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

2.8.6 · Chemistry › Chemical Kinetics

Wo cheez jo hum dhundhne ki koshish kar rahe hain

Neeche sab kuch $n$ ko measurable data se nikalne ki strategy hai.

Method 1 — Initial Rates (Isolation / Differential method)

KAISE (first principles se derivation):

Rate law $r = k [A]^{n}$ lo. Do experiments run karo initial concentrations $[A]_{1}, [A]_{2}$ ke saath aur initial rates $r_{1}, r_{2}$ measure karo (initial, isliye koi product interfere nahi karta aur concentration exactly pata hoti hai):

$\frac{r _{2}}{r _{1}} = \frac{k [ A ] _{2}^{n}}{k [ A ] _{1}^{n}} = (\frac{[ A ] _{2}}{[ A ] _{1}})^{n}$

Dono sides ka log lo — KYUN? $n$ ko exponent se neeche kheenchne ke liye:

$n = \frac{l n ( r _{2} / r _{1} )}{l n ([ A ] _{2} / [ A ] _{1} )}$

Worked example Do-experiment case

Expt	$[A]$ (M)	initial rate (M/s)
1	0.10	$2.0 \times 1 0^{- 3}$
2	0.20	$8.0 \times 1 0^{- 3}$

Step 1: rates ka ratio $= 8.0/2.0 = 4$ . Kyun? Concentration change ke effect ko isolate karta hai. Step 2: concentrations ka ratio $= 0.20/0.10 = 2$ . Step 3: $2^{n} = 4 \Rightarrow n = 2$ . Kyun? Kyunki $ln 4/ ln 2 = 2$ .

Reaction $A$ mein second order hai.

Worked example Multiple reactants (isolation logic)

$A + B \to$ products ke liye data, $r = k [A]^{a} [B]^{b}$ :

Expt	$[A]$	$[B]$	rate
1	0.1	0.1	1.0
2	0.2	0.1	2.0
3	0.1	0.2	4.0

1→2 compare karo: $[B]$ fixed, $[A]$ double, rate double → $2^{a} = 2 \Rightarrow a = 1$ . B ko fix kyun kiya? Taaki sirf $A$ ka exponent dikhे. 1→3 compare karo: $[A]$ fixed, $[B]$ double, rate ×4 → $2^{b} = 4 \Rightarrow b = 2$ . Overall order $= a + b = 3$ .

Method 2 — Integrated Rate Law Method (graphical fitting)

Teeno integrated laws derive karo $A \to$ products ke liye, $r = - \frac{d [ A ]}{d t} = k [A]^{n}$ .

Zero order ( $n = 0$ ): $- \frac{d [ A ]}{d t} = k$ . Integrate karo: $\int_{[A]_{0}}^{[A]} d [A] = - k \int_{0}^{t} d t$ $[A] = [A]_{0} - k t \Rightarrow plot [A] vs t linear hai.$

First order ( $n = 1$ ): $- \frac{d [ A ]}{d t} = k [A] \Rightarrow \int \frac{d [ A ]}{[ A ]} = - k \int d t$ $ln [A] = ln [A]_{0} - k t \Rightarrow plot ln [A] vs t linear hai.$

Second order ( $n = 2$ ): $- \frac{d [ A ]}{d t} = k [A]^{2} \Rightarrow \int \frac{d [ A ]}{[ A ] ^{2}} = - k \int d t$ $\frac{1}{[ A ]} = \frac{1}{[ A ] _{0}} + k t \Rightarrow plot \frac{1}{[ A ]} vs t linear hai.$

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

Method 3 — Half-Life Method

Har order ke liye $t_{1/2}$ derive karo (integrated laws mein $[A] = [A]_{0} /2$ set karo):

Zero order: $\frac{[ A ] _{0}}{2} = [A]_{0} - k t_{1/2} \Rightarrow t_{1/2} = \frac{[ A ] _{0}}{2 k}$ — proportional $[A]_{0}$ se.

First order: $ln \frac{[ A ] _{0} /2}{[ A ] _{0}} = - k t_{1/2} \Rightarrow - ln 2 = - k t_{1/2} \Rightarrow t_{1/2} = \frac{ln 2}{k} = \frac{0.693}{k}$ — $[A]_{0}$ se independent.

Second order: $\frac{2}{[ A ] _{0}} = \frac{1}{[ A ] _{0}} + k t_{1/2} \Rightarrow t_{1/2} = \frac{1}{k [ A ] _{0}}$ — $[A]_{0}$ se inversely proportional.

Exponent rule ki general derivation: $n \neq = 1$ ke liye, integrated law deta hai $t_{1/2} \propto [A]_{0}^{1 - n}$ . Do experiments ka log lo: $n = 1 - \frac{l n ( t _{1/2, 2} / t _{1/2, 1} )}{l n ([ A ] _{0, 2} / [ A ] _{0, 1} )}$

Common Mistakes

Flashcards

Order of reaction determine hota hai

experiment se, stoichiometry se nahi

Do initial-rate experiments se order ka formula

n = \frac{ln ( r _{2} / r _{1} )}{ln ([ A ] _{2} / [ A ] _{1} )}

Integrated method mein, first order ke liye kaun sa plot linear hota hai?

ln [A]

t

Second order ke liye kaun sa plot linear hota hai?

1/ [A]

t

Zero order ke liye kaun sa plot linear hota hai?

[A]

t

ln [A]

t

ka slope barabar hota hai

- k

First-order reaction ki half-life

t_{1/2} = 0.693/ k

(

[A]_{0}

se independent)

Zero-order reaction ki half-life

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

(∝

[A]_{0}

)

Second-order reaction ki half-life

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

(∝

1/ [A]_{0}

)

Concentration par general half-life dependence

t_{1/2} \propto [A]_{0}^{1 - n}

Isolation method mein, A ka order nikalne ke liye aap

baaki sabhi reactants ki concentrations constant rakhte ho

Agar

[A]

double karne se rate ×4 ho jaaye, toh A mein order hai

Agar

[A]_{0}

double karne se

t_{1/2}

unchanged rahe, toh order hai

Recall Feynman: 12-saal ke bacche ko samjhao

Ek car imagine karo jiska speed tank mein kitna fuel hai uske upar depend karta hai. Order bas yeh batata hai ki speed fuel par kitni strongly depend karti hai.

Initial rates: do tanks ko alag-alag bhar do, dekho har car kitni tezi se shuru hoti hai. Agar double fuel se 4× tez ho jaaye, yeh strong (order-2) dependence hai.
Integrated method: time ke saath fuel drop hote dekhte raho aur teen alag graph papers try karo — jo perfectly straight line banaye woh rule bata deta hai.
Half-life: time karo ki tank aadha khaali hone mein kitna waqt lagta hai. Agar woh time kabhi nahi badlta chahe kitna bhi bhar ke shuru karo (order 1), ya chhota/bada hota hai (order 2 ya 0), toh aapne pattern pakad liya. Teeno tricks ko ek hi answer point karna chahiye.

Connections

Rate Law and Rate Constant — woh equation jo yeh methods extract karte hain.
Integrated Rate Equations — poori derivations jo yahan reuse hoti hain.
Half-Life of Reactions
Elementary vs Complex Reactions — kyun order ≠ molecularity.
Arrhenius Equation — jab $k$ mil jaaye, temperature dependence.
Radioactive Decay — classic first-order constant-half-life case.

Order	Linear plot	Slope
0	$[A]$ vs $t$	$- k$
1	$ln [A]$ vs $t$	$- k$
2	$1/ [A]$ vs $t$	$+ k$

Order	$t_{1/2}$	$[A]_{0}$ par dependence
0	$[A]_{0} /2 k$	$[A]_{0}$ badhne par ↑
1	$0.693/ k$	constant
2	$1/ k [A]_{0}$	$[A]_{0}$ badhne par ↓