Heterogeneous equilibria — pure solids - liquids excluded

2.6.6 · Chemistry › Equilibrium

Core Concept

Mathematical Wajah: Exclusion Rule Derive Karna

First Principles se Shuru Karte Hain

Kisi bhi equilibrium ke liye, thermodynamic equilibrium constant hai:

$K=\prod \frac{(\text{activities of products})}{(\text{activities of reactants})}$

Activity kya hoti hai? Activity $a_i$ woh "effective concentration" hai jo non-ideal behavior ko account karti hai:

$a_i={\gamma }_i\cdot \frac{[i]}{c^{\circ }}$

jahan ${\gamma }_i$ activity coefficient hai, $[i]$ concentration/pressure hai, aur $c^{\circ }$ standard state hai.

Alag-Alag Phases ke Liye:

1. Gases: $a_i\approx \frac{P_i}{P^{\circ }}$ jahan $P^{\circ }=1$ bar
2. Aqueous solutions: $a_i\approx \frac{[i]}{1\text{ M}}$ (dilute approximation)
3. Pure solids/liquids: $a_i\approx 1$ har waqt!

Pure substances ke liye $a=1$ kyun hota hai?

Pure solid ya liquid ke liye, standard state ko khud woh pure substance choose kiya jaata hai. Activity uss standard state ke ratio ke roop mein define hoti hai:

$a_{\text{pure}}=\frac{[\text{actual concentration of the pure phase}]}{[\text{concentration of the pure phase in its standard state}]}$

Lekin "actual" pure phase hi standard state hai. Uska molar concentration fixed rehta hai:

$[\text{pure substance}]=\frac{\text{density}}{\text{molar mass}}=\text{constant (given T, P par)}$

Kyunki numerator aur denominator same fixed quantity hain:

$a_{\text{pure solid/liquid}}=\frac{\text{(density/molar mass)}}{\text{(density/molar mass)}}=1$

Zyada solid add karne se total amount badhti hai lekin density ya molar mass nahi badhta, isliye activity exactly 1 hi rehti hai.

Exclusion Ka Practical Use

Consider karo: ${\text{CaCO}}_3(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_2(g)$

Poora expression: $K=\frac{a_{\text{CaO}}\cdot a_{{\text{CO}}_2}}{a_{{\text{CaCO}}_3}}$

Kyunki $a_{\text{CaO}}=1$ aur $a_{{\text{CaCO}}_3}=1$:

$K_p=\frac{1\cdot P_{{\text{CO}}_2}}{1}=P_{{\text{CO}}_2}$

Include karo:

• Gases: $K_p$ ke liye partial pressures $(P_i)$ use karo ya $K_c$ ke liye concentrations
• Aqueous ions/molecules: molar concentrations $[i]$ use karo

Exclude karo:

• Pure solids
• Pure liquids (including dilute aqueous solutions mein paani)
• Solvents jo bahut zyada excess mein hain

Relationship: $K_p=K_c(RT{)}^{\Delta n_g}$ jahan $\Delta n_g=\sum n_{\text{gas products}}-\sum n_{\text{gas reactants}}$

Worked Examples with Step Reasoning

Is equilibrium ke liye $K_p$ likho.

Step 1: Phases identify karo

• ${\text{CaCO}}_3$: solid → exclude
• $\text{CaO}$: solid → exclude
• ${\text{CO}}_2$: gas → include

Yeh step kyun? Sirf CO₂ ka pressure independently vary kar sakta hai. Solids ki "concentrations" unki densities se fixed hoti hain.

Step 2: Sirf included species ke saath expression likho $K_p=P_{{\text{CO}}_2}$

Physical meaning: Ek given temperature par, limestone tab tak decompose hota rahega jab tak CO₂ ek specific pressure tak na pahunch jaaye. Agar CO₂ pump out kar do, aur zyada limestone decompose hoga. Agar $K_p$ se zyada CO₂ add kar do, toh CaO aur CO₂ react karke CaCO₃ banate hain.

Numerical: Approximately 1100 K par, $K_p\approx 1.0$ atm. (Calcium carbonate sirf high temperatures par appreciably decompose hota hai, roughly 900 °C ≈ 1173 K, jahan $P_{{\text{CO}}_2}$ 1 atm tak pahunchta hai.) Iska matlab hai limestone spontaneously decompose hota hai jab surrounding $P_{{\text{CO}}_2}$ us temperature par equilibrium value se neeche ho.

$K_{sp}$ (solubility product) likho.

Step 1: Phases identify karo

• $\text{AgCl}$: solid → exclude
• ${\text{Ag}}^{+}$: aqueous → include
• ${\text{Cl}}^{-}$: aqueous → include

Yeh step kyun? Undissolved AgCl pure solid hai. Sirf dissolved ions ke concentrations variable hote hain.

Step 2: Expression likho $K_{sp}=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]$

Physical interpretation: $K_{sp}$ humein bataata hai ki saturation par ion concentrations ka maximum product kya ho sakta hai. Chahe kitna bhi solid AgCl paani mein daalo, product $[{\text{Ag}}^{+}][{\text{Cl}}^{-}]$ kabhi $K_{sp}$ se zyada nahi ho sakta.

Numerical: $K_{sp}=1.8\times 10^{-10}$ at 25°C. Pure water mein, agar $[{\text{Ag}}^{+}]=s$ ho, toh $[{\text{Cl}}^{-}]=s$, isliye $s^2=K_{sp}$, jo deta hai $s=1.3\times 10^{-5}$ M.

High temperature par $K_p$ likho.

Step 1: Phases identify karo

• $\text{C}$: solid (graphite) → exclude
• ${\text{H}}_2\text{O}$, $\text{CO}$, ${\text{H}}_2$: sab gases → include

Yeh step kyun? High T par, paani steam (gas) hota hai, liquid nahi. Solid carbon ki activity 1 hai.

Step 2: Expression likho $K_p=\frac{P_{\text{CO}}\cdot P_{{\text{H}}_2}}{P_{{\text{H}}_2\text{O}}}$

Key insight: Aur graphite add karne se equilibrium shift nahi hota—sirf gas partial pressures change karne se farq padta hai.

Numerical reasoning: Agar $K_p=10$ at 1000 K aur initially $P_{{\text{H}}_2\text{O}}=2$ atm, $P_{\text{CO}}=P_{{\text{H}}_2}=0$:

• Maano $x$ = extent of reaction
• Equilibrium par: $P_{{\text{H}}_2\text{O}}=2-x$, $P_{\text{CO}}=P_{{\text{H}}_2}=x$
• $10=\frac{x^2}{2-x}$ → $x^2+10x-20=0$ → $x=\frac{-10+\sqrt{180}}{2}\approx 1.71$ atm

Yeh step kyun? Hum positive root lete hain kyunki pressure positive honi chahiye; $\sqrt{180}\approx 13.42$, isliye $x\approx (-10+13.42)/2\approx 1.71$ atm.

Kya iron catalyst $K_p$ mein appear karta hai?

Answer: Nahi. Catalysts dono forward aur reverse reactions ko equally speed up karte hain—yeh equilibrium position shift nahi karte. Chahe Fe solid ho, yeh reactant ya product bhi nahi hai, isliye yeh doubly excluded hai.

$K_p=\frac{(P_{{\text{NH}}_3}{)}^2}{P_{{\text{N}}_2}\cdot (P_{{\text{H}}_2}{)}^3}$

Yeh kyun important hai: Equilibrium tak pahunchne ki speed badhane ke liye tum tons catalyst add kar sakte ho, lekin final pressures sirf $K_p$, T, aur initial amounts par depend karte hain.

Common Mistakes aur Steel-Manning

Kyun sahi lagta hai: Roz ki zindagi mein, zyada starting material → zyada product. Confusion isliye hoti hai kyunki hum solids ko gases ki tarah "use up" hone wala samajhte hain.

Sach: Jab tak koi bhi solid present hai, aur zyada add karne se equilibrium position nahi badlti. Solid ki activity pehle se hi apne maximum (1) par hai. Aise sochiye: solid surface reaction ke liye sites provide karta hai, lekin saari available sites pehle se hi use ho rahi hain. Zyada solid = zyada sites, lekin har site par same rate, isliye same equilibrium.

Sahi: Equilibrium par CO₂ ki amount sirf temperature par depend karti hai (jo $K_p$ determine karta hai), nahi iss par ki tumne kitna CaCO₃ liya tha—jab tak kuch CaCO₃ unreacted raha ho.

Caveat: Agar tum itna thoda CaCO₃ se start karo ki woh poora decompose ho jaaye, toh tumhare paas equilibrium nahi hai (solid phase bacha hi nahi). Yeh ek alag scenario hai—system ab heterogeneous nahi raha.

Kyun sahi lagta hai: Zyaatar aqueous chemistry mein (acids, bases, salts dissolving), paani solvent hota hai bahut zyada excess mein, isliye uski activity ≈ 1 rehti hai.

Sach: Paani tabhi exclude hota hai jab woh pure liquid ya solvent bahut zyada excess mein ho. Agar paani:

• Gas (steam) ho: include karo! Example: ${\text{H}}_2(g)+{\text{CO}}_2(g)\rightleftharpoons {\text{H}}_2\text{O}(g)+\text{CO}(g)$ → $K_p=\frac{P_{{\text{H}}_2\text{O}}\cdot P_{\text{CO}}}{P_{{\text{H}}_2}\cdot P_{{\text{CO}}_2}}$
• Non-aqueous solution mein ho jahan uska concentration significantly vary karta hai: include karo!
• Apni initial quantity ke relative significant amount mein consume/produce ho raha ho: include karo!

Decision rule: Agar substance ki amount itni change hoti hai ki uski concentration/activity meaningfully affect ho, toh include karo. Agar yeh ek ocean hai jo barely ripple karta hai, toh exclude karo.

Kyun sahi lagta hai: Hum ratios ke aadi hain—agar main numerator double karunga, ratio change ho jaayega.

Sach: $K$ sirf temperature par depend karta hai. Yeh ek fundamental thermodynamic property hai. Solid add karne se na forward na reverse rate constants ($k_f,k_r$) change hote hain, aur kyunki $K=k_f/k_r$, K same rehta hai.

Jo actually change hota hai: Equilibrium tak pahunchne ka time change ho sakta hai (zyada surface area → faster), lekin equilibrium position (final pressures/concentrations) identical hoti hai.

Analogy: $K$ ek "finish line" jaisa hai jo temperature set karta hai. Tum alag-alag raaste le sakte ho (alag initial amounts), lekin hamesha same jagah pahunchte ho.

Active Recall Practice

Us saturated paani mein, dissolved sugar ki concentration same hoti hai chahe neeche 1 extra cube baitha ho ya 100 extra cubes. Paani har case mein "full" hai.

Chemical reactions ke liye, pure solids aur liquids un extra undissolved cubes ki tarah hain. Woh wahin hain, reaction ka hissa hain, lekin aur add karne se solution mein dissolved ya gas ke roop mein floating jo hai woh nahi badalta. Equilibrium "balance" sirf un cheezein ki care karta hai jo freely move kar sakti hain—gases aur dissolved stuff—nahi un bade solid chunks ki jo neeche baithe hain.

Isliye jab hum equilibrium math formula (jise $K$ kehte hain) likhte hain, hum sirf un cheezein ko include karte hain jinki amounts actually change hoti hain aur matter karti hain: gases aur dissolved chemicals. Solids aur pure liquids hamesha "full strength" par hain, isliye hum unhe formula mein nahi likhte, taaki yeh simple aur accurate rahe.

Connections and Extensions

Prerequisites:

• Chemical Equilibrium Fundamentals
• Equilibrium Constant Expressions
• Le Chatelier's Principle
• Activity vs Concentration

Related Concepts:

• Solubility Product (Ksp)
• Precipitation Reactions
• Phase Diagrams
• Gibbs Free Energy and Equilibrium

Applications:

• Metallurgy: Ore Reduction
• Cement Production
• Limestone Caves and Stalactites
• Buffer Solutions (when water matters)

Advanced:

• Ellingham Diagrams for Metal Oxides
• Activity Coefficients in Concentrated Solutions
• Multi-phase Equilibria

#flashcards/chemistry

Pure solids aur liquids ko equilibrium expressions se kyun exclude kiya jaata hai? :: Unki activity constant (≈1) rehti hai chahe kitni bhi amount present ho, kyunki pure substance ke liye concentration = density/molar mass nahi badlta. K mein ek constant ko include karna redundant hai—use equilibrium constant ki value mein absorb kar lete hain.

Is equilibrium ke liye expression likho: CaCO₃(s) ⇌ CaO(s) + CO₂(g) :: $K_p=P_{{\text{CO}}_2}$ (solids excluded; sirf gas partial pressure include hoti hai)

C(s) + H₂O(g) ⇌ CO(g) + H₂(g) ke liye Kp likho :: $K_p=\frac{P_{\text{CO}}\cdot P_{{\text{H}}_2}}{P_{{\text{H}}_2\text{O}}}$ (solid carbon excluded; paani yahan gas hai, isliye included)

Heterogeneous equilibrium ke liye Kp aur Kc relate karo :: $K_p=K_c(RT{)}^{\Delta n_g}$ jahan $\Delta n_g$ = (gas products ke moles) − (gas reactants ke moles). R = 0.0821 L·atm/(mol·K) ya 8.314 J/(mol·K).

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