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

2.8.4 · Chemistry › Chemical Kinetics

Why Half-Life Matters

Half-life sirf ek arbitrary metric nahi hai — yeh reaction speeds compare karne aur concentration decay predict karne ka sabse intuitive tarika hai:

Medicine mein: drug clearance times
Nuclear chemistry mein: radioactive decay
Environmental science mein: pollutant persistence
Manufacturing mein: shelf-life predictions

Deriving Half-Life for Each Order

Zero-Order Reactions (Rate = k)

Derivation of t₁/₂:

Integrated rate law se shuru karke, half-life par t = t₁/₂ par, hamare paas [A]_t = [A]₀/2 hai:

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

Yeh step kyun? Hum final concentration ko initial value ki exactly half set kar rahe hain — yahi half-life ki definition hai.

Rearrange karne par: $k t_{1/2} = [A]_{0} - \frac{[ A ] _{0}}{2} = \frac{[ A ] _{0}}{2}$

Physical meaning: Zero-order kinetics tab hoti hai jab reaction rate saturate ho jaati hai (jaise enzyme fully saturated ho, ya surface catalysis mein full coverage ho). Rate constant rehti hai chahe kitna bhi reactant bacha ho, isliye ek badi amount ki half consume karne mein proportionally zyada time lagta hai.

First-Order Reactions (Rate = k[A])

Derivation of t₁/₂:

t = t₁/₂ par, [A]_t = [A]₀/2:

$ln (\frac{[ A ] _{0}}{2}) = ln [A]_{0} - k t_{1/2}$

Yeh step kyun? Half-life condition ko integrated rate law mein substitute kar rahe hain.

Logarithm expand karne par: $ln [A]_{0} - ln 2 = ln [A]_{0} - k t_{1/2}$

Yeh step kyun? Logarithm property ln(a/b) = ln(a) - ln(b) use kar rahe hain.

ln[A]₀ terms cancel ho jaate hain: $- ln 2 = - k t_{1/2}$

Physical meaning: First-order reactions mein rate hamesha current concentration ke proportional hoti hai. Chahe aap 100 M se start karo ya 0.01 M se, half karne mein utna hi time lagta hai kyunki rate amount ke saath proportionally scale karti hai.

Second-Order Reactions (Rate = k[A]²)

Derivation of t₁/₂:

t = t₁/₂ par, [A]_t = [A]₀/2:

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

Yeh step kyun? Integrated rate law mein half-life condition apply kar rahe hain.

Left side simplify karne par: $\frac{2}{[ A ] _{0}} = \frac{1}{[ A ] _{0}} + k t_{1/2}$

Yeh step kyun? Algebra se 1/(a/2) = 2/a hota hai.

t₁/₂ ke liye solve karne par: $k t_{1/2} = \frac{2}{[ A ] _{0}} - \frac{1}{[ A ] _{0}} = \frac{1}{[ A ] _{0}}$

Physical meaning: Second-order reactions mein do molecules ka collide karna zaroori hota hai. Zyada concentration par collisions zyada frequent hote hain, isliye reaction faster proceed karti hai. Jaise-jaise concentration girti hai, collisions rare ho jaate hain, aur remaining amount ko half karne mein increasingly zyada time lagta hai.

Worked example Second-Order Half-Life Calculation

Ek second-order reaction mein k = 0.54 M⁻¹s aur [A]₀ = 0.20 M hai.

(a)* Pehli half-life calculate karo. (b)* Pehli half-life ke baad concentration kya hogi? (c)* Doosri half-life calculate karo.

Solution: (a)* Pehli half-life: $t_{1/2, 1} = \frac{1}{k [ A ] _{0}} = \frac{1}{( 0.54 M ^{- 1} s ^{- 1} ) ( 0.20 M )} = \frac{1}{0.108 s ^{- 1}} = 9.26 s$

Yeh step kyun? Second-order half-life formula mein direct substitution.

(b)* Pehli half-life ke baad: $[A] = \frac{[ A ] _{0}}{2} = \frac{0.20 M}{2} = 0.10 M$

Yeh step kyun? Half-life ki definition ke anusaar.

(c) Doosri half-life (0.10 M se start karke): $t_{1/2, 2} = \frac{1}{k [ A ]} = \frac{1}{( 0.54 M ^{- 1} s ^{- 1} ) ( 0.10 M )} = \frac{1}{0.054 s ^{- 1}} = 18.5 s$

Yeh step kyun? Doosri half-life ke liye "nayi" initial concentration 0.10 M hai, 0.20 M nahi.

Key insight: Doosri half-life (18.5 s) pehli (9.26 s) se exactly double hai kyunki concentration half ho gayi. Second-order reactions mein har successive half-life double hoti hai.

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

Comparison Table

| Order | Integrated Rate Law | Half-Life Formula | Dependence on [A]₀ | Half-Life Trend | |-------|---------------------|----------------|---------------| | 0 | [A]ₜ = [A]₀ - kt | $\frac{[ A ] _{0}}{2 k}$ | Proportional | Decreases | | 1 | ln[A]ₜ = ln[A]₀ - kt | $\frac{0.693}{k}$ | Independent | Constant | | 2 | 1/[A]ₜ = 1/[A]₀ + kt | $\frac{1}{k [ A ] _{0}}$ | Inversely proportional | Increases |

Common mistake Common Error: Yeh Assume Karna Ki Saari Half-Lives Constant Hoti Hain

Galat soch: "Agar ek reaction ki half-life 10 seconds hai, toh concentration half hone mein hamesha 10 seconds lagenge." Yeh sahi kyun lagta hai: First-order reactions (jaise radioactive decay) ki half-lives constant hoti hain, aur introductory courses mein inhi ke examples sabse zyada discuss hote hain.

Reality: Sirf first-order reactions ki half-lives constant hoti hain. Zero-order reactions mein, jaise [A] decrease hota hai, t₁/₂ bhi decrease karta hai. Second-order reactions mein, jaise [A] decrease hota hai, t₁/₂ increase karta hai.

Fix: Half-life behavior ke baare mein predictions karne se pehle hamesha reaction order check karo. Agar successive half-lives measure karo aur woh change ho rahi hain, toh aap first-order kinetics deal nahi kar rahe.

Worked example Half-Life Se Reaction Order Ka Experimental Determination

Aap ek reaction ko [A]₀ se [A]₀/2 tak, phir [A]₀/2 se [A]₀/4 tak jaane mein lagaa time measure karte ho:

Experiment 1: [A]₀ = 0.80 M, pehli t₁/₂ = 50 s, doosri t₁/₂ = 50 s
Experiment 2: [A]₀ = 0.40 M, pehli t₁/₂ = 50 s Reaction order kya hai? Solution: Half-life concentration se independent constant hai. Yeh first-order kinetics ka diagnostic hai.

Yeh step kyun? Constant t₁/₂ first-order reactions ka unique signature hai.

Hum k calculate kar sakte hain: $k = \frac{0.693}{t _{1/2}} = \frac{0.693}{50 s} = 0.0139 s^{- 1}$

Check: Agar zero-order hoti, toh [A]₀ ko 0.40 se 0.80 M tak double karne par t₁/₂ bhi double hoti (25 s se 50 s). Agar second-order hoti, toh [A]₀ double karne par t₁/₂ half hoti (100 s se 50 s). Dono hi hamare observation se match nahi karte.

The "Number of Half-Lives" Concept

Sirf first-order reactions ke liye, hum n half-lives ke baad remaining fraction predict kar sakte hain:

1 half-life ke baad: 1/2 bachta hai 2 half-lives ke baad: 1/4 bachta hai 3 half-lives ke baad: 1/8 bachta hai n half-lives ke baad: (1/2)ⁿ bachta hai

Total time: t = n × t₁/₂

Worked example Drug Clearance (First-Order)

Ek drug ki first-order elimination half-life 4 hours hai. Agar ek patient ko 200 mg dose milti hai, toh 12 hours baad kitna bachega?

Solution: Half-lives ki number: n = 12 h / 4 h = 3

Yeh step kyun? Count kar rahe hain ki elapsed time mein kitne half-life periods fit hote hain.

Remaining fraction: (1/2)³ = 1/8

Remaining amount: 200 mg × 1/8 = 25 mg

Yeh step kyun? Har half-life amount ko half kar deti hai, isliye teen half-lives mein 1/2 se teen baar multiply hota hai.

Alternative approach using the integrated rate law: $[A]_{t} = [A]_{0} e^{- k t}$

Pehle k find karo: $k = \frac{0.693}{t _{1/2}} = \frac{0.693}{4 h} = 0.173 h^{- 1}$ Phir: $[A]_{12} = 200 mg \times e^{- (0.173) (12)} = 200 \times e^{- 2.08} = 200 \times 0.125 = 25 mg$

Dono methods same answer dete hain, jo hamare approach ko validate karta hai.

Recall Feynman Explanation (12-saal ke bachche ko samjhao)

Socho tum ek badi bowl mein candy kha rahe ho, aur hum jaanna chahte hain ki aadhi khaane mein kitna time lagta hai.

Zero-order (constant rate): Tum exactly 10 candies har minute khaate ho, chahe kitni bhi baachi hon. Agar 100 candies se start karo, 50 khaane mein 5 minute lagte hain. Agar 200 se start karo, 100 khaane mein 10 minute. Zyada candies = aadhi khaane mein zyada time. "Half-time" depend karta hai kitni se start ki.

First-order (proportional rate): Tum hamesha jo bacha hai uska 10% khaate ho. Agar 100 hain, toh 10 khaate ho. Agle minute 90 hain, toh 9 khaate ho. Agle minute 81, toh 8.1. Cool baat yeh hai? Aadha khaane mein HAMESHA utna hi time lagta hai — 100 se 50 jaane mein lagbhag 7 minute, 50 se 25 jaane mein bhi 7 minute, aur 25 se 12.5 jaane mein bhi 7 minute. "Half-time" constant hai.

Second-order (squared rate): Yeh depend karta hai candies ke ek doosre se milne par (jaise tumhe do candies ka ek doosre se takrana chaahiye taki tum unhe notice karo aur khao). Jab bahut saari candies hoti hain, woh baar baar takraati hain, isliye tum fast khaate ho. Jaise khatam hoti jaati hain, takraav rare ho jaata hai, isliye khaana slow ho jaata hai. "Half-time" jaise-jaise khatam hoti hain, lambi hoti jaati hai.

Connections

Integrated rate laws — derivation and graphical analysis
Rate constant k — units and temperature dependence
Reaction order determination from experimental data
Radioactive decay — first-order kinetics
Pharmacokinetics — drug elimination
Method of initial rates vs. integrated rate laws

Flashcards

#flashcards/chemistry

Chemical kinetics mein half-life (t₁/₂) ki definition kya hai? :: Woh time jisme kisi reactant ki concentration apni initial value ki exactly half ho jaaye.

Zero-order reaction ke liye half-life formula kya hai?

t₁/₂ = [A]₀/(2k), jo dikhata hai ki half-life initial concentration ke directly proportional hai.

First-order reaction ke liye half-life formula kya hai?

t₁/₂ = 0.693/k = ln(2)/k, jo dikhata hai ki half-life initial concentration se independent hai.

Second-order reaction ke liye half-life formula kya hai?

t₁/₂ = 1/(k[A]₀), jo dikhata hai ki half-life initial concentration ke inversely proportional hai.

Zero-order reaction mein jaise reaction proceed karti hai, half-life kaise change hoti hai?

Half-life decrease hoti hai jaise reaction proceed karti hai (jaise [A] decrease hota hai).

First-order reaction mein jaise reaction proceed karti hai, half-life kaise change hoti hai?

Half-life poori reaction mein constant rehti hai.

Second-order reaction mein jaise reaction proceed karti hai, half-life kaise change hoti hai?

Half-life increase hoti hai jaise reaction proceed karti hai (jaise [A] decrease hota hai).

Second-order reaction ke liye rate constant k ki units kya hoti hain?

M⁻¹s⁻¹ ya L·mol⁻¹·s⁻¹

Agar ek first-order reaction mein k = 0.0462 min⁻¹ hai, toh uski half-life kya hai?

t₁/₂ = 0.693/0.0462 = 15.0 min

Ek second-order reaction [A]₀ = 0.50 M aur k = 0.20 M⁻¹s⁻¹ se start hoti hai. Pehli half-life kya hai?

t₁/₂ = 1/(0.20 × 0.50) = 10s

Pichle card ki usi second-order reaction mein doosri half-life kya hai?

20 s (pehli half-life se double, kyunki [A] ab 0.25 M hai)

First-order reaction ki 3 half-lives ke baad original reactant ka kitna fraction bachta hai?

(1/2)³ = 1/8 ya 12.5%

Agar experimentally measure karo ki t₁/₂ initial concentration se independent constant hai, toh reaction order kya hai?

First-order

Agar experimentally measure karo ki initial concentration half karne par t₁/₂ double ho jaati hai, toh reaction order kya hai?

Second-order

First-order half-life calculations mein use hone wali ln(2) ki value kya hai? :: 0.693