S^{5}$

Jab solution equilibrium par na ho, to ion concentrations ka yahi product ionic product $Q_{s p}$ kehlata hai. Agar $Q_{s p} < K_{s p}$ hai, to aur solid dissolve ho sakta hai. Agar $Q_{s p} = K_{s p}$ hai, to solution saturated hai. Agar $Q_{s p} > K_{s p}$ hai, to solid tab tak precipitate hota hai jab tak $Q_{s p}$ girkar $K_{s p}$ na ho jaaye (Fig. 6.20).

2/3 $Q_{s p} = K_{s p}$ : dissolve hona aur precipitate hona balance hain.

Fig. 6.20

Q_{s p} < K_{s p}

: solid dissolve hota hai;

Q_{s p} = K_{s p}

: saturated;

Q_{s p} > K_{s p}

: solid precipitate hota hai.

Ek beaker jiski tali par solid silver chloride hai aur solution mein silver aur chloride ions hain. Step 1: kam ions aur solid se upar jaate do arrows, unsaturated likha hai. Step 2: zyada ions aur dono taraf ek-ek arrow, saturated likha hai. Step 3: extra chloride ions aur solid ke chhote naye flakes, silver chloride comes out likha hai.

Example 6.26NCERT Problem 6.26 · p. 205

Har formula mein paanch ions wale salt ki molar solubility

Calcium phosphate, $C a_{3} (P O_{4})_{2}$ , jo haddiyon aur daanton ka mineral hai, ka $K_{s p} = 2.0 \times 10^{- 33}$ hai. Pure paani mein iski molar solubility nikalo, yeh maankar ki ions paani se react nahi karte.

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

Example 6.27NCERT Problem 6.27 · p. 205

Chhote Ksp ka matlab hamesha kam soluble nahi hota

Paani mein kaun zyada soluble hai: $M g (O H)_{2}$ ( $K_{s p} = 1.8 \times 10^{- 11}$ ) ya AgCl ( $K_{s p} = 1.8 \times 10^{- 10}$ )?

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

6.13.2 Ionic Salts ki Solubility par Common Ion Effect

Kisi bhi ion ka concentration badhao to $Q_{s p}$ $K_{s p}$ se zyada ho jaata hai, isliye kuch salt tab tak precipitate hota hai jab tak $Q_{s p} = K_{s p}$ phir na ho jaaye; ek ion kam karo to aur salt dissolve hota hai. Soluble salts bhi aise hi behave karte hain, molarities ki jagah activities ke saath. Saturated brine mein HCl gas bubble karne se bahut pure sodium chloride precipitate hota hai. Gravimetric analysis mein precipitate karne wale ion ki extra amount precipitation ko completion tak le jaati hai, jaise chloride ko AgCl ke roop mein, sulphate ko $B a S O_{4}$ ke roop mein aur iron(III) ko uske hydrated oxide ke roop mein.

Example 6.28NCERT Problem 6.28 · p. 206

Common ion ke saath solubility

0.010 M NaCl mein AgCl ( $K_{s p} = 1.8 \times 10^{- 10}$ ) ki molar solubility nikalo, aur use pure paani se compare karo.

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

pH bhi matter karta hai. Weak acids ke salts, jaise phosphates, carbonates aur fluorides, acid mein zyada dissolve hote hain. $H^{+}$ anion $X^{-}$ ke kuch hisse ko HX mein badal deta hai, jisse $[X^{-}]$ kam hota hai, isliye $[M^{+}] [X^{-}] = K_{s p}$ banaye rakhne ke liye aur salt dissolve hota hai. 1 : 1 salt MX ke liye, dissolved anion ka jo fraction $X^{-}$ ke roop mein bachta hai woh $f = K_{a} / (K_{a} + [H^{+}])$ hai, isliye $K_{s p} = S \cdot f S$ aur

Weak acid ke salt ki solubility

S = \sqrt{K_{s p} \frac{K_{a} + [H^{+}]}{K_{a}}}

(6.46)

Kam pH (bada $[H^{+}]$ ) bada S deta hai. Isiliye acidic drinks tooth enamel par bhi asar karte hain, jo calcium phosphate mineral hai.

Recap

Summary aur key points

Equilibrium dynamic hai: closed system mein ulte processes barabar rates par chalte hain, isliye properties constant rehti hain (Section 6.1, Section 6.2).
Har physical equilibrium ki given temperature par ek fixed quantity hoti hai: melting point, vapour pressure, solubility ya Henry's-law ratio (Table 6.1).
$K_{c}$ products over reactants hai, har concentration uske coefficient ki power ke saath; yeh sirf temperature par aur equation kaise likhi gayi hai is par depend karta hai (Section 6.3).
$K_{p} = K_{c} (R T)^{Δ n}$ ; pure solids aur liquids K mein nahi aate (Section 6.4.1, Section 6.5).
Bada K: zyaadatar products; $Q < K$ : net forward; ICE tables compositions deti hain; $Δ G^{\circ} = - R T \ln K$ (Section 6.6, Section 6.7).
Le Chatelier: jode gaye substances use ho jaate hain, pressure kam gas moles ko favour karta hai, heat endothermic side ko favour karti hai, catalysts sirf rate badalte hain (Section 6.8).
Arrhenius, Brønsted–Lowry (conjugate pairs) aur Lewis (electron pairs) acids aur bases (Section 6.10).
298 K par $K_{w} = 10^{- 14}$ ; pH + pOH = 14 (Section 6.11.1, Section 6.11.2).
$K_{a} = c α^{2} / (1 - α)$ , $K_{a} K_{b} = K_{w}$ ; common ions ionization ko dabate hain; salts hydrolyse hote hain (Section 6.11).
Buffers $p H = p K_{a} + \log ([salt] / [acid])$ follow karte hain; $K_{s p}$ se molar solubility milti hai, aur $Q_{s p} > K_{s p}$ ka matlab precipitation (Section 6.12, Section 6.13).

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 6.1	Practice Q1
NCERT Exercises 6.2, 6.6	Practice Q2
NCERT Exercises 6.3	Practice Q3
NCERT Exercises 6.4	Practice Q4
NCERT Exercises 6.5, 6.10	Practice Q5
NCERT Exercises 6.7, 6.13	Practice Q6
NCERT Exercises 6.8, 6.33	Practice Q7
NCERT Exercises 6.9	Practice Q8
NCERT Exercises 6.11	Practice Q9
NCERT Exercises 6.12, 6.21	Practice Q10
NCERT Exercises 6.14, 6.15	Practice Q11
NCERT Exercises 6.16, 6.22	Practice Q12
NCERT Exercises 6.17	Practice Q13
NCERT Exercises 6.18	Practice Q14
NCERT Exercises 6.19, 6.34	Practice Q15
NCERT Exercises 6.20	Practice Q16
NCERT Exercises 6.23	Practice Q17
NCERT Exercises 6.24	Practice Q18
NCERT Exercises 6.25, 6.26	Practice Q19
NCERT Exercises 6.27	Practice Q20
NCERT Exercises 6.28	Practice Q21
NCERT Exercises 6.29	Practice Q22
NCERT Exercises 6.30	Practice Q23
NCERT Exercises 6.31	Practice Q24
NCERT Exercises 6.32	Practice Q25
NCERT Exercises 6.35, 6.37, 6.38, 6.39	Practice Q26
NCERT Exercises 6.36, 6.40	Practice Q27
NCERT Exercises 6.41, 6.42, 6.55, 6.56	Practice Q28
NCERT Exercises 6.43, 6.44	Practice Q29
NCERT Exercises 6.45	Practice Q30
NCERT Exercises 6.46, 6.53, 6.59	Practice Q31
NCERT Exercises 6.47, 6.50, 6.60	Practice Q32
NCERT Exercises 6.48, 6.49, 6.57, 6.58	Practice Q33
NCERT Exercises 6.51, 6.52, 6.54	Practice Q34
NCERT Exercises 6.61, 6.62, 6.63, 6.64	Practice Q35
NCERT Exercises 6.65	Practice Q36
NCERT Exercises 6.66	Practice Q37
NCERT Exercises 6.67, 6.68	Practice Q38
NCERT Exercises 6.69, 6.71, 6.73	Practice Q39
NCERT Exercises 6.70	Practice Q40
NCERT Exercises 6.72	Practice Q41

Quick check: MCQs

6.1.5 Physical Processes wale Equilibria ki General Characteristics3 questions
6.2 Chemical Processes mein Equilibrium – Dynamic Equilibrium3 questions
6.3 Law of Chemical Equilibrium aur Equilibrium Constant3 questions
6.4.1 Gaseous Systems mein Equilibrium Constant2 questions
6.5 Heterogeneous Equilibria2 questions
6.6.3 Equilibrium Concentrations Calculate karna3 questions
6.7 Equilibrium Constant K, Reaction Quotient Q aur Gibbs Energy G ke beech Relationship3 questions
6.8.5 Catalyst ka Effect2 questions
6.10.3 Lewis Acids aur Bases3 questions
6.11.2 pH Scale2 questions
6.11.9 Salts ka Hydrolysis aur unke Solutions ka pH3 questions
6.12.1 Buffer Solution Design karna3 questions
6.13.2 Ionic Salts ki Solubility par Common Ion Effect3 questions

Glossary · 15

Key terms (English + Hindi)

buffer solutionबफर विलयनbuffer vilayan: Weak acid aur uska conjugate base, ya weak base aur uska conjugate acid, jo pH change ka virodh karta hai.Chapter mein pehli baar yahan aaya →

common ion effectसम-आयन प्रभावsam-aayan prabhav: Equilibrium mein pehle se maujood ion jodne se ionization ya solubility ka dab jaana.
conjugate acid–base pairसंयुग्मी अम्ल-क्षारक युग्मsanyugmi amla-ksharak yugm: Ek acid aur ek base jinmein ek proton ka fark hota hai.

degree of ionizationआयनन की मात्राaayanan ki matra: Weak electrolyte ka woh fraction $α$ jo ionized hai.Chapter mein pehli baar yahan aaya →
dynamic equilibriumगतिक साम्यgatik saamya: Closed system ki aisi state jismein ulte processes barabar rates par chalte hain, isliye koi measurable property nahi badalti.Chapter mein pehli baar yahan aaya →

electrolyteविद्युत्-अपघट्यvidyut-apaghatya: Aisa substance jiska aqueous solution conduct karta hai kyunki usmein ions hote hain.Chapter mein pehli baar yahan aaya →
equilibrium constantसाम्य स्थिरांकsaamya sthirank: Given temperature par fixed value: products over reactants, har concentration (ya pressure) uske coefficient ki power ke saath.Chapter mein pehli baar yahan aaya →

heterogeneous equilibriumविषमांगी साम्यvishamangi saamya: Ek se zyada phase wala equilibrium.Chapter mein pehli baar yahan aaya →
hydrolysis of saltsलवणों का जल-अपघटनlavanon ka jal-apghatan: Salt ke ions ka paani se reaction jo solution ko acidic ya basic bana deta hai.Chapter mein pehli baar yahan aaya →

ionic product of waterजल का आयनिक गुणनफलjal ka aayanik gunanphal: $K_{w} = [H^{+}] [O H^{-}]$ , 298 K par $10^{- 14}$ .Chapter mein pehli baar yahan aaya →

Le Chatelier’s principleला-शातलिए सिद्धांतla-shatliye siddhant: Concentration, pressure ya temperature ke change se disturb hua equilibrium is tarah shift hota hai ki change kuch had tak undo ho jaaye.Chapter mein pehli baar yahan aaya →
Lewis acidलूइस अम्लlewis amla: Electron-pair acceptor; Lewis base donor hota hai.Chapter mein pehli baar yahan aaya →

pH: Hydrogen ion activity ka negative base-10 logarithm, lagbhag $- \log [H^{+}]$ .Chapter mein pehli baar yahan aaya →

reaction quotientअभिक्रिया भागफलabhikriya bhagphal: Kisi bhi pal par nikala gaya K expression; Q ko K se compare karne par net reaction ki direction milti hai.Chapter mein pehli baar yahan aaya →

solubility product constantविलेयता गुणनफलvileyata gunanphal: Sparingly soluble salt aur uske ions ke beech ka equilibrium constant $K_{s p}$ .Chapter mein pehli baar yahan aaya →

6.1 Physical Processes mein Equilibrium

6.1.1 Solid–Liquid Equilibrium

6.1.2 Liquid–Vapour Equilibrium

6.1.3 Solid–Vapour Equilibrium

6.1.4 Liquids mein Solids ya Gases ke Dissolution wala Equilibrium

6.1.5 Physical Processes wale Equilibria ki General Characteristics

6.2 Chemical Processes mein Equilibrium – Dynamic Equilibrium

6.3 Law of Chemical Equilibrium aur Equilibrium Constant

6.4 Homogeneous Equilibria

6.4.1 Gaseous Systems mein Equilibrium Constant

6.5 Heterogeneous Equilibria

6.6 Equilibrium Constants ke Applications

6.6.1 Reaction ka Extent Predict karna

6.6.2 Reaction ki Direction Predict karna

6.6.3 Equilibrium Concentrations Calculate karna

6.7 Equilibrium Constant K, Reaction Quotient Q aur Gibbs Energy G ke beech Relationship

6.8 Equilibria ko Affect karne wale Factors

6.8.1 Concentration Change ka Effect

6.8.2 Pressure Change ka Effect

6.8.3 Inert Gas Addition ka Effect

6.8.4 Temperature Change ka Effect

6.8.5 Catalyst ka Effect

6.9 Solution mein Ionic Equilibrium

6.10 Acids, Bases aur Salts

6.10.1 Acids aur Bases ka Arrhenius Concept

6.10.2 Brönsted–Lowry Acids aur Bases

6.10.3 Lewis Acids aur Bases

6.11 Acids aur Bases ka Ionization

6.11.1 Water ka Ionization Constant aur uska Ionic Product

6.11.2 pH Scale

6.11.3 Weak Acids ke Ionization Constants

6.11.4 Weak Bases ka Ionization

6.11.5 Ka aur Kb ke beech Relation

6.11.6 Di- aur Polybasic Acids aur Di- aur Polyacidic Bases

6.11.7 Acid Strength ko Affect karne wale Factors

6.11.8 Acids aur Bases ke Ionization mein Common Ion Effect

6.11.9 Salts ka Hydrolysis aur unke Solutions ka pH

6.12 Buffer Solutions

6.12.1 Buffer Solution Design karna

6.13 Sparingly Soluble Salts ke Solubility Equilibria

6.13.1 Solubility Product Constant

6.13.2 Ionic Salts ki Solubility par Common Ion Effect

Apna NCERT exercise question dhoondo

Quick check: MCQs

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