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

2.8.3 · D5 · Chemistry › Chemical Kinetics › Differential rate equations for 0, 1st, 2nd order — derivati

Yeh ek misconception hunt hai. Neeche har line ek trap hai jo yeh topic set karta hai. Prompt padho, answer zor se bolo reveal karne se pehle, aur — sabse zaroori — check karo ki tumhara reasoning justification se match karta hai, sirf haan/nahi se nahi. Agar tumhe sirf verdict sahi mila lekin reason galat tha, toh tumne kuch nahi seekha.

Machinery mein naye ho? Pehle isse parent note se rebuild karo, aur Rate Laws and Reaction Order, Half-Life Calculations, aur Integrated Rate Laws Applications ko doosre tab mein khula rakho.

True or false — justify

Zero-order reaction ki rate kabhi bhi bilkul nahi badlti, chahe reactant khatam ho jaaye.

False — rate

k

tab tak rahti hai jab tak reactant present hai; jaise hi

[A]

zero ho jaata hai reaction achanak ruk jaati hai, isliye constant-rate line ka ek hard floor

[A] = 0

par hota hai.

First-order reaction ki half-life reaction aage badhne ke saath chhoti hoti jaati hai.

False — first-order half-life

t_{1/2} = 0.693/ k

hai, jisme koi concentration term nahi hai, isliye har successive half-life bilkul same rehti hai chahe kitna bhi bacha ho.

Second-order reaction ke liye, 1.0 M se 0.5 M tak girne ka time 0.5 M se 0.25 M tak girne ke time se kam hota hai.

True — second-order half-life

t_{1/2} = 1/ (k [A]_{0})

badhti hai jaise

[A]_{0}

giraata hai, isliye har successive halving mein zyada time lagta hai; dilute reaction bahut dheere chalti hai.

Rate constant

k

reaction ki ek property hai jo concentration girne ke saath change hoti hai.

False —

k

temperature (aur catalyst) par depend karta hai, concentration par nahi; yahi exactly woh cheez hai jo fixed rehti hai jab concentration change hoti hai, isliye hum rate law ko integrate kar paate hain.

[A]

t

plot par ek straight line guarantee karta hai ki reaction zero-order hai.

True tested reactant ke liye — sirf zero-order hi

[A]

t

mein linear deta hai; first- aur second-order us axis par curve karte hain, isliye yeh specific linear plot zero-order ki pehchaan hai.

- d [A] / d t = k [A]^{n}

mein negative sign ka matlab hai ki rate ek negative number hai.

False —

d [A] / d t

khud negative hota hai (concentration girti hai), isliye minus sign usse flip karta hai taaki rate ek positive quantity bane, jaise rate honi chahiye.

[A]

ko double karne se second-order reaction mein rate double hoti hai.

False — rate

\propto [A]^{2}

hai, isliye concentration double karne se rate chaar guna ho jaati hai; doubling sirf first-order ke liye hoti hai.

k

ki units teeno orders ke liye same hoti hain.

False — dimensional consistency

k = M^{1 - n} s^{- 1}

force karti hai, isliye zero-order

M s^{- 1}

hai, first-order

s^{- 1}

hai, second-order

M^{- 1} s^{- 1}

hai.

Spot the error

"First-order ke liye maine

\int d [A] / [A] = ln [A]

integrate kiya, toh integrated law hai

[A]_{t} = [A]_{0} - k t

Integral sahi hai lekin final form galat hai — logs subtract karte hain, jo

ln [A]_{t} - ln [A]_{0} = - k t

deta hai, yaani

[A]_{t} = [A]_{0} e^{- k t}

; linear-difference form zero-order ka hai.

"Second-order integrate hokar

\frac{1}{[ A ] _{t}} = \frac{1}{[ A ] _{0}} - k t

deta hai kyunki rate law mein minus sign hai."

Galat sign — antiderivative

- [A]^{- 1}

ko limits ke through carry karne par cheezein flip ho jaati hain isliye correct law

\frac{1}{[ A ] _{t}} = \frac{1}{[ A ] _{0}} + k t

hai; reciprocal time ke saath badhta hai kyunki concentration shrink hoti hai.

"Maine

ln [A]

t

plot kiya, straight line mili, toh reaction second-order hai."

Straight

ln [A]

plot first-order signal karta hai; second-order tab hi linearize hota hai jab tum

1/ [A]

t

plot karo — integrated law ka left side hamesha tumhara y-axis hota hai.

"Reaction

2 A \to P

coefficient 2 ki wajah se second-order honi chahiye."

Stoichiometric coefficient order set nahi karta — order experimentally determine hota hai;

2 A \to P

likhi reaction apne mechanism ke hisaab se zero-, first-, ya second-order ho sakti hai.

"Ammonia-on-platinum decomposition ke liye

k

ka value

2.5 \times 1 0^{- 4} s^{- 1}

hai."

Exponent theek hai lekin units galat hain — zero-order

k

M s^{- 1}

carry karta hai, isliye yeh

2.5 \times 1 0^{- 4} M s^{- 1}

hona chahiye; units khud hi order encode karte hain.

"Kyunki rate =

- \frac{1}{a} \frac{d [ A ]}{d t}

, main

\frac{1}{a}

factor jab chaahun drop kar sakta hun."

Sirf tab agar tumhe kisi specific species ke saath

k

match karne ki parwah nahi —

1/ a

, A ke disappearance rate ko reaction rate mein convert karta hai, isliye isse ignore karne par

k

silently

a

factor se rescale ho jaata hai.

"Zero-order ki half-life

0.693/ k

hai, first-order jaisi same idea."

Nahi —

[A]_{t} = [A]_{0} - k t

mein

[A]_{t} = [A]_{0} /2

rakho toh

t_{1/2} = [A]_{0} / (2 k)

milta hai; zero-order half-life initial concentration par depend karti hai, aur

0.693

kabhi nahi aata.

Why questions

Zero-order rate concentration par depend karne se kyun inkaar karti hai, jabki reactant bahut zyada hai?

Rate kisi cheez se limited hoti hai aur kisi reactant abundance se nahi — usually ek saturated catalyst surface ya fixed enzyme/photon supply — isliye extra reactant bas line mein wait karta hai aur cheezein speed up nahi kar sakta.

Natural logarithm specifically first-order law mein kyun appear karta hai aur kahin nahi?

Kyunki

\int d [A] / [A] = ln [A]

— log

1/ [A]

ka unique antiderivative hai, isliye yeh exactly tab appear hone par majboor hota hai jab rate concentration ki first power carry kare.

Second-order rate physically

[A]^{2}

par kyun depend karti hai, sirf algebraically nahi?

Second-order step ke liye do A molecules ko milna padta hai; meeting ki probability (number of A)

\times

(number of A) ke hisaab se scale hoti hai, isliye A double karne par encounters aur hence rate chaar guna ho jaati hai.

First-order half-life dating aur dosing ke liye itni useful kyun hai?

Kyunki yeh concentration-independent hai — fixed time per lost fraction constant hoti hai, isliye "kitna bacha hai" sirf elapsed half-lives ki sankhya par depend karta hai, jo ek clean exponential clock deta hai.

k

ki units order ke saath bilkul kyun badalni chahiye?

Rate hamesha

M s^{- 1}

carry karta hai, lekin

k

[A]^{n}

ko multiply karta hai jo

M^{n}

carry karta hai; product ko

M s^{- 1}

par rakhne ke liye,

k

M^{1 - n}

absorb karna padta hai, isliye uski units automatically order track karti hain.

Integrate karne se pehle variables separate karne ki zahmaat kyun uthate hain?

Integration ke liye

[A]

-wali sab cheezein ek side par aur

t

-wali cheezein doosri side par chahiye; separation hi ek equation jo dono ko link karta hai ko do independent integrals mein todta hai jinhein hum evaluate kar sakte hain.

Wohi reaction kabhi kabhi first-order lagti kyun hai jab woh actually second-order hai?

Agar ek reactant bahut zyada excess mein ho toh uski concentration barely change hoti hai, aur

k

mein pseudo-constant ban ke fold ho jaati hai — yeh "pseudo-first-order" trick ek deliberate simplification hai, true mechanism mein koi change nahi (dekho Rate Laws and Reaction Order).

Edge cases

Zero-order

[A]

-vs-

t

line us moment mein kya karti hai jab reactant exhaust ho jaata hai?

Yeh time

t = [A]_{0} / k

par

[A] = 0

hit karti hai aur rukni padti hai — mathematical line negative ho jaati, lekin concentration nahi ho sakti, isliye physical model wahan truncate hota hai.

Kya first-order concentration kabhi finite time mein exactly zero reach kar sakti hai?

Nahi —

[A]_{t} = [A]_{0} e^{- k t}

zero ki taraf asymptotically approach karta hai; exponential actually kabhi zero touch nahi karta, isliye first-order reactions technically kabhi poori tarah "finish" nahi hoti.

Second-order half-life ka kya hoga agar tum bahut dilute start karo,

[A]_{0} \to 0

ke paas?

t_{1/2} = 1/ (k [A]_{0})

infinity ki taraf blow up karta hai — almost koi molecules nahi hain jo collide karain, reaction vanishingly slow ho jaati hai, jo collision picture se match karta hai.

Agar kisi bhi order ke liye

k = 0

ho — kya reaction ab bhi inhi laws se describe hoti hai?

Haan trivially — har integrated law

[A]_{t} = [A]_{0}

collapse ho jaata hai, matlab koi reaction nahi;

k = 0

"kuch nahi hota" wala degenerate case hai jo sabhi orders mein common hai.

Ek reaction ke liye jo observation time ke comparison mein genuinely instantaneous hai, kaunsa order model apply hota hai?

Koi cleanly nahi — yeh differential laws ek measurable finite rate assume karti hain; essentially instantaneous step diffusion- ya mixing-limited hota hai aur concentration-driven framework se bahar hai.

t = 0

par, kya teeno integrated laws agree karte hain?

Haan — har mein

t = 0

plug karo aur right-hand extra term vanish ho jaata hai, jo

[A]_{t} = [A]_{0}

deta hai (ya

ln [A]_{0}

, ya

1/ [A]_{0}

); yeh sirf time ke saath diverge karte hain, jo exactly wahi cheez hai jo humein orders alag karne deti hai.

Recall Jaane se pehle ek-line self-test

Zor se bolo: har order ko straighten karne ke liye y-axis par kaunsi quantity jaati hai, aur har half-life $[A]_{0}$ par kaise depend karti hai? Answer ::: y-axis hai $[A]$ (0th), $ln [A]$ (1st), $1/ [A]$ (2nd); half-life $\propto [A]_{0}$ (0th), independent (1st), $\propto 1/ [A]_{0}$ (2nd).