Hess's law — enthalpy is a state function; enthalpy cycles

2.5.8 · Chemistry › Thermodynamics (Chemical)

Hess's law exist KYUN karta hai?

$H$ state function KYUN hai? Kyunki $H = U + P V$ hai, aur $U$ (internal energy), $P$ , aur $V$ — ye sab state functions hain. State functions ka sum bhi ek state function hota hai.

ISKA kya matlab hai: agar aap reactants se products tak jaate ho, toh change $Δ H = H_{final} - H_{initial}$ fixed hota hai. Chahe aap ek violent step mein wahan jaao ya das gentle steps mein, endpoints wahi rehte hain, isliye difference bhi wahi rahta hai.

Hess's law ko STATE kaise karein

Do operational rules (usi fact se derived)

Reaction reverse karo ⟹ $Δ H$ ka sign flip karo. Kyun? $Δ H_{Z \to A} = H_{A} - H_{Z} = - (H_{Z} - H_{A}) = - Δ H_{A \to Z}$ .
Reaction ko $n$ se multiply karo ⟹ $Δ H$ ko $n$ se multiply karo. Kyun? Enthalpy extensive hai — cheez ki amount double karo, enthalpy content double hoga, change double hoga.

Figure — Hess's law — enthalpy is a state function; enthalpy cycles

WORKED EXAMPLE 1 — Reactions ki classic algebra

Goal: $C (s) + \frac{1}{2} O_{2} (g) \to C O (g)$ ke liye $Δ H$ nikalna (seedha measure karna mushkil hai — CO hamesha over-oxidize ho jaata hai).

Diya gaya hai:

(i) $C (s) + O_{2} (g) \to C O_{2} (g)$ , $Δ H_{1} = - 393.5 kJ$
(ii) $C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$ , $Δ H_{2} = - 283.0 kJ$

Target = (i) $-$ (ii): (ii) ko reverse karo aur (i) mein add karo.

Step	Reaction	$Δ H$
(i)	$C + O_{2} \to C O_{2}$	$- 393.5$
(ii) reversed	$C O_{2} \to C O + \frac{1}{2} O_{2}$	$+ 283.0$
Sum	$C + \frac{1}{2} O_{2} \to C O$	$- 110.5$

(ii) ko reverse KYUN kiya? Humein $C O$ product ke roop mein chahiye, lekin (ii) mein yeh reactant hai. Reverse karne par yeh product side par aa jaata hai aur sign flip ho jaata hai. $C O_{2}$ kyun gayab ho gaya? Yeh (i) mein product ke roop mein aur reversed (ii) mein reactant ke roop mein aata hai — same species, same state, isliye cancel ho jaata hai (telescoping).

$Δ H = - 393.5 + 283.0 = - 110.5 kJ$

WORKED EXAMPLE 2 — Formation enthalpies

Methane jalana: $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$ Data ( $Δ_{f} H^{\circ}$ , kJ/mol): $C H_{4} = - 74.8, C O_{2} = - 393.5, H_{2} O (l) = - 285.8, O_{2} = 0$ .

$Δ_{r} H^{\circ} = [(- 393.5) + 2 (- 285.8)] - [(- 74.8) + 2 (0)]$ Water par coefficient 2 KYUN hai? Extensive property × stoichiometric coefficient. $= (- 965.1) - (- 74.8) = - 890.3 kJ/mol$

WORKED EXAMPLE 3 — Born–Haber style cycle (mini)

Nikalo $Δ H$ for $A \to D$ agar: $A \to B (+ 50)$ , $B \to C (- 120)$ , $D \to C (+ 30)$ .

Path: $A \to B \to C \to D$ . Note karo $C \to D = - (D \to C) = - 30$ . $Δ H = (+ 50) + (- 120) + (- 30) = - 100 kJ$ Last one flip KYUN kiya? Diya gaya data $D \to C$ jaata hai; humaare path ko $C \to D$ chahiye, isliye sign reverse karo.

Common mistakes

Feynman check

Recall 12-saal ke bachche ko samjhao (click to reveal)

Socho tumhare paas ₹100 hain. Chahe tum ek ₹100 ki note mein kharcho ya thodi-thodi coins mein, tum phir bhi ₹100 kam ho jaate ho. Jitna lose hota hai woh sirf start aur end money par depend karta hai, na ki kharch karne ke tarike par. Chemistry ki "money" enthalpy (heat content) hai. Kisi mushkil reaction mein kitni heat nikalti hai yeh jaanne ke liye, hum aasaan reactions lete hain aur unhe money receipts ki tarah add/subtract karte hain — aur hum exactly same total par pahunch jaate hain.

Active recall

Enthalpy state function KYUN hai?

Kyunki

H = U + P V

aur

U, P, V

sab state functions hain; state functions ka combination bhi ek state function hota hai, isliye

Δ H

sirf initial/final states par depend karta hai.

Hess's law bolo.

Kisi reaction ka total

Δ H

un kisi bhi steps ke

Δ H

ke sum ke barabar hota hai jo milkar usse banaate hain:

Δ H_{r x n} = \sum Δ H_{i}

Reaction reverse karne par

Δ H

ka kya hota hai?

Uska sign flip ho jaata hai:

Δ H_{r e v} = - Δ H_{f w d}

(kyunki

Δ H = H_{f ina l} - H_{ini t ia l}

Reaction ko

n

se multiply karne par

Δ H

ka kya hota hai?

Yeh

n

se multiply ho jaata hai (enthalpy extensive hai).

Formation enthalpies se

Δ_{r} H^{\circ}

ka formula?

Δ_{r} H^{\circ} = \sum ν_{p} Δ_{f} H_{p r o d}^{\circ} - \sum ν_{r} Δ_{f} H_{r e a c t}^{\circ}

Apni standard state mein element ka

Δ_{f} H^{\circ}

zero KYUN hota hai?

Element ko apne aap se form karna koi change nahi hai; endpoints identical hain ⟹

Δ H = 0

C + \frac{1}{2} O_{2} \to C O

ke liye

Δ H

compute karo, diya gaya

C + O_{2} \to C O_{2} (- 393.5)

aur

C O + \frac{1}{2} O_{2} \to C O_{2} (- 283.0)

- 393.5 - (- 283.0) = - 110.5

kJ.

Hum directly CO banane ka

Δ H

KYUN measure nahi kar sakte?

CO aasaani se aage

C O_{2}

mein oxidize ho jaata hai, isliye pure single-step data milna mushkil hai — Hess's law se hum ise indirectly nikaate hain.

Hess's law — enthalpy is a state function; enthalpy cycles

Hess's law exist KYUN karta hai?

Hess's law ko STATE kaise karein

Do operational rules (usi fact se derived)

WORKED EXAMPLE 1 — Reactions ki classic algebra

WORKED EXAMPLE 2 — Formation enthalpies

WORKED EXAMPLE 3 — Born–Haber style cycle (mini)

Common mistakes

Feynman check

Active recall

Connections

Concept Map