Integrated rate laws — half-life t₁ - ₂ for each order: Question bank

2.8.4 · D5 · Chemistry › Chemical Kinetics › Integrated rate laws — half-life t₁ - ₂ for each order

Shuru karne se pehle, ek shared vocabulary reminder taaki koi bhi symbol undefined na rahe:

Definition Jo quantities kaam mein aati hain (aur unke units)

$[A]_{0}$ ::: reactant A ki shuruwati concentration stopwatch ke zero par, molar mein measure hoti hai, $M$ (moles per litre).
$[A]_{t}$ ::: time $t$ ke baad bacha hua concentration, yeh bhi $M$ mein.
$[A]_{start}$ ::: concentration us half-life interval ki shuruwat mein jo abhi time ho raha hai. Pehli half-life ke liye yeh $[A]_{0}$ ke barabar hai; doosri ke liye $[A]_{0} /2$ ; aur aage bhi aise hi. Yeh hamesha $[A]_{0}$ nahi hoti — yahi distinction neeche kai traps exploit karti hai.
$t_{1/2}$ ::: half-life — woh time jisme $[A]$ us interval ke $[A]_{start}$ se exactly aadha ho jaata hai.
$k$ ::: rate constant. Iske units order ke saath change hote hain: zero-order $k$ ka unit $M s^{- 1}$ hai, first-order $k$ ka $s^{- 1}$ hai, second-order $k$ ka $M^{- 1} s^{- 1}$ hai. Dekho Rate constant k — units and temperature dependence.
$t_{1/2}^{(0)}$ , $t_{1/2}^{(1)}$ , $t_{1/2}^{(2)}$ mein superscript sirf reaction order (0, 1, ya 2) ka label hai. Yeh koi power nahi hai — " $t_{1/2}^{(2)}$ " ko zor se padho "second-order half-life."

Formula Teeno half-lives ek card par — aur har ek aisa kyun dikhta hai

Har formula us order ki integrated rate law mein $[A]_{t} = [A]_{start} /2$ rakhne se aata hai. Har ek ka ek-line "kyun" yahan hai: $t_{1/2}^{(0)} = \frac{[ A ] _{0}}{2 k} (k : M s^{- 1})$ Zero-order fixed speed $k$ par ek seedhi line mein girta hai: $[A]_{0} - k t$ . Left side ko start ka aadha rakho, $\frac{[ A ] _{0}}{2} = [A]_{0} - k t_{1/2}$ , toh $k t_{1/2} = \frac{[ A ] _{0}}{2}$ — time proportional hai is baat se ki tumne kitne se shuru kiya. $t_{1/2}^{(1)} = \frac{l n 2}{k} = \frac{0.693}{k} (k : s^{- 1})$ First-order exponentially decay karta hai: $[A]_{0} e^{- k t}$ . $e^{- k t_{1/2}} = \frac{1}{2}$ rakhne aur phir exponential ko undo karne ke liye logarithm lene par $- k t_{1/2} = ln \frac{1}{2} = - ln 2$ milta hai, toh $t_{1/2} = \frac{l n 2}{k}$ — $[A]_{0}$ gayab ho gaya, sirf ek constant bachi. $t_{1/2}^{(2)} = \frac{1}{k [ A ] _{start}} (k : M^{- 1} s^{- 1})$ Second-order ka reciprocal linearly badhta hai: $\frac{1}{[ A ] _{t}} = \frac{1}{[ A ] _{0}} + k t$ . Left side par $\frac{1}{[ A ] _{0} /2} = \frac{2}{[ A ] _{0}}$ rakhne par $k t_{1/2} = \frac{2}{[ A ] _{0}} - \frac{1}{[ A ] _{0}} = \frac{1}{[ A ] _{0}}$ milta hai — interval ki shuruwat mein concentration se inversely proportional. Ise yahan rakho — tumhe kabhi page chhodke check nahi karna padega ki kaun se $k$ units kis order se belong karte hain. Poori derivations Integrated rate laws — derivation and graphical analysis mein hain.

Neeche ke do figures in teeno "kyun" ko visual banate hain — captions padho, phir reveals argue karo.

Figure 1 — decay ki teen shapes. Teeno curves ek hi concentration se shuru hoti hain ( $1.0 M$ , top-left). Red line zero-order hai: ek seedhi diagonal jo finite time par zero par crash karti hai. Blue curve first-order hai: ek exponential jo mودti hai aur kabhi zero tak nahi pahunchti. Green curve second-order hai: ek lamba lazy tail jo shuruaat mein sabse fast decay karta hai (crowded molecules) phir reng ta hai. Dashed yellow line "start ki aadhi" mark karti hai — jahan har curve ise cross karti hai woh us reaction ki pehli half-life hai.

Figure — Integrated rate laws — half-life t₁ - ₂ for each order

Figure 2 — successive half-life widths. Har coloured block $[A]$ ki ek successive halving hai (block 1 tumhe full se half tak le jaata hai, block 2 half se quarter tak, aur aage bhi). Widths padho: red (zero-order) blocks shrink hote hain, blue (first-order) blocks equal rehte hain, aur green (second-order) blocks har round mein double hote hain. Block width = us half-life mein laga time, toh yeh picture literally orders 0, 1, 2 ke liye "Down, Flat, Up" dikhata hai.

True or false — justify

Har reaction ki ek fixed half-life hoti hai jo ek number mein bata sako

False. Sirf first-order reactions ki ek constant half-life hoti hai. Zero- aur second-order half-lives is par depend karti hain ki abhi kitna reactant bacha hai.

First-order reaction ke liye,

[A]_{0}

double karne par half-life double ho jaati hai

False. First-order

t_{1/2} = \frac{ln 2}{k}

mein

[A]_{0}

hai hi nahi, toh start double karne se kuch nahi badalta.

Zero-order reaction ke liye, zyada reactant se shuru karna matlab lambi half-life

True.

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

directly

[A]_{0}

ke proportional hai; rate

k

par fixed hai, toh bade pile ka aadha karna proportionally zyada time leta hai.

Second-order reaction ke liye, doosri half-life pehli se lambi hoti hai

True. Har interval ki half-life

\frac{1}{k [ A ] _{start}}

hai, jahan

[A]_{start}

us interval ki shuruwat ka concentration hai; kyunki yeh har round mein aadha ho jaata hai, agla

t_{1/2}

double ho jaata hai.

Radioactive decay ki constant half-life prove karti hai ki yeh first-order hai

True. Amount se independent constant

t_{1/2}

first-order kinetics ka unique fingerprint hai — dekho Radioactive decay — first-order kinetics.

Teen successive first-order half-lives ke baad

[A]_{0} /6

bachta hai

False. Teen baar halving karne par

(\frac{1}{2})^{3} = \frac{1}{8}

milta hai, yani

[A]_{0} /8

, na ki

[A]_{0} /6

. Half-lives multiply hoti hain, denominator mein add nahi hoti.

Zero-order reaction ke liye,

[A]

finite time ke baad exactly zero ho jaata hai aur ruk jaata hai

True. Straight-line drop

[A]_{t} = [A]_{0} - k t

zero par

t = [A]_{0} / k

mein pahunchti hai aur reaction abruptly ruk jaata hai — first-order ke unlike, jo zero ko sirf approach karta rehta hai hamesha.

First-order reaction ke liye,

[A]

do half-lives ke baad zero ho jaata hai

False. Do first-order half-lives ke baad

[A]_{0} /4

bachta hai, aur halvings ki koi finite number kabhi zero tak nahi pahunchti — exponential decay sirf ise approach karti hai.

Bada

k

matlab hamesha choti half-life

Har order ke liye True, kyunki

k

teeno half-life expressions ke denominator mein hai (

\frac{[ A ] _{0}}{2 k}

\frac{l n 2}{k}

\frac{1}{k [ A ] _{0}}

). Faster constant, halving mein kam time.

Spot the error

"Second-order reaction ke liye

t_{1/2} = \frac{0.693}{k}

Galat — yeh first-order formula hai. Second-order hai

t_{1/2} = \frac{1}{k [ A ] _{start}}

, jo interval ki shuruwat ke concentration par bhi depend karta hai.

k = 0.5 s^{- 1}

, second-order, toh

t_{1/2} = \frac{1}{k [ A ] _{start}}

Units se pakad aata hai: second-order

k

ka unit

M^{- 1} s^{- 1}

hona chahiye. Akela

s^{- 1}

matlab kisi ne tumhe first-order constant diya hai, toh yeh formula apply nahi hota.

"First-order

t_{1/2}

derive karte waqt:

[A]_{t} = [A]_{0} /2

rakhne par

ln \frac{1}{2} = - k t_{1/2}

milta hai, toh

t_{1/2} = - \frac{ln 2}{k}

— ek negative time."

Sign slip yeh hai ki

ln \frac{1}{2} = - ln 2

drop ho gaya. Do minus signs cancel hote hain:

- ln 2 = - k t_{1/2} \Rightarrow t_{1/2} = + \frac{ln 2}{k}

. Time positive aata hai, jaisa hona chahiye.

"Kyunki first-order ki half-life constant hai, concentration har half-life mein constant amount se girti hai."

Fraction aur amount mein confusion hai. Har half-life ek constant fraction (jo bacha hai uska aadha) remove karta hai, toh removed amount shrink hoti hai har round mein:

[A]_{0} \to \frac{[ A ] _{0}}{2} \to \frac{[ A ] _{0}}{4}

mein pehle

\frac{[ A ] _{0}}{2}

phir

\frac{[ A ] _{0}}{4}

remove hota hai.

"Zero-order

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

, toh zyada reactant matlab faster halving."

Ulta hai. Zyada

[A]_{0}

t_{1/2}

ko bada (lamba) banata hai, kyunki fixed rate ko proportionally bada amount chew karna padta hai.

"Maine do half-lives measure kiye, dono 40 s, toh main ise second-order kahunga."

Successive half-lives constant hona matlab first-order hai, second nahi. Second-order ke liye doosri half-life pehli se double hoti.

"Second-order ke liye,

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

har successive half-life ke liye kaam karta hai."

Sirf pehli ke liye. Har baad ke interval mein

[A]_{start}

use karna padega — us interval ki shuruwat ka concentration — original

[A]_{0}

nahi; doosri half-life

[A]_{0} /2

use karti hai, double time deti hai.

t_{1/2} = \frac{1}{k [ A ] _{start}}

, toh agar

[A]_{start} = 0

ho toh half-life infinite hai — reaction kabhi hoti hi nahi."

Agar

[A]_{start} = 0

hai toh react karne ko kuch hai hi nahi; "zero ka aadha" instantly hota hai aur formula ka division-by-zero sirf signal hai ki question meaningless hai, time infinite nahi.

Why questions

First-order half-life concentration se independent kyun hai, jabki baaki orders ki nahi?

Kyunki first-order rate

[A]

ke proportional hai: jaise concentration girti hai, rate bhi saath-saath girta hai, toh fraction-halving time kabhi nahi badlta. Zero-order ki rate fixed hai (shrinking amount se mismatched); second-order ki rate

[A]

se bhi faster girti hai.

ln 2

first-order half-life mein kyun aata hai par kahin aur nahi?

Kyunki first-order decay exponential hai, aur exponential ko undo karne ke liye logarithm chahiye;

[A]_{t} = [A]_{0} /2

rakhne par

ln \frac{1}{2} = - ln 2

aata hai, aur woh akela minus sign flip karke positive

t_{1/2} = ln 2/ k

deta hai. Derivation dekho Integrated rate laws — derivation and graphical analysis mein.

Second-order half-lives reaction ke sath-sath kyun double hoti rehti hain?

Second-order reaction ke liye do molecules ko milna padta hai; jaise

[A]

girta hai, encounters roughly square ke saath rare hote jaate hain, toh har successive halving mein double time lagta hai.

Reaction order dhundhne ke liye successive half-lives measure karna achha tarika kyun hai?

Trend (constant, decreasing, ya increasing) ek aisa fingerprint hai jo koi aur order share nahi karta, toh do stopwatch readings curve-fitting ke bina orders distinguish kar deti hain — dekho Reaction order determination from experimental data.

Pharmacologists ko first-order half-life itni kyun pasand hai?

Kyunki constant half-life se woh kisi bhi time par drug concentration predict kar sakte hain ek number se, dose se independent — yeh Pharmacokinetics — drug elimination ki backbone hai.

Zero- ya second-order reactions ke liye hum "number of half-lives" shortcuts kyun use nahi kar sakte?

Kyunki un orders ke liye half-life har interval mein change hoti hai, toh "3 half-lives =

\frac{1}{8}

" tabhi kaam karta hai jab har half-life identical ho, yani sirf first-order mein.

Edge cases

Zero-order reaction ki half-life completion ke paas aate-aate kya hoti hai?

Yeh baaki bache

[A]

ke saath shrink hoti rehti hai, aur poora reaction us pal dead ruk jaata hai jis pal reactant zero ho jaata hai (rate

k

par rehta hai bilkul us wall tak).

Kya first-order reaction kabhi sach mein

[A] = 0

tak pahunchti hai?

Nahi — exponential decay hamesha halve karti rehti hai aur sirf zero approach karti hai. Practice mein hum ~7 half-lives ke baad "done" bolte hain (

< 1%

bacha), par mathematically yeh kabhi zero nahi chhuti.

Second-order reaction ke liye, jab

[A]_{start}

bahut bada ho toh half-life kya hoti hai?

Bahut choti:

t_{1/2} = \frac{1}{k [ A ] _{start}}

[A]_{start}

badhne par shrink hoti hai, kyunki crowded molecules hamesha collide karte hain. Opposite limit (tiny

[A]_{start}

) enormous half-life deta hai.

Agar tum observe karo ki pehli half-life doosre interval par double ho rahi hai, par teesre par triple, kya yeh abhi bhi simple second-order hai?

Nahi — clean second-order har baar double hoti hai (ek fixed geometric pattern). Shifting pattern ek mixed ya higher-order mechanism ka signal hai, textbook second-order ka nahi.

Kaun sa order aisa half-life deta hai jo har half-life ki bajaye har unit time mein constant amount removed ho?

Zero-order: reactant steady

k

mol per litre per second par disappear hota hai, ek straight-line drop, toh "amount removed per second" fixed hai chahe half-life khud shrink ho rahi ho.

Kya do reactions ka same

k

ho sakta hai par wildly different half-lives?

Haan — agar woh different orders ke hain (ya orders 0 aur 2 ke liye different

[A]_{0}

), kyunki half-life formula har order ke liye

k

[A]_{0}

se alag tarike se mix karta hai.

Recall Ek-line self-test

Kaun sa ek word har order ki half-life trend describe karta hai jaise reaction chalta hai? ::: Zero-order = decreasing, first-order = constant, second-order = increasing.