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

2.8.4 · D4 · Chemistry › Chemical Kinetics › Integrated rate laws — half-life t₁ - ₂ for each order

Definition Woh symbols jo hum har jagah use karenge

$t$ — time, reaction shuru hone ke moment se measure kiya jaata hai ( $t = 0$ ), seconds ( $s$ ), minutes, hours, ya days mein. Is page ka har formula track karta hai ki concentration kaise badalta hai jaise $t$ badhta hai.
$[A]_{0}$ — reactant A ki starting concentration (hum kitne se shuru karte hain, yaani $t = 0$ par iski value), mol per litre mein, $M$ likha jaata hai.
$[A]_{t}$ — time $t$ ke baad bacha hua concentration (subscript $t$ sirf yeh kehta hai "time $t$ par measure kiya gaya").
$k$ — rate constant, woh number jo speed set karta hai. Iske units order par depend karte hain (dekho Rate constant k — units and temperature dependence).
$t_{1/2}$ — half-life, $t$ ki woh special value jis par concentration exactly aadha ho jaata hai jo pehle tha.
$ln$ — natural logarithm, woh tool jo exponential $e^{- k t}$ ko "undo" karta hai. Hum ise sirf first order ke liye use karte hain kyunki sirf first order hi exponential decay deta hai.

Level 1 — Recognition

Goal: kya tum order ko uski fingerprint se naam de sakte ho, aur sahi formula padh sakte ho?

Exercise 1.1

Ek reaction ko monitor kiya jaata hai aur uski successive half-lives measure ki jaati hain: 40 s, 40 s, 40 s, 40 s. Reaction order kya hai?

Recall Solution

Ek half-life jo reaction chalte waqt kabhi nahi badlti woh first order ki unique signature hai.

Zero order kyun nahi? Zero-order half-lives time ke saath shrink hoti hain (har ek current concentration par depend karti hai, jo gir rahi hai).
Second order kyun nahi? Second-order half-lives time ke saath badhti hain (har ek current concentration ke inversely proportional hoti hai).
Sirf first order $t_{1/2} = \frac{0.693}{k}$ ko fixed rakhta hai — isme koi concentration nahi hai, toh badlne ke liye kuch nahi.

Answer: first order.

Exercise 1.2

Zero-order reaction ke liye, half-life formula likho aur ek sentence mein batao ki agar tum bade $[A]_{0}$ se shuru karo toh $t_{1/2}$ ka kya hoga.

Recall Solution

$t_{1/2} = \frac{[ A ] _{0}}{2 k}$ Kyunki $[A]_{0}$ upar baitha hai, badi starting concentration ek badi half-life deti hai — woh directly proportional hain. Physically: ek zero-order reaction reactant ko fixed rate $k$ se consume karti hai, isliye do guna zyaada reactant ko aadha karne mein do guna zyaada waqt lagta hai.

Exercise 1.3

Ek first-order reaction ka $k = 0.0250 s^{- 1}$ hai. Sirf $k$ ke units se, order confirm karo, phir $t_{1/2}$ nikalo.

Recall Solution

$k$ ke units $s^{- 1}$ hain (sirf inverse time). Yeh exactly first-order unit hai — toh yeh first order hai. $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0250 s ^{- 1}} = 27.7 s$

Level 2 — Application

Goal: numbers ko sahi se plug in karo, units ko honest rakhte hue.

Exercise 2.1

Ek first-order decomposition ka $k = 1.15 \times 1 0^{- 3} s^{- 1}$ hai. Iska half-life seconds aur minutes mein nikalo.

Recall Solution

$t_{1/2} = \frac{0.693}{1.15 \times 1 0 ^{- 3} s ^{- 1}} = 602.6 s$ Minutes mein: $602.6 s \div 60 = 10.04 min$ .

Exercise 2.2

Ek zero-order surface reaction ka $k = 0.020 M s^{- 1}$ aur $[A]_{0} = 0.50 M$ hai. $t_{1/2}$ nikalo, aur check karo ki units cancel hokar seconds mein aate hain.

Recall Solution

$t_{1/2} = \frac{[ A ] _{0}}{2 k} = \frac{0.50 M}{2 ( 0.020 M s ^{- 1} )} = \frac{0.50 M}{0.040 M s ^{- 1}} = 12.5 s$ Unit check: $\frac{M}{M s ^{- 1}} = \frac{1}{s ^{- 1}} = s$ . ✓ M cancel ho jaate hain, seconds bachte hain.

Exercise 2.3

Ek second-order reaction ka $k = 0.35 M^{- 1} s^{- 1}$ aur $[A]_{0} = 0.80 M$ hai. Pehla half-life nikalo.

Recall Solution

$t_{1/2} = \frac{1}{k [ A ] _{0}} = \frac{1}{( 0.35 M ^{- 1} s ^{- 1} ) ( 0.80 M )} = \frac{1}{0.28 s ^{- 1}} = 3.57 s$ Unit check: $M^{- 1} s^{- 1} \times M = s^{- 1}$ ; uska ulta $s$ hai. ✓

Level 3 — Analysis

Goal: successive half-lives ka behaviour samjho, aur unhe backwards padho.

Exercise 3.1

Ek second-order reaction $[A]_{0} = 0.40 M$ se shuru hoti hai aur $k = 0.50 M^{- 1} s^{- 1}$ hai. (a) Pehla half-life? (b) Uske baad concentration? (c) Doosra half-life? (d) (doosra half-life)/(pehla half-life) ka ratio kya hai?

Recall Solution

(a) $t_{1/2, 1} = \frac{1}{k [ A ] _{0}} = \frac{1}{( 0.50 ) ( 0.40 )} = \frac{1}{0.20} = 5.0 s$ (b) Ek half-life ke baad, $[A] = \frac{0.40}{2} = 0.20 M$ . (c) Ab $0.20 M$ ko nayi starting concentration maano: $t_{1/2, 2} = \frac{1}{k [ A ]} = \frac{1}{( 0.50 ) ( 0.20 )} = \frac{1}{0.10} = 10.0 s$ (d) Ratio $= 10.0/5.0 = 2$ . Har successive second-order half-life double hoti hai, kyunki $\frac{1}{k [ A ]}$ ke andar concentration aadhi ho gayi.

Neeche wali figure kya dikhati hai (alt text: ek concentration-vs-time decay curve black mein, time axis par do red double-headed horizontal arrows jo pehle half-life stretch ko 0 se 5 s tak aur doosre ko 5 se 15 s tak mark karte hain; dashed guide lines concentration levels 0.40, 0.20, 0.10 M se curve tak girte hain): black curve $[A]$ hai jo time $t$ ke saath girta hai. Do red double-headed arrows time axis par do half-life stretches ko mark karte hain. Notice karo ki doosra red arrow ( $0.20 \to 0.10 M$ wala leg) pehle se do guna chauda hai ( $0.40 \to 0.20 M$ wala leg): jaise concentration patli hoti hai, molecules kam often collide karte hain, isliye har successive halving mein zyaada waqt lagta hai. Yeh visually us ratio of 2 ko confirm karta hai jo tumne abhi compute kiya.

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

Exercise 3.2

Ek first-order reaction ka $t_{1/2} = 25 s$ hai. $1.00 M$ se shuru karke, 100 s ke baad kya concentration bachti hai? Calculator ke bina half-lives count karke jawab do.

Recall Solution

$100 s \div 25 s per half-life = 4$ half-lives. Har half-life concentration ko $\frac{1}{2}$ se multiply karti hai: $1.00 \to 0.50 \to 0.25 \to 0.125 \to 0.0625 M$ 4 half-lives ke baad, $[A] = 1.00 \times (\frac{1}{2})^{4} = \frac{1}{16} = 0.0625 M$ .

Exercise 3.3

Ek zero-order reaction diya gaya hai jisme $[A]_{0} = 1.0 M$ hai, pehla half-life 20 s hai. Doosra half-life nikalo ( $0.5 M$ se $0.25 M$ tak jaane ka waqt).

Recall Solution

Pehle half-life se $k$ nikalo: $t_{1/2} = \frac{[ A ] _{0}}{2 k} \Rightarrow k = \frac{[ A ] _{0}}{2 t _{1/2}} = \frac{1.0}{2 ( 20 )} = 0.025 M s^{- 1}$ . Doosra half-life nayi starting concentration $0.5 M$ use karta hai: $t_{1/2, 2} = \frac{[ A ]}{2 k} = \frac{0.5}{2 ( 0.025 )} = \frac{0.5}{0.05} = 10 s$ Zero-order doosra half-life pehle ka aadha hai — half-lives shrink hoti hain. Isko second order se contrast karo, jahan woh badhti hain.

Level 4 — Synthesis

Goal: order-diagnosis ko calculation ke saath combine karo, aur tools cross karo.

Exercise 4.1

Usi reaction par do experiments:

Exp 1: $[A]_{0} = 1.00 M$ , pehla $t_{1/2} = 100 s$ .
Exp 2: $[A]_{0} = 2.00 M$ , pehla $t_{1/2} = 50 s$ . Order determine karo aur phir rate constant $k$ nikalo.

Recall Solution

Diagnose: $[A]_{0}$ double karna (×2) ne $t_{1/2}$ half kar diya (×½). Yeh inverse relationship $t_{1/2} \propto 1/ [A]_{0}$ second order ki signature hai.

(Zero order ne $t_{1/2}$ double kiya hota; first order ise unchanged chhod deta.) $k$ nikalo kisi bhi experiment se. Exp 1 se: $t_{1/2} = \frac{1}{k [ A ] _{0}} \Rightarrow k = \frac{1}{t _{1/2} [ A ] _{0}} = \frac{1}{( 100 ) ( 1.00 )} = 0.0100 M^{- 1} s^{- 1}$ Exp 2 se check karo: $k = \frac{1}{( 50 ) ( 2.00 )} = 0.0100 M^{- 1} s^{- 1}$ . ✓ Consistent. Yeh order nikalne ka half-life method hai.

Exercise 4.2

Ek radioactive isotope first-order kinetics se decay karta hai jisme half-life 8.0 days hai. Ek patient ko dose diya jaata hai; 24 days ke baad kya fraction bachta hai, aur $k$ $day^{- 1}$ mein kya hai?

Recall Solution

$24 days \div 8.0 = 3$ half-lives, toh bacha hua fraction $= (\frac{1}{2})^{3} = \frac{1}{8} = 0.125$ . Rate constant: $k = \frac{0.693}{t _{1/2}} = \frac{0.693}{8.0 day} = 0.0866 day^{- 1}$ Yahi reasoning drug elimination schedules ke peeche hai.

Exercise 4.3

Ek first-order reaction 60 minutes mein 75% complete hai. $k$ aur half-life nikalo. (Hint: 75% complete ka matlab $[A]_{t} = 0.25 [A]_{0}$ .)

Recall Solution

$75%$ chala gaya toh $25% = \frac{1}{4}$ bachta hai, yaani do half-lives (har halving: $1 \to \frac{1}{2} \to \frac{1}{4}$ ). Toh $2 t_{1/2} = 60 min \Rightarrow t_{1/2} = 30 min$ . $k = \frac{0.693}{30 min} = 0.0231 min^{- 1}$ Integrated law se cross-check $ln [A]_{t} = ln [A]_{0} - k t$ : $k = \frac{1}{t} ln \frac{[ A ] _{0}}{[ A ] _{t}} = \frac{1}{60} ln 4 = \frac{1.386}{60} = 0.0231 min^{- 1} ✓$

Level 5 — Mastery

Goal: full mixed reasoning, degenerate cases, aur unit thinking.

Exercise 5.1

Ek enzyme reaction zero order mein chalti hai jab substrate zyaada hai, $k = 0.10 M s^{- 1}$ aur $[A]_{0} = 1.0 M$ ke saath. Yeh zero order tab tak rehti hai jab tak $[A]$ gir kar $0.20 M$ na ho jaaye, jiske neeche yeh first order mein switch ho jaati hai nayi rate constant $k^{'} = 0.50 s^{- 1}$ ke saath (prime symbol $^{'}$ sirf ise second-phase rate constant ke roop mein label karta hai, zero-order $k$ se alag; note karo iske units $s^{- 1}$ confirm karte hain ki yeh first order hai). $0.10 M$ tak pahunchne mein total kitna waqt lagega?

Recall Solution

Phase 1 (zero order, $1.0 \to 0.20 M$ ). Hume zero-order integrated law chahiye, toh pehle ise build karte hain.

Zero order ka matlab hai rate constant hai: concentration har second $k$ units of M kho deta hai. Kyun: zero order ki definition se rate $[A]$ par depend nahi karti, toh yeh sirf fixed number $k$ hai.
Constant loss rate ka matlab hai concentration straight line mein girta hai: time $t$ ke baad humne $k \times t$ remove kar liya hai, toh $[A]_{t} = [A]_{0} - k t$ (Yeh wahi law hai jo integrated rate law note mein quoted hai.)
$[A]_{0}$ se target $[A]_{t}$ tak girne ke waqt ke liye rearrange karo: subtract karo, phir $k$ se divide karo: $t = \frac{[ A ] _{0} - [ A ] _{t}}{k} = \frac{1.0 - 0.20}{0.10} = 8.0 s$ Phase 2 (first order, $0.20 \to 0.10 M$ ): yeh exactly ek halving hai ( $0.20 \to 0.10$ ), toh hum $k^{'}$ ke saath first-order half-life use karte hain: $t_{2} = t_{1/2} = \frac{0.693}{k ^{'}} = \frac{0.693}{0.50} = 1.386 s$ Total: $t = 8.0 + 1.386 = 9.39 s$ . Yeh mixed behaviour isliye hai ki order ek hi reaction ke course mein change hota dikh sakta hai.

Exercise 5.2

Ek first-order reaction ke liye, prove karo ki initial value ka $\frac{1}{n}$ tak girne ka waqt $t = \frac{ln n}{k}$ hai, aur ise use karo woh waqt nikalne ke liye jab $k = 0.020 s^{- 1}$ par $10%$ bacha ho.

Recall Solution

Shuru karo first-order integrated law $ln [A]_{t} = ln [A]_{0} - k t$ se. Kyun: yeh first order ke liye concentration ko time se link karta hai. Set karo $[A]_{t} = [A]_{0} / n$ (target "one $n$ -th" condition): $ln \frac{[ A ] _{0}}{n} = ln [A]_{0} - k t$ Log-of-a-quotient rule apply karo $ln (\frac{a}{b}) = ln a - ln b$ (logarithm division ko subtraction mein convert karta hai — yahi exactly wajah hai ki logs yahan sahi tool hain). $a = [A]_{0}$ , $b = n$ ke saath: $ln [A]_{0} - ln n = ln [A]_{0} - k t$ Dono sides se $ln [A]_{0}$ subtract karo. Kyun: yeh dono sides par identically appear karta hai, toh ise remove karne se woh terms isolate ho jaate hain jo actually $t$ aur $n$ par depend karte hain: $- ln n = - k t$ Dono sides ko $- 1$ se multiply karo (minus signs clear karne ke liye), phir $k$ se divide karo ( $t$ isolate karne ke liye): $k t = ln n \Rightarrow t = \frac{l n n}{k}$ Sanity check: $n = 2$ set karne par $t = \frac{ln 2}{k}$ milta hai, jaana-pehchana half-life. ✓ Numerical application — $10%$ remaining tak pahunchne ka waqt. " $10%$ remaining" ka matlab $[A]_{t} = \frac{1}{10} [A]_{0}$ , toh $n = 10$ : $t = \frac{l n 10}{k} = \frac{l n 10}{0.020 s ^{- 1}} = \frac{2.3026}{0.020} = 115.1 s$

Exercise 5.3 (degenerate / limiting case)

Ek zero-order reaction jisme $[A]_{0} = 0.60 M$ aur $k = 0.15 M s^{- 1}$ hai. (a) Reactant completely kab khatam hota hai ( $[A] = 0$ )? (b) Explain karo ki second-order aur first-order laws $[A]$ ko exactly zero kyun nahi pahunchne dete, lekin zero order deta hai.

Recall Solution

(a) $[A]_{t} = [A]_{0} - k t = 0 \Rightarrow t = \frac{[ A ] _{0}}{k} = \frac{0.60}{0.15} = 4.0 s$ . Note karo yeh $2 t_{1/2}$ ke barabar hai kyunki $t_{1/2} = [A]_{0} /2 k = 2.0 s$ . (b) First order $[A]_{t} = [A]_{0} e^{- k t}$ deta hai; exponential $e^{- k t}$ 0 ki taraf shrink karta hai lekin finite $t$ ke liye kabhi exactly 0 nahi hota — toh first order approach karta hai, kabhi empty nahi pahunchta. Second order $1/ [A]_{t} = 1/ [A]_{0} + k t$ deta hai, jo infinity tak sirf $t \to \infty$ par jaata hai, toh $[A]_{t} \to 0$ sirf infinite-time limit mein hota hai. Zero order exception hai: iska straight-line decay $[A]_{0} - k t$ finite time $[A]_{0} / k$ par zero hit karta hai, phir model stop hona chahiye (concentration negative nahi ho sakti).

Neeche wali figure kya dikhati hai (alt text: teen decay curves ek hi point se shuru hoti hain; do black curves — ek solid first-order exponential aur ek dashed second-order curve — dono bend karke time axis ki taraf flatten hoti hain lekin kabhi touch nahi karti; ek red straight line zero order ke liye slope karte hue axis se milti hai ek marked red dot par jisme $t = [A]_{0} / k$ label hai): do black curves (first order, solid; second order, dashed) dono bend aur flatten hoti hain, time-axis ke paas aati hain lekin forever approach karti hain bina touch kiye — visual proof ki woh kabhi truly empty nahi hoti. Red straight line (zero order) ek finite point par axis mein crash karti hai, ek red dot se mark ki gayi $t = [A]_{0} / k$ par: woh sahi "finish line" hai jo sirf zero order ke paas hai.

Exercise 5.4

Ek drug first-order kinetics se eliminate hoti hai jisme $t_{1/2} = 4.0 h$ hai. Effective rehne ke liye, plasma level peak dose ke $25%$ se neeche nahi girna chahiye. Agla dose lene se pehle maximum safe dosing interval kya hai?

Recall Solution

$25% = \frac{1}{4} = (\frac{1}{2})^{2}$ , yaani exactly do half-lives (har halving: $100% \to 50% \to 25%$ ). $t = 2 \times t_{1/2} = 2 \times 4.0 = 8.0 h$ Toh drug ko zyaada se zyaada har 8.0 ghante par redose karna chahiye. Yahi everyday logic real dosing schedules ke peeche hai.

Common mistake L5 trap — "

$[A] \to 0$ " ko sabhi orders ke liye same samajhna Galat soch: "Har reaction eventually kaafi half-lives ke baad $[A] = 0$ pahunch jaati hai." Yeh sahi kyun lagta hai: Hamesha halving karna lagta hai ki eventually kuch nahi bachega. Haqeekat: $1 \to \frac{1}{2} \to \frac{1}{4} \to \dots$ halving karna finite steps mein literally 0 nahi pahunchta — isliye first aur second order sirf $t \to \infty$ par zero approach karte hain. Zero order genuinely alag hai: yeh straight line mein decrease karta hai aur finite time $[A]_{0} / k$ par zero hit karta hai, jiske baad rate law break down ho jaata hai (yeh impossible negative concentrations predict karta). Fix: Zero order ko woh ek case recognise karo jisme sahi finish line hai; first/second order ke liye, "gone" ka matlab "practically negligible" hai, mathematically zero nahi.

Recall Self-test — answers chhupa lo

Constant half-life ka matlab order ::: first order Zero-order half-life formula ::: $t_{1/2} = [A]_{0} /2 k$ Second-order half-life formula ::: $t_{1/2} = 1/ (k [A]_{0})$ Agar $[A]_{0}$ double karne se $t_{1/2}$ half ho jaaye, toh order hai ::: second order "75% complete" ka matlab bacha hua fraction hai ::: 0.25 (ek chauthai) First order ke liye $1/ n$ tak pahunchne ka waqt ::: $t = ln n / k$ Woh akela order jiska $[A]$ finite time par exactly zero pahunchta hai ::: zero order