Le Chatelier's principle — pressure, temperature, concentration, catalyst effects

2.6.5 · Chemistry › Equilibrium

WHY aisa hota hai? Equilibrium par, forward aur reverse rates equal hoti hain. Jab tum system ko disturb karte ho, tum temporarily yeh balance tod dete ho. System response karta hai us direction ko favor karke jo naye conditions mein equilibrium re-establish kare.

Core Definition

Key term: Stress = koi bhi external change jo $Q$ ka ratio $K$ se door le jaaye.

Derivation from First Principles

Starting point: Reaction $a A + b B ⇌ c C + d D$ ke liye, equilibrium constant hai:

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

Reaction quotient $Q$ ka same form hai lekin current concentrations use karta hai (equilibrium wali nahi):

$Q = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}_{current}$

Shift rule:

Agar $Q < K$ : Bahut zyada reactant → forward shift (zyada products banana)
Agar $Q > K$ : Bahut zyada product → backward shift (zyada reactants banana)
Agar $Q = K$ : Equilibrium par, koi net shift nahi

WHY? Kyunki $Δ G = Δ G^{\circ} + R T ln Q$ . Jab $Q \neq = K$ , toh $Δ G \neq = 0$ , isliye reaction us direction mein spontaneous hai jo $Q$ ko $K$ ke paas laaye.

Figure — Le Chatelier's principle — pressure, temperature, concentration, catalyst effects

Effect 1: Concentration Changes

Derivation: $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ ke liye, agar tum $N_{2}$ add karo:

$Q = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$

$[N_{2}]$ badhane se denominator bada ho jaata hai, isliye $Q$ decrease ho jaata hai. Kyunki ab $Q < K$ hai, system forward shift karta hai added $N_{2}$ consume karne ke liye aur $[N H_{3}]$ badhane ke liye jab tak $Q = K$ nahi ho jaata.

Stress: $0.2 M$ $S O_{2}$ add karo.

Step 1 — Q calculate karo: $Q = \frac{[ S O _{3} ] ^{2}}{[ S O _{2} ] ^{2} [ O _{2} ]} = \frac{( 0.8 ) ^{2}}{( 0.7 ) ^{2} ( 0.3 )} = \frac{0.64}{0.147} = 4.35$

Yeh step kyun? Naya $[S O_{2}] = 0.5 + 0.2 = 0.7 M$ . Humein dekhna hai ki $Q$ , $K$ ke relative kitna change hua.

Step 2 — Agar $K = 5.0$ : Kyunki $Q = 4.35 < K = 5.0$ , system forward shift karta hai.

WHY? Naye conditions mein reactants ke relative zyada product nahi hai.

Step 3 — Prediction: $[S O_{3}]$ badhega, $[S O_{2}]$ aur $[O_{2}]$ tab tak ghatengy jab tak $Q = K$ nahi ho jaata.

Effect 2: Pressure and Volume Changes

WHY? Gas-phase reactions ke liye, $K_{p}$ partial pressures se relate karta hai. Ideal gas law se, constant $T$ par $P \propto n / V$ . $V$ change karne se saari concentrations proportionally change hoti hain, lekin equilibrium expression mein har species ke alag powers hote hain.

Derivation: $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ ke liye (4 moles reactant gas → 2 moles product gas):

$K_{p} = \frac{P _{N H_{3}}^{2}}{P _{N_{2}} \cdot P _{H_{2}}^{3}}$

Agar tum compress karo (volume decrease karo), toh saare partial pressures increase ho jaate hain. Lekin denominator ka total exponent zyada hai ( $1 + 3 = 4$ vs numerator ka $2$ ), isliye $Q$ , $K$ se zyada decrease hota hai. Thus $Q < K$ → forward shift.

Moles count karo:

Reactants: $1 + 3 = 4$ moles gas
Products: $2$ moles gas
Compression products ko favor karta hai (fewer moles)

Stress: $1 L$ tak compress karo (volume half, pressure double).

Step 1 — Moles count karo:

Reactant: 1 mole $N_{2} O_{4}$
Product: 2 moles $N O_{2}$
Product side par zyada moles hain

Yeh step kyun? Zyada moles wali side pressure changes ke liye zyada sensitive hoti hai.

Step 2 — Shift predict karo: Pressure increase fewer moles ko favor karta hai → backward shift (toward $N_{2} O_{4}$ ).

WHY? $N_{2} O_{4}$ banana total gas molecules ko reduce karta hai, pressure stress ko partially relieve karta hai.

Key insight: Agar dono sides par equal moles hain, toh pressure change se no shift hoga.

Effect 3: Temperature Changes

WHY temperature special hai? Concentration ya pressure stresses se alag, temperature change karna actually $K$ ki value ko hi change karta hai (van't Hoff equation ke through). Yahi ek stress hai jo $K$ ko modify karta hai.

Derivation from van 't Hoff:

$\frac{d l n K}{d T} = \frac{Δ H ^{\circ}}{R T ^{2}}$

Exothermic ke liye ( $Δ H^{\circ} < 0$ ): Jaise $T$ badhta hai, $\frac{d l n K}{d T} < 0$ , isliye $K$ decrease hota hai. System ko naye, chhote $K$ ko match karne ke liye backward shift karna padta hai.

Endothermic ke liye ( $Δ H^{\circ} > 0$ ): Jaise $T$ badhta hai, $K$ badhta hai, isliye system forward shift karta hai.

Stress: Temperature $400° C$ se $500° C$ tak badhao.

Step 1 — Reaction type identify karo: $Δ H < 0$ → exothermic → heat ek "product" hai

Yeh step kyun? Humein jaanna hai ki kaun si direction heat absorb ya release karti hai.

Step 2 — Heat ko product ki tarah apply karo: $T$ badhana "product add karne" jaisa hai → backward shift (reactants ki taraf)

WHY? System added heat consume karne ki koshish karta hai reverse (endothermic) direction ko favor karke.

Step 3 — Predict karo: $[N_{2}]$ aur $[H_{2}]$ badhenge, $[N H_{3}]$ ghategaa. Saath hi, $K$ decrease hoga (van 't Hoff se verified).

Practical note: Isliye Haber process moderate temperatures par chalaaya jaata hai, zyada yield chahne ke bawajood—high $T$ reaction speed karta hai lekin equilibrium ko unfavorably shift karta hai.

Effect 4: Catalyst

WHY? Catalyst dono directions ke liye $E_{a}$ usi amount se ghata deta hai. Kyunki $K = \frac{k _{f}}{k _{r}}$ , aur dono rate constants same factor se increase hote hain, unka ratio (aur thus $K$ ) unchanged rehta hai.

$K = \frac{k _{f}}{k _{r}} = \frac{A _{f} e ^{- E_{a, f} / R T}}{A _{r} e ^{- E_{a, r} / R T}} = constant$

Catalyst change karta hai: $E_{a, f}$ aur $E_{a, r}$ dono $Δ E_{a}$ se decrease hote hain

Naya ratio: $\frac{A _{f} e ^{- (E_{a, f} - Δ E_{a}) / R T}}{A _{r} e ^{- (E_{a, r} - Δ E_{a}) / R T}} = \frac{A _{f} e ^{- E_{a, f} / R T}}{A _{r} e ^{- E_{a, r} / R T}} \cdot \frac{e ^{Δ E_{a} / R T}}{e ^{Δ E_{a} / R T}} = K$ (same!)

Stress: $M n O_{2}$ catalyst add karo.

Step 1 — Rates ka kya hoga?

Forward rate: increase hoti hai (maano, 1000×)
Reverse rate: increase hoti hai (same factor, 1000×)

Yeh step kyun? Catalyst alternate pathway provide karta hai jisme $E_{a}$ kam hota hai.

Step 2 — Equilibrium ka kya hoga?

Equilibrium position: unchanged
Equilibrium tak pahunchne ka time: bahut jaldi

WHY? $K = \frac{k _{f}}{k _{r}}$ constant rehta hai jab dono proportionally change hote hain.

Practical: Catalysts yield nahi badhate, lekin reactions ko industrially viable banate hain time years se seconds tak reduce karke.

Summary Table of Stress Effects

Stress	Change	System Response	$K$ Changes?
Reactant add karo	$Q ↓$	Forward shift	Nahi
Product remove karo	$Q ↓$	Forward shift	Nahi
Pressure badhao	(fewer moles favored)	Fewer moles ki taraf shift	Nahi
$T$ badhao (exo)	$Δ H < 0$	Backward shift	Haan ( $K ↓$ )
$T$ badhao (endo)	$Δ H > 0$	Forward shift	Haan ( $K ↑$ )
Catalyst add karo	—	No shift (sirf jaldi)	Nahi

Kyun sahi lagta hai: Catalysts reactions speed karte hain, aur hum product jaldi banta dekhte hain.

Steel-man: Yeh intuition ki "jaldi forward = zyada product" sahi hoti agar reverse reaction exist nahi karti. Ek one-way reaction mein, catalyst per unit time product zaroor badhata hai.

Fix: Equilibrium mein, reverse reaction bhi equally speed ho jaati hai. Dono rates same factor se scale hote hain, isliye ratio (jo equilibrium position determine karta hai) unchanged rehta hai. Catalyst kinetics (time) change karta hai, thermodynamics (position) nahi.

Test karo: Agar catalyst equilibrium shift karta, toh hum equilibrium systems se infinite energy nikal sakte—yeh thermodynamics violate karta hai!

Kyun sahi lagta hai: Bahut saare industrial reactions mein (Haber, contact process), high pressure actually products ko favor karta hai. Students in examples se overgeneralize kar lete hain.

Steel-man: Un reactions ke liye jahan products mein fewer moles hain (jaise $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ ), yeh actually sahi hai. Galti yeh assume karna hai ki yeh hamesha sach hai.

Fix: Pressure fewer total moles of gas wali side ko favor karta hai. $N_{2} O_{4} ⇌ 2 N O_{2}$ ke liye (1 mole → 2 moles), pressure reactants ko favor karta hai, products ko nahi. Pehle hamesha moles count karo.

Recall Ek 12-saal ke bachhe ko explain karo

Socho tum ek seesaw par apne dost ke saath perfectly balanced ho. Yahi equilibrium hai—dono sides equal hain.

Ab, agar koi tumhare dost ki side ko neeche push kare? Seesaw ab balanced nahi hai. Lekin yahan cool part hai: seesaw naturally khud ko balance karne ki koshish karta hai, tumhari side thoda neeche jaake aur tumhare dost ki side thoda upar aakar.

Le Chatelier's principle bilkul us seesaw jaisi hai. Jab tum ek chemical reaction ko "push" karte ho (zyada chemicals add karke, use squeeze karke, ya heat karke), reaction automatically doosri taraf push karti hai balanced hone ke liye.

Zyada reactant add kiya? Reaction zyada product banati hai use use karne ke liye.
Heat kiya? Agar reaction normally heat release karti hai, toh woh slow ho jaati hai cheezein cool karne ke liye.
Squeeze kiya? Agar reaction fewer gas molecules banati hai, toh woh us taraf shift ho jaati hai squeeze relieve karne ke liye.

Reaction hamesha balance wapas laane ki koshish karti hai, bilkul ek seesaw ki tarah!

Pressure mnemonic: "Fewer Favored when Flattened" (compression fewer moles ko favor karta hai)

Temperature mnemonic: "Endo Eats heat, Exo Expels heat" → heat add karna endothermic direction ki taraf shift karta hai

Connections

Chemical Equilibrium — Le Chatelier explain karta hai kaise equilibrium disturbances ke response mein react karta hai
Equilibrium Constant — $K$ target hai; $Q$ batata hai kis direction mein shift karna hai
Reaction Quotient Q — $Q$ vs $K$ compare karna shift direction predict karta hai
Haber Process — industrial application jo pressure aur temperature use karke yield maximize karta hai
van 't Hoff Equation — quantify karta hai kaise $K$ temperature ke saath change karta hai
Activation Energy — kyun catalysts dono directions ko equally speed karte hain
Gibs Free Energy — $Δ G = R T ln (Q / K)$ dikhata hai kyun systems shift karte hain jab $Q \neq = K$ ho

#flashcards/chemistry

Le Chatelier's principle kya hai? :: Agar ek system equilibrium par ek stress (concentration, pressure, volume, ya temperature mein change) ke under aata hai, toh equilibrium us stress ko counteract karne ke liye shift karta hai aur partially equilibrium restore karta hai.

Jab tum equilibrium mein zyada reactant add karte ho, toh kis taraf shift hoga?

Forward (products ki taraf), kyunki reactant add karne se Q, K se neeche chala jaata hai, isliye system zyada product banakar compensate karta hai.

Reaction

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

ke liye, pressure badhane par kya hoga?

Forward shift hoga (products/ammonia ki taraf) kyunki products mein fewer moles of gas hain (2) reactants se (4), aur high pressure fewer moles ko favor karta hai.

Exothermic reaction mein temperature badhane se backward shift kyun hota hai?

Kyunki exothermic reactions mein heat ek product jaisi hoti hai; heat badhana (T increase karna) product add karne jaisa hai, isliye system use consume karne ke liye backward shift karta hai. Saath hi, exothermic reactions ke liye T badhne par K decrease ho jaata hai.

Kya catalyst equilibrium position change karta hai?

Nahi. Catalyst forward aur reverse dono reactions ko equally speed karta hai, isliye K = kf/kr unchanged rehta hai. Yeh sirf equilibrium tak jaldi pahunchne mein help karta hai.

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

ke liye, pressure badhane par equilibrium kis taraf shift hoga?

Backward (toward

N_{2} O_{4}

) kyunki reactants mein fewer moles hain (1) products se (2), aur pressure fewer moles ko favor karta hai.

Q aur K use karke shift ki direction kaise predict karte hain?

Agar Q < K, forward shift karo (zyada products banao). Agar Q > K, backward shift karo (zyada reactants banao). Agar Q = K, koi shift nahi (equilibrium par hai).

Temperature hi ek aisa stress kyun hai jo K ko khud change karta hai?

Kyunki K, ΔG° par depend karta hai jo van 't Hoff equation ke through temperature par depend karta hai: d(ln K)/dT = ΔH°/(RT²). Baaki stresses concentrations/pressures change karte hain lekin thermodynamic equilibrium constant nahi.

Endothermic equilibrium mein temperature ghataane par kya hoga?

Backward shift hoga (reactants ki taraf) kyunki endothermic reactions mein heat ek reactant ki tarah kaam karta hai; heat remove karna reactant remove karne jaisa hai, jo products se door shift karta hai.

Agar ek reaction mein dono sides par equal moles of gas hain, toh pressure change karne par kya hoga?

Equilibrium position mein koi shift nahi hoga, kyunki jab mole counts same hote hain toh pressure changes dono sides ko equally affect karte hain.