Gibbs-Helmholtz equation: Exercises — Hinglish deep dive

2.4.4 · D4 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Gibbs-Helmholtz equation

Shuru karne se pehle, ek shared picture. Hum quantity $Δ G / T$ (ek reaction ki free-energy change, absolute temperature $T$ se divide ki gayi) ko $1/ T$ (temperature ka ulta — ek "thanda-pan" axis) ke against plot karenge. Parent note ne prove kiya ki us straight line ki slope $Δ H$ hai, yaani enthalpy change. Is graph ko apne dimag mein rakho analysis problems ke liye.

Level 1 — Recognition

Goal: pehchano ki tum kaun sa form dekh rahe ho, aur har symbol ka matlab kya hai.

L1.1 In mein se kaun sa Gibbs–Helmholtz equation ka woh form hai jisme straight-line plot $Δ H$ ko slope ke roop mein deta hai? (a) $(\frac{\partial ( G / T )}{\partial T})_{P} = - \frac{H}{T ^{2}}$ (b) $(\frac{\partial ( Δ G / T )}{\partial ( 1/ T )})_{P} = Δ H$ (c) $(\frac{\partial G}{\partial T})_{P} = - S$

Recall Solution L1.1

Answer hai (b).

Har ek kya hai. (a) temperature form hai; iska slope constant nahi hai, isliye ye straight line nahi deta. (c) Gibbs–Helmholtz bilkul nahi hai — ye definition $(\partial G / \partial T)_{P} = - S$ hai, yaani entropy relation.
Kyun (b). Agar tum horizontal axis par $x = 1/ T$ aur vertical axis par $y = Δ G / T$ rakh do, toh $\frac{d y}{d x} = Δ H$ . Ek derivative jo ek constant ke barabar hai, woh ek straight line ki slope hai. Toh $Δ G / T$ ko $1/ T$ ke against plot karo aur slope napo, seedha $Δ H$ mil jaata hai.

L1.2 Equation $(\partial (G / T) / \partial T)_{P} = - H / T^{2}$ mein subscript $P$ ka kya matlab hai, aur woh wahan kyun hai?

Recall Solution L1.2

Matlab. Subscript $P$ ka matlab hai "temperature change karte waqt aur derivative lete waqt pressure ko constant rakho". Partial derivative $\partial / \partial T$ tabhi meaningful hoti hai jab tum bolo ki kya fixed rakha ja raha hai.
Yahan kyun. Gibbs energy $G$ naturally dono $T$ aur $P$ par depend karta hai ( $d G = - S d T + V d P$ se). Agar pressure drift karne dete, toh extra $V d P$ term ghus aata aur clean result toot jaata. Constant $P$ us term ko khatam kar deta hai, sirf temperature effect bachta hai.

L1.3 Har symbol ko uske plain-English naam se match karo: $G$ , $H$ , $S$ , $A$ , $U$ .

Recall Solution L1.3

$G$ = Gibbs free energy (Gibbs free energy) — "kya yeh constant $T, P$ par hoga?" quantity.
$H$ = enthalpy (Enthalpy) — constant pressure par heat content.
$S$ = entropy (Entropy) — spread/disorder ka measure.
$A$ = Helmholtz free energy (Helmholtz free energy) — $G$ ka constant-volume cousin.
$U$ = internal energy — total microscopic energy, $H$ ka constant-volume cousin.

Level 2 — Application

Goal: numbers daalo aur carefully calculate karo.

L2.1 Ek reaction mein $Δ G^{\circ} = - 30.0$ kJ/mol at $T_{1} = 280$ K aur $Δ G^{\circ} = - 27.0$ kJ/mol at $T_{2} = 310$ K hai. Maano $Δ H$ roughly constant hai, $Δ H$ nikalo.

Recall Solution L2.1

Hum kya use karte hain, aur kyun integrate hota hai. $1/ T$ form kehta hai ki $Δ G / T$ ka rate of change $1/ T$ ke respect mein $Δ H$ ke barabar hai: $\frac{d ( Δ G / T )}{d ( 1/ T )} = Δ H .$ Hum integrate kyun kar sakte hain. Agar $Δ H$ (maana gaya) constant hai, toh yeh exactly "ek quantity jiska slope ek fixed number hai" — ek straight line hai. Dono taraf ko chhoti step $d (1/ T)$ se multiply karo aur state 1 se state 2 tak add karo (integrate karo): $\int_{1}^{2} d (\frac{Δ G}{T}) = \int_{1}^{2} Δ H d (\frac{1}{T}) = Δ H \int_{1}^{2} d (\frac{1}{T}) .$ Left integral bas $Δ G / T$ ka total change hai; right mein $Δ H$ times $1/ T$ ka total change hai. Toh $\frac{Δ G _{2}}{T _{2}} - \frac{Δ G _{1}}{T _{1}} = Δ H (\frac{1}{T _{2}} - \frac{1}{T _{1}}) .$ Left side (units kJ·mol⁻¹·K⁻¹): $\frac{- 27.0}{310} - \frac{- 30.0}{280} = - 0.087097 + 0.107143 = + 0.020045.$ $1/ T$ ka difference (units K⁻¹): $\frac{1}{310} - \frac{1}{280} = 0.0032258 - 0.0035714 = - 3.4562 \times 1 0^{- 4} .$ Divide karo: $Δ H = \frac{0.020045}{- 3.4562 \times 1 0 ^{- 4}} = - 58.0 kJ/mol .$ Sign check. $Δ G$ $T$ badhne par kam negative hua → reaction garmi mein kam favourable hai → yeh exothermic hai, toh $Δ H < 0$ . ✓

L2.2 $T = 298$ K par ek system mein $G = - 12.0$ kJ aur slope $(\partial G / \partial T)_{P} = - 0.15$ kJ/K hai. $H$ aur $S$ nikalo.

Recall Solution L2.2

Formula kahan se aata hai. Definition $G = H - T S$ se start karo. Hum yeh bhi jaante hain (constant $P$ par $d G = - S d T + V d P$ se) ki $(\partial G / \partial T)_{P} = - S$ , yaani $S = - (\partial G / \partial T)_{P}$ . Is $S$ ko definition mein substitute karo: $G = H - T = S [- (\frac{\partial G}{\partial T})_{P}] = H + T (\frac{\partial G}{\partial T})_{P} .$ $H$ ke liye rearrange karne par " $(⋆)$ form" milta hai: $H = G - T\left(\frac{\partial G}{\partial T}\right)_P. \tag{$\star$}$ Toh yeh koi naya law nahi hai — yeh $G = H - T S$ hai jismein $S$ ko $G$ ki slope ke roop mein likhaa gaya hai. Entropy. $S = - (\partial G / \partial T)_{P} = - (- 0.15) = + 0.15$ kJ/K. Enthalpy via $(⋆)$ : $H = - 12.0 - 298 (- 0.15) = - 12.0 + 44.7 = + 32.7 kJ .$ Cross-check $H = G + T S = - 12.0 + 298 (0.15) = - 12.0 + 44.7 = + 32.7$ kJ se. ✓

L2.3 Forms ke beech convert karo: agar $(\partial (G / T) / \partial T)_{P} = - H / T^{2}$ hai, toh woh step dikhao jo isse $(\partial (G / T) / \partial (1/ T)))_{P} = H$ mein badalta hai.

Recall Solution L2.3

Hum kya karte hain. Variable ko $T$ se $β \equiv 1/ T$ mein change karo. Kyun. Hum chahte hain ki slope constant ho taaki plot ek straight line ho. Kaise. Kyunki $β = 1/ T = T^{- 1}$ , differentiate karo: $d β = - T^{- 2} d T$ , yaani $d T = - T^{2} d β$ . Phir chain rule se $\frac{\partial ( G / T )}{\partial ( 1/ T )} = \frac{\partial ( G / T )}{\partial T} \cdot \frac{d T}{d β} = (- \frac{H}{T ^{2}}) (- T^{2}) = H .$ Dono minus signs aur $T^{2}$ sab cancel ho jaate hain, sirf $H$ bachta hai.

Level 3 — Analysis

Goal: slopes, signs, aur har case ke baare mein reason karo — hot, cold, exo-, endo-.

L3.1 Tum $Δ G / T$ (vertical) ko $1/ T$ (horizontal) ke against plot karte ho aur positive slope wali straight line paate ho. Kya reaction exothermic hai ya endothermic? $Δ H$ ki sign kya hai?

Recall Solution L3.1

Reasoning. Slope $Δ H$ ke barabar hai ( $1/ T$ form). Positive slope ⇒ $Δ H > 0$ ⇒ endothermic (heat absorb karta hai). Figure s01 dekho: amber line ki positive slope hai, aur arrow "slope $= Δ H$ " mark karta hai. Physical cross-check. Endothermic reactions ke liye, $T$ badhana (smaller $1/ T$ ki taraf left move karna) $Δ G$ ko zyada negative banata hai — heat unki madad karta hai. Consistent.

L3.2 $(Δ H, Δ S)$ ke charon sign combinations sketch/describe karo aur batao ki temperature badhne par har case mein $Δ G$ kaise behave karta hai. ( $Δ G = Δ H - T Δ S$ use karo.)

Recall Solution L3.2

Kyunki $Δ G = Δ H - T Δ S$ hai, $Δ G$ vs $T$ ki slope $- Δ S$ hai. Chaar cases:

$Δ H < 0, Δ S > 0$ : $Δ G < 0$ har $T$ par aur garmi mein aur zyada negative hota hai. Hamesha spontaneous.
$Δ H > 0, Δ S < 0$ : $Δ G > 0$ har $T$ par, garmi mein aur positive. Kabhi spontaneous nahi.
$Δ H < 0, Δ S < 0$ : spontaneous sirf thandi mein ( $T < Δ H /Δ S$ ); $- T Δ S$ term eventually jeet jaata hai.
$Δ H > 0, Δ S > 0$ : spontaneous sirf garmi mein ( $T > Δ H /Δ S$ ). Charon lines aur unke crossovers ke liye figure s02 dekho.

L3.3 (degenerate case). Agar $Δ H = 0$ ho (ek "thermoneutral" reaction) toh plot $Δ G / T$ vs $1/ T$ ka kya hoga? Aur agar $Δ S = 0$ ho?

Recall Solution L3.3

$Δ H = 0$ : slope zero hai → line flat (horizontal) hai. Phir $Δ G / T = - Δ S$ = constant, toh $Δ G$ linearly $T$ ke saath scale karta hai aur equilibrium kabhi temperature ke saath shift nahi karta (van 't Hoff: $d ln K / d (1/ T) = 0$ , $K$ $T$ -independent hai).
$Δ S = 0$ : ab $Δ G = Δ H$ (constant) hai, toh $Δ G / T = Δ H / T = Δ H \cdot (1/ T)$ . $1/ T$ ke against plot karne par origin se guzarti straight line milti hai jiska slope $Δ H$ hai — sabse cleanest possible case.

Level 4 — Synthesis

Goal: Gibbs–Helmholtz ko doosre laws ke saath combine karo.

L4.1 $Δ G^{\circ} = - R T ln K$ se shuru karke (jahan $R = 8.314$ J·mol⁻¹·K⁻¹ gas constant hai aur $K$ equilibrium constant hai), Gibbs–Helmholtz use karke van 't Hoff equation derive karo.

Recall Solution L4.1

$T$ se divide karo: $\frac{Δ G ^{\circ}}{T} = - R ln K .$ $1/ T$ form apply karo $\frac{\partial ( Δ G ^{\circ} / T )}{\partial ( 1/ T )} = Δ H^{\circ}$ : $\frac{\partial ( - R l n K )}{\partial ( 1/ T )} = Δ H^{\circ} .$ Partial ordinary derivative kyun ban jaati hai. Standard state ke fixed standard pressure par, ek hi variable bachta hai jis par $Δ G^{\circ} / T$ (isliye $ln K$ ) depend karta hai, woh hai temperature. Jab koi quantity sirf ek variable par depend kare, toh partial derivative $\partial / \partial (1/ T)$ aur ordinary derivative $d / d (1/ T)$ ek hi cheez hain. Toh hum $d ln K / d (1/ T)$ likh sakte hain. Kyunki $R$ constant hai, use bahar nikalo: $- R \frac{d l n K}{d ( 1/ T )} = Δ H^{\circ} \Rightarrow \frac{d ln K}{d ( 1/ T )} = - \frac{Δ H ^{\circ}}{R} .$ Iska matlab: $ln K$ vs $1/ T$ ka plot $- Δ H^{\circ} / R$ ka slope rakhta hai. Van 't Hoff equation hai hi Gibbs–Helmholtz jo equilibrium-constant ka costume pehne hua hai.

L4.2 Ek reaction ke liye $ln K = 4.00$ at $T_{1} = 300$ K aur $ln K = 2.50$ at $T_{2} = 350$ K hai. $Δ H^{\circ}$ nikalo (constant maano). $R = 8.314$ J·mol⁻¹·K⁻¹ use karo.

Recall Solution L4.2

Integrated van 't Hoff: $ln K_{2} - ln K_{1} = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) .$ Numbers. $ln K_{2} - ln K_{1} = 2.50 - 4.00 = - 1.50$ . $\frac{1}{350} - \frac{1}{300} = 0.00285714 - 0.00333333 = - 4.7619 \times 1 0^{- 4}$ K⁻¹. $Δ H^{\circ}$ ke liye solve karo: $Δ H^{\circ} = \frac{( l n K _{2} - l n K _{1} ) R}{- ( 1/ T _{2} - 1/ T _{1} )} = \frac{( - 1.50 ) ( 8.314 )}{- ( - 4.7619 \times 1 0 ^{- 4} )} = \frac{- 12.471}{4.7619 \times 1 0 ^{- 4}} = - 26190 J/mol .$ Toh $Δ H^{\circ} \approx - 26.2$ kJ/mol. Sign check: $T$ badhne par $K$ gira → garmi mein kam favourable → exothermic, $Δ H^{\circ} < 0$ . ✓

L4.3 Helmholtz (constant-volume) analogue likho aur symbol by symbol explain karo ki $U$ ne $H$ ki jagah kyun li.

Recall Solution L4.3

Twin equation: $(\frac{\partial ( A / T )}{\partial ( 1/ T )})_{V} = U .$ Kyun. $G = H - T S$ constant-pressure life se aaya, jahan $H$ natural energy hai. Constant volume par hum Helmholtz free energy $A = U - T S$ use karte hain — same algebra mein $H \to U$ aur constant- $P$ → constant- $V$ . Parent note ki derivation iske baad identical hai, kyunki $d A = - S d T - P d V$ deta hai $(\partial A / \partial T)_{V} = - S$ , jo $(\partial G / \partial T)_{P} = - S$ ko mirror karta hai.

Level 5 — Mastery

Goal: ek correction banao, woh case handle karo jo easy formula ignore karta hai.

L5.1 Two-point formula maanta hai ki $Δ H$ constant hai. Reality mein $Δ H$ Kirchhoff's law (thermochemistry) ke zariye temperature ke saath drift karta hai: $\frac{d Δ H}{d T} = Δ C_{p}$ . Maano $Δ C_{p} = - 40$ J·mol⁻¹·K⁻¹ constant hai, aur $Δ H (300 K) = - 58.0$ kJ/mol. $Δ H$ at $310$ K nikalo.

Recall Solution L5.1

Kirchhoff, constant $Δ C_{p}$ ke saath integrate karke: $Δ H (T_{2}) = Δ H (T_{1}) + Δ C_{p} (T_{2} - T_{1}) .$ Numbers. $Δ H (310) = - 58000 + (- 40) (310 - 300) = - 58000 - 400 = - 58400$ J/mol $= - 58.4$ kJ/mol. Iska matlab: negative $Δ C_{p}$ reaction ko garm hone par thoda aur exothermic banata hai — easy constant- $Δ H$ assumption $10$ K mein $0.4$ kJ/mol se off thi, chhota par real.

L5.2 (Synthesis + Mastery). Jab $Δ H (T) = Δ H_{0} + Δ C_{p} (T - T_{0})$ ho (constant $Δ C_{p}$ ), full integrated Gibbs–Helmholtz result derive karo, $Δ G (T) / T$ ko $1/ T$ ke function ke roop mein dete hue.

Recall Solution L5.2

Start karo $\frac{d ( Δ G / T )}{d T} = - \frac{Δ H ( T )}{T ^{2}}$ se aur $Δ H (T) = (Δ H_{0} - Δ C_{p} T_{0}) + Δ C_{p} T$ substitute karo. $a \equiv Δ H_{0} - Δ C_{p} T_{0}$ likho taaki $Δ H = a + Δ C_{p} T$ . Term by term integrate karo $T_{0}$ se $T$ tak: $\frac{Δ G ( T )}{T} - \frac{Δ G ( T _{0} )}{T _{0}} = - \int_{T_{0}}^{T} \frac{a + Δ C _{p} T ^{'}}{T ^{'2}} d T^{'} = - a (- \frac{1}{T ^{'}})_{T_{0}}^{T} - Δ C_{p} ln T^{'}_{T_{0}}^{T} .$ Result: $\frac{Δ G ( T )}{T} = \frac{Δ G ( T _{0} )}{T _{0}} + a (\frac{1}{T} - \frac{1}{T _{0}}) - Δ C_{p} ln \frac{T}{T _{0}},$ jahan $a = Δ H_{0} - Δ C_{p} T_{0}$ . Matlab. Pehla correction term $1/ T$ mein linear hai (constant- $Δ H$ line); $ln (T / T_{0})$ term woh curvature hai jo Kirchhoff introduce karta hai. Agar $Δ C_{p} = 0$ ho, toh dono extra pieces gum ho jaate hain aur tum straight two-point line recover karte ho. Yeh wide temperature ranges ke liye honest formula hai.

L5.3 (numeric). L5.2 ka full formula use karte hue, $Δ G (T_{0}) = - 30.0$ kJ/mol at $T_{0} = 280$ K, $Δ H_{0} = - 58.0$ kJ/mol ( $T_{0}$ par enthalpy change), aur constant $Δ C_{p} = - 40$ J·mol⁻¹·K⁻¹ ke saath, $Δ G$ at $T = 310$ K nikalo.

Recall Solution L5.3

Poore calculation mein J/mol mein kaam karo (kJ aur J kabhi mix mat karo). Step 1 — constant piece $a$ . $a = Δ H_{0} - Δ C_{p} T_{0} = - 58000 - (- 40) (280) = - 58000 + 11200 = - 46800$ . Step 2 — L5.2 formula ke teen terms evaluate karo $Δ G (310) /310$ ke liye. $\frac{Δ G ( 310 )}{310} = \frac{- 30000}{280} + (- 46800) (\frac{1}{310} - \frac{1}{280}) - (- 40) ln \frac{310}{280} .$

Pehla term: $\frac{- 30000}{280} = - 107.1429$ .
Doosra term: $\frac{1}{310} - \frac{1}{280} = 0.0032258 - 0.0035714 = - 3.4562 \times 1 0^{- 4}$ ; times $- 46800 = + 16.175$ .
Teesra term: $- (- 40) ln (310/280) = 40 ln (1.107143) = 40 (0.101783) = 4.0713$ . Step 3 — add karo aur $T$ se multiply karo. Sum $= - 107.1429 + 16.175 + 4.0713 = - 86.897$ J·mol⁻¹·K⁻¹. $Δ G (310) = - 86.897 \times 310 = - 26938 J/mol \approx - 26.9 kJ/mol .$ Sanity: $Δ G$ $T$ badhne par $- 30.0$ se $- 26.9$ kJ/mol ho gayi — ek exothermic reaction garmi mein kam favourable hoti ja rahi hai. ✓

Connections

Gibbs free energy — woh quantity jo har problem differentiate karta hai.
Enthalpy — woh prize jo har exercise nikaalti hai.
Entropy — woh term jo cancel ho jaata hai; L3.2 mein dhyan rakho.
van 't Hoff equation — L4 poori tarah is bridge ke baare mein hai.
Kirchhoff's law (thermochemistry) — L5 corrections ko power deta hai.
Helmholtz free energy — L4.3 mein constant-volume twin.
Maxwell relations — same identity family $(\partial G / \partial T)_{P} = - S$ ke peeche.