Visual walkthrough — Differential rate equations for 0, 1st, 2nd order — derivations

2.8.3 · D2 · Chemistry › Chemical Kinetics › Differential rate equations for 0, 1st, 2nd order — derivati

Yeh page teen differential rate equations ko parent note sirf molecules ke gayab hone ki ek tasveer se dobara banata hai. Hum uss ek sachche sawaal se shuru karte hain — "abhi iss waqt cheezein kitni tezi se khatam ho rahi hain?" — aur har step pictures se karte hain. Koi bhi symbol tab tak nahin aata jab tak usse draw na kar liya ho.

Step 1 — "Rate" ka matlab kya hota hai: ek shrinking-crowd graph par ek slope

KYA HAI. Ek jar mein reactant imagine karo. Uski per litre miqdar ko concentration kaho, likho $[\text{A}]$ — padhte hain "concentration of A". Square brackets ka matlab sirf yeh hai ki "per litre kitna A hai", measure hota hai $\text{M}$ mein (molar = ${\text{mol L}}^{-1}$). Jaise jaise time $t$ aage badhta hai, $[\text{A}]$ girta hai.

KYUN. Yeh poochhne se pehle ki "kitni tezi se", humein ek picture chahiye ki kya change ho raha hai. Figure mein $[\text{A}]$ ko neechey jaate dikhaya gaya hai jaise $t$ aage badhta hai.

PICTURE. Blue curve ko downhill slide karte dekho. Do kareeb ke instants chuno. Concentration ek choti si miqdar $d[\text{A}]$ se girta hai ek choti si time $dt$ mein. Us choti si drop ki steepness hi derivative $\frac{d[\text{A}]}{dt}$ hai — woh tool jo yeh jawab deta hai "ek quantity mein ek doosri quantity ki choti si change per kitni change hoti hai?". Hum derivative use karte hain aur saada division nahin kyunki slope curve ke saath-saath badalta rehta hai; derivative ek ek instant par slope hota hai.

$\text{Rate}=\underset{\text{crowd shrinks, so make it positive}}{\underbrace{-}}\underset{\begin{array}{c}\text{slope of the blue curve:}\\ \text{drop in }[\text{A}]\text{ per tiny second}\end{array}}{\underbrace{\frac{d[\text{A}]}{dt}}}$

Minus sign isliye hai kyunki $[\text{A}]$ gir raha hai, toh $\frac{d[\text{A}]}{dt}$ ek negative number hai; minus use karke hum isse positive bana dete hain taaki rate kabhi negative na ho.

Step 2 — Ek design choice: slope ka bheed par kya depend karta hai?

KYA HAI. Ab hum guess karte hain ki rate current crowd size ko kisi power tak uthane se control hoti hai: $-\frac{d[\text{A}]}{dt}=k[\text{A}{]}^n$

KYUN. Aage ki har cheez sirf yahi ek equation hai $n=0$, $1$, ya $2$ ke saath. Toh har symbol ko dhyan se padhna zaroori hai.

PICTURE. Teen choti bheeden draw ki gayi hain, aur unke saath ek arrow jis ki length rate hai. Left bheed mein arrow bheed ki size ignore karta hai; middle mein woh bheed ke saath badhta hai; right mein woh square ke saath badhta hai.

$-\frac{d[\text{A}]}{dt}=\underset{\begin{array}{c}\text{rate constant:}\\ \text{fixed for this reaction}\\ \text{at this temperature}\end{array}}{\underbrace{k}}\underset{\begin{array}{c}\text{crowd size}\\ \text{raised to the}\\ \text{order }n\end{array}}{\underbrace{[\text{A}{]}^n}}$

• $k$ = rate constant. Yeh tab tak nahin badalta jab reaction chal rahi ho — dekho yeh sirf temperature ke saath badalta hai.
• $n$ = order, experimentally measure kiya hua number. Yahi poori kahaani hai.

Step 3 — Universal trick: do worlds ko alag karo, phir integrate karo

KYA HAI. Teeno derivations mein se har ek yehi do moves use karti hai. Move 1: saara $[\text{A}]$ wala stuff left side par dhakelo, saara $t$ wala stuff right side par. Move 2: integrate karo — shuru se time $t$ tak saare chote slices ko joodo.

KYUN. Equation $[\text{A}]$ aur $t$ ko ek derivative ke andar mila deti hai. Hum $[\text{A}]$ solve nahin kar sakte jab tak woh ulajhe hue hain. Unhe alag karne se har side ko apne aap sum kiya ja sakta hai. Integration woh tool hai jo yeh jawab deta hai "agar mujhe har instant par choti change pata hai, toh kul mila-julakar change kya hai?" — yeh Step 1 ke derivative ka ulta hai.

PICTURE. Figure mein ek curve ke neeche ka area patli strips mein kata hua aur stack kiya hua dikhaya gaya hai: yahi stacking hai jo integral sign $\int$ ka matlab hai.

$\underset{\begin{array}{c}\text{sum over all}\\ \text{concentrations from}\\ \text{start }[\text{A}{]}_0\text{ to now }[\text{A}{]}_t\end{array}}{\underbrace{{\int }_{[\text{A}{]}_0}^{[\text{A}{]}_t}}}f([\text{A}])d[\text{A}]=-k\underset{\begin{array}{c}\text{sum over all}\\ \text{time from }0\text{ to }t\end{array}}{\underbrace{{\int }_0^t}}dt$

Yahan $[\text{A}{]}_0$ starting whistle par concentration hai ($t=0$) aur $[\text{A}{]}_t$ clock reading $t$ par concentration hai. Left side par $f([\text{A}])$ ki shape hi ek maatra cheez hai jo teeno orders mein alag hai. Ab hum ek-ek karke karte hain.

Step 4 — Zero order ($n=0$): seedha ramp neeche ki taraf

KYA HAI. $n=0$ rakho. Kyunki kuch bhi power $0$ se $1$ hota hai, crowd size equation se gayab ho jaati hai: $-\frac{d[\text{A}]}{dt}=k[\text{A}{]}^0=k$ Alag karo aur integrate karo (yahan $f([\text{A}])=1$): ${\int }_{[\text{A}{]}_0}^{[\text{A}{]}_t}d[\text{A}]=-k{\int }_0^{t}dt⟹[\text{A}{]}_t-[\text{A}{]}_0=-kt$

KYUN. Rate constant $k$ se chipki hone ke saath, bheed har second ek hi miqdar se girती hai — ek bilkul seedhi downhill line.

PICTURE. Seedhi pink line fixed steepness se girti hai. Uska slope $-k$ hai; $t=0$ par uski height $[\text{A}{]}_0$ hai.

$\underset{\text{abhi}}{\underbrace{[\text{A}{]}_t}}=\underset{\begin{array}{c}\text{y-intercept:}\\ \text{starting height}\end{array}}{\underbrace{[\text{A}{]}_0}}-\underset{\begin{array}{c}\text{slope size:}\\ \text{drop per second}\end{array}}{\underbrace{k}}t$

Degenerate case (ek asli trap). Seedha ramp zero se neechey nahin ja sakta — lekin formula bade $t$ ke liye khushi se negative concentration de deta. Physically reactant bas $t_{\text{end}}=[\text{A}{]}_0/k$ par khatam ho jaata hai, aur line ruk jaati hai. Figure mein dekho kahan pink line axis se milti hai: us point ke aage yeh law laagu nahin hota.

Step 5 — First order ($n=1$): ek curve jo kabhi zero tak nahin pahunchti

KYA HAI. $n=1$ rakho, toh $f([\text{A}])=1/[\text{A}]$: $-\frac{d[\text{A}]}{dt}=k[\text{A}]⟹\frac{d[\text{A}]}{[\text{A}]}=-kdt$ $1/[\text{A}]$ ko integrate karne se natural logarithm $\ln$ milta hai: $\ln [\text{A}{]}_t-\ln [\text{A}{]}_0=-kt⟹\ln \frac{[\text{A}{]}_t}{[\text{A}{]}_0}=-kt$

LOGARITHM KYUN? $1/x$ ka integral $\ln x$ hota hai — yahi woh specific tool hai jo "crowd ke ek upar" ko undo karta hai. Natural log ka jawab hai "$e\approx 2.718$ ko kis power par uthaya jaaye taaki yeh number mile?". Log ko undo karne par (dono sides par $e$ uthao) ek clean exponential milta hai: $[\text{A}{]}_t=[\text{A}{]}_0{e}^{-kt}$

PICTURE. Left panel: yellow exponential curve pehle tezi se jhukti hai, phir flat hoti jaati hai, zero ke paas aati hai par kabhi chhooti nahin. Right panel: $\ln [\text{A}]$ plot karne par yeh $-k$ slope ki ek seedhi line ban jaati hai.

$\underset{\text{abhi}}{\underbrace{[\text{A}{]}_t}}=\underset{\text{start}}{\underbrace{[\text{A}{]}_0}}\underset{\begin{array}{c}\text{shrink factor:}\\ \text{1 se kam se multiply karo}\\ \text{jaise time badhta hai}\end{array}}{\underbrace{e^{-kt}}}$

Khaas feature. Half-life $\ln 2/k$ mein $[\text{A}{]}_0$ nahin hai — har halving mein utna hi time lagta hai, chahe shuru mein kitna bhi ho. Isliye radioactive dating kaam karti hai.

Step 6 — Second order ($n=2$): reciprocal seedhi line mein upar chadhta hai

KYA HAI. $n=2$ rakho, toh $f([\text{A}])=1/[\text{A}{]}^2=[\text{A}{]}^{-2}$: $-\frac{d[\text{A}]}{dt}=k[\text{A}{]}^2⟹[\text{A}{]}^{-2}d[\text{A}]=-kdt$ $[\text{A}{]}^{-2}$ ko integrate karne se $-[\text{A}{]}^{-1}$ milta hai (power mein ek jodo, nayi power se divide karo): $[-[\text{A}{]}^{-1}{]}_{[\text{A}{]}_0}^{[\text{A}{]}_t}=-kt⟹\frac{1}{[\text{A}{]}_t}-\frac{1}{[\text{A}{]}_0}=kt$

KYUN. Yahan integration ordinary power rule use karta hai, log nahin, kyunki power $-2$ hai jo $-1$ nahin hai. Woh natural variable jo seedhi nikalta hai woh $1/[\text{A}]$ hai, yaani reciprocal — "crowd divided by ek".

PICTURE. Left: blue $[\text{A}]$ curve tail mein first order se bhi zyada dheemi hoti jaati hai. Right: $1/[\text{A}]$ plot karne par yeh $+k$ slope ki ek upar jaati seedhi line ban jaati hai (note: positive).

$\underset{\begin{array}{c}\text{reciprocal abhi}\\ \text{(y-axis)}\end{array}}{\underbrace{\frac{1}{[\text{A}{]}_t}}}=\underset{\begin{array}{c}\text{y-intercept:}\\ \text{start par reciprocal}\end{array}}{\underbrace{\frac{1}{[\text{A}{]}_0}}}+\underset{\begin{array}{c}\text{slope}\\ \text{(positive!)}\end{array}}{\underbrace{k}}t$

Worked check (parent ke ${\text{NO}}_2$ example se). $[\text{A}{]}_0=0.0100\text{M}$, $k=0.543{\text{M}}^{-1}{\text{s}}^{-1}$, $t=300\text{s}$ ke saath: $\frac{1}{[\text{A}{]}_t}=100+0.543\times 300=262.9\Rightarrow [\text{A}{]}_t=3.80\times 10^{-3}\text{M}$

Step 7 — Graphs kyun kabhi jhooth nahin bolte: seedhi line se order padhna

KYA HAI. Teen integrated laws mein se har ek ek seedhi line ban jaati hai jab tum y-axis par sahi cheez plot karo. Integrated law ka left-hand side hi tera y-axis hai.

KYUN. Seedhi line ek maatra shape hai jise insaani aankhein turant judge kar sakti hain. Toh hum har law ko $y=(\text{intercept})+(\text{slope})t$ mein mod dete hain.

PICTURE. Teeno linear plots ek ke upar ek: $[\text{A}]$ (zero), $\ln [\text{A}]$ (first), $1/[\text{A}]$ (second). Sirf matching transform seedhi hoti hai; baaki dono curve karte hain.

Ek-tasveer summary

Neeche, teeno worlds ek saath side by side hain: same starting height, same clock, teen taqdeer. Pink zero-order line seedhi zero tak chalti hai aur ruk jaati hai; yellow first-order curve equal steps mein forever halve hoti hai; blue second-order curve rengti hai, uski tail sabse lambi.

Retrieval check