Methods to determine order — initial rates, integrated method, half-life method: Worked examples

2.8.6 · D3 · Chemistry › Chemical Kinetics › Methods to determine order — initial rates, integrated metho

The scenario matrix

"Find the order" ka har problem inhi cells mein se kisi ek mein fit hota hai. Table ke neeche, examples E1–E9 mein se har ek announce karta hai ki woh kaun si cell cover karta hai. Milake yeh har row fill karte hain.

Cell	Method	Situation jo stressed hai	Covered by
C1	Initial rates	ek reactant, clean integer order	E1
C2	Initial rates	do reactants, isolation logic	E2
C3	Initial rates	fractional / non-integer order	E3
C4	Integrated	raw data se linear plot chunna	E4
C5	Integrated	degenerate: zero order (constant rate)	E5
C6	Half-life	degenerate: first order ( $t_{1/2}$ constant)	E6
C7	Half-life	second order, $t_{1/2}$ grows/shrinks	E7
C8	Any	real-world word problem (drug clearance)	E8
C9	Any	exam twist: $k$ ki units order pin karti hain	E9

E1 — Ek reactant, clean integer order (cell C1)

Worked example Statement

$A \to$ products ke liye rate law $r = k [A]^{n}$ hai. Do experiments:

Expt	$[A]$ (M)	initial rate (M/s)
1	0.050	$1.5 \times 1 0^{- 4}$
2	0.150	$1.35 \times 1 0^{- 3}$

$n$ find karo.

Forecast: $[A]$ teen guna hua. Guess: kya rate ×3, ×9, ya ×27 hogi?

Step 1 — rate ratio. $\frac{r _{2}}{r _{1}} = \frac{1.35 \times 1 0 ^{- 3}}{1.5 \times 1 0 ^{- 4}} = 9$ . Yeh step kyun? Ratio unknown $k$ ko cancel kar deta hai, sirf concentration effect reh jaata hai.

Step 2 — concentration ratio. $\frac{[ A ] _{2}}{[ A ] _{1}} = \frac{0.150}{0.050} = 3$ . Kyun? Yeh cause hai; rate ratio effect hai.

Step 3 — $3^{n} = 9$ solve karo. $n = \frac{ln ( r _{2} / r _{1} )}{ln ([ A ] _{2} / [ A ] _{1} )} = \frac{ln 9}{ln 3} = 2$ use karte hue. Log kyun? Logs $n$ ko exponent se neeche kheench laate hain taaki hum use directly read kar sakein.

Answer: $n = 2$ , second order.

Verify: $n = 2$ se $r_{2}$ predict karo: $r_{2} = r_{1} \cdot 3^{2} = 1.5 \times 1 0^{- 4} \times 9 = 1.35 \times 1 0^{- 3}$ M/s. ✓ Table se exactly match karta hai.

E2 — Do reactants, isolation logic (cell C2)

Worked example Statement

$A + B \to$ products, $r = k [A]^{a} [B]^{b}$ .

Expt	$[A]$	$[B]$	rate (M/s)
1	0.10	0.10	$2.0 \times 1 0^{- 3}$
2	0.30	0.10	$6.0 \times 1 0^{- 3}$
3	0.10	0.30	$1.8 \times 1 0^{- 2}$

$a$ , $b$ , aur overall order find karo.

Forecast: kaun sa reactant zyada matter karta hai — woh jahan teen guna karne se ×3 hua, ya woh jahan teen guna karne se ×9 hua?

Step 1 — 1→2 use karke $a$ find karo. Yahaan $[B]$ fixed hai, sirf $[A]$ teen guna hua. $\frac{r _{2}}{r _{1}} = \frac{6.0 \times 1 0 ^{- 3}}{2.0 \times 1 0 ^{- 3}} = 3 = 3^{a} \Rightarrow a = 1$ . $[B]$ fix kyun kiya? Taaki rate change sirf $[A]$ ki wajah se ho — yeh $a$ ko isolate karta hai.

Step 2 — 1→3 use karke $b$ find karo. Ab $[A]$ fixed, $[B]$ teen guna hua. $\frac{r _{3}}{r _{1}} = \frac{1.8 \times 1 0 ^{- 2}}{2.0 \times 1 0 ^{- 3}} = 9 = 3^{b} \Rightarrow b = 2$ . Kyun? Same isolation trick, ab roles swap ho gaye.

Step 3 — overall order $= a + b = 1 + 2 = 3$ . Jodna kyun? Overall order $r = k [A]^{a} [B]^{b}$ mein individual exponents ka sum hota hai.

Answer: $a = 1, b = 2$ , overall third order.

Verify: $k = \frac{r _{1}}{[ A ] ^{a} [ B ] ^{b}} = \frac{2.0 \times 1 0 ^{- 3}}{( 0.10 ) ^{1} ( 0.10 ) ^{2}} = \frac{2.0 \times 1 0 ^{- 3}}{1 0 ^{- 3}} = 2.0$ . Predict $r_{3} = k (0.10) (0.30)^{2} = 2.0 \times 0.10 \times 0.09 = 1.8 \times 1 0^{- 2}$ . ✓

E3 — Fractional (non-integer) order (cell C3)

E4 — Integrated method, linear plot chunna (cell C4)

Worked example Statement

$A$ ka decomposition. Neeche data hai. Order decide karo yeh test karke ki $[A]$ , $ln [A]$ , $1/ [A]$ mein se kaun $t$ mein linear hai, phir $k$ nikalo.

$t$ (s)	$[A]$ (M)
0	1.00
100	0.50
200	0.25
300	0.125

Forecast: $[A]$ har 100 s mein half hoti hai, level chahe kuch bhi ho. Kaun sa order constant halving time deta hai?

Step 1 — zero order test karo ( $[A]$ vs $t$ ). Drops $0.50, 0.25, 0.125$ — equal steps nahi hain. Linear nahi. Kyun? Zero order har interval mein ek fixed amount girti hai, jo yeh nahi hai.

Step 2 — first order test karo ( $ln [A]$ vs $t$ ). $ln$ values: $0, - 0.693, - 1.386, - 2.079$ . Har baar $- 0.693$ ke steps — bilkul even! Linear. ✓ Kyun? Equal times mein equal $ln$ steps first order ki pehchaan hai.

Step 3 — $k$ nikalo. Slope $= \frac{- 0.693 - 0}{100 - 0} = - 6.93 \times 1 0^{- 3}$ s⁻¹. Kyunki slope $= - k$ , $k = 6.93 \times 1 0^{- 3}$ s⁻¹. Minus kyun? Concentration girti hai, toh line neeche slope karti hai; $k$ magnitude hai.

Answer: first order, $k = 6.93 \times 1 0^{- 3}$ s⁻¹. Neeche teen-plot comparison dekho.

Verify: first-order law $[A] = [A]_{0} e^{- k t}$ : $t = 200$ par, $[A] = 1.00 e^{- 6.93 \times 1 0^{- 3} \times 200} = e^{- 1.386} = 0.250$ M. ✓

Figure — Methods to determine order — initial rates, integrated method, half-life method

E5 — Degenerate case: ZERO order (cell C5)

Worked example Statement

Ek saturated catalyst surface par reaction. Data:

$t$ (s)	$[A]$ (M)
0	0.80
100	0.60
200	0.40
300	0.20

Order aur $k$ find karo.

Forecast: drop har 100 s mein $0.20$ M hai — hamesha same amount, same fraction nahi. Kaun sa order yeh karta hai?

Step 1 — constant drop spot karo. $Δ [A] = - 0.20$ M per $100$ s, har interval. Kyun care karna? Fixed time mein fixed drop matlab rate $- \frac{d [ A ]}{d t}$ constant hai — yeh zero order ki definition hai, $r = k [A]^{0} = k$ .

Step 2 — confirm karo ki $[A]$ vs $t$ straight line hai. Points $0.80, 0.60, 0.40, 0.20$ line par hain. Zero order confirmed. Yeh plot kyun? $[A] = [A]_{0} - k t$ se, sirf $[A]$ khud ko plot karna $n = 0$ ke liye straight line deta hai.

Step 3 — $k$ nikalo. $k = - slope = - \frac{0.20 - 0.80}{300 - 0} = - \frac{- 0.60}{300} = 2.0 \times 1 0^{- 3}$ M/s. Units M/s kyun? Zero order ke liye $k$ ke units rate ke units hi hote hain.

Answer: zero order, $k = 2.0 \times 1 0^{- 3}$ M/s.

Verify (limiting behaviour): zero-order reactions khatam hoti hain — $[A]$ , $t = [A]_{0} / k = 0.80/2.0 \times 1 0^{- 3} = 400$ s par 0 ho jaati hai. Uske baad rate 0 ho jaani chahiye (koi $A$ nahi bachi); law sirf tab valid hai jab $[A] > 0$ ho. Sanity ✓.

E6 — Degenerate case: FIRST order via half-life (cell C6)

Worked example Statement

Ek reaction ke liye, $[A]_{0} = 0.40$ M se $t_{1/2} = 200$ s milta hai. Jab tum $[A]_{0} = 0.80$ M se rerun karte ho, $t_{1/2}$ phir bhi $200$ s hai. Order aur $k$ find karo.

Forecast: half-life ne starting concentration ko ignore kar diya. Kaun sa ek order yeh karta hai? (Hint: radioactivity isi tarah behave karti hai.)

Step 1 — fingerprint padho. $t_{1/2}$ independent of $[A]_{0}$ $\Rightarrow$ order $= 1$ . Kyun? Saare orders mein, sirf $n = 1$ deta hai $t_{1/2} = \frac{0.693}{k}$ , jisme $[A]_{0}$ nahi hai.

Step 2 — general check $n = 1 - \frac{ln ( t _{1/2, 2} / t _{1/2, 1} )}{ln ([ A ] _{0, 2} / [ A ] _{0, 1} )} = 1 - \frac{ln ( 200/200 )}{ln ( 0.80/0.40 )} = 1 - \frac{ln 1}{ln 2} = 1 - 0 = 1$ se. Yeh formula kyun? Yeh how-t-half-changes-with-concentration slope ko kisi bhi $n \neq = 1$ ke liye order mein convert karta hai, aur gracefully $1$ return karta hai jab koi change nahi hota.

Step 3 — $k$ nikalo. $k = \frac{0.693}{t _{1/2}} = \frac{0.693}{200} = 3.47 \times 1 0^{- 3}$ s⁻¹.

Answer: first order, $k = 3.47 \times 1 0^{- 3}$ s⁻¹.

Verify: ek half-life ke baad, $[A]_{0} = 0.40$ se $[A]$ $0.20$ M honi chahiye: $[A] = 0.40 e^{- 3.47 \times 1 0^{- 3} \times 200} = 0.40 e^{- 0.693} = 0.40 \times 0.500 = 0.200$ M. ✓

E7 — Second order, $t_{1/2}$ changes with $[A]_{0}$ (cell C7)

Worked example Statement

$[A]_{0} = 0.10$ M se $t_{1/2} = 250$ s milta hai. $[A]_{0} = 0.25$ M se $t_{1/2} = 100$ s milta hai. Order aur $k$ find karo.

Forecast: starting concentration $2.5 \times$ badha, aur $t_{1/2}$ ghata. Inverse relation — order 2 ki smell aa rahi hai.

Step 1 — general formula. $n = 1 - \frac{ln ( t _{1/2, 2} / t _{1/2, 1} )}{ln ([ A ] _{0, 2} / [ A ] _{0, 1} )} = 1 - \frac{ln ( 100/250 )}{ln ( 0.25/0.10 )}$ . Kyun? Hum abhi "yeh inverse lagta hai" par trust nahi karte; formula exact $n$ deta hai.

Step 2 — calculate karo. $ln (0.4) = - 0.9163$ , $ln (2.5) = 0.9163$ . Toh $n = 1 - \frac{- 0.9163}{0.9163} = 1 - (- 1) = 2$ . Exactly 2 kyun aata hai? Kyunki $t_{1/2} \propto [A]_{0}^{1 - n} = [A]_{0}^{- 1}$ , toh iska ratio concentration ratio ka reciprocal hai — reciprocals ke logs exact negatives hote hain.

Step 3 — $k$ nikalo $t_{1/2} = \frac{1}{k [ A ] _{0}}$ se: pehle data point use karte hue, $k = \frac{1}{t _{1/2} [ A ] _{0}} = \frac{1}{250 \times 0.10} = 0.040$ M⁻¹s⁻¹. Units M⁻¹s⁻¹ kyun? Second-order $k$ mein yeh units honi chahiye taaki $k [A]^{2}$ M/s mein aaye.

Answer: second order, $k = 0.040$ M⁻¹s⁻¹.

Verify: $[A]_{0} = 0.25$ ke liye second half-life predict karo: $t_{1/2} = \frac{1}{0.040 \times 0.25} = \frac{1}{0.010} = 100$ s. ✓ Table se match karta hai.

E8 — Real-world word problem: drug clearance (cell C8)

Worked example Statement

Ek patient ko drug di jaati hai. Blood tests dikhate hain ki iska concentration har $4$ ghante mein half ho jaata hai, chahe starting dose kuch bhi ho. $80$ mg/L se start karke, yeh $10$ mg/L kab drop karega, aur elimination constant $k$ kya hai?

Forecast: "har 4 ghante mein half, dose se chahe kuch bhi ho" — yeh fingerprint ek example pehle dekha hai. Kaun sa order?

Step 1 — order identify karo. Constant $t_{1/2}$ independent of dose $\Rightarrow$ first order (typical drug pharmacokinetics). Kyun? Same reasoning as E6 — sirf $n = 1$ mein dose-free half-life hoti hai.

Step 2 — $10$ mg/L tak pahunchne ke liye half-lives count karo. $80 \to 40 \to 20 \to 10$ : yeh $3$ halvings hain. Time $= 3 \times 4 = 12$ h. Halvings kyun count karo? First order har half-life mein ek fixed fraction (½) girti hai, toh successive halvings multiplicatively chain hote hain.

Step 3 — $k$ nikalo. $k = \frac{0.693}{t _{1/2}} = \frac{0.693}{4} = 0.173$ h⁻¹. Kyun? First-order half-life relation, $k$ ke liye solve karke.

Answer: first order, $12$ h baad $10$ mg/L tak pahunchta hai, $k = 0.173$ h⁻¹.

Verify: $[A] = 80 e^{- 0.173 \times 12} = 80 e^{- 2.079} = 80 \times 0.125 = 10.0$ mg/L. ✓

E9 — Exam twist: $k$ ki units order reveal karti hain (cell C9)

Worked example Statement

Ek exam sirf yeh deta hai: ek single-reactant reaction ke liye, $k = 3.0 \times 1 0^{- 2} mol L^{- 1} s^{- 1}$ . Order batao — koi rate data nahi diya.

Forecast: tumhare paas koi concentrations nahi, koi rates nahi... lekin $k$ ki units ek hidden fingerprint hain. Kya tum unhe decode kar sakte ho?

Step 1 — units rule yaad karo. $r = k [A]^{n}$ se, rate ke units hamesha $mol L^{- 1} s^{- 1}$ hote hain, toh $k$ ke units $(mol L^{- 1})^{1 - n} s^{- 1}$ hote hain. Kyun? $[A]^{n}$ $(mol L^{- 1})^{n}$ carry karta hai; product ko rate ke units ke barabar karne ke liye, $k$ ko $(mol L^{- 1})^{1 - n}$ supply karna hota hai.

Step 2 — match karo. Diye gaye $k$ mein $mol L^{- 1}$ power $+ 1$ par hai. Set karo $1 - n = 1 \Rightarrow n = 0$ . Yeh pin kyun karta hai? Har order concentration unit par ek unique power deta hai: $n = 0 \to (M)^{+ 1}$ , $n = 1 \to (M)^{0}$ (sirf s⁻¹), $n = 2 \to (M)^{- 1}$ .

Answer: zero order.

Verify (unit table sanity):

$n$	power $1 - n$	units of $k$
0	$+ 1$	mol L⁻¹ s⁻¹
1	$0$	s⁻¹
2	$- 1$	L mol⁻¹ s⁻¹

Diye gaye units $n = 0$ row mein hain. ✓ Rate Law and Rate Constant convention se consistent.

Recall check

Recall Kaun si cells degenerate orders ko stress karti hain?

Zero order (C5, E9) aur first order (C6, E8) — do "special" cases jahan behaviour unusually simple lagta hai. C5/E5 shows ::: ek constant drop per unit time (rate independent of $[A]$ ). C6/E6 shows ::: ek constant half-life independent of $[A]_{0}$ .

Recall Sabse fast single-clue shortcuts

Koi data nahi, sirf $k$ ki units? ::: concentration unit ki power match karo — $n = 1 - (that power)$ . $t_{1/2}$ dose se nahi badalti? ::: first order. Har equal time mein constant amount loss? ::: zero order. $[A]$ double karne se rate ×4? ::: second order.