Pseudo-first-order kinetics: Worked examples — Hinglish deep dive

2.8.5 · D3 · Chemistry › Chemical Kinetics › Pseudo-first-order kinetics

Yeh page pseudo-first-order kinetics ka drill ground hai. Parent note ne idea build kiya; yahan hum ensure karte hain ki exam chahe kuch bhi phenke — koi ajeeb fraction remaining ho, half-life question ho, "kya yeh pseudo-first-order bhi hai?" wala trap ho, units ki trick ho, ya ek limiting case jahan poori approximation toot jaaye — tumne pehle se uska ek fully worked twin dekha hua hai.

Sab kuch parent se ek equation par tikaa hai, toh shuru karne se pehle chalein har symbol ko dobara earn karein.

The scenario matrix

Har pseudo-first-order problem jo tum kabhi bhi miloge, in cells mein se kisi ek mein aata hai. Neeche ke examples us cell ke label ke saath hain jo woh cover karta hai.

#	Cell class	Kya mushkil banaata hai	Example
C1	Given $t$ → find $[A]$	pick exponential form	Ex 1
C2	Given fraction/percent → find $t$	pick log form, convert percent right	Ex 2
C3	Recover true $k$ from measured $k^{'}$	divide by $[B]_{0}$ , watch units	Ex 3
C4	Half-life & multiples ( $t_{1/2}$ , $t_{3/4}$ …)	half-life independent of $[A]_{0}$	Ex 4
C5	Graphical / slope reading	slope $= - k^{'}$ , curved vs straight	Ex 5
C6	Degenerate: excess too small	approximation breaks, quantify error	Ex 6
C7	Limiting: $t \to 0$ and $t \to \infty$	sanity endpoints, $[A] \to [A]_{0}$ and $\to 0$	Ex 7
C8	Real-world word problem + exam twist	translate words, two-excess trap	Ex 8

Trig topic mein jo cells "signs/quadrants" hote hain, woh kinetics mein inequality ki direction ( $[B]_{0} ≫ [A]_{0}$ vs comparable) aur time ke endpoints ( $0$ , finite, $\infty$ ) ban jaate hain. Hum inhe sabko cover karte hain.

Worked examples

Worked example Example 1 — Cell C1: given time, find concentration

Sucrose inversion mein $k^{'} = 6.0 \times 1 0^{- 5} s^{- 1}$ hai, starting concentration $[A]_{0} = 0.50 M$ . 2 hours baad $[A]$ kya hai?

Forecast: Do ghante yahan bahut kam hain considering $k^{'}$ kitna dheere kaam karta hai. Guess: abhi bhi zyaadatar bacha hoga — kahin $0.3$ – $0.35 M$ ke aas paas.

Time ko seconds mein convert karo: $t = 2 \times 3600 = 7200 s$ . Yeh step kyun? $k^{'}$ mein units $s^{- 1}$ hain, toh exponent $k^{'} t$ tabhi dimensionless hoga jab $t$ seconds mein ho.
Humein $t$ diya hai aur $[A]$ chahiye, toh exponential form use karo: $[A] = [A]_{0} e^{- k^{'} t}$ . Yeh step kyun? Unknown $[A]$ pehle se hi left side par akele hai — koi algebra nahi chahiye.
Exponent: $k^{'} t = (6.0 \times 1 0^{- 5}) (7200) = 0.432$ . Yeh step kyun? Yeh number hai "kitne decay-lengths guzar gaye"; $0.432$ matlab hum ek se kaafi neeche hain, toh bahut kam react hua — humara forecast match karta hai.
$[A] = 0.50 \times e^{- 0.432} = 0.50 \times 0.6492 = 0.325 M$ .

Verify: Fraction left $= 0.325/0.50 = 0.65$ , toh $35%$ 2 ghante mein react hua — forecast ke saath consistent ki "zyaadatar" bacha hai. Units: $M \times (dimensionless) = M$ . ✓

Worked example Example 2 — Cell C2: given percent completion, find time

Ethyl acetate hydrolysis, $k^{'} = 2.5 \times 1 0^{- 4} s^{- 1}$ . 75% completion ke liye kitna time chahiye?

Forecast: 75% gone ek half-life (50%) aur do (75% exactly do half-lives hai — kyunki $1 - \frac{1}{4} = \frac{3}{4}$ ) ke beech mein hai. Toh roughly $2 \times t_{1/2}$ expect karo.

"75% completion" matlab 75% consume hua, toh 25% bacha hai: $[A] = 0.25 [A]_{0}$ , yaani $[A] / [A]_{0} = 0.25$ . Yeh step kyun? Equation sirf remaining ratio se matlab rakhti hai, kabhi bhi absolute start value se nahi — toh humein $[A]_{0}$ ki zaroorat hi nahi.
Fraction diya hai, time chahiye → log form use karo: $ln \frac{[ A ]}{[ A ] _{0}} = - k^{'} t$ . Yeh step kyun? Logarithm exponential ko undo karta hai aur $t$ ko exponent se bahar ek aisi jagah laata hai jahan hum solve kar sakein.
$ln (0.25) = - 1.3863$ , toh $- 1.3863 = - k^{'} t$ .
$t = \frac{1.3863}{2.5 \times 1 0 ^{- 4}} = 5545 s \approx 92.4 min$ . Yeh step kyun? $t$ isolate karne ke liye $k^{'}$ se divide karo; minutes mein convert karna ( $/60$ ) scale ko human-friendly banata hai.

Verify: $t_{1/2} = ln 2/ k^{'} = 0.6931/2.5 \times 1 0^{- 4} = 2773 s$ . Two half-lives $= 5545 s$ — exactly humara answer, forecast confirm karta hai ki 75% = two half-lives. ✓

Worked example Example 3 — Cell C3: recover the true rate constant

Wahi hydrolysis, $k^{'} = 2.5 \times 1 0^{- 4} s^{- 1}$ , water excess mein $[H_{2} O] = 55.5 M$ par. True second-order $k$ nikalo.

Forecast: Ek chhoti number ko ek badi number (55.5) se divide karna → answer bahut chhota, around $1 0^{- 6}$ .

Yaad karo $k^{'} = k [B]_{0}$ jahan $[B]_{0} = [H_{2} O] = 55.5 M$ . Yeh step kyun? Water woh reactant hai jo $k^{'}$ ke andar chhupa hua hai; wahi excess species $B$ hai.
$k$ ke liye solve karo: $k = k^{'} / [B]_{0} = \frac{2.5 \times 1 0 ^{- 4}}{55.5}$ . Yeh step kyun? Us multiplication ko undo karo jisne $[B]_{0}$ ko $k^{'}$ mein bundle kiya tha.
$k = 4.50 \times 1 0^{- 6} M^{- 1} s^{- 1}$ .

Verify — units hi saari baat hai: $\frac{s ^{- 1}}{M} = M^{- 1} s^{- 1}$ , correct second-order unit. Agar tumne divide karna bhool gaye, toh $2.5 \times 1 0^{- 4} s^{- 1}$ report karte — ek first-order unit — jo parent note ki classic Mistake 2 hai. ✓

Worked example Example 4 — Cell C4: half-life and its multiples

Ek pseudo-first-order reaction ke liye $k^{'} = 0.0231 s^{- 1}$ hai: (a) $t_{1/2}$ nikalo; (b) woh time nikalo jab concentration apne start ka one-eighth ho jaaye.

Forecast: One-eighth $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$ = teen half-lives hai. Toh (b) $3 \times$ (a) hona chahiye.

Half-life: $ln ([A] / [A]_{0}) = - k^{'} t$ mein $[A] / [A]_{0} = \frac{1}{2}$ set karo: $ln \frac{1}{2} = - k^{'} t_{1/2}$ . Yeh step kyun? Half-life define hi woh time hai jab ratio $\frac{1}{2}$ ho jaaye.
$t_{1/2} = \frac{ln 2}{k ^{'}} = \frac{0.6931}{0.0231} = 30.0 s$ . Yeh step kyun? $ln 2$ isliye aata hai kyunki halving har baar ek fixed fraction hoti hai — yahi reason hai first-order half-life $[A]_{0}$ se independent hoti hai.
One-eighth ke liye: $[A] / [A]_{0} = \frac{1}{8}$ , toh $ln \frac{1}{8} = - 3 ln 2$ , jo deta hai $t = 3 ln 2/ k^{'} = 3 t_{1/2}$ .
$t = 3 \times 30.0 = 90.0 s$ .

Verify: Plug karo $t = 90$ ko $[A] / [A]_{0} = e^{- k^{'} t} = e^{- 0.0231 \times 90} = e^{- 2.079} = 0.125 = \frac{1}{8}$ mein. ✓ Forecast se match karta hai (teen half-lives).

Worked example Example 5 — Cell C5: reading the graph (geometric)

Do experiments $ln [A]$ ko $t$ ke against track karte hain. Experiment 2 ( $B$ $100 \times$ excess mein) ek seedhi line deta hai $(0, - 4.605)$ aur $(100, - 9.605)$ se hote hue. $k^{'}$ nikalo, aur true $k$ nikalo agar $[B]_{0} = 1.0 M$ hai.

Forecast: Seedhi neeche jaati line matlab pseudo-first-order confirm. Jitna steep fall, utna bada $k^{'}$ . 100 s mein value 5 units giri → slope ka magnitude $5/100 = 0.05$ .

$ln [A]$ vs $t$ ka slope hai $\frac{Δ ln [ A ]}{Δ t} = \frac{- 9.605 - ( - 4.605 )}{100 - 0} = \frac{- 5.000}{100} = - 0.0500 s^{- 1}$ . Yeh step kyun? Integrated law $ln [A] = ln [A]_{0} - k^{'} t$ basically $y = c - k^{'} t$ hai: iska slope literally $- k^{'}$ hai.
Toh $k^{'} = 0.0500 s^{- 1}$ (minus hata do — $k^{'}$ positive hai; minus "decay" direction mein rehta hai).
True $k$ recover karo: $k = k^{'} / [B]_{0} = 0.0500/1.0 = 0.0500 M^{- 1} s^{- 1}$ . Yeh step kyun? Ex 3 jaisi wahi division — excess concentration ko wapis strip karo.

Verify: Intercept $ln [A]_{0} = - 4.605 \Rightarrow [A]_{0} = e^{- 4.605} = 0.010 M$ — stated start. Figure mein, red curved line (Experiment 1, comparable concentrations) seedhi nahi hai, toh woh pseudo-first-order NAHI hai — exactly isliye hum excess vary karte hain. ✓

Worked example Example 6 — Cell C6 (degenerate): the approximation breaks

Ek student sirf $10 \times$ excess use karta hai: $[A]_{0} = 0.10 M$ , $[B]_{0} = 1.0 M$ . Jab A ka aadha consume ho jaata hai, tab $[B]$ practically kitne percent se badla hai, aur kya ise "constant" kehna honest hai?

Forecast: Sirf $10 \times$ excess parent note ka warning zone hai. A ka aadha consume karna = $0.05 M$ , aur B bhi utna hi kho deta hai. $0.05$ out of $1.0$ matlab $5%$ — noticeable hai.

50% par consume hua A ka amount: $Δ = 0.50 \times 0.10 = 0.050 M$ . Yeh step kyun? Stoichiometry $A + B \to$ products matlab har A consume hone par ek B khatam hota hai.
Naya $[B] = 1.0 - 0.050 = 0.950 M$ .
B mein percent change: $\frac{0.050}{1.0} \times 100 = 5.0%$ . Yeh step kyun? Yahi woh error hai jo tum inject karte ho jab $[B]$ ko $[B]_{0}$ par fixed pretend karte ho.

Verify: Ek proper $100 \times$ excess se compare karo ( $[A]_{0} = 0.10$ , $[B]_{0} = 10$ ): B sirf $0.05/10 = 0.5%$ change hota. Toh $10 \times$ se $100 \times$ jaane par error das guna kam hoti hai — yahi reason hai parent $[B]_{0} / [A]_{0} \geq 50$ – $100$ demand karta hai. $5%$ par, Ex 5 ki "line" visibly bend kar jaayegi. ✓

Worked example Example 7 — Cell C7 (limiting): the two endpoints of time

$[A] = [A]_{0} e^{- k^{'} t}$ use karke kisi bhi positive $k^{'}$ ke saath, physical endpoints evaluate karo: (a) $t \to 0$ ; (b) $t \to \infty$ . Confirm karo ki model kabhi misbehave nahi karta.

Forecast: Bilkul shuru mein kuch react nahi hua → $[A]$ ko $[A]_{0}$ equal hona chahiye. Infinite time ke baad sab consume ho jaata hai → $[A] \to 0$ . Ek achha model exactly yahi dena chahiye.

Jab $t \to 0$ : $e^{- k^{'} \cdot 0} = e^{0} = 1$ , toh $[A] \to [A]_{0} \times 1 = [A]_{0}$ . Yeh step kyun? $e^{0} = 1$ exponential ka anchor hai — yeh guarantee karta hai ki curve sahi height par start hoti hai.
Jab $t \to \infty$ : $- k^{'} t \to - \infty$ , aur $e^{- \infty} = 0$ , toh $[A] \to 0$ . Yeh step kyun? Positive $k^{'}$ exponent ko $- \infty$ ki taraf march karaata hai; concentration decay hoti hai (lekin kabhi poori tarah zero nahi pahunchti) — match karta hai "A eventually khatam ho jaata hai."
Degenerate case $k^{'} = 0$ (koi reaction nahi, e.g. catalyst absent): $e^{0} = 1$ sab $t$ ke liye, toh $[A] = [A]_{0}$ hamesha — ek flat line. Model sahi kehta hai "kuch nahi hota."

Verify: Numerically $[A]_{0} = 1, k^{'} = 0.05$ ke saath: $t = 0$ par, $[A] = 1.000$ ; $t = 1000 s$ par, $[A] = 1 \cdot e^{- 50} \approx 1.9 \times 1 0^{- 22} \approx 0$ . Dono endpoints hit ho gaye. ✓

Worked example Example 8 — Cell C8: real-world word problem with an exam twist

Ek pollutant $P$ ek lake mein dissolved oxygen ke saath reaction karke degrade hota hai: $P + O_{2} \to$ products, truly second-order, $k = 3.0 M^{- 1} s^{- 1}$ . Lake mein $[O_{2}] = 2.0 \times 1 0^{- 4} M$ hai, effectively constant (air se continuously resupplied), jabki $[P]_{0} = 1.0 \times 1 0^{- 8} M$ . Twist: exam tumhe $k$ deta hai (na ki $k^{'}$ ) aur poochta hai pollutant ko apne start ka 10% tak girne mein kitna time lagega.

Forecast: Oxygen ( $2 \times 1 0^{- 4}$ ) pollutant ( $1 0^{- 8}$ ) se $2 \times 1 0^{4}$ factor se zyada hai — massively pseudo-first-order, approximation rock-solid. Pehle humein khud $k^{'}$ build karna hoga, phir time ke liye solve karna hoga. "10% remaining" teen half-lives ( $12.5%$ exactly teen hain) ke thoda baad hai, toh $t$ thoda $3 t_{1/2}$ se zyada expect karo.

Excess confirm karo: $[O_{2}] / [P]_{0} = 2 \times 1 0^{- 4} /1 0^{- 8} = 2 \times 1 0^{4} \geq 100$ . ✓ Yeh step kyun? Pseudo-first-order trust karne se pehle tumhe excess check karna hi hoga — assume kabhi mat karo.
Pseudo constant build karo: $k^{'} = k [O_{2}] = 3.0 \times 2.0 \times 1 0^{- 4} = 6.0 \times 1 0^{- 4} s^{- 1}$ . Yeh step kyun? Twist ne true $k$ diya; hum constant excess ko fold in karte hain taaki woh $k^{'}$ mile jo hamare time-equation ko chahiye (units: $M^{- 1} s^{- 1} \times M = s^{- 1}$ ✓).
"10% remaining": $[P] / [P]_{0} = 0.10$ , log form use karo: $ln (0.10) = - k^{'} t$ . Yeh step kyun? Fraction diya hai, time chahiye → logarithm $t$ isolate karta hai.
$ln (0.10) = - 2.3026$ , toh $t = 2.3026/6.0 \times 1 0^{- 4} = 3838 s \approx 64 min$ .

Verify: $t_{1/2} = ln 2/ k^{'} = 0.6931/6 \times 1 0^{- 4} = 1155 s$ ; teen half-lives $= 3466 s$ deta hai $12.5%$ . Humara $3838 s$ thoda zyada hai → $12.5%$ se kam bacha hai, yaani exactly $10%$ — forecast ke saath consistent. Plug back karo: $e^{- 6 \times 1 0^{- 4} \times 3838} = e^{- 2.303} = 0.100$ . ✓

Recall Quick self-test

Percent completion diya ho toh kaun sa form use karte ho aur kyun? ::: $ln$ form, kyunki logarithm exponential ko undo karta hai aur unknown time $t$ isolate karta hai. $t_{3/4}$ (75% done) exactly $2 t_{1/2}$ kyun hai? ::: Kyunki 25% remaining $= \frac{1}{2} \times \frac{1}{2}$ hai, yaani do successive halvings, aur first-order half-life constant hoti hai. Tumne $k^{'} = 0.05 s^{- 1}$ measure kiya $[B]_{0} = 2.0 M$ ke saath. True $k$ ? ::: $k = k^{'} / [B]_{0} = 0.025 M^{- 1} s^{- 1}$ . $10 \times$ excess mein, jab A ka aadha khatam ho jaaye toh B kitna badal jaata hai? ::: Lagbhag 5% — itna kaafi hai "straight" line ko bend karne ke liye, toh $10 \times$ safe nahi hai.

Related: Integrated Rate Laws · Second-Order Reactions · Method of Isolation · Half-life · Enzyme Kinetics · back to Pseudo-first-order kinetics.