Born-Haber cycle revisited — calculating lattice energy

2.5.10 · Chemistry › Thermodynamics (Chemical)

Hume Born–Haber cycle ki ZAROORAT kyun hai?

Lattice energy KYA hai?

Ise directly measure kyun nahi kar sakte? Tum ek bench par gaseous Na $^{+}$ aur Cl $^{-}$ ions ki bottle nahi rakh sakte aur unhe calorimeter mein condense hone de sakte — woh free gaseous ions normal conditions mein exist hi nahi karte. Toh hum us unmeasurable step ko route around karte hain un steps se jo hum measure kar sakte hain.

KAISE: cycle ko first principles se build karna

Hum elements (standard states mein) se ionic compound MX(s) tak do paths compare karte hain.

Path 1 (direct): Elements $\to$ MX(s), ek step = enthalpy of formation $Δ H_{f}$ .

Path 2 (lamba rasta, gaseous ions ke zariye):

Metal ko atomise karo — solid/liquid $\to$ gaseous atoms: $Δ H_{s u b}$ (sublimation/atomisation). Endothermic.
Metal ko ionise karo — electron hatao: ionisation energy $I E$ . Endothermic.
Non-metal ko atomise karo — jaise $\frac{1}{2} Cl_{2} \to Cl$ todna: $\frac{1}{2} D$ (bond dissociation). Endothermic.
Non-metal mein electron add karo — electron affinity $E A$ . Usually exothermic (negative).
Gaseous ions ko lattice mein assemble karo — lattice energy $U$ . Strongly exothermic.

Figure — Born-Haber cycle revisited — calculating lattice energy

Worked Example 1 — NaCl ki Lattice Energy

Diye gaye (kJ mol⁻¹): $Δ H_{f} = - 411$ , $Δ H_{s u b} (Na) = + 108$ , $I E_{1} (Na) = + 496$ , $\frac{1}{2} D (Cl_{2}) = + 122$ , $E A (Cl) = - 349$ .

$U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - E A$ $U = (- 411) - (108) - (496) - (122) - (- 349)$

Last sign kyun flip hota hai? $E A$ already negative hai ( $- 349$ ); negative ko subtract karne se add hota hai $349$ .

$U = - 411 - 108 - 496 - 122 + 349 = - 788 kJ mol^{- 1}$

Negative aur bada kyun? Ions se lattice banana bahut stabilising hota hai — yahi magnitude NaCl ka high melting point aur stability drive karti hai.

Worked Example 2 — MgCl₂ (ek +2 cation, DO chlorides)

Diye gaye (kJ mol⁻¹): $Δ H_{f} = - 641$ , $Δ H_{s u b} (Mg) = + 148$ , $I E_{1} = + 738$ , $I E_{2} = + 1451$ , $D (Cl_{2}) = + 244$ (poora bond, kyunki hume 2 Cl chahiye), $E A (Cl) = - 349$ (per Cl).

$U = Δ H_{f} - Δ H_{s u b} - (I E_{1} + I E_{2}) - D - 2 E A$ $U = - 641 - 148 - (738 + 1451) - 244 - 2 (- 349)$ $U = - 641 - 148 - 2189 - 244 + 698 = - 2524 kJ mol^{- 1}$

NaCl se itna bada kyun? Zyada charge (+2) aur zyada paas, chhota ion → bahut zyada strong electrostatic attraction (yaad karo $U \propto \frac{q _{+} q _{-}}{r}$ ).

Worked Example 3 — U ke alawa koi doosra unknown dhundhna

Kabhi kabhi $U$ diya hota hai (theory se) aur tum, maano, $E A$ dhundh te ho. Same equation, alag unknown. KBr ke liye diya gaya: $U = - 671$ , $Δ H_{f} = - 394$ , $Δ H_{s u b} (K) = + 89$ , $I E (K) = + 419$ , $\frac{1}{2} D (Br_{2}, in c l . v a p or i s a t i o n) = + 112$ . $E A (Br)$ nikalo.

Master equation ko $E A$ ke liye rearrange karo: $E A = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - U$ $E A = - 394 - 89 - 419 - 112 - (- 671) = - 343 kJ mol^{- 1}$ Yeh kyun kaam karta hai: cycle mein ek equation, ek unknown hai — jo bhi term tumhe nahi pata, use isolate karo.

Active Recall

Recall Lattice energy directly measure kyun nahi kar sakte?

Kyunki free gaseous ions (M⁺, X⁻) isolable substances ke roop mein exist nahi karte jinhein calorimeter mein condense kiya ja sake — toh hum $U$ Hess's Law ke zariye indirectly compute karte hain.

Recall Master equation memory se likho.

$U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - E A$ .

Recall MgCl₂ ke liye NaCl se kya kya badalta hai?

Do ionisation energies ( $I E_{1} + I E_{2}$ ), poora $D (Cl_{2})$ , aur $2 \times E A$ .

Recall Feynman: ek 12 saal ke bachche ko samjhao

Socho LEGO bricks ek giant sculpture (salt crystal) mein chipke hue hain. Sculpture ko todte hue single floating bricks mein convert karne ke liye bahut bada pull chahiye — woh pull hi "lattice energy" hai. Hum woh rip directly measure nahi kar sakte, isliye hum doosre raaste se sculpture banane ke har doosre step ki cost count karte hain, phir missing rip cost figure out karte hain taaki total match ho. Energy, jaise paise count karna, ek hi total deti hai chahe tum koi bhi rasta lo.

Flashcards

Lattice energy ki definition

1 mol ionic solid apne gaseous ions se bane toh enthalpy change; strongly exothermic.

Born–Haber cycle ko kaun sa law valid banata hai

Hess's Law — enthalpy ek state function hai, path-independent.

U ke liye master Born–Haber equation

U = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - E A

Cycle mein Cl ke electron affinity ka sign

Negative (

- 349

kJ/mol), yeh energy release karta hai.

Standard data se NaCl lattice energy

\approx - 788

kJ/mol.

MgCl₂ ki lattice energy NaCl se bahut badi kyun hai

Cation ka zyada charge (+2) aur chhota ion → zyada strong

q_{+} q_{-} / r

attraction.

MgCl₂ ke liye kaun se terms double ya change hote hain

I E_{1} + I E_{2}

use karo, poora

D (Cl_{2})

, aur

2 E A

Lattice energy directly measure kyun nahi ho sakti

Free gaseous ions ko calorimeter mein isolate/condense nahi kiya ja sakta.

U \propto q_{+} q_{-} / r

hume kya batata hai

Zyada charges aur chhote ionic radii → zyada exothermic (badi) lattice energy.

Agar U pata ho aur EA unknown ho, kaise solve karo

Same equation rearrange karo:

E A = Δ H_{f} - Δ H_{s u b} - I E - \frac{1}{2} D - U

Connections

Hess's Law and state functions
Enthalpy of formation
Ionisation energy trends
Electron affinity
Bond dissociation enthalpy
Ionic bonding and Coulomb's law
Kapustinskii equation (theoretical lattice energy)
Solubility and enthalpy of hydration