Visual walkthrough — Pseudo-first-order kinetics

2.8.5 · D2 · Chemistry › Chemical Kinetics › Pseudo-first-order kinetics

Kisi bhi algebra se pehle, chalte hain ek picture par agree karte hain jo hum draw kar rahe hain.

---

Step 1 — Molecules ki do dheriyaan, aur "rate" ka matlab kya hai

KYA. Hamare paas ek reaction hai jisme A ke ek molecule ko B ke ek molecule se milna padta hai react karne ke liye: $A+B\to \text{Products}$ Socho do dheriyaan marbles ki. A-dher choti hai. B-dher bahut badi hai.

YE YAHAN KYU SHURU. Har rate law asal mein collisions ke baare mein ek baat hai (dekho Collision Theory). A tabhi khatam hota hai jab woh B se takraata hai. Toh reaction ki speed depend karni chahiye ki har dher kitni bheed bhari hai. "Bheed bhari" = concentration.

PICTURE. Figure dekho: baayi taraf choti laal A-dher, daayeen taraf badi kaali B-dher. "Reaction" wala arrow tabhi chalta hai jab ek laal marble ek kaale ko dhoondh leta hai.

---

Step 2 — Sachcha rate law: A aur B dono matter karte hain

KYA. Reaction ki sach mein speed hai $\underset{\text{kitni tezi se A khatam hota hai}}{\underbrace{-\frac{d[A]}{dt}}}=\underset{\text{collision-luck}}{\underbrace{k}}\underset{\text{A kitna bheed bhara hai}}{\underbrace{[A]}}\underset{\text{B kitna bheed bhara hai}}{\underbrace{[B]}}$

HAR PIECE KYU.

• $-\frac{d[A]}{dt}$ — $d[A]/dt$ A-vs-time curve ki slope hai (har instant $[A]$ kitna badalta hai). Yeh negative hai kyunki A gayab ho raha hai, toh hum aage minus sign lagaate hain taaki poori cheez ek positive speed bane.
• $[A][B]$ — dono crowd-densities ko multiply karo kyunki ek collision ke liye ek ek ki zaroorat hoti hai. A-dher double karo → chances double. B-dher double karo → chances phir double. Product, sum nahi.
• $k$ — rate constant: ek fixed number jo capture karta hai ki ek collision actually reaction lead karne ki kitni probability hai (temperature, orientation — dekho Activation Energy). Yahan iske units hain ${\text{M}}^{-1}{\text{s}}^{-1}$, algebra ke force se.

"Second-order" KYU. Do concentration factors milake total power $2$ banaate hain; reaction overall second-order hai (dekho Second-Order Reactions aur Integrated Rate Laws).

PICTURE. Figure collision frequency ko ek rectangle ke area ke roop mein dikhata hai jiske sides $[A]$ aur $[B]$ hain. Koi bhi side chota karo → area (rate) chota ho jaata hai.

---

Step 3 — B ko giant banao: excess condition

KYA. Hum jaanbujhkar flask mein A se bahut zyada B dalte hain — 50 se 100 guna zyada: $[B{]}_0\gg [A{]}_0$ ($\gg$ ka matlab hai "bahut zyada bada.")

KYU. Dekho ki ek poora reaction har dher ke saath kya karta hai. Jab har A khatam ho jaata hai, har A ke saath bilkul ek B consume hota hai — lekin itne saare B hain ki B-dher ko pata hi nahi chalta.

Concrete numbers: $[A{]}_0=0.01$ M aur $[B{]}_0=1.0$ M se shuru karo (ek $100\times$ excess). Saara A jalao: $[B{]}_{\text{final}}=1.0-0.01=0.99\text{ M}.$ B 1% gira. Essentially flat.

PICTURE. Do bars: laal A-bar zero tak drain hoti hai, jabki kaali B-bar muskil se thodi si neeche jaati hai. Dashed line "pehle" B-level dikhata hai jo almost "baad" B-level ko touch kar raha hai.

---

Step 4 — B ko freeze karo, aur use ek naye constant mein chhupaao

KYA. Kyunki B muskil se hilta hai, hum maanlete hain ki yeh ek fixed number hai: $[B]\approx [B{]}_0(\text{constant}).$ Ise honest rate law mein daalo: $-\frac{d[A]}{dt}=k[B{]}_0[A].$ Ab $k$ aur $[B{]}_0$ dono constants hain, saath mein baithe hain. Inhe ek naye symbol mein bundle karo: $\boxed{k^{\prime }=k[B{]}_0}⟹-\frac{d[A]}{dt}=k^{\prime }[A].$

KYU. $k^{\prime }$ (kaho "k-prime") pseudo-first-order rate constant hai. Iska kaam hai frozen B-concentration ko absorb karna taaki equation mein sirf ek variable bache, $[A]$. Note karo ki units apne aap theek ho gayi: $[k^{\prime }]=\underset{k}{\underbrace{{\text{M}}^{-1}{\text{s}}^{-1}}}\times \underset{[B{]}_0}{\underbrace{\text{M}}}={\text{s}}^{-1}.$ Woh ${\text{s}}^{-1}$ ek genuine first-order rate constant ki pehchaan hai — disguise poora ho gaya.

PICTURE. Figure Step 2 ka rectangle dikhata hai jiska $[B]$ side locked hai (ek padlock icon), toh sirf $[A]$ side ab slide kar sakta hai. Rate ab ek length par hi depend karta hai.

---

Step 5 — One-variable equation solve karo (kyun log aata hai)

KYA. Ab hamare paas hai $-\frac{d[A]}{dt}=k^{\prime }[A].$ Variables ko alag karo — $[A]$ wali sab cheez baayein, $t$ wali sab cheez daayein: $\frac{d[A]}{[A]}=-k^{\prime }dt.$ Dono sides ko add karo (integrate karo), shuru se ($[A{]}_0$ at $t=0$) kisi baad ke moment tak ($[A]$ at time $t$): ${\int }_{[A{]}_0}^{[A]}\frac{d[A]}{[A]}=-k^{\prime }{\int }_0^{t}dt.$

LOGARITHM KYU. Left side poochh raha hai: "kaun sa function hai jiska slope $1/[A]$ hai?" Jawaab hai $\ln [A]$ — natural logarithm woh ek tool hai jo bilkul isi $1/x$ shape ko undo karne ke liye bana hai. Isliye $\ln$ yahan aata hai aur kahin nahi; yeh antidote hai us rate ke liye jo amount ke proportional hai.

Sum carry out karo: $\underset{\text{left side}}{\underbrace{\ln [A]-\ln [A{]}_0}}=\underset{\text{right side}}{\underbrace{-k^{\prime }t}}.$

PICTURE. Figure plot karta hai $[A]$ vs $t$ (ek swooping decay curve, laal) aur saath mein $\ln [A]$ vs $t$ (ek straight kaali line). Log-transform curve ko ek ruler-straight line mein flatten kar deta hai jiska steepness hai $k^{\prime }$.

---

Step 6 — Straight line padho: yahi poora payoff hai

KYA. $\ln [A]=\ln [A{]}_0-k^{\prime }t$ ko schoolbook line $y=c+mx$ se compare karo:

• $y\leftrightarrow \ln [A]$
• $x\leftrightarrow t$
• intercept $c\leftrightarrow \ln [A{]}_0$
• slope $m\leftrightarrow -k^{\prime }$

KYU USEFUL HAI. Apna measured $\ln [A]$ $t$ ke against plot karo. Agar straight line mile, toh reaction pseudo-first-order behave kar raha hai, aur slope tumhe $k^{\prime }$ free mein de deta hai. Yeh method of isolation ka experimental heart hai: B se flood karo, ek line paao, $k^{\prime }$ padho, phir true $k$ ke liye $[B{]}_0$ se divide karo.

PICTURE. Figure line par slope triangle mark karta hai: $\Delta t$ ka ek "run" aur $\Delta \ln [A]$ ka ek "rise"; unka ratio (negative) equals $k^{\prime }$.

---

Step 7 — Edge case: agar B excess mein nahi hai toh?

KYA. Maano $[B{]}_0=[A{]}_0$ (bilkul bhi excess nahi). Ab B utni hi tezi se drain hota hai jitna A — Step 4 ka padlock gone hai, $[B]$ phir se ek moving variable hai, aur $-d[A]/dt=k[A][B]$ sach mein do-variabled rehta hai.

LINE KYU MARTA HAI. Sache second-order law ka sahi integration straight line deta hai sirf tab jab tum $\frac{1}{[A]}$ vs $t$ plot karo — $\ln [A]$ vs $t$ nahi. Toh agar tum yahan $\ln [A]$ vs $t$ plot karne par insist karo, tumhe ek curve milega, line nahi. Woh curvature yeh signal hai ki excess trick fail ho gayi.

Kitna excess kaafi hai? B ko freeze karne mein error roughly woh fraction hai jo B ka khatam ho jaata hai, $\approx [A{]}_0/[B{]}_0$:

Excess $[B{]}_0/[A{]}_0$	Completion par B consumed	Verdict
$1\times$	$100\%$	curve — truly 2nd order
$10\times$	$\approx 10\%$	noticeable bend
$50\times$	$\approx 2\%$	achha
$100\times$	$\approx 1\%$	excellent

PICTURE. Ek axis par do $\ln [A]$-vs-$t$ traces: laal curved wala ($1\times$, fail) aur straight wala ($100\times$, works). Same reaction — sirf excess alag hai.

---

Step 8 — Real numbers daalo (ester hydrolysis)

KYA. Water ultimate excess reactant hai: pure water $[H_{2}O]\approx 55.5$ M hai, jo ek $0.01$ M ester ko bahut peeche chhod deta hai. Ethyl-acetate hydrolysis ke liye $k^{\prime }=2.5\times 10^{-4}{\text{s}}^{-1}$ ke saath:

True $k$ recover karo (Step 4 ki bundling undo karo): $k=\frac{k^{\prime }}{[H_{2}O]}=\frac{2.5\times 10^{-4}}{55.5}=4.5\times 10^{-6}{\text{M}}^{-1}{\text{s}}^{-1}.$

$75\%$ completion ka time (A ka $25\%$ baca hai, toh $[A]=0.25[A{]}_0$). Step 5 ka log form use karo: $\ln (0.25[A{]}_0)=\ln [A{]}_0-k^{\prime }t\Rightarrow \ln (0.25)=-k^{\prime }t$ $t=-\frac{\ln 0.25}{k^{\prime }}=\frac{1.386}{2.5\times 10^{-4}}=5544\text{s}\approx 92.4\text{min}.$

Sucrose inversion ($k^{\prime }=6.0\times 10^{-5}{\text{s}}^{-1}$, $[A{]}_0=0.5$ M, $t=7200$ s ke baad). Exponential twin use karo — yeh seedha concentration deta hai: $[A]=0.5e^{-(6.0\times 10^{-5})(7200)}=0.5e^{-0.432}=0.5\times 0.649=0.325\text{M}.$ Bacha fraction $=0.325/0.5=65\%$, toh $35\%$ react hua — samajh mein aata hai.

PICTURE. Decay curve $[A]=0.5e^{-k^{\prime }t}$ jisme $t=7200$ s ka point laal mein $0.325$ M par mark hai, plus ester ke liye $75\%$-completion marker.

---

Ek-picture summary

Ek panel, poori kahani: true law $k[A][B]$ → B se flood karo (padlock) → bundle karo $k^{\prime }=k[B{]}_0$ → $\ln$ se integrate karo → slope $-k^{\prime }$ wali straight line.

Recall Feynman retelling — apne words mein wapas bolo

A ko react karne ke liye B se takrana padta hai, aur speed depend karti hai ki dono dheriyaan kitni bheed bhari hain. Yeh ek two-variable problem hai — annoying. Toh hum thodi cheating karte hain: hum B ka ek pahaad daal dete hain. Ab, chahe A kitna bhi react kare, B-pahaad muskil se sikundta hai, toh hum maanlete hain ki B ka crowd-level kabhi nahi badlata. Hum woh frozen B-number ko rate constant mein fold karte hain aur bundle ko $k^{\prime }$ kehte hain. Achanak speed sirf ek cheez par depend karta hai, $[A]$, bilkul first-order reaction ki tarah. Solve karne par ek logarithm milta hai (kyunki log woh ek function hai jiska slope $1/[A]$ hai), aur jab tum $\ln [A]$ ko time ke against plot karte ho toh tumhe ek bilkul seedhi line milti hai jiska steepness $k^{\prime }$ hai. Sach mein two-body constant dhundhne ke liye, sirf $k^{\prime }$ ko B-pahaad ke size se divide karo. Aur agar tum kaafi B nahi daalo? Line bend ho jaati hai — woh curvature reaction ka yeh kehna hai ki disguise slip ho gaya.

Recall Quick self-check

$\ln$ kyun aata hai aur, say, square root kyun nahi? ::: Kyunki rate $[A]$ ke hi proportional hai; woh function jiska slope $1/[A]$ hai woh $\ln [A]$ hai, toh $d[A]/[A]$ integrate karne par logarithm milta hai. $k^{\prime }$ ke units kya hain aur kyun? ::: ${\text{s}}^{-1}$; yeh equals hai $k[B{]}_0={\text{M}}^{-1}{\text{s}}^{-1}\times \text{M}$, M cancel ho jaate hain. Measured $k^{\prime }$ se true $k$ kaise recover karo? ::: Excess concentration se divide karo: $k=k^{\prime }/[B{]}_0$. Agar B excess mein nahi hai toh $\ln [A]$-vs-$t$ plot ka kya hota hai? ::: Yeh curve ho jaata hai, kyunki $[B]$ ab constant nahi hai aur reaction sach mein second-order rehta hai.

Yeh bhi dekho: Half-life · Enzyme Kinetics · Integrated Rate Laws · 2.8.05 Pseudo-first-order kinetics (Hinglish)