Gibbs-Helmholtz equation: Question bank — Hinglish deep dive

2.4.4 · D5 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Gibbs-Helmholtz equation

Definition Is page par use ki gayi notation (yahan banai gayi, kahin se borrow nahi ki)

Taaki ye page apne aap mein complete rahe:

$G$ = Gibbs free energy (Gibbs free energy), constant temperature $T$ aur pressure $P$ par "spend-worthiness budget". $G = H - T S$ se bana.
$H$ = enthalpy (Enthalpy), heat-content term.
$S$ = entropy (Entropy), disorder term; real substances ke liye hamesha positive.
$T$ = absolute temperature kelvin mein — hamesha positive, isliye $T^{2} > 0$ hamesha.
$P, V$ = pressure aur volume; kisi derivative par subscript $_{P}$ ya $_{V}$ ka matlab hai "jab $T$ change karo to use fixed rakho."
$A$ = Helmholtz free energy (Helmholtz free energy), $A = U - T S$ ; $U$ = internal energy.
Prefix $Δ$ ka matlab hai "ek process ke liye final minus initial" (jaise koi reaction). Toh $Δ G = G_{products} - G_{reactants}$ , aur usi tarah $Δ H, Δ S$ . Jo bhi $G$ ke liye sach hai wo $Δ G$ ke liye bhi sach hai kyunki $Δ$ sirf do states ka subtraction hai, aur derivatives seedhe iske through nikalti hain.
$K$ = equilibrium constant, $R$ = gas constant, $Δ G^{\circ} = - R T ln K$ se jude hue (the $^{\circ}$ = standard conditions).

The mechanism, shown once (so the answers can point at it)

$- S$ kyun cancel hota hai — explicitly carry out kiya gaya. $G = H - T S$ se shuru karo aur $T$ se divide karo: $\frac{G}{T} = \frac{H}{T} - S$ Ab constant $P$ par differentiate karo, $H / T$ par product/quotient rule use karte hue aur linked variation ko imandari se treat karte hue (yaad karo $d H = T d S + V d P$ , toh constant $P$ par, $\partial_{T} H = T \partial_{T} S$ ): $(\frac{\partial ( G / T )}{\partial T})_{P} = \frac{1}{T} = T \partial_{T} S \partial_{T} H - \frac{H}{T ^{2}} - \partial_{T} S = \partial_{T} S - \frac{H}{T ^{2}} - \partial_{T} S = - \frac{H}{T ^{2}}$ Do $\partial_{T} S$ pieces bilkul ek dusre ko khatam kar dete hain — yahi cancellation hai, likha hua. Neeche ka figure ise do arrows ke roop mein dikhata hai jo annihilate ho jaate hain.

$1/ T$ form sign kyun flip karta hai — chain rule ek line mein. Maano $u = 1/ T$ , toh $T = 1/ u$ aur $\frac{d T}{d u} = - \frac{1}{u ^{2}} = - T^{2}$ . Tab kisi bhi quantity $Q (T)$ ke liye: $\frac{\partial Q}{\partial ( 1/ T )} = \frac{\partial Q}{\partial u} = \frac{\partial Q}{\partial T} \cdot \frac{d T}{d u} = - T^{2} \frac{\partial Q}{\partial T}$ Ise $Q = G / T$ ke saath apply karo: $- T^{2}$ ne $- H / T^{2}$ ko multiply kiya aur tumhe $+ H$ mila. $- T^{2}$ minus aur $T^{2}$ dono ko khaa jaata hai. Doosra figure wo straight line dikhata hai jo ye produce karta hai.

True ya false — justify karo

True ya false:

(\frac{\partial G}{\partial T})_{P} = H

False.

d G = - S d T + V d P

se, constant

P

par tumhe

(\partial G / \partial T)_{P} = - S

(entropy) milta hai, nahi

H

. Enthalpy tab hi aata hai jab pehle

T

se divide karo, jaise mechanism section mein dikhaya gaya hai.

True ya false: differentiate karne se pehle

T

se divide karna sirf cosmetic hai;

G

akele se bhi

H

nikal sakte ho.

Spirit mein False. Tum

H = G - T (\partial G / \partial T)_{P}

likh sakte ho, lekin ye

G

aur uski slope ko awkwardly mix karta hai.

T

se divide karne par

\partial_{T} S

terms exactly cancel ho jaate hain (mechanism dekho), jo tumhein clean

- H / T^{2}

deta hai. Division real algebraic kaam karta hai.

True ya false: enthalpy ko directly

H = - T^{2} (\frac{\partial ( G / T )}{\partial T})_{P}

se nikaal sakte ho.

True. Pehli identity

\partial (G / T) / \partial T = - H / T^{2}

- T^{2}

se multiply karo: right side

+ H

ban jaati hai. Ye hi

1/ T

form hai disguise mein, kyunki

- T^{2} \partial_{T} = \partial_{(1/ T)}

upar ke chain rule se.

True ya false:

(\frac{\partial ( G / T )}{\partial T})_{P} = + \frac{H}{T ^{2}}

False. Mechanism section line by line

- H / T^{2}

derive karta hai. Plus sign sirf

1/ T

form mein aata hai,

\partial (G / T) / \partial (1/ T) = + H

, kyunki chain-rule factor

- T^{2}

minus ko cancel kar deta hai.

True ya false: Gibbs–Helmholtz equation ke liye

H

directly measure karna zaroori hai.

False, aur yahi poora point hai. Ye tumhein bina measure kiye

H

nikalne deta hai: kai

T

par

Δ G

measure karo (

K

Δ G^{\circ} = - R T ln K

ke zariye), aur uska

1/ T

-slope hi

Δ H

hai.

H

directly measure karna purpose ko defeat kar deta.

True ya false:

Δ G / T

1/ T

ke plot par slope

Δ H

ke saath minus sign ke saath hoga.

False. Chain rule se,

\partial (Δ G / T) / \partial (1/ T) = + Δ H

— slope

+ Δ H

hai, koi minus nahi. Isliye hi

1/ T

form preferred hai (straight-line figure dekho).

True ya false: constant-volume analogue mein

H

ki jagah

U

aur

P

ki jagah

V

aata hai.

True.

A = U - T S

ka shape

G = H - T S

jaisa hi hai; same division-and-cancel chalane par

(\partial (A / T) / \partial T)_{V} = - U / T^{2}

milta hai. Gibbs

\to

Helmholtz,

H \to U

, constant-

P \to

constant-

V

— teeno swaps ek saath.

True ya false: van 't Hoff equation ek alag law hai jise Gibbs–Helmholtz se independently yaad karna padta hai.

False. Van 't Hoff hi Gibbs–Helmholtz hai

Δ G^{\circ} = - R T ln K

par apply kiya hua: tab

Δ G^{\circ} / T = - R ln K

, aur

\partial (Δ G^{\circ} / T) / \partial (1/ T) = Δ H^{\circ}

seedha

d ln K / d (1/ T) = - Δ H^{\circ} / R

deta hai.

True ya false: two-point integrated formula assume karta hai ki

Δ H

temperature range mein constant hai.

True.

\frac{Δ G _{2}}{T _{2}} - \frac{Δ G _{1}}{T _{1}} = Δ H (\frac{1}{T _{2}} - \frac{1}{T _{1}})

\partial (Δ G / T) / \partial (1/ T) = Δ H

ko integrate karne se aata hai jab

Δ H

ko bahar kheencha jaata hai — valid tab hi jab

Δ H

barely drift kare (chhota

Δ C_{p}

, narrow range).

Error dhundho

Error: "Kyunki

G = H - T S

, constant

P

par

\partial / \partial T

lene se

\partial G / \partial T = - S - T \partial_{T} S

milta hai, isliye main

- T \partial_{T} S

term rakhta hoon."

Student ka product rule arithmetically correct hai lekin incomplete hai: unhone

H

ka variation bhool gaye. Poori derivative

\partial_{T} G = \partial_{T} H - S - T \partial_{T} S

hai, aur kyunki

\partial_{T} H = T \partial_{T} S

(constant

P

par

d H = T d S

se), pehla aur aakhri term cancel ho jaate hain:

\partial_{T} G = T \partial_{T} S - S - T \partial_{T} S = - S

\partial_{T} H

term drop karna overcount hai; use waapas daalte hi leftover hat jaata hai.

Error: "Main

Δ H

Δ G

(

Δ G / T

nahi) ko

T

ke against plot karke aur slope padh kar nikaalunga."

Wo slope

- Δ S

hai,

Δ H

nahi (ye sirf

\partial Δ G / \partial T = - Δ S

hai). Tumhe

Δ G / T

1/T

ke against plot karna hoga; tabhi

\partial_{T} S

cancellation hota hai aur slope

Δ H

ban jaata hai.

Error: "

Δ G

T

badhne par zyada negative ho gaya, toh reaction exothermic honi chahiye."

Sign ulta hai.

\partial (Δ G / T) / \partial (1/ T) = Δ H

T

badhna =

1/ T

ghatna; gar garam hone par zyada favourable ho, toh

Δ G / T

badhta hai jab

1/ T

ghatta hai, matlab positive slope

\Rightarrow Δ H > 0

(endothermic). Exothermic (

Δ H < 0

) reactions garam hone par kam favourable ho jaate hain.

Error: "

1/ T

form mein main safe rehne ke liye minus sign lagata hoon:

\partial (Δ G / T) / \partial (1/ T) = - Δ H

Koi minus nahi. Chain-rule factor

- T^{2}

pehle se

- Δ H / T^{2}

ka minus consume kar chuka hai:

(- T^{2}) (- Δ H / T^{2}) = + Δ H

1/ T

form sign-free waala hai.

Error: "

H = G - T (\partial G / \partial T)_{P}

, aur

(\partial G / \partial T)_{P}

positive hai, isliye

H > G

(\partial G / \partial T)_{P} = - S

usually negative hota hai (entropy positive), toh

- T (\partial G / \partial T)_{P} = + T S > 0

, deta hai

H = G + T S > G

. Sahi conclusion, galat raasta — derivative ka sign galat padha gaya.

Error: "Phase transition ke liye melting point par,

Δ G = 0

, isliye Gibbs–Helmholtz kehta hai

Δ H = 0

Galat. Gibbs–Helmholtz slope

\partial (Δ G / T) / \partial (1/ T) = Δ H

read karta hai, jo generally nonzero hota hai chahe value

Δ G = 0

ho.

Δ G

ka ek akela zero

Δ H

ke baare mein kuch nahi kehta.

Why questions

T

se divide karne par (multiply ya

T^{2}

se divide karne ki bajaye) entropy kyun cancel ho jaati hai?

Kyunki

G / T = H / T - S

: entropy ek bare

- S

ban jaata hai, jiska

\partial_{T} S

derivative exactly match ho jaata hai aur

H / T

\partial_{T} H = T \partial_{T} S

piece se cancel ho jaata hai (mechanism dekho). Koi bhi doosra divisor

S

-terms ko mismatched chhodta hai, toh wo cancel nahi hote.

Actual data analysis ke liye

1/ T

form kyun prefer kiya jaata hai?

Kyunki ye Gibbs–Helmholtz ko ek straight line banata hai:

Δ G / T

1/ T

ka constant slope

Δ H

hota hai (agar

Δ H

constant hai), toh linear fit se read off ho jaata hai.

\partial_{T}

form curve

- Δ H / T^{2}

deta hai, jo fit karna mushkil hai. Straight-line figure dekho.

Gibbs–Helmholtz bina calorimeter touch kiye equilibrium measurements se

Δ H

kyun recover kar sakta hai?

Kyunki

Δ G^{\circ} = - R T ln K

free energy ko measurable

K

se jodta hai. Temperatures ke across

K

dekhne se

Δ G^{\circ} (T)

milta hai, aur

Δ G^{\circ} / T

1/ T

ka slope hi

Δ H^{\circ}

hai — heat indirectly equilibrium ke through measure hoti hai.

Usi trick ka

U

ke saath Helmholtz twin kyun hai?

Kyunki

A = U - T S

structurally

G = H - T S

se identical hai.

T

se divide karne aur

\partial_{T} S

terms cancel karne par

H

ki jagah

U

bachta hai; sirf held-constant variable (

P

ki jagah

V

) alag hota hai.

\partial (G / T) / \partial T

mein

S

ko fixed treat karna valid kyun hai lekin

H - T S

ko naively differentiate karne mein nahi?

Tum

S

ko kabhi fixed treat nahi karte.

G / T = H / T - S

mein

- S

\partial_{T} S

genuinely present hai — ye sirf exactly cancel hota hai

\partial_{T} (H / T)

ke andar chuppe

\partial_{T} S

ke saath. Naive

H - T S

expansion

\partial_{T} H

drop karta hai aur term uncancelled chhod deta hai.

Wide temperature range par two-point formula apply karte waqt Kirchhoff's law kyun matter karta hai?

Kirchhoff's law kehta hai

Δ H

T

ke saath

Δ C_{p}

ki rate par drift karta hai. Wide range par ye drift two-point formula ki "constant

Δ H

" assumption todti hai, toh tum

Δ H (T)

correct karte ho ya narrow intervals tak restrict rehte ho.

Edge cases

Edge case: Gibbs–Helmholtz kya kehta hai jab

T \to 0^{+}

- H / T^{2} \to - \infty

(jab

H \neq = 0

ho), toh

G / T

infinitely steeply change karta hai. Lekin yahan third law

S \to 0

force karta hai aur quantum effects dominate karte hain, toh classical formula ek limiting idealization hai, literal measurement nahi.

Edge case:

T \to \infty

par kya hota hai?

G / T = H / T - S \to - S

, kyunki

H / T \to 0

. Toh bahut high

T

par plot ki value

- S

ke paas aati hai aur entropy term dominate karta hai —

G / T

mein enthalpy negligible ho jaata hai.

Δ G / T

-vs-

1/ T

plot par ye

1/ T \to 0

intercept hai, jo

- Δ S

ke barabar hai: graph literally ek end par entropy aur throughout slope mein enthalpy read out karta hai.

Edge case: agar kisi reaction ka

Δ H = 0

exactly ho, toh

1/ T

plot mein kya hota hai?

Slope zero hai, toh

Δ G / T

1/ T

flat hai:

Δ G / T = - Δ S

constant hai, matlab

Δ G = - T Δ S

purely entropic hai. Horizontal line ek "no heat" process ka fingerprint hai.

Edge case: kya

T

in formulas mein kabhi negative ho sakta hai?

Ordinary thermodynamics mein, nahi —

T

absolute (kelvin) aur positive hota hai, toh

T^{2} > 0

aur

- H / T^{2}

ka sign cleanly

- H

track karta hai. "Negative temperature" systems (bounded-energy spins) exotic hain aur is equation ke scope se bahar hain.

Edge case: agar kisi process ke liye

Δ S = 0

ho — kya Gibbs–Helmholtz simplify hota hai?

Tab

Δ G = Δ H

(temperature-independent), toh

\partial (Δ G / T) / \partial (1/ T) = Δ H = Δ G

: slope value ke barabar hai. Entropy term jo tum normally cancel karte wo pehle se zero tha, toh dono forms trivially agree karte hain.

Edge case: jis temperature par

Δ G

T

mein maximum hota hai, wahan

Δ S

ke saath kya ho raha hai?

Δ G

T

ka max/min matlab

(\partial Δ G / \partial T)_{P} = - Δ S = 0

, toh wahan

Δ S = 0

hai. Gibbs–Helmholtz tab bhi hold karta hai; us point par

Δ H = Δ G

kyunki

T Δ S

vanish ho jaata hai.

Edge case: kya Gibbs–Helmholtz ke liye process ka equilibrium mein hona zaroori hai?

Nahi. Ye function

G (T, P)

ke baare mein (ya do defined states ke beech

Δ G

) ek identity hai, equilibrium ki parwah kiye bina sach. Equilibrium tab aata hai jab tum isme

Δ G^{\circ} = - R T ln K

feed karte ho van 't Hoff paane ke liye.

Edge case: agar

T_{1} = T_{2}

ho toh two-point formula mein kya hota hai?

Dono sides

0 = Δ H \cdot 0

mein collapse ho jaate hain — uninformative. Slope dekhne ke liye tumhe do alag temperatures chahiye; ek temperature ek single point deta hai, line nahi.

Connections

Gibbs free energy — wo function jiska $T$ -dependence hum probe karte hain.
Enthalpy — wo quantity jise ye traps $- S$ se baar baar confuse karti hain.
Entropy — wo term jo cancel hota hai, aur zyaadatar sign errors ki source.
van 't Hoff equation — equilibrium constants ke liye Gibbs–Helmholtz.
Kirchhoff's law (thermochemistry) — kyun "constant $Δ H$ " eventually fail hota hai.
Helmholtz free energy — $U$ ke saath constant-volume twin.
Maxwell relations — sibling identities jahan sign discipline bhi matter karti hai.