Methods to determine order — initial rates, integrated method, half-life method: Visual walkthrough

2.8.6 · D2 · Chemistry › Chemical Kinetics › Methods to determine order — initial rates, integrated metho

Step 0 — Pehle words, koi symbols nahi

Ek bhi formula se pehle, har symbol ko samajhte hain.

Hum jaanna chahte hain: jab main zyada bhari tank se start karta hoon, toh kya aadha hone mein zyada waqt lagta hai, kam lagta hai, ya utna hi? Yeh ek simple haan/nahi/kitna waala sawaal hi poora method hai.

Step 1 — Tank khali hote hua draw karo

KYA. Ek tank ka picture banao jisme reactant drain ho raha ho. Starting height $[A]_{0}$ aur half-line $[A]_{0} /2$ mark karo. Ek se doosre tak jaane mein jo time lagta hai woh $t_{1/2}$ hai.

KYUN. "Order" ke baare mein jo bhi abstract hai woh sab visible ho jaata hai jab hum maan lete hain ki concentration = height aur half-life = "half-line tak pahunchne ka time."

PICTURE.

Step 2 — Woh ek equation jisse sab kuch aata hai

KYA. Rate law: woh rule jo kehta hai ki level kitni jaldi girta hai yeh current height par depend karta hai.

$speed the level drops - \frac{d [ A ]}{d t} = rate constant k current height, raised to power n [A]^{n}$

Yeh symbols kyun:

$\frac{d [ A ]}{d t}$ derivative hai — girte curve ki instantaneous slope, "kitni height per second." Hum derivative use karte hain na ki plain fraction isliye kyunki slope lagaataar badlati rehti hai jaise tank khali hota hai; sirf calculus ek chalti hui slope ko capture kar sakta hai. (Dekho Rate Law and Rate Constant.)
Minus sign isliye hai kyunki $[A]$ ghatta hai, toh $d [A] / d t$ negative hai; hum isse flip karte hain taaki left side ek positive speed ho.
$k$ rate constant hai — ek fixed number jo ek given temperature par reaction ke liye hota hai (yeh collisions kitni aasaan hain yeh chupaata hai; dekho Arrhenius Equation). Yeh speed ka overall scale set karta hai lekin dependence ki shape nahi.
$n$ order hai — show ka star. Yeh woh power hai jis par height uthaayi jaati hai.

PICTURE.

Step 3 — Rule ko level-vs-time law mein badlo (integrate karo)

KYA. Rate law sirf humein har instant ki slope batata hai. Baad ke kisi time par actual height paane ke liye hume un saari slopes ko add karna padega — yahi integration hai. Isse integrated rate law milta hai.

Integrate kyun karo, seedha plug in kyun nahi? Kyunki $[A]$ dono sides par hai (slope usi height par depend karta hai jo khud change ho rahi hai). Hum ise ek step mein solve nahi kar sakte; hum choti-choti drops accumulate karte hain. Integration exactly woh tool hai jo answer deta hai "slope har jagah given hai, toh curve kya hai?"

Hum yeh har order ke liye alag se karte hain.

n = 0 : n = 1 : n = 2 : [A] = [A]_{0} - k t ln [A] = ln [A]_{0} - k t \frac{1}{[ A ]} = \frac{1}{[ A ] _{0}} + k t

Har term padhna:

$[A]$ — height abhi.
$[A]_{0}$ — start par height (integration ka constant, "at $t = 0$ height $[A]_{0}$ hai" se fix hota hai).
$k t$ — kitna "draining budget" clock ne spend kiya hai. Har law kehta hai: starting value, minus (ya plus) $k t$ , jo bhi coordinate us order ko straight banata hai usme.
$ln$ (natural log) $n = 1$ ke liye aata hai kyunki $1/ [A]$ integrate karne par logarithm milta hai — log honest answer hai, koi trick nahi.
$1/ [A]$ $n = 2$ ke liye aata hai kyunki $1/ [A]^{2}$ integrate karne par $- 1/ [A]$ milta hai.

PICTURE.

Step 4 — Har law mein tank ko aadhe par laao

KYA. $[A] = [A]_{0} /2$ set karo aur time ke liye solve karo. Woh solved time $t_{1/2}$ hai.

KYUN. "Half" sabse clean checkpoint hai: yeh $[A]$ ke units ki parwah nahi karta (yeh hamesha 2 ka factor hota hai), toh answer mein sirf $k$ aur $[A]_{0}$ bachte hain — precisely yahi hum compare karna chahte hain.

Zero order. $\frac{[ A ] _{0}}{2} = [A]_{0} - k t_{1/2} \Rightarrow k t_{1/2} = \frac{[ A ] _{0}}{2} \Rightarrow t_{1/2} = \frac{[ A ] _{0}}{2 k}$ Term-by-term: $[A]_{0} - \frac{[ A ] _{0}}{2} = \frac{[ A ] _{0}}{2}$ woh height hai jo humein khoni thi; drain speed $k$ se divide karo toh time milta hai. Taller start ⇒ zyada khona hai ⇒ zyada time.

First order. $ln \frac{[ A ] _{0} /2}{[ A ] _{0}} = - k t_{1/2} \Rightarrow ln \frac{1}{2} = - k t_{1/2} \Rightarrow - ln 2 = - k t_{1/2} \Rightarrow t_{1/2} = \frac{ln 2}{k} = \frac{0.693}{k}$ Dekho kya cancel hua: ratio $\frac{[ A ] _{0} /2}{[ A ] _{0}} = \frac{1}{2}$ ne $[A]_{0}$ ko completely cancel kar diya. Starting height ka koi nishan nahi — dekho Half-Life of Reactions.

Second order. $\frac{1}{[ A ] _{0} /2} = \frac{1}{[ A ] _{0}} + k t_{1/2} \Rightarrow \frac{2}{[ A ] _{0}} - \frac{1}{[ A ] _{0}} = k t_{1/2} \Rightarrow t_{1/2} = \frac{1}{k [ A ] _{0}}$ Yahan $[A]_{0}$ denominator mein hai: zyada taller start se $t_{1/2}$ chota ho jaata hai.

PICTURE.

Step 5 — Fingerprint: teen tarike jisme red bar react karta hai

KYA. Ab ek picture mein decisive experiment karo: har order ka curve do baar draw karo, ek $[A]_{0}$ se aur ek $2 [A]_{0}$ se, aur dekho half-life bar ka kya hota hai.

KYUN. Yahi actual lab move hai — aap $n$ directly nahi dekh sakte, lekin aap starting concentration double kar sakte ho aur halving time measure kar sakte ho.

PICTURE.

Step 6 — Ek exponent mein compress karo

KYA. Teeno boxed results secretly ek hi formula hain. Har ek mein $[A]_{0}$ ki power notice karo:

$n = 0 : t_{1/2} \propto [A]_{0}^{+ 1}$
$n = 1 : t_{1/2} \propto [A]_{0}^{0}$ (koi dependence nahi)
$n = 2 : t_{1/2} \propto [A]_{0}^{- 1}$

Exponent $n = 0, 1, 2$ ke liye $1, 0, - 1$ hai — yani exactly $1 - n$ .

$t_{1/2} \propto [A]_{0}^{1 - n}$

Yeh kyun matter karta hai. Ek formula har order ko cover karta hai, fractional bhi. Aur yeh humein do experiments se $n$ directly solve karne ka seedha tarika deta hai.

Do starting heights $[A]_{0, 1}, [A]_{0, 2}$ lo jinki half-lives $t_{1/2, 1}, t_{1/2, 2}$ hain. Divide karo: $\frac{t _{1/2, 2}}{t _{1/2, 1}} = (\frac{[ A ] _{0, 2}}{[ A ] _{0, 1}})^{1 - n}$ Logarithm lo — woh tool jo exponent ko neeche ek multiplier mein khiinch laata hai taaki hum use isolate kar sakein: $n = 1 - \frac{ln ( t _{1/2, 2} / t _{1/2, 1} )}{ln ( [ A ] _{0, 2} / [ A ] _{0, 1} )}$ Har piece: upar wala log measure karta hai "halving-time kitna change hua," neeche wala log measure karta hai "start kitna change hua," unka ratio exponent $1 - n$ hai, aur $n$ free karne ke liye $1$ se subtract karte hain.

Step 7 — Edge aur degenerate cases

Reader ko koi unseen scenario mein mat chhodna.

Ek-picture summary

Recall Feynman: walkthrough plain words mein

Ek tank of stuff drain hote hua picture karo. Order bas yeh hai ki "drain sunti kitna hai tank kitna bhara hai." Agar woh fullness ignore kare, order 0; agar dhyaan se sune (proportional), order 1; agar fiercely sune (squared), order 2.

Hum order directly nahi dekh sakte. Toh hum time karte hain ki tank ko aadha khali hone mein kitna lagta hai — yahi half-life hai. Phir clever move: tank ko do guna bhar do aur phir time karo.

Agar halving time double ho jaaye, toh drain ne parwah ki kitna tha — order 0.
Agar halving time na hile, toh drain ne extra paani ki parwah nahi ki — order 1 (yeh radioactivity ki pehchaan hai).
Agar halving time aadhi ho jaaye, toh zyada bhara tank aur proportionally fast drain hota hai — order 2.

Package karo teeno ko ek sentence mein: $t_{1/2} \propto [A]_{0}^{1 - n}$ . Do experiments ka log lo aur $n$ bahar nikal aata hai. Yahi poora half-life method hai — kuch yaad nahi karna, bas ek red bar ko badhte, rukate, ya shrink hote dekho.

Recall

Half-life definition

time jisme

[A]

[A]_{0}

[A]_{0} /2

tak girta hai

Rate law integrate kyun karo

slope usi changing height par depend karta hai, toh hum saari slopes add karte hain

First order ke liye

[A]_{0}

kya cancel karta hai

ratio

\frac{[ A ] _{0} /2}{[ A ] _{0}} = \frac{1}{2}

log ke andar

Order 0 ke liye

[A]_{0}

double karne par

t_{1/2}

ka kya hota hai

doubles ho jaata hai

Order 1 ke liye

[A]_{0}

double karne par

t_{1/2}

ka kya hota hai

kuch nahi (unchanged)

Order 2 ke liye

[A]_{0}

double karne par

t_{1/2}

ka kya hota hai

aadhi ho jaati hai

General half-life dependence

t_{1/2} \propto [A]_{0}^{1 - n}

Order from two half-lives

n = 1 - \frac{ln ( t _{1/2, 2} / t _{1/2, 1} )}{ln ([ A ] _{0, 2} / [ A ] _{0, 1} )}

n

find karne ke liye log kyun lete hain

yeh exponent

1 - n

ko neeche kheench laata hai jahan hum use isolate kar sakein

Order	$t_{1/2}$	$[A]_{0}$ double karein $\Rightarrow t_{1/2}$
0	$[A]_{0} /2 k$	doubles ↑
1	$0.693/ k$	unchanged =
2	$1/ k [A]_{0}$	halves ↓