Differential rate equations for 0, 1st, 2nd order — derivations: Exercises

2.8.3 · D4 · Chemistry › Chemical Kinetics › Differential rate equations for 0, 1st, 2nd order — derivati

Yeh parent derivations ka self-test hai. Pehle har problem paper pe karo, PHIR collapsible solution kholo. Exercises ek ladder climb karti hain: L1 Recognition (order pehchano) → L2 Application (law mein plug karo) → L3 Analysis (data se $k$ nikalo) → L4 Synthesis (ideas combine karo) → L5 Mastery (design/prove karo).

Shuru karne se pehle, ek figure har symbol fix karti hai jo tum use karoge — ise ek baar padh lo taaki neeche kuch bhi surprise na ho.

Teen laws jinpe hum rely karte hain, sab parent mein derive kiye gaye hain:

Level 1 — Recognition

Exercise 1.1 (L1)

Ek reaction ka rate constant $s^{- 1}$ units mein hai. Iski order kya hai? Kaunsa plot straight line deta hai?

Recall Solution

HUM KYA POOCH RAHE HAIN: $k$ ki units ek fingerprint ki tarah kaam karti hain. Rate ki units hamesha $M s^{- 1}$ hoti hain, aur $k [A]^{n} = M s^{- 1}$ , isliye $k = \frac{M s ^{- 1}}{M ^{n}} = M^{1 - n} s^{- 1} .$ KYUN: $M^{1 - n}$ ko diye gaye $s^{- 1}$ (yaani $M^{0}$ ) se match karne pe $1 - n = 0$ force hota hai, toh $n = 1$ . Answer: first-order. Straight-line plot $ln [A]$ vs $t$ hai (slope $- k$ ).

Exercise 1.2 (L1)

Neeche concentration data ke teen sketches hain. Kaun sa order har ek se match karta hai? Neeche di gayi figure dekho.

Recall Solution

Trick yeh hai: har order sirf tab straight jaata hai jab y-axis ka sahi transform kiya jaaye.

Left panel — $[A]$ vs $t$ seedha girta hua line hai → zero-order.
Middle panel — $ln [A]$ vs $t$ straight hai → first-order.
Right panel — $\frac{1}{[ A ]}$ vs $t$ straight hai → second-order. KYUN: integrated law ka left side HI tumhara y-axis hai. Zero mein $[A]$ use hota hai, first mein $ln [A]$ , second mein $1/ [A]$ .

Level 2 — Application

Exercise 2.1 (L2)

Ek zero-order reaction mein $[A]_{0} = 0.500 M$ aur $k = 0.010 M s^{- 1}$ hai. $20 s$ baad $[A]$ kya hoga?

Recall Solution

$[A]_{t} = [A]_{0} - k t$ use karo. $[A]_{t} = 0.500 - (0.010) (20) = 0.500 - 0.200 = 0.300 M .$ KYUN subtraction: zero-order mein reactant ek fixed pace $k$ se consume hota hai, toh har second same chunk remove hota hai — seedha line neeche jaati hai.

Exercise 2.2 (L2)

Ek first-order reaction mein $k = 0.0693 s^{- 1}$ hai. $[A]_{0} = 1.00 M$ se start karke, $30 s$ baad $[A]$ kya hoga?

Recall Solution

$[A]_{t} = [A]_{0} e^{- k t}$ use karo. $k t = (0.0693) (30) = 2.079.$ $[A]_{t} = 1.00 e^{- 2.079} = 0.125 M .$ KYUN exponential (subtraction nahi): yahan pace proportional hai jo bacha hua hai, toh jaise jaise kam hota hai waise waise slow hota jaata hai — mathematically yeh "self-shrinking" decay exactly $e^{- k t}$ hai. Note karo $2.079 = 3 \times 0.693$ , toh yeh exactly 3 half-lives hai: $1 \to \frac{1}{2} \to \frac{1}{4} \to \frac{1}{8} = 0.125 M$ . Dekho Half-Life Calculations.

Exercise 2.3 (L2)

Ek second-order reaction mein $k = 0.50 M^{- 1} s^{- 1}$ aur $[A]_{0} = 0.20 M$ hai. $10 s$ baad $[A]$ kya hoga?

Recall Solution

$\frac{1}{[ A ] _{t}} = \frac{1}{[ A ] _{0}} + k t$ use karo. $\frac{1}{[ A ] _{t}} = \frac{1}{0.20} + (0.50) (10) = 5.0 + 5.0 = 10.0 M^{- 1} .$ $[A]_{t} = \frac{1}{10.0} = 0.10 M .$ KYUN reciprocals: second-order $1/ [A]$ mein linearise hota hai, toh tum reciprocal mein add karte ho, phir end mein flip karte ho.

Level 3 — Analysis

Exercise 3.1 (L3)

Ek first-order reaction mein $[A]$ $60 s$ mein $0.80 M$ se $0.20 M$ tak girta hai. $k$ aur $t_{1/2}$ nikalo.

Recall Solution

Step 1 — $k$ nikalo. $ln \frac{[ A ] _{t}}{[ A ] _{0}} = - k t$ use karo. $ln \frac{0.20}{0.80} = ln (0.25) = - 1.386 = - k (60) .$ $k = \frac{1.386}{60} = 0.0231 s^{- 1} .$ Step 2 — half-life. $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0231} = 30 s .$ Sanity check (elegant tarika): $0.80 \to 0.40 \to 0.20$ exactly 2 half-lives hai $60 s$ mein, toh $t_{1/2} = 30 s$ — match karta hai. Dekho Integrated Rate Laws Applications.

Exercise 3.2 (L3)

Ek zero-order reaction $[A]_{0} = 0.60 M$ se start hoti hai aur $120 s$ mein $0.00 M$ tak pahunchti hai. $k$ aur $t_{1/2}$ nikalo.

Recall Solution

Zero-order mein kya special hai: yeh actually ek finite time $t_{end} = [A]_{0} / k$ pe zero hit karta hai (first/second order ke ulta, jo sirf zero ko approach karte hain). $[A]_{t} = 0$ set karo: $0 = 0.60 - k (120) \Rightarrow k = \frac{0.60}{120} = 0.0050 M s^{- 1} .$ Half-life: $t_{1/2} = \frac{[ A ] _{0}}{2 k} = \frac{0.60}{2 ( 0.0050 )} = 60 s .$ KYUN yeh $t_{end}$ ka aadha hai: line constant speed se girti hai, toh halfway concentration tak exactly aadhe total time mein pahunchti hai. Clean geometry.

Level 4 — Synthesis

Exercise 4.1 (L4)

Ek first-order reaction $50 min$ mein $75%$ complete ho jaati hai. $99%$ complete hone mein kitna time lagega?

Recall Solution

"75% complete" ka matlab: $25%$ bacha hai, toh $[A]_{t} / [A]_{0} = 0.25$ . Step 1 — $k$ nikalo: $ln (0.25) = - k (50) \Rightarrow k = \frac{1.386}{50} = 0.02773 min^{- 1} .$ Step 2 — 99% complete ke liye time ( $1%$ bacha hai, ratio $= 0.01$ ): $ln (0.01) = - k t \Rightarrow t = \frac{4.605}{0.02773} = 166 min .$ Elegant check: $75%$ complete = 2 half-lives (kyunki $0.25 = (1/2)^{2}$ ), toh $t_{1/2} = 25 min$ . $99%$ complete matlab lagbhag $0.01 = (1/2)^{6.64}$ , yaani $6.64$ half-lives $= 6.64 \times 25 \approx 166 min$ . Dono routes agree karte hain.

Exercise 4.2 (L4)

Do reactions same $[A]_{0} = 0.10 M$ se start hoti hain. Reaction P second-order hai $k_{P} = 2.0 M^{- 1} s^{- 1}$ ke saath. Reaction Q first-order hai $k_{Q} = 0.20 s^{- 1}$ ke saath. Kiski pehli half-life shorter hai, aur kitni zyada?

Recall Solution

Reaction P (second-order): $t_{1/2} = \frac{1}{k _{P} [ A ] _{0}} = \frac{1}{( 2.0 ) ( 0.10 )} = 5.0 s .$ Reaction Q (first-order): $t_{1/2} = \frac{0.693}{k _{Q}} = \frac{0.693}{0.20} = 3.465 s .$ Answer: Q shorter hai $5.0 - 3.465 = 1.535 s \approx 1.54 s$ se. KYUN comparison matter karta hai: second-order half-life $[A]_{0}$ pe depend karti hai jabki first-order nahi — toh starting concentration change karo toh winner flip ho sakta hai. Yeh concentration-sensitivity Collision Theory ke collision picture se aati hai.

Level 5 — Mastery

Exercise 5.1 (L5)

Prove karo ki first-order reaction ke liye half-life starting concentration se independent hoti hai, lekin second-order reaction ke liye half-life double ho jaati hai har baar jab tum $[A]_{0}$ ko aadha karo.

Recall Solution

First-order. $ln ([A]_{t} / [A]_{0}) = - k t$ mein $[A]_{t} = \frac{1}{2} [A]_{0}$ set karo: $ln (\frac{1}{2}) = - k t_{1/2} \Rightarrow t_{1/2} = \frac{l n 2}{k} .$ KYUN concentration-free hai: $[A]_{0}$ ratio ke andar cancel ho gaya, sirf constants bache. Kuch bhi aadha karne mein same time lagta hai. Second-order. $\frac{1}{[ A ] _{t}} - \frac{1}{[ A ] _{0}} = k t$ mein $[A]_{t} = \frac{1}{2} [A]_{0}$ set karo: $\frac{2}{[ A ] _{0}} - \frac{1}{[ A ] _{0}} = k t_{1/2} \Rightarrow \frac{1}{[ A ] _{0}} = k t_{1/2} \Rightarrow t_{1/2} = \frac{1}{k [ A ] _{0}} .$ KYUN double hota hai: $t_{1/2} \propto 1/ [A]_{0}$ . $[A]_{0}$ aadha karo toh $t_{1/2}$ double ho jaata hai. Physically, kam molecules matlab kam collisions per second (rate $\propto [A]^{2}$ ), toh har successive half zyada time leta hai.

Exercise 5.2 (L5)

Ek student zero-order reaction measure karta hai: $[A]_{0} = 0.40 M$ , $k = 0.020 M s^{- 1}$ . (a) Exact time do jab reaction khatam hogi. (b) $t = 30 s$ pe $[A]$ kya hai? (c) $t = 25 s$ pe? Explain karo kyun ek answer carefully report karna chahiye.

Recall Solution

(a) Finish time. Reaction tab khatam hoti hai jab $[A] = 0$ : $t_{end} = \frac{[ A ] _{0}}{k} = \frac{0.40}{0.020} = 20 s .$ (b) $t = 30 s$ pe. Naive formula deta hai $0.40 - (0.020) (30) = 0.40 - 0.60 = - 0.20 M$ — impossible. Concentration negative nahi ho sakti. Jab $t > t_{end}$ , saara A khatam ho gaya hai, toh $[A] = 0 M$ . (c) $t = 25 s$ pe. Same reasoning: $25 > 20$ , toh $[A] = 0 M$ . KYUN carefully report karo: zero-order line $[A]_{0} - k t$ sirf tab valid hai jab reactant bacha ho ( $t \leq t_{end}$ ). Iske baad physics ise zero pe floor kar deti hai — ek degenerate case jo blind formula ignore karta hai.

Recall Self-Check Scoreboard

Pehli try mein kitne nikale bina jhankne ke? Level 1 correct matlab tum order identify kar sakte ho. Level 5 correct matlab tum laws prove aur defend kar sakte ho. Sabse weak level ::: parent note ka woh section revisit karo, phir sirf usi level ke problems dobara try karo.