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

2.8.4 · D3 · Chemistry › Chemical Kinetics › Integrated rate laws — half-life t₁ - ₂ for each order

Yeh page half-life topic ki practice lab hai. Parent note ne teen formulas banaye. Yahan hum har tarah ke question dhundhte hain jo un formulas mein wrap ho sakte hain — har order, har trick, har degenerate edge — aur har ek ko zor se solve karte hain.

Shuru karne se pehle, teen tools ko pin down karte hain jinhe hum baar baar use karenge, taaki koi bhi symbol unexplained na rahe.

Scenario matrix

Is table ko ek checklist ki tarah socho. Neeche har worked example us cell ke saath tagged hai jo woh fill karta hai. Jab saari cells tick ho jayein, tumne is topic ke har scenario ko dekh liya hoga.

Cell	Scenario class	Kya tricky banata hai isko
C1	Zero-order, forward calc	$t_{1/2}$ depends karta hai $[A]_{0}$ par — har step mein sahi starting value lo
C2	First-order, forward calc	yeh recognize karna padega ki $[A]_{0}$ irrelevant hai
C3	Second-order, successive half-lives	har half-life double hoti hai — nayi starting concentration track karo
C4	Zero-order degenerate edge	reaction expect se pehle khatam ho jaati hai — concentration zero hit kar sakti hai
C5	"Number of half-lives" (first-order)	fractional decay $(1/2)^{n}$ , seedha countdown nahi
C6	Half-life data se order diagnose karo	trend padho, single number nahi
C7	Units / sign sanity trap	galat-order $k$ units nonsense time dete hain
C8	Real-world word problem	drug elimination / decay prose mein dress ki hui
C9	Exam twist: $k$ ya $[A]_{0}$ ke liye back-solve karo	formula ko doosri taraf se rearrange karo

Worked examples

C1 — Zero-order forward calculation

Worked example Ek saturated surface reaction

Ek gas fully-covered catalyst surface par decompose hoti hai, isliye yeh zero-order chalta hai $k = 0.050 M s^{- 1}$ ke saath. $[A]_{0} = 2.0 M$ se shuru karke, pehla half-life nikalo, phir woh half-life jo us moment se shuru hoti hai jab $[A]$ $1.0 M$ tak pahunchta hai.

Forecast: zero-order ke liye, $t_{1/2}$ starting concentration ke proportional hota hai. Kyunki doosra stage aadhi concentration se shuru hota hai, andaza lagao ki doosra half-life aadha utna lamba hoga.

Pehla half-life. Yeh step kyun? $t_{1/2} = [A]_{0} / (2 k)$ use karo true start ke saath. $t_{1/2} = \frac{2.0}{2 ( 0.050 )} = \frac{2.0}{0.10} = 20 s$
Nayi starting concentration ab $[A] = 1.0 M$ hai. Yeh step kyun? Zero-order mein har stage ka half-life us stage ki starting value use karta hai. $t_{1/2}^{'} = \frac{1.0}{2 ( 0.050 )} = \frac{1.0}{0.10} = 10 s$ Verify: units $\frac{M}{M s ^{- 1}} = s$ ✓. Doosra half-life (10 s) pehle (20 s) ka exactly aadha hai — forecast se match karta hai. Zero-order half-lives ghatte hain.

C2 — First-order forward calculation

C3 — Second-order successive half-lives

Worked example Har half-life double hoti hai

Ek second-order reaction mein $k = 0.54 M^{- 1} s^{- 1}$ aur $[A]_{0} = 0.20 M$ hai. Pehle teen successive half-lives nikalo.

Forecast: second-order $t_{1/2}$ starting concentration ke inversely proportional hai. Har stage concentration ko aadha karta hai, isliye har half-life double honi chahiye.

Pehla half-life. Yeh step kyun? $t_{1/2} = 1/ (k [A]_{0})$ with $[A]_{0} = 0.20$ . $t_{1/2, 1} = \frac{1}{( 0.54 ) ( 0.20 )} = \frac{1}{0.108} = 9.26 s$
Nayi start $= 0.10 M$ . Yeh step kyun? Stage do aadhi concentration se shuru hota hai. $t_{1/2, 2} = \frac{1}{( 0.54 ) ( 0.10 )} = \frac{1}{0.054} = 18.5 s$
Nayi start $= 0.050 M$ . Yeh step kyun? Stage teen phir se aadhe se shuru hota hai. $t_{1/2, 3} = \frac{1}{( 0.54 ) ( 0.050 )} = \frac{1}{0.027} = 37.0 s$ Verify: $9.26 \to 18.5 \to 37.0$ — har ek pichle ka double hai ✓. Neeche ki figure teen orders ko side by side dikhati hai taaki tum trends dekh sako.

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

C4 — Zero-order degenerate edge: reaction khatam ho sakti hai

Worked example Agar khatam ho jaye toh?

C1 ki zero-order reaction ( $k = 0.050 M s^{- 1}$ , $[A]_{0} = 2.0 M$ ). Kitne time mein A bilkul khatam ho jaayega? Kya "teesra" ya "chautha" half-life compute karna valid hai?

Forecast: zero-order concentration straight line mein girta hai (constant slope), isliye yeh zero ko finite time par hit karta hai — first/second-order se alag jo sirf zero ke paas jaate hain. Half-lives ghatte rehte hain aur ruk jaane chahiye.

Saara A consume hone ka total time. Yeh step kyun? $[A]_{t} = 0$ set karo $[A]_{t} = [A]_{0} - k t$ mein. $0 = 2.0 - 0.050 t ⟹ t = \frac{2.0}{0.050} = 40 s$
Half-lives ke against check karo. Yeh step kyun? Successive zero-order half-lives hain $20, 10, 5, 2.5, \dots$ s; inki sum hai $20 + 10 + 5 + \dots \to 40$ s. $20 + 10 + 5 + 2.5 + \dots = 20 \cdot \frac{1}{1 - \frac{1}{2}} = 40 s$ Verify: shrinking half-lives ki infinite chain exactly finite completion time 40 s tak sum hoti hai ✓. Degenerate warning: jab $t > 40$ s ho jaata hai toh formula $[A]_{t} = [A]_{0} - k t$ negative concentration deta, jo physically impossible hai — concentration 0 par reh jaati hai. Figure mein mint straight line dekho: woh sach mein axis ko touch karti hai.

C5 — First-order ke liye "Number of half-lives"

C6 — Data se order diagnose karo

Worked example Trend padhna

Do experiments successive half-lives dete hain. Har ek mein order identify karo.

Set X: $[A]_{0} = 0.80 M$ : pehla $t_{1/2} = 50 s$ , doosra $t_{1/2} = 100 s$ .
Set Y: $[A]_{0} = 0.80 M$ : pehla $t_{1/2} = 50 s$ , doosra $t_{1/2} = 50 s$ .

Forecast: constant $\Rightarrow$ first-order; increasing $\Rightarrow$ second-order; decreasing $\Rightarrow$ zero-order. Toh X second-order lagta hai, Y first-order.

Set X: concentration aadhi hui toh half-life double ho gayi. Yeh step kyun? Doubling $t_{1/2} \propto 1/ [A]_{0}$ se match karta hai. $second-order ⟹ k = \frac{1}{t _{1/2} [ A ] _{0}} = \frac{1}{50 \times 0.80} = 0.025 M^{- 1} s^{- 1}$
Set Y: half-life nahi badi. Yeh step kyun? Constant $t_{1/2}$ first-order ki unique signature hai. $first-order ⟹ k = \frac{0.693}{50} = 0.01386 s^{- 1}$ Verify: X units $\frac{1}{s \cdot M} = M^{- 1} s^{- 1}$ ✓ (second-order); Y units $s^{- 1}$ ✓ (first-order). Method Reaction order determination from experimental data se match karta hai.

C7 — Units / sign sanity trap

C8 — Real-world word problem (drug elimination)

Worked example Blood se drug clear karna

Ek drug first-order kinetics se eliminate hoti hai jisme $t_{1/2} = 4.0 hours$ hai. Ek patient ko dose milti hai jisse blood concentration $80 mg L^{- 1}$ ho jaati hai. (a) Kab yeh $5 mg L^{- 1}$ se neeche jaayegi? (b) $k$ nikalo.

Forecast: $80 \to 40 \to 20 \to 10 \to 5$ mein chaar halvings hain, yaani $4 \times 4.0 = 16$ hours. Andaza 16 h.

Target tak pahunchne ke liye half-lives count karo. Yeh step kyun? Har halving ek $t_{1/2}$ hai; $80 \to 5$ mein $16 = 2^{4}$ se divide hota hai. $\frac{80}{5} = 16 = 2^{4} ⟹ n = 4 ⟹ t = 4 \times 4.0 = 16 h$
$k$ compute karo. Yeh step kyun? First-order $k$ aur $t_{1/2}$ ko directly link karta hai. $k = \frac{0.693}{4.0} = 0.173 h^{- 1}$ Verify: $[A] = 80 e^{- 0.173 \cdot 16} = 80 e^{- 2.772} = 80 \times 0.0625 = 5.0 mg L^{- 1}$ ✓. Yeh constant-half-life behaviour exactly isliye hai ki dosing schedules periodic hote hain (dekho Pharmacokinetics — drug elimination).

C9 — Exam twist: unknown ke liye back-solve karo

Har example ne kaun sa cell fill kiya?

Recall Quick self-test

Zero-order successive half-lives chote hote hain ya bade? ::: Chote (har stage mein aadhe ho jaate hain). Kaun sa order fraction $= (1/2)^{n}$ safely use karne deta hai? ::: Sirf first-order (constant $t_{1/2}$ ). Second-order $k$ ke units kya hote hain? ::: $M^{- 1} s^{- 1}$ . Agar do successive half-lives equal hain, toh order kya hai? ::: First-order. Zero-order reaction $[A] = 0$ reach kar sakti hai lekin first-order kyun nahi? ::: Straight-line decay axis ko finite time par hit karti hai; exponential sirf zero ke paas jaata hai.