Gibbs-Helmholtz equation: Worked examples — Hinglish deep dive

2.4.4 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Gibbs-Helmholtz equation

Scenario matrix

Har Gibbs–Helmholtz problem actually kuch gine-chune shapes mein se ek hoti hai. Yeh raha pura grid — har row ek "case class" hai, aur baad mein har worked example us cell ke saath tagged hai jise woh fill karta hai.

#	Case class	Kya special hai	Filled by
A	$Δ H < 0$ (exothermic), $Δ G$ $T$ ke saath badhta hai	$Δ G / T$ vs $1/ T$ ka slope negative hai	Ex 1
B	$Δ H > 0$ (endothermic), $Δ G$ $T$ ke saath girta hai	slope positive	Ex 2
C	Single $G (T)$ point + slope $(\partial G / \partial T)_{P}$	$(⋆)$ form use karo, do points nahi	Ex 3
D	$Δ H \approx 0$ (degenerate / flat)	$Δ G / T$ constant hai — line horizontal hai	Ex 4
E	Limiting behaviour: $T \to 0$ aur $T \to \infty$	$G / T$ aur $- H / T^{2}$ edges par kya karte hain	Ex 5
F	Equilibrium constant $K$ (van 't Hoff face)	plug karo $Δ G^{\circ} = - R T ln K$	Ex 6
G	Real-world word problem (battery / cell)	$Δ G = - n F E$ , temperature coefficient	Ex 7
H	Exam twist: wrong-form / sign trap	test karta hai ki tum $\partial T$ vs $\partial (1/ T)$ jaante ho	Ex 8

Neeche wala plot rows A, B aur D ke peeche ka mental picture hai ek saath: $Δ G / T$ ko $1/ T$ ke against plot karo toh ek straight line milti hai jiska slope $Δ H$ hai.

Case A — exothermic, $Δ G$ $T$ ke saath badhta hai

Worked example Ex 1 (cell A) — negative

$Δ H$ recover karo Ek reaction mein $Δ G^{\circ} = - 20.0$ kJ/mol at $T_{1} = 300$ K aur $Δ G^{\circ} = - 18.0$ kJ/mol at $T_{2} = 320$ K hai. Assume karo $Δ H$ constant hai. $Δ H^{\circ}$ find karo.

Forecast: $Δ G$ heat karne par less negative ho gaya — reaction kam favourable ho gayi jab garam ki. Cold isko favour karta hai, jo heat release karne ki pehchaan hai. Toh guess karo $Δ H < 0$ .

Two-point form likho. Yeh step kyun? Constant $Δ H$ matlab $\partial (Δ G / T) / \partial (1/ T) = Δ H$ integrate hokar do points ke beech straight line banta hai. $\frac{Δ G _{2}}{T _{2}} - \frac{Δ G _{1}}{T _{1}} = Δ H (\frac{1}{T _{2}} - \frac{1}{T _{1}})$
Left-hand side. Kyun: yeh $Δ G / T$ quantity mein change hai jo equation predict karta hai. $\frac{- 18.0}{320} - \frac{- 20.0}{300} = - 0.056250 + 0.066667 = + 0.010417 \frac{kJ}{mol\cdotK}$
$1/ T$ step. Kyun: yeh hamari line ka run hai. $\frac{1}{320} - \frac{1}{300} = - 2.0833 \times 1 0^{- 4} K^{- 1}$
Divide karo (rise over run = slope = $Δ H$ ). Kyun: $Δ G / T$ vs $1/ T$ ka slope $Δ H$ hai. $Δ H = \frac{0.010417}{- 2.0833 \times 1 0 ^{- 4}} = - 50.0 \frac{kJ}{mol}$

Verify: Sign negative hai — forecast se match karta hai (exothermic). Units: $(kJ\cdotmol^{- 1} K^{- 1}) / (K^{- 1}) = kJ\cdotmol^{- 1}$ . ✓

Case B — endothermic, $Δ G$ $T$ ke saath girta hai

Case C — ek temperature, slope use karke

Worked example Ex 3 (cell C) — single

$G (T)$ measurement $T = 298$ K par ek system mein $G = - 10.0$ kJ hai aur measured slope $(\partial G / \partial T)_{P} = - 0.100$ kJ/K hai. $H$ find karo.

Forecast: Slope $(\partial G / \partial T)_{P} = - S$ hai, isliye yahan $S = + 0.100$ kJ/K hai. Phir $H = G + T S = - 10 + (298) (0.1) \approx + 20$ kJ — positive, aur $G$ se bada kyunki hum entropy term wapas add karte hain.

Raw $(⋆)$ form use karo. Yeh step kyun? Sirf ek temperature hone par line slope karne ke liye koi line nahi hai — lekin hamare paas $G$ ka slope hai, jise identity $H = G - T (\partial G / \partial T)_{P}$ seedha $H$ mein convert kar deti hai. $H = G - T (\frac{\partial G}{\partial T})_{P}$
Substitute karo. Kyun: numbers plug karo. $H = - 10.0 - (298) (- 0.100) = - 10.0 + 29.8 = + 19.8 kJ$
$S$ se cross-check karo. Kyun: slope term mein sign slip pakadta hai. $S = - (\partial G / \partial T)_{P} = + 0.100$ kJ/K, toh $H = G + T S = - 10.0 + (298) (0.100) = + 19.8 kJ$

Verify: Dono routes $+ 19.8$ kJ dete hain. ✓ Units: kJ. Forecast se match karta hai.

Case D — degenerate flat line ( $Δ H \approx 0$ )

Worked example Ex 4 (cell D) — zero enthalpy

Ek reaction $Δ G^{\circ} = - 8.00$ kJ/mol at $T_{1} = 300$ K aur $Δ G^{\circ} = - 9.60$ kJ/mol at $T_{2} = 360$ K deti hai. $Δ H^{\circ}$ find karo aur interpret karo.

Forecast: $Δ G / T$ ratio dekho. 300 K par yeh $- 8/300$ hai; 360 K par yeh $- 9.6/360$ hai. Dono $- 0.02 \overline{6}$ ke barabar hain! Agar $Δ G / T$ change nahi karta, toh $1/ T$ ke against uska slope zero hai, isliye $Δ H = 0$ . Yeh purely entropy-driven process hai.

Dono $Δ G / T$ values compute karo. Kyun: equation sirf $Δ G / T$ ki parwah karta hai, $Δ G$ ki nahi. $\frac{- 8.00}{300} = - 0.026667, \frac{- 9.60}{360} = - 0.026667$
Difference lo. Kyun: woh difference $Δ H Δ (1/ T)$ hai. $Δ (\frac{Δ G}{T}) = 0$
Conclude karo. Kyun: $0 = Δ H \cdot (nonzero Δ (1/ T)) \Rightarrow Δ H = 0$ . $Δ H^{\circ} = 0 \frac{kJ}{mol}$

Verify: Kyunki $Δ G = Δ H - T Δ S$ aur $Δ H = 0$ , hume milta hai $Δ S = - Δ G / T = + 0.026667$ kJ·mol⁻¹·K⁻¹ dono temperatures par — constant entropy, koi enthalpy nahi. ✓ $s 01$ figure par yeh horizontal orange line hai.

Case E — limits ( $T \to 0$ , $T \to \infty$ )

Worked example Ex 5 (cell E) — edges par kya hota hai

Ek fixed exothermic system lo jisme $H = - 40$ kJ ho ( $H$ roughly constant maano). $G / T$ aur uske slope $- H / T^{2}$ ka describe karo jab $T \to 0^{+}$ aur $T \to \infty$ .

Forecast: $- H / T^{2} = + 40/ T^{2}$ . Jab $T \to 0$ toh yeh $+ \infty$ tak explode karta hai; jab $T \to \infty$ toh yeh $0$ tak die karta hai. Isliye $G / T$ ka curve cold hone par bahut steep hoga aur hot hone par almost flat.

Cold limit $T \to 0^{+}$ . Yeh step kyun? $- H / T^{2} = + 40/ T^{2} \to + \infty$ : $G / T$ vs $T$ ka slope infinite ho jaata hai. Physically, ek finite $G$ ko tiny $T$ se divide karne par $G / T$ enormous ho jaata hai. $T = 1$ K par evaluate karo: $- H / T^{2} = + 40$ kJ/K².
Hot limit $T \to \infty$ . Kyun: $- H / T^{2} \to 0$ . $G / T$ curve flat ho jaata hai; enthalpy ki slope par pakad fade ho jaati hai kyunki denominator mein $T^{2}$ jeet jaata hai. $T = 1000$ K par: $- H / T^{2} = + 40/1 0^{6} = + 4.0 \times 1 0^{- 5}$ kJ/K².
$G / T$ ka apna sanity check. Kyun: $G = H - T S$ use karke, $G / T = H / T - S$ . Jab $T \to \infty$ , $H / T \to 0$ isliye $G / T \to - S$ ; jab $T \to 0^{+}$ , $H / T$ dominate karta hai aur $G / T \to \pm \infty$ $H$ ke sign ke saath (yahan $- \infty$ ).

Verify: $T = 1$ par: $- H / T^{2} = 40$ . $T = 1000$ par: $- H / T^{2} = 4 \times 1 0^{- 5}$ . Ratio $= 1 0^{6} = T^{2}$ -ratio. ✓ $1/ T^{2}$ law ke consistent — koi scenario unshown nahi: cold steep, hot flat.

Case F — van 't Hoff face (equilibrium $K$ )

Worked example Ex 6 (cell F) — do

$K$ values se $Δ H^{\circ}$ nikalo Ek equilibrium mein $K_{1} = 10.0$ at $T_{1} = 298$ K aur $K_{2} = 50.0$ at $T_{2} = 328$ K hai. $R = 8.314$ J·mol⁻¹·K⁻¹ lo. $Δ H^{\circ}$ find karo.

Forecast: $K$ temperature ke saath badha, isliye heating reaction ko aage push karti hai — endothermic, $Δ H^{\circ} > 0$ .

Integrated van 't Hoff form use karo (Gibbs–Helmholtz with $Δ G^{\circ} = - R T ln K$ ). Yeh step kyun? Substitute karne par $Δ G^{\circ} / T = - R ln K$ milta hai, aur $\partial (Δ G^{\circ} / T) / \partial (1/ T) = Δ H^{\circ}$ become karta hai $d ln K / d (1/ T) = - Δ H^{\circ} / R$ , jo integrate hokar deta hai: $ln \frac{K _{2}}{K _{1}} = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$
Left side. Kyun: $ln K$ ka rise. $ln \frac{50.0}{10.0} = ln 5 = 1.60944$
$1/ T$ run. Kyun: denominator. $\frac{1}{328} - \frac{1}{298} = - 3.0704 \times 1 0^{- 4} K^{- 1}$
$Δ H^{\circ}$ solve karo. Kyun: step 1 rearrange karo. $Δ H^{\circ} = - R \frac{l n ( K _{2} / K _{1} )}{( 1/ T _{2} - 1/ T _{1} )} = - 8.314 \cdot \frac{1.60944}{- 3.0704 \times 1 0 ^{- 4}} = + 4.358 \times 1 0^{4} \frac{J}{mol}$

Verify: $Δ H^{\circ} \approx + 43.6$ kJ/mol, positive — endothermic forecast se match karta hai. Units: J·mol⁻¹K⁻¹·(dimensionless)/K⁻¹ = J·mol⁻¹. ✓

Case G — real-world word problem (ek galvanic cell)

Worked example Ex 7 (cell G) — battery reaction ki enthalpy

Ek galvanic cell $n = 2$ moles of electrons drive karta hai. Uska cell voltage $E = 1.100$ V at $298$ K hai aur uska temperature coefficient $(\partial E / \partial T)_{P} = - 5.00 \times 1 0^{- 4}$ V/K hai. Faraday constant $F = 96485$ C/mol. Cell reaction ka $Δ H$ find karo.

Forecast: $Δ G = - n F E$ negative hai (cell kaam karta hai), around $- 212$ kJ. Negative temperature coefficient matlab voltage heat hone par drop karta hai — isliye entropy lost ho rahi hai, aur $H$ , $G$ se bhi zyada negative hoga. Guess karo $Δ H < Δ G < 0$ .

$Δ G$ nikalo. Yeh step kyun? Electrical work Gibbs energy se linked hai: $Δ G = - n F E$ . $Δ G = - (2) (96485) (1.100) = - 212267 J = - 212.27 kJ$
Slope $(\partial G / \partial T)_{P}$ nikalo. Kyun: Gibbs–Helmholtz $(⋆)$ ko uski zaroorat hai. $Δ G = - n F E$ differentiate karo: $(\frac{\partial Δ G}{\partial T})_{P} = - n F (\frac{\partial E}{\partial T})_{P} = - (2) (96485) (- 5.00 \times 1 0^{- 4}) = + 96.485 \frac{J}{K}$
$(⋆)$ form apply karo. Kyun: yeh $G$ aur uske slope ko seedha $H$ mein turn karta hai. $Δ H = Δ G - T (\frac{\partial Δ G}{\partial T})_{P} = - 212267 - (298) (96.485) = - 241000 J$ $Δ H \approx - 241.0 kJ$

Verify: $Δ H = - 241.0$ kJ, $Δ G = - 212.3$ kJ se zyada negative hai, bilkul forecast ke according girte voltage ke liye. Entropy se cross-check: $Δ S = (\partial Δ G / \partial T)_{P} \cdot (- 1) = - 96.485$ J/K, aur $Δ H = Δ G + T Δ S = - 212267 + (298) (- 96.485) = - 241000$ J. ✓ Units: J. ✓

Case H — exam sign trap

Worked example Ex 8 (cell H) — wrong form spot karo aur fix karo

Ek student likhta hai: "Kyunki $\partial (G / T) / \partial T = - H / T^{2}$ hai, plot se $Δ H$ nikalne ke liye main $Δ G / T$ ko $T$ ke against plot karta hoon aur slope $Δ H$ hai." $T = 400$ K par ek system mein $Δ G / T = - 30.0$ J/(mol·K) hai aur numerically measured slope of $Δ G / T$ vs $T$ wahan $- 1.25 \times 1 0^{- 4}$ J·mol⁻¹·K⁻² hai. Student ne kya galat kiya, aur $Δ H$ kya hai?

Forecast: $Δ G / T$ vs $T$ ka slope $Δ H$ nahi hai — yeh $- Δ H / T^{2}$ hai. Isliye student ke slope ko $Δ H$ recover karne ke liye $- T^{2}$ se multiply karna hoga.

Do legitimate forms name karo. Yeh step kyun? Trap $\partial T$ aur $\partial (1/ T)$ ko confuse karna hai.

vs $T$ : $\frac{\partial ( Δ G / T )}{\partial T} = - \frac{Δ H}{T ^{2}}$ (slope $- Δ H / T^{2}$ hai, $Δ H$ nahi).
vs $1/ T$ : $\frac{\partial ( Δ G / T )}{\partial ( 1/ T )} = Δ H$ (slope hi $Δ H$ hai — $1/ T$ ke against plot karo, $T$ ke nahi).

vs- $T$ slope se $Δ H$ recover karo. Kyun: sahi relation invert karo. $Δ H = - T^{2} (\frac{\partial ( Δ G / T )}{\partial T})_{P} = - (400)^{2} (- 1.25 \times 1 0^{- 4}) = + 20.0 \frac{J}{mol}$

Verify: Student ka slope tiny tha ( $- 1.25 \times 1 0^{- 4}$ ); agar woh " $Δ H$ " hota toh absurdly small hota. $- T^{2} = - 1.6 \times 1 0^{5}$ se multiply karne par ek sensible $Δ H = + 20.0$ J/mol milta hai. ✓ Units: K²·(J·mol⁻¹·K⁻²)=J·mol⁻¹. ✓ Enthalpy aur Gibbs free energy notes dekho ki kyun $Δ H$ ek normal energy scale par hona chahiye.

Active recall

Recall Kaun sa cell, kaun sa tool?

Do $Δ G$ values at do $T$ diye hain → kaun sa formula? ::: two-point integrated form (cells A, B, D) Ek $G$ aur $(\partial G / \partial T)_{P}$ diye hain → kaun sa formula? ::: $(⋆)$ form $H = G - T (\partial G / \partial T)_{P}$ (cell C) Do $K$ values diye hain → kaun sa formula? ::: integrated van 't Hoff $ln (K_{2} / K_{1}) = - \frac{Δ H ^{\circ}}{R} (1/ T_{2} - 1/ T_{1})$ (cell F) $Δ G / T$ ka $T$ across change na hona matlab? ::: $Δ H = 0$ , entropy-driven (cell D) Jab $T \to 0^{+}$ , $- H / T^{2}$ kya karta hai? ::: diverge karta hai (steep); jab $T \to \infty$ toh $\to 0$ (flat) (cell E) $Δ G / T$ vs $T$ ka slope kya hai? ::: $- Δ H / T^{2}$ , $Δ H$ nahi (cell H trap)

Connections

Parent: Gibbs–Helmholtz equation — woh identity jo har example use karta hai.
Gibbs free energy — $Δ G$ jo hum feed karte hain (Ex 1–4, 7, 8).
Enthalpy — $Δ H$ jo hum har jagah extract karte hain.
Entropy — woh term jo disappear hoti hai; Ex 4 aur 7 cross-checks mein wapas aati hai.
van 't Hoff equation — Ex 6, Gibbs–Helmholtz hai apna $K$ mask pahanke.
Kirchhoff's law (thermochemistry) — fix karta hai jab Ex 1, 2 ki "constant $Δ H$ " assumption wide $T$ par toot jaati hai.
Helmholtz free energy — swap karo $G \to A$ , $H \to U$ constant-volume twin ke liye.
Maxwell relations — exact thermodynamic derivatives ki same family.