2.8.3 · D3Differential rate equations for 0, 1st, 2nd order — derivati
Yeh page parent derivations ka stress-test hai. Hum sirf "formula mein plug karo" wali baat repeat nahi karte. Hum har tarah ke sawal enumerate karte hain jo ek rate law throw kar sakta hai — har order, har direction (forward: concentration dhundho; backward: time ya k
Agar tumne abhi tak teen integrated laws nahi padhe, pehle parent note padho. Yahan sab kuch sirf in teen lines par depend karta hai, toh chaliye inhe wahan pin kar dete hain jahan nazar seedhi padti ho.
Definition Square brackets aur "M" ka matlab kya hai[ A ] concentration of A " padha jaata hai — matlab ek litre solution mein packed substance A ke moles ki tadaad. Iska unit hai mol L − 1 M molar ). Toh "0.10 M 0.10 − 1 0.10 M mol L − 1
t 1/2 Half-life t 1/2 aadhi ho jaati hai: [ A ] → 2 1 [ A ] 1/2 0th: t 1/2 = 2 k [ A ] 0 1st: t 1/2 = k l n 2 = k 0.693 2nd: t 1/2 = k [ A ] 0 1
Shuru karne se pehle teeno ki pictures ek saath dekho — woh alag dikhti hain , aur wahi shape tumhe batati hai ki kaun sa law kaam kar raha hai.
Intuition Upar wali figure mein kya notice karna haiHar curve ko usi starting height se trace karo. White line (0th) perfectly constant tilt par girta hai — har second same drop, aur eventually axis se crash karta hai. Cyan curve (1st) start mein sabse steep hai aur phir hamesha gentle hota jaata hai, lekin zero ko kabhi touch nahi karta — sirf approach karta hai. Amber curve (2nd) sabse pehle sabse fast start karta hai, phir cyan wale se bhi zyada lazy tail mein flatten ho jaata hai. Is poore page ka game bas yeh hai: shape padho, order batao. Constant slope ⇒ 0th; ek fixed fraction se hamesha shrink hona ⇒ 1st; ek stubborn long tail jo khatam hone ka naam nahi leta ⇒ 2nd.
Intuition Time ke units ke baare mein ek baatCells A–F, H aur I mein hum seconds (s) mein kaam karte hain, kyunki rate constants waise hi diye gaye hain. Drug problem (Cell G) mein data hours (h) mein aata hai, toh hum k h − 1 har time ko hours mein — s aur h ko ek hi equation mein mix karna waise hi hai jaise metres ko miles mein add karo. Rule yeh hai: k t k t
Har kinetics problem is grid ke exactly ek cell mein fit hoti hai. Last column us worked example ka naam batata hai jo use cover karta hai.
Cell
Order
Kya unknown hai / kya weird hai
Covered by
A
0
forward: [ A ] t
Ex 1
B
0
degenerate : reactant t khatam ho jaata hai (formula negative ho jaayega!)Ex 2
C
1
backward: target tak pahunchne ka time dhundho
Ex 3
D
1
limiting behaviour: t → ∞
Ex 4
E
2
forward: [ A ] t
Ex 5
F
2
backward: do concentrations se k
Ex 6
G
any
word problem (real world), units of k Ex 7
H
any
exam twist : order chhupa hua hai — data se decide karo, phir answer doEx 8
I
0 vs 1 vs 2
half-life comparison — same numbers, teen alag t 1/2
Ex 9
Hum inhe order mein chalte hain. Har example pehle tumse Forecast maangta hai (rough guess lagao — yeh intuition train karta hai), phir solve karta hai, phir Verify karta hai.
Worked example Ex 1 — Ammonia on hot platinum2 NH 3 → N 2 + 3 H 2 [ NH 3 ] 0 = 0.10 M k = 2.5 × 1 0 − 4 M s − 1 t = 300 s [ NH 3 ]
Forecast: Zero order mein concentration straight line mein girta hai. 300 s mein hum k ⋅ t = 2.5 × 1 0 − 4 × 300 = 0.075 M 0.10 0.025 M
Step 1. Law chunno. Zero order → [ A ] t = [ A ] 0 − k t Yeh step kyun? Kyunki rate concentration par depend nahi karta; remove ki gayi amount sirf (rate)×(time) hai, yaani constant slope.
Step 2. Substitute karo.
[ NH 3 ] t = 0.10 − ( 2.5 × 1 0 − 4 ) ( 300 ) = 0.10 − 0.075 = 0.025 M Yeh step kyun? Direct plug-in; subtraction hi physics hai — toll booth ne har second fixed number of molecules pass kiye.
Verify: 0.025 M 0.10 M − ( M s − 1 ) ( s ) = M − M = M
Yeh woh trap hai jo almost sab miss kar dete hain. Zero order mein line same slope par girti rehti hai. Ek line ko nahi pata ki zero par ruk jaaye. Toh blindly ek bada t negative concentration aati hai, jo physically impossible hai.
t end zero-order reactant completely use ho jaata hai: t end [ A ] t 0 derive karenge.
Intuition Upar wali figure mein kya notice karna haiDashed grey line bare formula 0.10 − k t negative territory mein chali jaati hai — woh jagah jahan concentration kabhi nahi ja sakta. Solid cyan line honest concentration hai: yeh grey line ko exactly tab tak follow karta hai jab tak axis se meet nahi karta, phir [ A ] = 0 t end tool hai jiska valid domain hota hai , aur amber dot us domain ka edge hai.
Worked example Ex 2 — Formula kab toot jaata hai?Same reaction: [ NH 3 ] 0 = 0.10 M k = 2.5 × 1 0 − 4 M s − 1 t = 500 s t end
Forecast: 500 s mein hum 2.5 × 1 0 − 4 × 500 = 0.125 M shuru se zyada hai. Toh honest answer hai "500 s se pehle khatam." Reactant 500 s se pehle kahi zero hit karega.
Step 1. Naively substitute karo.
[ NH 3 ] 500 = 0.10 − 0.125 = − 0.025 M ( impossible! ) Yeh step kyun? Failure dekhne ke liye. Negative concentration matlab model apne valid region se bahar nikal gaya — reaction pehle hi khatam ho gayi.
Step 2. Run-out time t end [ A ] t = 0 0 = [ A ] 0 − k t end ⇒ t end = k [ A ] 0 = 2.5 × 1 0 − 4 0.10 = 400 s Yeh step kyun? Definition se t end [ A ] = 0 t x [ A ] = 0 0
Step 3. Real answer piecewise batao.
[ NH 3 ] t = { 0.10 − k t 0 t ≤ t end = 400 s t > 400 s Yeh step kyun? Single formula 0.10 − k t jab tak ammonia bachi ho . t end = 0 t end
Verify: t = 400 0.10 − 2.5 × 1 0 − 4 × 400 = 0.10 − 0.10 = 0 400 2 × t 1/2 t 1/2 = [ A ] 0 /2 k = 200 amount khoata hai, toh do half-lives mein completely khatam ho jaata hai. ✓
Common mistake The zero-order cliffFirst aur second order zero ko asymptotically approach karte hain — woh kabhi truly zero nahi pahunchte. Sirf zero order actually finite time t end Number trust karne se pehle hamesha check karo ki k t > [ A ] 0
Worked example Ex 3 — Target tak pahunchne mein kitna time?Ek first-order isomerisation ka k = 3.0 × 1 0 − 3 s − 1 [ A ] 0 = 0.80 M [ A ] t = 0.20 M
Forecast: Hum start ke one-quarter tak gir rahe hain. One-quarter = do halvings, toh roughly do half-lives . Half-life = 0.693/ k ≈ 231 460 – 470
Step 1. Log form chunno aur t ln [ A ] 0 [ A ] t = − k t ⇒ t = − k 1 ln [ A ] 0 [ A ] t Yeh step kyun? Time exponential ke andar trapped hai; natural log woh tool hai jo e − k t undo karke t ln log 10 e
Step 2. Substitute karo.
t = − 3.0 × 1 0 − 3 1 ln 0.80 0.20 = − 3.0 × 1 0 − 3 1 ln ( 0.25 ) ln ( 0.25 ) = − 1.386 t = − 3.0 × 1 0 − 3 − 1.386 = 462 s Yeh step kyun? ln ( 0.25 ) − k
Verify: Back plug karo: [ A ] t = 0.80 e − 3.0 × 1 0 − 3 × 462 = 0.80 e − 1.386 = 0.80 × 0.25 = 0.20 M 462 ≈ 2 × 231
Worked example Ex 4 — Bachi hui fraction14 C t 1/2 = 5730 4 half-lives ke baad kaunsi fraction bachi hai? (b) t → ∞
Forecast: Har half-life 2 1 2 1 × 2 1 × 2 1 × 2 1 = 16 1 t → ∞ e − k t → 0
Step 1. k = ln 2/ t 1/2 k = 5730 0.693 = 1.21 × 1 0 − 4 yr − 1 Yeh step kyun? First order ke liye half-life formula t 1/2 = ln 2/ k k
Step 2. Char half-lives matlab t = 4 × 5730 = 22920 [ A ] 0 [ A ] t = e − k t = e − 1.21 × 1 0 − 4 × 22920 = e − 2.773 = 0.0625 = 16 1 Yeh step kyun? Hum ( 1/2 ) 4 e − 4 l n 2
Step 3. Limit. Jab t → ∞ − k t → − ∞ e − k t → 0 Yeh step kyun? Yeh first order ki defining feature hai — asymptotic approach, kabhi hard stop nahi (Cell B se contrast karo).
Verify: ( 1/2 ) 4 = 0.0625 e − 2.773 = 0.0625 k yr − 1 k t
2 2 NO 2 → N 2 O 4 2 [ NO 2 ] 0 = 0.0100 M k = 0.543 M − 1 s − 1 t = 300 s [ NO 2 ] t
Forecast: Second order pehle fast decay karta hai, phir crawl karta hai . Kyunki 1/ [ A ] 0 = 100 k t = 163 k t × 1 0 − 3 M
Step 1. Reciprocal law.
[ NO 2 ] t 1 = [ NO 2 ] 0 1 + k t Yeh step kyun? Second order 1/ [ A ] [ A ]
Step 2. Substitute karo.
[ NO 2 ] t 1 = 0.0100 1 + ( 0.543 ) ( 300 ) = 100 + 162.9 = 262.9 M − 1 Yeh step kyun? 1/0.0100 = 100 ( 0.543 ) ( 300 ) = 162.9 1/ [ A ] linearly grow karta hai.
Step 3. Flip back karo.
[ NO 2 ] t = 262.9 1 = 3.80 × 1 0 − 3 M Yeh step kyun? Law reciprocal space mein rehta hai; final flip humein real concentration par wapas laata hai.
Verify: 1/ ( 3.80 × 1 0 − 3 ) = 263 ≈ 262.9 M − 1 + ( M − 1 s − 1 ) ( s ) = M − 1 ≈ 38% k
Worked example Ex 6 — Do data points sek [ A ] 0 = 0.050 M t = 250 s [ A ] t = 0.020 M k
Forecast: Reciprocal space mein hum 1/0.050 = 20 1/0.020 = 50 30 M − 1 k ≈ 30/250 = 0.12 M − 1 s − 1
Step 1. Law ko k k = t 1 ( [ A ] t 1 − [ A ] 0 1 ) Yeh step kyun? k 1/ [ A ] t slope hai; slope = (rise in 1/ [ A ] t
Step 2. Substitute karo.
k = 250 1 ( 0.020 1 − 0.050 1 ) = 250 1 ( 50 − 20 ) = 250 30 = 0.12 M − 1 s − 1 Yeh step kyun? 1/0.020 = 50 1/0.050 = 20 30 M − 1 250 hi k
Verify: Is k 1/ [ A ] t = 20 + 0.12 × 250 = 20 + 30 = 50 [ A ] t = 0.020 M k ( M − 1 ) / ( s ) = M − 1 s − 1
Worked example Ex 7 — Body mein drug clearanceEk drug first-order kinetics se eliminate hoti hai. Ek patient ko 500 mg 6.0 hours 125 mg k ≤ 5%
Forecast: 500 → 125 1/16 = 6.25%
Part (a) — k
Step 1. Amount concentration ke proportional hai, toh ratio law masses par directly kaam karta hai.
ln 500 125 = − k ( 6.0 ) ⇒ ln ( 0.25 ) = − 1.386 = − 6.0 k Yeh step kyun? First-order sirf amounts ke ratio par depend karta hai; volume cancel ho jaata hai, toh hum M ki jagah mg use kar sakte hain. Yahan sab times hours mein hain, toh k
Step 2. k k = 6.0 1.386 = 0.231 h − 1 Yeh step kyun? Units h − 1 hours , data ke hours se match karta hai) kyunki first-order k
Part (b) — half-life dhundho.
Step 3. First-order half-life formula use karo.
t 1/2 = k 0.693 = 0.231 0.693 = 3.0 h Yeh step kyun? First-order half-life ln 2/ k k h − 1 hours mein aata hai — koi conversion nahi chahiye kyunki k 500 → 125
Part (c) — 5% tak pahunchne ka time.
Step 4. Remaining fraction ko 0.05 ln ( 0.05 ) = − k t ⇒ t = − k 1 ln ( 0.05 ) = − 0.231 − 2.996 = 12.97 h ≈ 13 h Yeh step kyun? Same "time ke liye solve karo" move jaise Cell C mein — natural log exponential ko undo karke t 0.05 [ A ] t / [ A ] 0 = 0.05
Verify: Forecasts confirm hue: t 1/2 = 3.0 500 e − 0.231 × 6 = 500 × 0.25 = 125 500 e − 0.231 × 12.97 = 500 × 0.0500 = 25 = 5% 500 12.97 6.25%
Sabse nasty exam questions order nahi batate . Tumhe ise data se decide karna hota hai, phir answer dena hota hai. Sabse clean tell: check karo ki equal time steps mein konsi quantity constant amount change karti hai.
Intuition Upar wali figure mein kya notice karna haiAmber dots raw data ko 1/ [ A ] karte hain. Notice karo woh bilkul straight line par gir rahe hain — har equal 100 s step exactly same 1.00 M − 1 1/ [ A ] 1/ [ A ] t = 400 [ A ] ln [ A ]
Worked example Ex 8 — Yeh konsa order hai?Ek reactant monitor kiya jaata hai:
t 0
100
200
300
[ A ] 1.00
0.50
0.333
0.25
Order identify karo, k t = 400 s [ A ]
Forecast: Pehle 100 s mein concentration aadhi ho gayi, lekin 200 s tak phir aadhi nahi hui (yeh 0.25 nahi 0.333 par gayi). Aisa shrinking fall second order ka signature hai (fast then slow).
Step 1. Zero order test karo: kya Δ [ A ] 0.50 , 0.167 , 0.083 constant nahi . Zero order reject.
Yeh step kyun? Zero order straight line mein girta hai (equal drops per equal time), toh unequal drops ise turant rule out karte hain.
Step 2. First order test karo: kya har step ka ratio constant hai? 0.50/1.00 = 0.50 0.333/0.50 = 0.667 constant nahi . First order reject.
Yeh step kyun? First order har equal interval mein same factor se multiply karta hai; changing factor ise rule out karta hai.
Step 3. Second order test karo: kya 1/ [ A ] constant amount se badh raha hai?
[ A ] 1 : 1.00 , 2.00 , 3.00 , 4.00 1.00 M − 1 Second order confirmed.
Yeh step kyun? Second order woh hai jo 1/ [ A ]
Step 4. Slope se k k = Δ t Δ ( 1/ [ A ]) = 100 1.00 = 1.0 × 1 0 − 2 M − 1 s − 1 Yeh step kyun? Second order ke liye k 1/ [ A ] t 1.00 M − 1 100
Step 5. t = 400 s 1/ [ A ] = 1.00 + ( 0.01 ) ( 400 ) = 5.00 [ A ] 400 = 5.00 1 = 0.20 M Yeh step kyun? Jab line ka slope aur intercept pata hai, ek step aur extrapolate karna sirf t = 400
Verify: Table reproduce karo: 1/ [ A ] = 1 + 0.01 t t = 200 1/ [ A ] = 3 ⇒ 0.333 t = 300 = 4 ⇒ 0.25 0.20
Sabse illuminating comparison: same [ A ] 0 same numeric k t 1/2
Intuition Upar wali figure mein kya notice karna haiTeen bars, same input numbers, teen alag heights . White (0th) bar sabse lamba hai — ek badi starting stock fixed removal rate par burn through hone mein zyada time leti hai. Amber (2nd) bar sabse chota hai — yahan ek bada starting concentration matlab abundant collision partners, toh pehla halving sabse jaldi hota hai. Cyan (1st) bar beech mein hai aur, critically, bilkul nahi hilega agar tum [ A ] 0 k [ A ] 0 dekhte ho: proportional, independent, inverse.
Worked example Ex 9 — Ek start, teen fates[ A ] 0 = 2.0 M k = 0.10 t 1/2
Forecast: Zero order "slow to start" lagni chahiye kyunki yeh fixed amount khoata hai; second order yahan sabse fast half tak pahunchna chahiye kyunki bada [ A ] 0 [ A ] 0
Step 1. Zero order: t 1/2 = 2 k [ A ] 0 = 2 ( 0.10 ) 2.0 = 10 s Yeh step kyun? Constant removal rate par aadha material — bada start matlab longer half-life ([ A ] 0
Step 2. First order: t 1/2 = k 0.693 = 0.10 0.693 = 6.93 s Yeh step kyun? Exponential decay ek fixed time mein half hota hai regardless of [ A ] 0 — woh ek constant half-life.
Step 3. Second order: t 1/2 = k [ A ] 0 1 = ( 0.10 ) ( 2.0 ) 1 = 5.0 s Yeh step kyun? Yahan ek bada start actually pehle half-life ko chota karta hai — shuruaat mein collide karne ke liye plenty of partners hote hain ([ A ] 0
Verify: 10 , 6.93 , 5.0 [ A ] 0
Recall Quick self-testZero-order NH3 [ A ] 0 = 0.10 k = 2.5 × 1 0 − 4 − 1 t end = [ A ] 0 / k = 400 1/ [ A ] [ A ] 0 k t 1/2 M mol L − 1
Mnemonic Pick-the-law reflex"Amount, Ratio, Reciprocal." Har step constant amount khoye → 0th. Har step constant ratio → 1st. Reciprocal mein constant jump → 2nd.
