Chemical Kinetics: Class 12 Chemistry Chapter 3 Hinglish mein

Chemistry Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

90 min
44 practice questions
19 MCQs

Is chapter mein aap kya seekhenge

Reaction ka average aur instantaneous rate define aur calculate karo, aur balanced equation ke coefficients use karke ek rate ko kisi bhi reactant ya product ke terms mein likho
Rate law likho, experimental data se reaction ka order nikaalo aur uske rate constant ki units nikaalo
Elementary aur complex reactions mein, aur molecularity aur order mein fark samjho, aur simple mechanism ka rate-determining step pehchaano
Zero- aur first-order reactions ki integrated rate equations aur half-lives derive aur use karo, pseudo first-order reactions aur total-pressure data samet
Arrhenius equation se activation energy aur pre-exponential factor nikaalo, aur do temperatures par rate constants ko relate karo
Energy profiles aur energy distributions se samjhao ki temperature aur catalysts reaction ka rate kaise badalte hain
Collision theory samjhao, jismein collision frequency, activation energy aur steric factor shaamil hain

NCERT Chemistry Class 12, Chapter 3, pp. 61–88 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Table par rakhi newspaper ki sheet, chemistry ki nazar se, oxygen mein padi ek fuel hai. Use jalane se bahut saari energy nikalegi, aur iska $Δ G$ bada aur negative hai, phir bhi paper saalon tak waise hi pada reh sakta hai. Hydrogen aur oxygen ka mixture bhi aisa hi hai: woh bina badle intezaar karta hai, jab tak ek spark use dhamake ke saath shuru na kar de. React karne ki ijaazat hona aur sach mein noticeable speed se react karna, do alag baatein hain.

Thermodynamics batata hai ki koi change allowed hai ya nahi, aur equilibrium batata hai ki woh kitni door tak ja sakta hai; dono mein time nahi hai. Jo sawaal ek cook, engineer ya doctor aksar sabse pehle poochta hai, kitni tezi se?, woh chemical kinetics (रासायनिक बलगतिकी) ka hai. Naam Greek shabd kinesis, yaani movement, se aaya hai. Kinetics hi tay karta hai ki cold box ke bahar vaccine kitni der tak potent rehti hai, cement floor par kitni jaldi chal sakte hain, aur harmful gases ban hone ke baad ozone layer kitni jaldi theek hoti hai.

Yeh chapter pehle flask mein rates naapta hai aur unke laws dhoondhta hai, phir single collisions par zoom karke dekhta hai ki temperature aur catalysts itna farak kyun dalte hain.

3.1 Rate of a Chemical Reaction

Kuch reactions palak jhapakte hi khatam ho jaati hain; kuch mein poori zindagi lag jaati hai. Acid ko alkali ke saath milane par, ya firecracker jalane par, ek second se bhi kaafi kam mein products ban jaate hain. Kata hua apple minutes mein brown ho jaata hai, aur doodh ko dahi banne mein kai ghante lagte hain. Chamakti copper statue ko hara hone mein saal lagte hain. Fig. 3.1 inhe ek hi time line par dikhata hai.

3/3Copper surface ko hara hone mein saal lagte hain.

Fig. 3.1Reaction times das se zyada powers of ten tak phaile hain; axis logarithmic hai, isliye har tick pichhle wale se kai guna hai.NCERT p. 62

Ek horizontal time axis jis par 1 second, 1 minute, 1 hour, 1 day, 1 year aur 100 years par ticks hain, powers of ten mein barabar doori par. Step 1, fast: left end par do markers, acid plus alkali ek millisecond se kam par aur firecracker second ke kuch sauvein hisse par, fast label wale shaded band ke andar. Step 2, moderate: kate apple ke brown hone ka marker lagbhag pandrah minute (quarter hour) par, aur dahi jamne ka lagbhag chhe ghante par, moderate label wale band ke andar. Step 3, slow: copper statue ke lagbhag bees saal baad hara hone ka marker, slow label wale band ke andar, aur saare markers ab bhi dikh rahe hain.

Reactions ko compare karne ke liye humein ek number chahiye. Socho ek water tank bucket mein khaali ho raha hai: tum bata sakte ho ki woh kitni tezi se khaali ho raha hai, ya to is se ki tank har minute kitne litre khota hai, ya is se ki bucket kitne litre paata hai. Reaction ko bhi aise hi describe karte hain: reactant ka concentration kitni tezi se girta hai ya product ka kitni tezi se badhta hai.

Reactant ko R aur product ko P kaho, taaki reaction $R \overset{}{\to} P$ ho, har R se ek P, aur volume fixed rakho. Square brackets ka matlab hai moles per litre. Maan lo time $t_{1}$ par reading $[R]_{1}$ deti hai aur baad mein $t_{2}$ par reading $[R]_{2}$ deti hai. Gap $Δ t = t_{2} - t_{1}$ mein concentration $Δ [R] = [R]_{2} - [R]_{1}$ se badla hai, aur product ka $Δ [P]$ se. Time se divide karne par average rate milta hai:

R ke gayab hone ka average rate

r_{a v} = - \frac{Δ [R]}{Δ t}

(3.1)

P ke banne ka average rate

r_{a v} = + \frac{Δ [P]}{Δ t}

(3.2)

$Δ [R]$ negative hai kyunki R use ho raha hai, isliye (3.1) ka minus sign use positive rate bana deta hai. Dono expressions average rate $r_{a v}$ dete hain: poore interval ke liye ek hi number, chahe uske andar speed kitni bhi badli ho.

Definition

Rate of a reaction

Koi reactant kitni tezi se use hota hai ya product kitni tezi se banta hai, jise per unit time molar concentration ke positive change ke roop mein naapte hain aur balanced equation mein us species ke coefficient se divide karte hain.

Example $R \overset{}{\to} P$ ke liye, rate $= - Δ [R] / Δ t = Δ [P] / Δ t$ .

Rate ki units. Rate concentration ka change divided by time hai, isliye uski unit concentration unit per time unit hoti hai: zyadatar problems mein $m o l L^{- 1} s^{- 1}$ , ya slow reactions ke liye per minute ya per hour. Gas reactions ko aksar pressure se follow karte hain, jisse $a t m s^{- 1}$ ya $b a r m i n^{- 1}$ jaisi units milti hain.

3/4 $Δ t$ ko zero tak chhota karo to chord $t$ par tangent ban jaati hai: $r_{i n s t} = - d [R] / d t = d [P] / d t$ .

Fig. 3.2Average rate chord ka slope hai; instantaneous rate ek pal par tangent ka slope hai.NCERT Fig. 3.1 · p. 63

Concentration vs time ka graph. Reactant curve R left mein upar se shuru hota hai aur aur bhi dheere girta jaata hai; product curve P zero se shuru hokar mirror image ki tarah badhta hai. Step 1 sirf do curves dikhata hai. Step 2 R curve ke do points, times t1 aur t2, ko jodne wali seedhi chord jodta hai, horizontal gap par delta t aur vertical girawat par delta R mark hai; iska slope average rate deta hai. Step 3 chord ki jagah ek tangent laata hai jo R curve ko ek hi time t par chhooti hai, aur P curve par bhi ek matching tangent; inke slopes instantaneous rate dete hain. Step 4 baad ke time par ek doosri tangent jodta hai jo kaafi flat hai, jisse dikhta hai ki reaction aage badhne par rate girta hai.

Average rate yeh baat chhupa leta hai ki zyadatar reactions reactants use hone ke saath dheemi hoti jaati hain. Ek pal ka rate nikaalne ke liye interval chhota karo. Jaise $Δ t$ zero ki taraf jaata hai, concentration–time curve ke do points ke beech ki chord ek single point par tangent ban jaati hai, aur changes ka ratio derivative ban jaata hai:

Instantaneous rate

r_{i n s t} = - \frac{d [R]}{d t} = \frac{d [P]}{d t}

(3.3)

Graph mein, time $t$ par instantaneous rate $[P]$ – $t$ curve par $t$ wali tangent ka slope hai, ya $[R]$ – $t$ curve par tangent ke slope ka minus (Fig. 3.2). Reaction ki shuruaat mein tangents steep hoti hain; baad mein flat ho jaati hain. Section 3.3.3 ki rate lab mein tum ek hi curve par chord aur tangent bana sakte ho aur interval chhota karne par dono ko match hote dekh sakte ho.

Table 3.1Basic solution mein crystal violet ka fade hona (ek run)NCERT Table 3.1 · p. 64

$t$ / s	$[C V^{+}]$ / $10^{- 6} m o l L^{- 1}$
0	20.0
60	16.7
120	14.0
180	11.7
240	9.7
360	6.8
480	4.7
600	3.3

Example 3.1NCERT Example 3.1 · p. 63

Data table se average rates

Crystal violet ( $C V^{+}$ ) ek purple dye hai. Basic solution mein yeh hydroxide ions ke saath dheere-dheere react karke colourless product deta hai, isliye colorimeter iska concentration follow kar sakta hai. Table 3.1 ek run dikhata hai. Har interval mein dye ke ghatne ka average rate nikaalo, aur batao ki reaction aage badhne par rate kaise badalta hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Jab coefficients alag hon. $N_{2} (g) + 3 H_{2} (g) \overset{}{\to} 2 N H_{3} (g)$ mein, react karne wala har nitrogen molecule apne saath teen hydrogen molecules leta hai aur do ammonia molecules banata hai. Isliye kisi bhi interval mein $[H_{2}]$ , $[N_{2}]$ se teen guna tezi se girta hai, aur $[N H_{3}]$ do guna tezi se badhta hai. Ek hi reaction ko teen alag numbers describe karenge. Ek single rate paane ke liye har species ke rate of change ko balanced equation mein uske coefficient se divide karo (Fig. 3.3):

\begin{aligned} Rate & = - \frac{Δ [N_{2}]}{Δ t} \\ = - \frac{1}{3} \frac{Δ [H_{2}]}{Δ t} \\ = + \frac{1}{2} \frac{Δ [N H_{3}]}{Δ t} \end{aligned}

3/3Har per-second change ko uske coefficient se divide karo: teeno $0.010 m o l L^{- 1} s^{- 1}$ dete hain.

Fig. 3.3Alag species alag speed se badalti hain, lekin coefficients se divide karne par reaction ka ek hi rate milta hai.NCERT p. 65

Reaction N2 plus 3 H2 se 2 NH3 mein nitrogen, hydrogen aur ammonia ke liye teen bars saath-saath khade hain. Step 1, shuruaat mein: nitrogen 0.50, hydrogen 1.50 aur ammonia 0 moles per litre. Step 2, das second baad: nitrogen bar 0.40 tak, hydrogen bar 1.20 tak sikud gaya hai aur ammonia bar 0.20 tak badh gaya hai, aur unke upar changes minus 0.10, minus 0.30 aur plus 0.20 likhe hain. Step 3: har bar ke neeche per-second change ko coefficient 1, 3 aur 2 se divide kiya gaya hai, aur teeno same rate dete hain, 0.010 moles per litre per second.

Yahi rule kisi bhi reaction $a A + b B \to c C + d D$ par kaam karta hai: har species ka rate of change lo, use uske apne coefficient se divide karo, aur reactants ke aage minus sign lagao.

\begin{aligned} Rate & = - \frac{1}{a} \frac{Δ [A]}{Δ t} \\ = - \frac{1}{b} \frac{Δ [B]}{Δ t} \\ = + \frac{1}{c} \frac{Δ [C]}{Δ t} \\ = + \frac{1}{d} \frac{Δ [D]}{Δ t} \end{aligned}

Gases mein ek suvidha aur hai. Kyunki $p = c R T$ , fixed temperature par partial pressure bas bhes badla hua concentration hai, isliye upar ke har rate ko $Δ c$ ki jagah $Δ p$ se likh sakte hain.

Example 3.2NCERT Example 3.2 · p. 66

Ek reaction, kai rates

Nitrosyl bromide gas phase mein decompose hota hai: $2 N O B r (g) \overset{}{\to} 2 N O (g) + B r_{2} (g)$ . Ek experiment mein $[N O B r]$ , $25.0 m i n$ mein $0.0450$ se $0.0390 m o l L^{- 1}$ tak gira. Reaction ka average rate per minute, per hour aur per second nikaalo, aur woh rate bhi jis par NO bana.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 2 questions

Practice: reaction rates (Intext Questions 3.1–3.2): answers ke saath

1
NumericalEasyNCERT Intext Question 3.1
Ek reaction $R \overset{}{\to} P$ mein $[R]$ , $15 m i n$ mein $0.048$ se $0.036 m o l L^{- 1}$ tak girta hai.
1. Average rate $m o l L^{- 1} m i n^{- 1}$ mein nikaalo.
  mol L⁻¹ min⁻¹
  Solution dikhao
  $r_{a v} = \frac{0.048 - 0.036}{15} = 8.0 \times 10^{- 4}$
  $m o l L^{- 1} m i n^{- 1}$ mein.
2. Ise $m o l L^{- 1} s^{- 1}$ mein likho.
  mol L⁻¹ s⁻¹
  Solution dikhao
  60 s per minute se divide karo: $8.0 \times 10^{- 4} / 60 = 1.3 \times 10^{- 5} m o l L^{- 1} s^{- 1}$ .
2
NumericalEasyNCERT Intext Question 3.2
Reaction $3 A \overset{}{\to} p r o d u c t s$ ke liye $[A]$ , $12 m i n$ mein $0.90$ se $0.72 m o l L^{- 1}$ tak girta hai.
1. A ke gayab hone ka average rate $m o l L^{- 1} m i n^{- 1}$ mein nikaalo.
  mol L⁻¹ min⁻¹
  Solution dikhao
  $- \frac{Δ [A]}{Δ t} = \frac{0.18}{12} = 0.015$
2. Reaction ka average rate $m o l L^{- 1} m i n^{- 1}$ mein nikaalo.
  mol L⁻¹ min⁻¹
  Solution dikhao
  A ka coefficient 3 hai, isliye reaction ka rate $0.015 / 3 = 5.0 \times 10^{- 3} m o l L^{- 1} m i n^{- 1}$ hai.

3.2 Factors Influencing Rate of a Reaction

Pisi hui tablet poori tablet se jaldi fizz karke kyun khatam ho jaati hai, aur fridge mein doodh zyada der tak kyun tikta hai? Kisi reaction ka rate uski conditions par depend karta hai: reactants ke concentrations (gases ke liye unke pressures), temperature, catalyst ki maujoodgi, aur jab koi solid shaamil ho to woh kitna barik pisa hai. Yeh section concentration ki baat karta hai; temperature aur catalysts Section 3.4 mein aate hain.

Try karke dekho

Fizz karti tablet ki race

Tumhe chahiye

ek hi brand ki chaar effervescent antacid ya vitamin C tablets
paani ke chaar ek jaise glass: do thande, do garam
ek chammach aur ek saaf plate
phone ka stopwatch

Pehle guess karo

Kaunse glass mein fizzing sabse pehle rukegi?

thanda paani, poori tabletthanda paani, pisi tabletgaram paani, poori tabletgaram paani, pisi tablet

Do tablets ko saaf plate par pees kar powder bana lo; baaki do ko poora rakho.
Ek poori tablet thande paani ke glass mein daalo aur stopwatch shuru karo. Fizzing khatam hone par use roko.
Thande paani mein pisi tablet ke saath dohrao, phir garam paani mein ek poori aur ek pisi tablet ke saath.
Chaaron times table mein likho.

Table 3.2

Glass	Fizzing rukne ka time (s)
Thanda paani, poori tablet
Thanda paani, pisi tablet
Garam paani, poori tablet
Garam paani, pisi tablet

Tumhari entries apne aap save hoti hain.

3.2.1 Dependence of Rate on Concentration

Example 3.1 ka data pehle hi pattern dikhata hai: jaise-jaise dye use hoti hai, reaction dheemi hoti hai. Wahi reaction zyada dye se shuru karo to shuruaati rate zyada hota hai. Zyadatar reactions mein zyada concentrated reactants ka matlab hai tez reaction.

Chemists is dependence ko ek rate law (वेग नियम) mein pakadte hain, jise rate equation ya rate expression bhi kehte hain: ek formula jiske ek side rate hai aur doosri side reactant concentrations.

Definition

Rate law

Experiment se nikaali gayi equation jo rate ko reactants ke concentrations se jodti hai, aur har concentration ki apni power hoti hai.

Example $2 N O + 2 H_{2} \overset{}{\to} N_{2} + 2 H_{2} O$ ke liye $Rate = k [N O]^{2} [H_{2}]$ .

3.2.2 Rate Expression and Rate Constant

General reaction $a A + b B \to c C + d D$ ke liye experiments aam taur par dikhate hain ki rate har reactant ke concentration ki kisi power ke proportional hota hai:

Rate \propto [A]^{x} [B]^{y}

(3.4)

Proportionality sign ki jagah constant $k$ lagane par milta hai

Rate = k [A]^{x} [B]^{y}

(3.4a)

aur reactant R ke ghatne ke instantaneous rate se likhne par,

Differential rate equation

- \frac{d [R]}{d t} = k [A]^{x} [B]^{y}

(3.4b)

(3.4b) wale form ka rate law ek differential rate equation hai. Constant $k$ rate constant (वेग स्थिरांक) hai. Diye gaye temperature par reaction ke liye iski ek hi value hoti hai, concentrations chahe jo hon. (3.4a) mein har concentration $1 m o l L^{- 1}$ rakhne se iska matlab samajh aata hai: $k$ unit concentrations par rate hai.

Definition

Rate constant

Rate law mein proportionality constant $k$ . Yeh reaction aur temperature par depend karta hai, concentrations par nahi.

Powers $x$ aur $y$ ka coefficients $a$ aur $b$ ke barabar hona zaroori nahi; inhe sirf experiment se nikaala ja sakta hai. Ek simple plan, method of initial rates, milaane ke turant baad rate naapta hai, jab tak concentrations zyada nahi badle. Runs ki ek series mein ek starting concentration badla jaata hai aur baaki same rakhe jaate hain (Fig. 3.4).

3/3 $[H_{2}]$ teen guna karne se yeh 3 guna tez hoti hai: $3^{y} = 3$ , isliye $y = 1$ .

Fig. 3.4Ek baar mein ek hi starting concentration badlo aur dekho ki reaction kitne guna tez shuru hoti hai.NCERT Table 3.2 · p. 67

Teen vertical bars initial rates dikhate hain. Step 1: sirf run 1, reference, jismein dono starting concentrations 0.010 moles per litre hain; iska bar ek unit lamba hai. Step 2: run 2 aata hai, nitrogen monoxide double aur hydrogen same; iska bar chaar units lamba hai, us par times 4 likha hai, jo 2 squared ke barabar hai, isliye reaction nitrogen monoxide mein second order hai. Step 3: run 3 aata hai, nitrogen monoxide wapas 0.010 par aur hydrogen teen guna; iska bar teen units lamba hai, us par times 3 likha hai, jo 3 to the power 1 ke barabar hai, isliye reaction hydrogen mein first order hai.

Table 3.3Ek high temperature par

2 N O + 2 H_{2} \overset{}{\to}

N_{2} + 2 H_{2} O

ke initial rates (concentrations

m o l L^{- 1}

mein, rates

m o l L^{- 1} s^{- 1}

mein)

Run	$[N O]$	$[H_{2}]$	Initial rate
1	0.010	0.010	$2.4 \times 10^{- 6}$
2	0.020	0.010	$9.6 \times 10^{- 6}$
3	0.010	0.030	$7.2 \times 10^{- 6}$
4	0.030	0.020	$4.32 \times 10^{- 5}$

ExampleNCERT Table 3.2 · p. 67

Initial rates se rate law nikaalna

High temperature par nitrogen monoxide hydrogen ke saath react karta hai: $2 N O (g) + 2 H_{2} (g) \overset{}{\to}$ $N_{2} (g) + 2 H_{2} O (g)$ . Table 3.3 ke chaar runs se rate law aur rate constant nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Dhyaan do ki $[H_{2}]$ ki power 1 hai, jabki uska coefficient 2 hai. Balanced equation atoms ka overall hisaab-kitaab batata hai. Woh yeh nahi batata ki rate concentration par kaise depend karta hai.

Table 3.4Rate laws coefficients se nahi nikalteNCERT p. 68

Reaction	Rate law (naapa gaya)
$2 N O + 2 H_{2} \overset{}{\to}$ $N_{2} + 2 H_{2} O$	$k [N O]^{2} [H_{2}]$
$H_{2} + B r_{2} \overset{}{\to}$ $2 H B r$	$k [H_{2}] [B r_{2}]^{1 / 2}$ (shuruaat mein)
$N O_{2} + C O \overset{}{\to}$ $N O + C O_{2}$	$k [N O_{2}]^{2}$ (lagbhag 500 K se neeche)

Har row mein powers coefficients se alag hain; aakhri wali mein CO rate law mein aata hi nahi. Rate law balanced equation dekh kar nahi likha ja sakta; ise experiment se hi nikaalna padta hai.

3.2.3 Order of a Reaction

Rate law ki powers batati hain ki har concentration rate ko kitni zor se chalata hai. $Rate = k [A]^{x} [B]^{y}$ mein $x$ A ke respect mein order hai aur $y$ B ke respect mein order. Inka sum $x + y$ reaction ka overall order of the reaction (अभिक्रिया की कोटि) hai.

Definition

Order of a reaction

Measured rate law mein concentration terms ke exponents ka total.

Example $Rate = k [N O]^{2} [H_{2}]$ : NO mein second order, $H_{2}$ mein first order, overall third order.

Kyunki order naapa jaata hai, uska whole number hona zaroori nahi. Aam taur par yeh 1, 2 ya 3 hota hai, lekin yeh zero, fraction, ya negative bhi ho sakta hai jab koi substance apni hi reaction ko dheema kare. Kisi reactant ke respect mein zero order ka matlab hai ki uske concentration ka koi asar nahi: use double karne se kuch nahi badalta.

Example 3.3NCERT Example 3.3 · p. 68

Rate law se overall order

Har reaction ka overall order uske measured rate law se nikaalo. (a) $2 N O + 2 H_{2} \overset{}{\to} N_{2} + 2 H_{2} O$ , $Rate = k [N O]^{2} [H_{2}]$ . (b) shuruaat mein $H_{2} + B r_{2} \overset{}{\to} 2 H B r$ , $Rate = k [H_{2}] [B r_{2}]^{1 / 2}$ . (c) $2 O_{3} \overset{}{\to} 3 O_{2}$ , $Rate = k [O_{3}]^{2} [O_{2}]^{- 1}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Powers coefficients se alag kyun hon? Jawaab us cheez mein hai jo balanced equation chhupa leta hai. Zyadatar reactions ek jhatke mein nahi hotin: woh simple steps ki seedhi chadhti hain, aur un steps ki list reaction ka mechanism hai. Jis reaction ki equation aisa ek single step bilkul waise hi describe kare jaise woh hota hai, woh elementary reaction (प्राथमिक अभिक्रिया) hai; jise kai steps chahiye, woh complex reaction hai.

Definition

Elementary reaction

Aisi reaction jo ek hi step mein hoti hai, jismein reacting species bilkul waise hi takraati hain jaise uski equation mein dikhaya hai.

Complex reactions kai patterns follow karti hain. Consecutive reactions mein ek step ka product agle step ka reactant hota hai: liver mein ethanol oxidise hokar ethanal banta hai, aur phir ethanal oxidise hokar ethanoic acid banta hai. Reversible reactions mein products phir se reactants bana dete hain, jaise $N_{2} + 3 H_{2} \overset{}{⇌} 2 N H_{3}$ mein. Side (parallel) reactions mein reactants ka ek set ek saath kai products deta hai: methane ka chlorination $C H_{3} C l$ ke saath $C H_{2} C l_{2}$ , $C H C l_{3}$ aur $C C l_{4}$ bhi deta hai.

Rate constant ki units. $k$ ki unit overall order $n = x + y$ par depend karti hai. (3.4a) se $k = Rate / ([A]^{x} [B]^{y})$ , yaani rate divided by concentration to the power $n$ . Concentration $m o l L^{- 1}$ mein aur time seconds mein ho to yeh milta hai:

Order n ke liye rate constant ki unit

unit of k = (m o l L^{- 1})^{1 - n} s^{- 1}

Table 3.5Common orders ke liye

k

ki units (time seconds mein)NCERT Table 3.3 · p. 69

Overall order $n$	$k$ ki unit
0	$m o l L^{- 1} s^{- 1}$
1	$s^{- 1}$
3/2	$L^{1 / 2} m o l^{- 1 / 2} s^{- 1}$
2	$L m o l^{- 1} s^{- 1}$
3	$L^{2} m o l^{- 2} s^{- 1}$

Sirf first-order rate constant ki unit mein concentration nahi hota. Isliye uski value same rehti hai, chahe tum koi bhi concentration unit use karo, aur isi se first-order constants ko compare karna khaas taur par aasaan hai.

Example 3.4NCERT Example 3.4 · p. 69

k ki unit se order

Us reaction ka overall order kya hai jiska rate constant (i) $4.1 \times 10^{- 3} m o l L^{- 1} m i n^{- 1}$ , (ii) $6.5 \times 10^{- 2} L^{2} m o l^{- 2} s^{- 1}$ , (iii) $2.7 h^{- 1}$ hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3.2.4 Molecularity of a Reaction

Elementary step ke liye hum order se bhi tez sawaal pooch sakte hain: step hone ke liye ek saath kitne particles ka milna zaroori hai? Yeh ginti us step ki molecularity (आण्विकता) hai.

Definition

Molecularity

Kitne particles, chahe molecules hon, atoms ya ions, ko ek saath takraana padta hai taaki elementary step ho sake.

Example $N O + O_{3} \overset{}{\to} N O_{2} + O_{2}$ bimolecular hai.

Unimolecular: ek species apne aap rearrange hoti hai ya tootti hai, jaise cyclopropane khul kar propene banta hai, $C_{3} H_{6} \overset{}{\to} C H_{3} C H = C H_{2}$ .
Bimolecular: do species takraati hain, jaise atmosphere mein upar nitrogen monoxide ozone se milta hai, $N O + O_{3} \overset{}{\to} N O_{2} + O_{2}$ .
Termolecular: teen species ek saath milti hain. Ozone tab banti hai jab ek oxygen atom aur ek $O_{2}$ molecule takraate hain aur ek teesra molecule M, aksar $N_{2}$ , extra energy le jaata hai: $O + O_{2} + M \overset{}{\to} O_{3} + M$ .

Teen particles ka ek hi jagah ek hi pal mein pahunchna pehle se hi mushkil hai, isliye termolecular steps kam hote hain aur slow hote hain; chaar to itna unlikely hai ki teen se zyada molecularity wala koi elementary step nahi jaana jaata. Acid mein permanganate dwara oxalate ions ke oxidation ki balanced equation mein left side 23 ions hain ( $2 M n O_{4}^{-}$ , $5 C_{2} O_{4}^{2 -}$ aur $16 H^{+}$ ), isliye yeh zaroor kai steps mein hoti hai.

Jab reaction ke kai steps hon, to speed kaun tay karta hai? Ek dukaan socho jahan customers shelves se saamaan jaldi utha lete hain aur phir ek hi slow billing counter par line lagaate hain. Har minute nikalne wale customers ki sankhya billing rate ke barabar hoti hai, shelves chahe kitni bhi jaldi khaali hon. Isi tarah sabse slow elementary step, yaani rate-determining step (वेग निर्धारक पद), poori reaction ki speed tay karta hai.

Definition

Rate-determining step

Mechanism ka sabse slow elementary step; overall reaction is step se tez nahi chal sakti.

Ek asli example hai lagbhag 500 K se neeche nitrogen dioxide aur carbon monoxide ki reaction (Fig. 3.5). Saboot do bimolecular steps ki taraf ishara karte hain:

Step 1 (slow): $N O_{2} + N O_{2} \overset{}{\to} N O_{3} + N O$
Step 2 (fast): $N O_{3} + C O \overset{}{\to} N O_{2} + C O_{2}$

Steps jodne par $N O_{3}$ aur ek $N O_{2}$ cancel ho jaate hain, aur bachta hai $N O_{2} + C O \overset{}{\to} N O + C O_{2}$ . $N O_{3}$ ek intermediate hai: yeh ek step mein banta hai aur baad ke step mein use ho jaata hai, isliye overall equation mein kabhi nahi dikhta. Step 1 slow hai, isliye reaction step 1 ke rate par chalti hai, $Rate = k [N O_{2}]^{2}$ . Carbon monoxide slow step ke baad hi judta hai, isiliye woh Table 3.4 ke rate law mein nahi hai.

3/3Intermediate cancel ho jaata hai. Rate slow step ka hi hai: $Rate = k [N O_{2}]^{2}$ .

Fig. 3.5Slow pehla step bottleneck hai, isliye akela wahi rate law tay karta hai.NCERT p. 70

Left se right: nitrogen dioxide molecules ka ek pool, step 1 slow label wala patla gate, beech ka box jismein intermediate NO3 hai, step 2 fast label wala chauda gate jahan carbon monoxide aata hai, aur ek products box. Step 1: NO2 pairs ki line patle gate par intezaar karti hai jabki ek pair andar se guzar kar NO3 plus NO ban jaata hai. Step 2: NO3 chaude gate par ek CO molecule se milta hai aur turant NO2 aur CO2 ban jaata hai, isliye wahan koi line nahi banti. Step 3: overall equation NO2 plus CO se NO plus CO2 dikhti hai, beech ke box ka NO3 cancel hone wale intermediate ke roop mein label hai, aur rate law k times NO2 concentration ka square hai, jo slow gate tay karta hai.

Order aur molecularity mein aasaani se confusion ho jaata hai. Neeche ka comparison dono ko saath-saath rakhta hai.

Compare

Order vs molecularity

Order

experiment se, rate law se milta hai
zero, fraction ya negative ho sakta hai
elementary aur complex dono reactions ke liye defined
jab slow step pehle aaye, to us step ki molecularity ke barabar

Molecularity

ek elementary step ki equation se gina jaata hai
hamesha whole number: 1, 2 ya, kabhi-kabhi, 3
poori complex reaction ke liye iska koi matlab nahi
elementary step ke liye, uske order ke barabar

Practice · 2 questions

Practice: order (Intext Questions 3.3–3.4): answers ke saath

Quick answerEasyNCERT Intext Question 3.3
Reaction $A + B \overset{}{\to} p r o d u c t s$ ka rate law $Rate = k [A]^{1 / 2} [B]$ hai. Iska overall order kya hai?
Solution dikhao
Overall order $= \frac{1}{2} + 1 = \frac{3}{2}$ hai.
Quick answerEasyNCERT Intext Question 3.4
X ka Y mein conversion X mein second order hai. Agar $[X]$ ko uski value ka ek-tihaai kar diya jaaye, to Y ke banne ke rate ka kya hota hai?
ek-tihaai reh jaata haiek-nauvaan reh jaata haiek-chhatha reh jaata hainau guna badh jaata hai
Solution dikhao
Rate $= k [X]^{2}$ . Naya rate $= k ([X] / 3)^{2} = \frac{1}{9} k [X]^{2}$ : rate ek-nauvaan reh jaata hai.

3.3 Integrated Rate Equations

Tangents se kinetics karna jhanjhat ka kaam hai. Smooth curve, sthir haath aur ruler chahiye, aur slope ki chhoti si galti order mein badi galti ban jaati hai. Behtar plan hai rate law ko khud concentration ke formula mein badalna, taaki concentration aur time ki seedhi readings ko direct test kiya ja sake. Integration yeh ek hi step mein kar deta hai, aur result order par depend karta hai. Syllabus mein do cases chahiye: zero order aur first order.

3.3.1 Zero Order Reactions

Zero-order reaction mein rate reactant ke concentration par bilkul depend nahi karta. $R \overset{}{\to} P$ ke liye $Rate = k [R]^{0} = k$ . Reactant ek steady speed se use hota hai, us candle ki tarah jo har ghante utni hi lambai mein jalti hai, chahe woh lambi ho ya lagbhag khatam.

DerivationZero order ke liye integrated rate equation

Zero order wali differential rate equation se shuru karo:

- \frac{d [R]}{d t} = k [R]^{0} = k

Variables alag karo, $d [R] = - k d t$ , aur dono sides integrate karo, integration constant ko $I$ kaho:

[R] = - k t + I

(3.5)

Jab $t = 0$ , $[R]$ apni starting value $[R]_{0}$ ke barabar hai, isliye $I = [R]_{0}$ aur

Zero-order integrated rate equation

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

(3.6)

Rate constant ke liye rearrange karne par:

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

(3.7)

Equation (3.6) ka form $y = m x + c$ hai. $[R]$ vs $t$ ka plot ek straight line hai jiska slope $- k$ aur intercept $[R]_{0}$ hai (Fig. 3.6). Line $t = [R]_{0} / k$ par zero tak pahunchti hai, isliye zero-order reaction ek tay time mein khatam ho jaati hai.

Zero order rule nahi, exception hai. Yeh tab dikhta hai jab reactant ke concentration ke alawa koi aur cheez reaction ko rokti hai, jaise catalyst par sites ki sankhya. Jab high pressure par garam tungsten wire par phosphine decompose hota hai, $4 P H_{3} (g) \overset{}{\to} P_{4} (g) + 6 H_{2} (g)$ , to metal surface poori tarah $P H_{3}$ molecules se dhak jaati hai. Aur gas daalne se surface par aur molecules nahi ghus sakte, isliye rate surface sites ki sankhya se fixed ho jaata hai. Jab har enzyme molecule busy ho, to enzyme reactions bhi aise hi behave karti hain. Low pressure par surface sirf thodi dhaki hoti hai, aur wahi reaction first order ban jaati hai.

3/3Isliye $[R]$ constant rate se girta hai: slope $- k$ wali straight line.

Fig. 3.6Jab har surface site busy ho, to pressure ke saath rate nahi badh sakta: reaction zero order hai aur [R] straight line mein girta hai.NCERT Fig. 3.3 · p. 72

Left mein chhe marked sites wali garam tungsten ki patti gas ke neeche hai. Step 1, low pressure: sirf kuch gas molecules hain aur chhe mein se do sites par ek-ek molecule hai; zyada gas aur sites bhar degi, isliye rate pressure ke saath chalta hai. Step 2, high pressure: gas bhari hui hai aur chhe ki chhe sites bhari hain, ek label kehta hai ki surface full hai; extra gas cheezon ko tez nahi kar sakti, isliye rate constant hai. Step 3: right mein reactant concentration vs time ka graph ek straight line dikhata hai jo vertical axis par R zero se girti hai, uske slope par minus k mark hai, aur woh ek tay time par zero tak pahunchti hai.

Acid mein propanone ka iodination lab ka ek badhiya example hai: $C H_{3} C O C H_{3} + I_{2} \overset{}{\to} C H_{3} C O C H_{2} I + H I$ . Iska rate $[I_{2}]$ par bilkul depend nahi karta. Isliye jab propanone aur acid bahut zyada excess mein hon, to iodine ka brown colour ek steady speed se halka hota jaata hai jab tak gayab na ho jaaye.

Example

Ek zero-order reaction

Propanone aur acid bahut excess mein rakh kar iodination run mein $[I_{2}]$ , $2.0 \times 10^{- 3} m o l L^{- 1}$ se shuru hota hai aur constant $4.0 \times 10^{- 6} m o l L^{- 1} s^{- 1}$ par girta hai. (a) $150 s$ baad $[I_{2}]$ kitna hai? (b) Iodine ka colour kab gayab hota hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3.3.2 First Order Reactions

First-order reaction ka rate reactant ke concentration ke saath seedha badalta hai: concentration aadha karo to rate bhi aadha ho jaata hai. Isliye reaction lagataar dheemi hoti jaati hai aur, theory mein, kabhi poori tarah khatam nahi hoti.

DerivationFirst order ke liye integrated rate equation

First-order kinetics wale $R \overset{}{\to} P$ ke liye,

- \frac{d [R]}{d t} = k [R]

Variables alag karo, $d [R] / [R] = - k d t$ , aur integrate karo. $d [R] / [R]$ ka integral $\ln [R]$ hai, isliye

\ln [R] = - k t + I

(3.8)

$t = 0$ par $[R] = [R]_{0}$ , jisse $I = \ln [R]_{0}$ milta hai:

\ln [R] = \ln [R]_{0} - k t

(3.9)

Logarithms ko ek saath rakho aur $k$ ke liye solve karo:

k = \frac{1}{t} \ln \frac{[R]_{0}}{[R]}

(3.10)

(3.8) ko do times $t_{1}$ aur $t_{2}$ par likhne se starting concentration use kiye bina $I$ hat jaata hai:

\ln [R]_{1} = - k t_{1} + I

(3.11)

\ln [R]_{2} = - k t_{2} + I

(3.12)

(3.11) mein se (3.12) ghataane par:

k = \frac{1}{t_{2} - t_{1}} \ln \frac{[R]_{1}}{[R]_{2}}

(3.13)

(3.9) ka antilogarithm lene par khud concentration milta hai, ek exponential decay:

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

(3.14)

Aakhir mein, $\ln x = 2.303 \log x$ , isliye (3.10) ko aksar common logarithms ke saath use karte hain:

First-order integrated rate equation

k = \frac{2.303}{t} \log \frac{[R]_{0}}{[R]}

(3.15)

Equation (3.9) bhi ek straight line hai: $\ln [R]$ vs $t$ ke plot ka slope $- k$ aur intercept $\ln [R]_{0}$ hai. Isi tarah, $\log ([R]_{0} / [R])$ vs $t$ origin se guzarti straight line hai jiska slope $k / 2.303$ hai (Fig. 3.7). Yeh plots order ka sabse tez test hain: data ka jo plot straight aaye, wahi order batata hai.

2/3 $\ln [R]$ vs $t$ slope $- k$ wali straight line hai.

Fig. 3.7First-order data [R] vs t ke plot par curve banata hai, lekin logarithms plot karne par straight lines par aa jaata hai.NCERT Fig. 3.4 · p. 74

Axes ka ek set jismein time horizontal hai. Step 1: R ka concentration vs time ek curve hai jo pehle tezi se aur phir aur dheere-dheere zero ki taraf girta hai. Step 2: vertical axis R ka natural logarithm ban jaata hai aur wahi data ab ek straight line par hai jo ln R zero se girti hai, uske slope par minus k mark hai. Step 3: vertical axis log of R zero over R ban jaata hai, aur data origin se shuru hokar upar jaati straight line par hai, jiska slope k divided by 2.303 hai.

First-order kinetics aam hai. Cyclopropane first-order process se propene mein isomerise hota hai. Har radioactive decay first order hai: iodine-131, jo thyroid ki bimaari ke ilaaj mein use hota hai, har 8.0 din mein apne aadhe atoms kho deta hai, amount chahe jitna ho. Kai dawaiyaan bhi blood se first-order processes se nikalti hain, isiliye doctors kisi drug ki fixed half-life bata paate hain.

Example 3.5NCERT Example 3.5 · p. 74

Do concentrations se rate constant

Cyclopropane first-order reaction se propene mein isomerise hota hai. Ek high temperature par iska concentration $15.0 m i n$ mein $0.0500$ se $0.0300 m o l L^{- 1}$ tak gira. Rate constant calculate karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Constant volume par gas-phase reaction ke liye aksar total pressure follow karna sabse aasaan hota hai. Ek first-order decomposition $A (g) \overset{}{\to} B (g) + C (g)$ lo jo pressure $p_{i}$ par pure A se shuru hota hai.

DerivationTotal pressure se first-order rate constant

Maan lo time $t$ tak A ka partial pressure $x$ se girta hai. React karne wala har A ek B aur ek C deta hai:

Table 3.6

	$p_{A}$	$p_{B}$	$p_{C}$
$t = 0$	$p_{i}$	0	0
time $t$	$p_{i} - x$	$x$	$x$

Total pressure $p_{t} = (p_{i} - x) + x + x = p_{i} + x$ hai, isliye $x = p_{t} - p_{i}$ aur $p_{A} = p_{i} - x = 2 p_{i} - p_{t}$ . Partial pressure concentration ke proportional hai, isliye (3.15) ban jaata hai

k = \frac{2.303}{t} \log \frac{p_{i}}{2 p_{i} - p_{t}}

(3.16)

Agar ek A alag sankhya mein gas molecules deta hai, to hisaab badal jaata hai lekin method nahi: pehle $p_{t}$ se $p_{A}$ nikaalo, phir $k = (2.303 / t) \log (p_{i} / p_{A})$ use karo.

Example 3.6NCERT Example 3.6 · p. 75

Pressure se reaction follow karna

Di-tert-butyl peroxide, $(C H_{3})_{3} C O O C (C H_{3})_{3}$ , gas phase mein first-order reaction se decompose hota hai: har molecule propanone ke do molecules aur ethane ka ek molecule deta hai. Fixed volume ka ek sealed flask $0.400 a t m$ par pure peroxide se shuru hua. $20.0 m i n$ baad total pressure $0.520 a t m$ tha. Rate constant nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3.3.3 Half-Life of a Reaction

Reaction ki half-life (अर्धायु), $t_{1 / 2}$ , woh time hai jismein reactant ka concentration shuruaati value ka aadha reh jaata hai. Reaction kitni tez hai, yeh batane ke liye yeh ek kaam ka single number hai.

Definition

Half-life

Reactant ke aadhe hisse ko use hone mein lagne wala time, taaki uska concentration shuruaati value ka 50 % reh jaaye.

Zero order. (3.7) mein $t = t_{1 / 2}$ par $[R] = [R]_{0} / 2$ rakho:

Half-life, zero order

t_{1 / 2} = \frac{[R]_{0}}{2 k}

To zero-order half-life starting concentration ke saath badhti hai aur $k$ badhne par ghatti hai. Reaction aage badhne par har agli half-life pichhli wali se aadhi lambi hoti hai.

First order. (3.15) mein $[R] = [R]_{0} / 2$ rakho. Ratio $[R]_{0} / [R]$ 2 ho jaata hai, aur $\log 2 = 0.301$ :

Half-life, first order

t_{1 / 2} = \frac{2.303 \log 2}{k} = \frac{0.693}{k}

(3.17)

Starting concentration cancel ho gaya. First-order reaction ko 1.0 se 0.5 tak girne mein utna hi time lagta hai jitna 0.5 se 0.25 tak, ya 0.02 se 0.01 tak (Fig. 3.8). $n$ half-lives ke baad reactant ka $(1 / 2)^{n}$ hissa bachta hai. Constant half-life first-order reaction ki pehchaan hai, aur iska matlab hai ki $k$ aur $t_{1 / 2}$ ko seedha ek doosre mein convert kar sakte hain.

4/4Same pehli half-life wali ek zero-order reaction: har baad wali half-life aadhi lambi hai.

Fig. 3.8First order: har half-life barabar lambi hai. Zero order: har half-life pichhli wali ki aadhi hai.NCERT p. 77

Bache hue reactant ke fraction vs time ka graph. Step 1: ek first-order decay curve 1 se shuru hota hai aur jahan woh one half tak pahunchta hai, yaani ek half-life par, wahan mark hai. Step 2: doosra mark jahan woh one quarter tak pahunchta hai, theek ek aur half-life baad. Step 3: teesri barabar half-life ke baad one eighth par teesra mark, aur teeno barabar time steps bracket mein. Step 4: usi point se shuru hoti ek straight zero-order line judti hai; yeh one half tak usi pehle time par pahunchti hai, lekin one quarter tak sirf aadhe aur time mein, aur one eighth tak chauthaai time mein, isliye iski half-lives chhoti hoti jaati hain.

Example 3.7NCERT Example 3.7 · p. 77

Rate constant se half-life

Ek dawai bloodstream se first-order process se hatti hai, jismein $k = 3.85 \times 10^{- 5} s^{- 1}$ . Iski half-life hours mein kitni hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3.8NCERT Example 3.8 · p. 77

Half-lives mein completion times

Dikhao ki first-order reaction ko 87.5 % completion tak pahunchne mein theek teen half-lives lagti hain, aur nikaalo ki 99 % completion mein kitni half-lives lagti hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Table 3.7Zero- aur first-order reactions ki tulnaNCERT Table 3.4 · p. 77

	Zero order	First order
Differential rate law	$- d [R] / d t = k$	$- d [R] / d t = k [R]$
Integrated rate law	$[R] = [R]_{0} - k t$	$[R] = [R]_{0} e^{- k t}$
Straight-line plot	$[R]$ vs $t$	$\ln [R]$ vs $t$
Us line ka slope	$- k$	$- k$
Half-life	$[R]_{0} / 2 k$	$0.693 / k$
$k$ ki unit	$m o l L^{- 1} s^{- 1}$	$s^{- 1}$

InteractiveRate lab

0 se 100 seconds tak reactant concentration vs time ka graph. Buttons se zero-order ya first-order reaction chunte hain, aur sliders starting concentration ko 0.2 se 1.0 moles per litre tak aur rate constant ko set karte hain. Ek time slider, 0 se 80 seconds tak, curve par ek marker chalata hai, aur ek interval slider delta t set karta hai. Graph t se t plus delta t tak chord aur t par tangent banata hai, aur readouts t par concentration, interval ka average rate aur instantaneous rate dikhate hain, taaki delta t chhota karne par dono rates ko match hote dekha ja sake. Ek check box curve par lagataar half-lives mark karta hai aur unki lambaai list karta hai: first order ke liye barabar, zero order ke liye aadhi hoti hui. Buttons ka doosra set wahi data concentration, concentration ka natural log, ya log of R zero over R vs time ke roop mein dobara plot karta hai, aur ek line batati hai ki woh plot straight hai ya nahi.

Fig. 3.9Zero ya first order chuno, starting concentration aur

k

set karo, phir chord ko tangent se compare karo, half-lives mark karo aur test karo ki kaunsa plot straight line hai.

Key idea

Pseudo first-order reactions

Do reactants wali kuch reactions aise behave karti hain jaise woh first order hon. Trick hai unmein se ek ko bahut zyada excess mein rakhna. Example 3.1 ke crystal-violet fading mein hydroxide ions aam taur par $0.10 m o l L^{- 1}$ par hote hain jabki dye $2.0 \times 10^{- 5} m o l L^{- 1}$ par. Saari dye react ho jaane ke baad bhi $[O H^{-}]$ sirf 0.02 % girta hai.

Asli rate law $Rate = k [C V^{+}] [O H^{-}]$ hai, overall second order. $[O H^{-}]$ lagbhag constant hone se use ek naye constant $k^{'} = k [O H^{-}]$ mein mila sakte hain, aur bachta hai $Rate = k^{'} [C V^{+}]$ . Dye ka concentration bilkul first-order reactant ki tarah decay karta hai.

Aisi reaction ko pseudo first-order reaction (छद्म प्रथम कोटि अभिक्रिया) kehte hain. Peroxodisulfate dwara iodide ka oxidation, $S_{2} O_{8}^{2 -} + 2 I^{-} \overset{}{\to} 2 S O_{4}^{2 -} + I_{2}$ , ek aur example hai: iodide bahut excess mein ho to sirf $[S_{2} O_{8}^{2 -}]$ rate control karta hai. Koi bhi reaction jismein paani reactant bhi ho aur solvent bhi, aise hi behave karti hai, kyunki paani ka concentration, lagbhag $55.5 m o l L^{- 1}$ , mushkil se badalta hai.

Practice · 2 questions

Practice: first-order reactions (Intext Questions 3.5–3.6): answers ke saath

NumericalMediumNCERT Intext Question 3.5
Ek first-order reaction ka $k = 2.0 \times 10^{- 3} s^{- 1}$ hai. Reactant ke $6.0 g$ ko $4.0 g$ tak girne mein kitna time lagta hai?
s
Solution dikhao
$t = \frac{2.303}{2.0 \times 10^{- 3}} \log \frac{6.0}{4.0}$
$\log 1.5 = 0.1761$ , isliye $t = 1151.5 \times 0.1761 = 203 s$ (lagbhag 3.4 min).
NumericalEasyNCERT Intext Question 3.6
Ek organic peroxide first-order kinetics se decompose hota hai, half-life $45 m i n$ hai. Rate constant $s^{- 1}$ mein nikaalo.
s⁻¹
Solution dikhao
$k = \frac{0.693}{45 \times 60 s} = 2.57 \times 10^{- 4} s^{- 1}$

3.4 Temperature Dependence of the Rate of a Reaction

Garmi mein kitchen shelf par ek din mein khatta hone wala doodh fridge mein kai din chalta hai. Pressure cooker dal jaldi pakata hai kyunki andar paani $100^{\circ} C$ ki jagah lagbhag $120^{\circ} C$ par ubalta hai. Lagbhag har reaction garam karne par tez ho jaati hai. Ek kaam ka rule of thumb: room temperature ke aas-paas $10 K$ badhne se rate constant lagbhag double ho jaata hai.

Swedish chemist Svante Arrhenius ne 1889 mein temperature ke asar ko ek equation mein rakha, Dutch chemist J. H. van 't Hoff ke pehle suggest kiye form par aage badhte hue; Arrhenius ne iske peeche ki physical picture jodi. Arrhenius equation hai

Arrhenius equation

k = A e^{- E_{a} / R T}

(3.18)

jahan

$A$: pre-exponential (frequency) factor, reaction ke liye ek constant jiski unit $k$ jaisi hi hai
$E_{a}$: activation energy, $J m o l^{- 1}$ mein
$R$: gas constant, $8.314 J K^{- 1} m o l^{- 1}$
$T$: absolute temperature, K mein

Constant $A$ ko pre-exponential factor (आवृत्ति गुणांक) kehte hain, ise frequency factor ya Arrhenius factor bhi kehte hain. $E_{a}$ activation energy hai, jise dhyaan se dekhna padega.

Sirf kuch molecules hi react kyun karein? Ek reaction socho jismein atom A, molecule B–C se atom B le leta hai: $A + B - C \to A - B + C$ . Jaise A paas aata hai, B–C bond khinchta hai aur naya A–B bond banna shuru hota hai. Raaste mein ek bheed bhari, high-energy arrangement $A \dots B \dots C$ aata hai jismein dono bonds aadhe bane hote hain. Yeh pal bhar ka arrangement activated complex (सक्रियित संकुल) hai. Yeh lagbhag ek molecular vibration jitna tikta hai, phir toot kar products ban jaata hai ya wapas reactants.

Reaction ko ek pahaadi ke upar se chalne jaisa socho (Fig. 3.10). Horizontal axis, yaani reaction coordinate, ke saath atoms step by step reactants se products mein rearrange hote hain; height potential energy hai. Activated complex choti par baitha hai. Reactants se choti tak ki chadhaai activation energy (सक्रियण ऊर्जा), $E_{a}$ , hai. Is example mein safar shuruaat se neeche khatam hota hai, isliye reaction energy deti hai. Enthalpy change $Δ H$ sirf do end points se tay hota hai, pahaadi chahe kitni bhi oonchi ho.

Definition

Activation energy

Activated complex aur reactants ke beech ka energy difference: takraate reactant molecules ko react karne ke liye kam se kam jitni extra energy chahiye.

4/4 $Δ H$ sirf dono ends par depend karta hai. Reverse reaction ko badi chadhaai, $E_{a} + ∣ Δ H ∣$ , chadhni padti hai.

Fig. 3.10Activated complex energy hill ki choti par hai; Ea reactants se chadhaai hai aur ΔH dono ends ka difference.NCERT Fig. 3.7 · p. 79

Potential energy ko reaction coordinate ke against plot kiya gaya hai, aur curve ke upar atoms ke chhote sketches hain: left mein A aur B–C, choti par A dot dot B dot dot C aur right mein A–B aur C. Step 1: sirf do flat levels, left mein reactants aur right mein neeche products. Step 2: ek pahaadi unhe jodti hai, aur uski choti par activated complex mark hai. Step 3: reactant level se pahaadi ki choti tak ek double-headed arrow par E a likha hai. Step 4: reactant level se product level tak neeche jaate arrow par delta H likha hai, jo is exothermic reaction ke liye negative hai, aur products se pahaadi ki choti tak ek doosre arrow par reverse reaction ka E a likha hai, jo zyada bada hai.

Barrier chapter ki shuruaat wale newspaper ko samjhata hai. Jalne se bahut energy nikalti hai, lekin molecules ko pehle $E_{a}$ ke upar uthana padta hai. Match ek jagah par woh energy deti hai; uske baad jalne se nikli heat naye molecules ko barrier ke upar uthaati rehti hai.

Gas kabhi ek jaise molecules ki bheed nahi hoti. Collisions energy ko idhar-udhar baantte rehte hain, isliye kisi bhi pal kuch molecules lagbhag ruke hote hain, kuch daud rahe hote hain, aur zyadatar beech mein kahin hote hain. Ludwig Boltzmann aur James Clerk Maxwell ne is phailaav ki statistics nikaali. Iska graph (Fig. 3.11) zero se tezi se uthkar ek single peak tak jaata hai, jo most probable kinetic energy hai, aur phir high energies ki taraf ek lambi tail mein dheere-dheere girta hai.

Ab energy axis par $E_{a}$ mark karo. Sirf tail mein, $E_{a}$ se aage ke molecules hi react kar sakte hain, aur typical activation energies bahut aage hoti hain jahan curve exponentially girta hai. Gas garam karne se poora hump right ki taraf khisakta hai aur flat ho jaata hai; kyunki curve har molecule ka hisaab rakhta hai, iske neeche ka total area nahi badal sakta. Lekin $E_{a}$ se aage ka patla sa area bahut badh jaata hai. Lagbhag $50 k J m o l^{- 1}$ wale $E_{a}$ ke liye $300 K$ se $310 K$ jaane par yeh lagbhag double ho jaata hai, aur das-degree wala rule yahin se aata hai.

3/3Zyada $T_{2}$ par curve flat hokar right khisakta hai: shaded tail lagbhag double ho jaati hai.

Fig. 3.11Temperature badhaane se energies phail jaati hain, isliye molecules ke kahin bade hisse ke paas kam se kam Ea hota hai.NCERT Fig. 3.9 · p. 80

Molecules ke fraction ko kinetic energy ke against plot kiya gaya hai. Step 1: temperature T1 ke liye ek single hump jaisa curve, jo zero se uthta hai, most probable energy par peak karta hai, jo mark hai, aur right mein lambi tail mein girta hai. Step 2: ek vertical dashed line tail mein bahut aage activation energy E a mark karti hai, aur uske aage curve ke neeche ka chhota area shaded hai aur react kar sakne wale molecules ke roop mein label hai. Step 3: zyada temperature T2 ke liye doosra curve judta hai; yeh neecha, chauda aur right khisaka hua hai, same total area gherta hai, aur E a se aage ka iska shaded area lagbhag do guna hai.

Arrhenius equation mein factor $e^{- E_{a} / R T}$ yahi hissa naapta hai: kaafi had tak yeh un collisions ka fraction hai jinki energy kam se kam $E_{a}$ hai. Factor $A$ phir yeh sambhaalta hai ki collisions kitni baar hote hain. (3.18) ka natural logarithm lene par milta hai

\ln k = \ln A - \frac{E_{a}}{R T}

(3.19)

Yeh chhupi hui straight line hai: $\ln k$ vs $1 / T$ plot karo to slope $- E_{a} / R$ aur intercept $\ln A$ hai (Fig. 3.12). Kai temperatures par $k$ naap kar yeh Arrhenius plot banana activation energy nikaalne ka standard tareeka hai. Bada $E_{a}$ zyada steep line deta hai, matlab rate temperature ke liye zyada sensitive hai.

2/3Yeh slope $- E_{a} / R$ wali line par hain: slope naapo, $- R$ se multiply karo, aur tumhare paas $E_{a}$ hai.

Fig. 3.12ln k vs 1/T slope −Ea/R wali straight line hai.NCERT Fig. 3.10 · p. 80

Vertical axis k ka natural logarithm hai aur horizontal axis one over T hai, jo right ki taraf badhta hai, isliye zyada temperatures left mein hain. Step 1: paanch naape gaye points upar-left se neeche-right ki taraf girte hain. Step 2: unke beech se ek straight line khinchi gayi hai, aur ek right-angled slope triangle par slope equals minus E a over R likha hai. Step 3: usi beech wale point se guzarti doosri, zyada steep line par larger E a likha hai, jo dikhata hai ki aisi reaction temperature ke utne hi badhne par zyada tez hoti hai.

DerivationDo temperatures par rate constants

(3.19) ko do temperatures par likho. $A$ dono par same hai:

\ln k_{1} = \ln A - \frac{E_{a}}{R T_{1}}

(3.20)

\ln k_{2} = \ln A - \frac{E_{a}}{R T_{2}}

(3.21)

(3.21) mein se (3.20) ghataao; $\ln A$ cancel ho jaata hai:

\ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

Common logarithms mein badlo aur bracket ko common denominator par rakho:

\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} \cdot \frac{T_{2} - T_{1}}{T_{1} T_{2}}

(3.22)

Do temperatures par $k$ diya ho to (3.22) se $E_{a}$ milta hai. $E_{a}$ aur ek temperature par $k$ diya ho to kisi bhi doosre temperature par $k$ milta hai.

Example 3.9NCERT Example 3.9 · p. 81

Do rate constants se activation energy

Ek gas-phase decomposition ka rate constant $400 K$ par $3.0 \times 10^{- 3} s^{- 1}$ aur $450 K$ par $9.6 \times 10^{- 2} s^{- 1}$ hai. $E_{a}$ aur $A$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3.10NCERT Example 3.10 · p. 81

Naye temperature par rate constant

Dawaiyon ki shelf life test karne ke liye unhe garam rakha jaata hai, jisse unka decomposition tez hota hai. Ek active ingredient first-order reaction se tootta hai, $25^{\circ} C$ par $k = 1.5 \times 10^{- 8} s^{- 1}$ aur $E_{a} = 85.0 k J m o l^{- 1}$ . $45^{\circ} C$ par $k$ nikaalo aur dono temperatures par half-lives compare karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

InteractiveArrhenius lab

Sliders temperature ko 250 se 500 kelvin tak aur activation energy ko 20 se 150 kilojoules per mole tak set karte hain, aur ek check box catalyst jodta hai jo activation energy ko 30 kilojoules per mole kam karta hai. 1 se neeche 10 to the minus 30 tak jaata ek logarithmic scale un collisions ka fraction mark karta hai jinke paas kam se kam activation energy hai, yaani e to the minus Ea over RT, abhi ka aur, ek dashed ring ke roop mein, 300 kelvin ka. Readouts Ea over RT, woh fraction, A ko 10 to the 11 per second par fixed rakh kar rate constant, aur first-order half-life dikhate hain. ln k vs 1000 over T ka graph chuni hui activation energy ki straight line dikhata hai, catalyst on hone par ek doosri dashed line, aur abhi ka point. Record abhi ke point ko graph mein jodta hai; do ya zyada points hone par best-fit line banti hai aur uske slope ko activation energy mein badla jaata hai, Ea equals minus R times slope. Clear points hata deta hai. Ek status line batati hai ki kitne collisions mein se kitne safal hote hain aur reaction 300 kelvin ke mukaable kitni tez ya dheemi hai.

Fig. 3.13Temperature aur activation energy badlo, catalyst on karo, aur rate constants record karke apne Arrhenius plot se

E_{a}

naapo.

3.4.1 Effect of Catalyst

Kuch substances reaction ko tez karte hain aur phir bhi usmein se bina badle nikal aate hain; aisa substance catalyst (उत्प्रेरक) hai. Jaise hydrogen peroxide apne aap bahut dheere decompose hota hai, $2 H_{2} O_{2} \overset{}{\to} 2 H_{2} O + O_{2}$ , lekin chutki bhar manganese dioxide ya khoon ki ek boond daalne par zor se jhaag banata hai; khoon mein enzyme catalase hota hai. Jo substance reaction ko dheema kare, woh catalyst nahi hai; use inhibitor kehte hain.

Definition

Catalyst

Aisa substance jo reaction ko tez karta hai aur reaction khatam hone par chemically bina badle wapas mil jaata hai.

ExampleHydrogen peroxide ke decomposition mein manganese dioxide.

Yeh kaam kaise karta hai? Intermediate complex theory catalyst ko ek temporary partner ki tarah dekhti hai. Yeh thodi der ke liye reactant se bond karta hai, aur yeh combination, jise intermediate complex kehte hain, akele reactant se zyada aasaani se react karta hai. Jab complex tootta hai to products deta hai aur catalyst wapas lauta deta hai, jo phir se shuru karne ke liye free hai. Catalysed raasta ek alag mechanism hai jiski activation energy kam hai (Fig. 3.14). Kyunki (3.18) mein $E_{a}$ exponent mein hai, thodi si kami bhi bahut farak dalti hai: $298 K$ par $E_{a}$ ko $20 k J m o l^{- 1}$ kam karne se $k$ 3000 guna se zyada ho jaata hai, agar $A$ same rahe.

3/3Dono raaste same levels par shuru aur khatam hote hain: $Δ H$ aur $Δ G$ nahi badalte, aur reverse barrier bhi utna hi girta hai.

Fig. 3.14Catalyst unhi reactants aur products ke beech kam energy wala raasta khol deta hai.NCERT Fig. 3.11 · p. 82

Ek reaction ke liye potential energy vs reaction coordinate. Step 1: uncatalysed raasta reactant level se neeche product level tak ek single oonchi pahaadi hai, aur uski activation energy mark hai. Step 2: unhi do levels ke beech ek dashed catalysed raasta judta hai; isme do neeche humps hain aur unke beech ek dip hai jahan intermediate complex banta hai, aur iski chhoti activation energy mark hai. Step 3: ek dashed line reactant level ko products tak le jaati hai aur ek neeche jaata arrow delta H mark karta hai, jo dono raaston ke liye same hai; products se peeche jaane par catalysed humps utne hi neeche hain, isliye reverse barrier bhi girta hai.

Is picture se kai baatein nikalti hain:

Dono raaste reactants se shuru hokar products par khatam hote hain, isliye catalyst na $Δ H$ badalta hai na $Δ G$ . Isliye woh non-spontaneous reaction ( $Δ G > 0$ ) ko nahi kara sakta; woh sirf un changes ko tez karta hai jo pehle se feasible hain.
Reaction kisi bhi taraf chale, barrier utna hi girta hai, isliye forward aur reverse reactions barabar tez hoti hain. Equilibrium constant nahi badalta; equilibrium bas jaldi aa jaata hai. Haber process mein iron catalyst ammonia ki yield nahi badhata; woh plant ko itni jaldi equilibrium tak pahunchne deta hai ki use chalana faayde ka ho.
Har cycle mein catalyst wapas ban jaata hai, isliye thoda sa bhi bahut kaam karta hai: car ke catalytic converter mein kuch grams platinum-group metals saalon tak exhaust saaf karte hain.

Try karke dekho

Aaloo mein catalyst

Tumhe chahiye

chemist ki dukaan se 3 % hydrogen peroxide (lagbhag 30 mL)
do chhote glass
kachche aaloo ka ek slice aur uble aaloo ka ek slice
dish-wash liquid ki kuch boondein
safety glasses

Pehle guess karo

Kachcha aaloo peroxide mein daalne par kya hoga?

kuch dikhega nahiek gas ke lagataar bubblesliquid neela ho jaayegaaaloo ghul jaayega

Har glass mein lagbhag 10 mL peroxide aur dish-wash liquid ki kuch boondein daalo.
Kachche aaloo ka slice ek glass mein aur uble aaloo ka slice doosre glass mein daalo.
Do minute dekho aur jhaag compare karo. Phir kachcha slice nikaalo, dho lo, aur naye peroxide mein daalo.

Table 3.8

Glass	Tumhe kya dikhta hai
Kachcha aaloo
Ubla aaloo
Kachcha slice, dobara use kiya

Tumhari entries apne aap save hoti hain.

3.5 Collision Theory of Chemical Reactions

Arrhenius equation temperature ka asar describe karti hai, lekin yeh nahi batati ki $A$ ka matlab kya hai. Collision theory, jise Max Trautz aur William Lewis ne 1916 aur 1918 ke beech develop kiya, ek molecular picture deti hai. Iska starting point kinetic theory of gases se liya gaya hai: molecules chhoti-chhoti sakht gendein hain, aur jab tak do molecules sach mein ek doosre se takraate nahi, kuch nahi hota.

Mixture ke har unit volume mein har second kitne collisions hote hain, yeh gino, to tumhare paas collision frequency, $Z$ , hai. Yeh bahut badi hoti hai: room conditions par hawa mein har molecule ko ek second mein kai billion baar takkar lagti hai. Agar har collision se reaction hoti, to har gas reaction pal bhar mein khatam ho jaati. Sirf kaafi energy wale collisions hi ginti mein aate hain, isliye bimolecular elementary reaction $A + B \overset{}{\to} p r o d u c t s$ ke liye

Rate = Z_{A B} e^{- E_{a} / R T}

(3.23)

Yahan $Z_{A B}$ A aur B ke beech collisions ki frequency hai, aur $e^{- E_{a} / R T}$ un collisions ka fraction hai jinki energy kam se kam $E_{a}$ hai. Ise (3.18) ke saath rakho: pre-exponential factor $A$ collision frequency ka role nibhata hai.

(3.23) ko experiment se test karo to ek pattern dikhta hai. Jab atoms ya chhote molecules react karte hain to yeh achchha kaam karta hai, lekin bade molecules ke liye measured rate aksar prediction se bahut kam hota hai, kabhi-kabhi kai powers of ten kam. Sirf energy kaafi nahi: molecules ko sahi direction mein bhi milna padta hai, taaki tootne wala bond aur banne wala bond sahi jagah par hon. Kaafi energy, yaani threshold energyThreshold energy woh kam se kam total energy hai jo takraate molecules ke paas react karne ke liye honi chahiye: unki pehle se maujood energy plus activation energy., aur sahi orientation wale collisions effective collisions (प्रभावी संघट्ट) hain.

Definition

Effective collision

Aisa collision jo sach mein products banata hai: molecules kam se kam threshold energy ke saath aur is tarah line up hokar milte hain ki sahi bond toot sake aur naya ban sake.

Chlorine atom ki nitrosyl chloride ke saath reaction lo, $C l + N O C l \overset{}{\to} C l_{2} + N O$ (Fig. 3.15). Naya Cl–Cl bond tabhi ban sakta hai jab aane wala Cl atom NOCl ke chlorine end se takraaye, aur kaafi zor se takraaye. Oxygen end par takkar bas uchhal jaati hai, chahe usmein kitni bhi energy ho.

3/3Sahi end aur kaafi energy: ek effective collision $C l_{2}$ aur NO deta hai.

Fig. 3.15Collision se products tabhi bante hain jab usmein kaafi energy aur sahi orientation dono hon.NCERT Fig. 3.12 · p. 83

Ek mude hue nitrosyl chloride molecule, jismein upar oxygen, beech mein nitrogen aur neeche-right chlorine hai, ki taraf ek single chlorine atom aata hai. Step 1: chlorine atom upar se oxygen end se takraata hai aur ek mude arrow ke saath wapas uchhal jaata hai, label hai wrong end, no reaction. Step 2: chlorine atom chlorine end ko chhoota hai lekin bas halke se, aur uchhal jaata hai, label hai right end, too little energy. Step 3: chlorine atom chlorine end se zor se takraata hai; dono chlorine atoms saath milkar Cl2 molecule ke roop mein nikalte hain aur ek alag NO molecule door chala jaata hai, label hai right end and enough energy: products.

Collision theory orientation ko ek aur number se sambhaalti hai, steric factor $P$ , jise probability factor bhi kehte hain: energetic collisions ka woh fraction jo react karne ke liye theek se line up hai.

Steric factor ke saath collision theory

Rate = P Z_{A B} e^{- E_{a} / R T}

To reaction ka rate teen cheezon se tay hota hai: molecules kitni baar takraate hain ( $Z_{A B}$ ), collisions ka kitna fraction kaafi energy le jaata hai ( $e^{- E_{a} / R T}$ ), aur unmein se kitna fraction sahi nishaane par hai ( $P$ ). Molecules ko bina features wali gendein maanna hi theory ki kamzori bhi hai: yeh asli molecules ki shapes aur bonds se $P$ predict nahi kar sakti. Reaction rates ki zyada poori theories aage ki padhai mein aati hain.

Practice · 3 questions

Practice: temperature aur energy (Intext Questions 3.7–3.9): answers ke saath

Short answerEasyNCERT Intext Question 3.7
Ek typical reaction ka rate constant temperature badhne par kaise respond karta hai, aur kaunsi equation is change ko describe karti hai?
Solution dikhao
AnswerYeh tezi se badhta hai (room temperature ke paas har 10 K par lagbhag double), jaisa Arrhenius equation describe karti hai.
Rate constant badhta hai, room temperature ke paas har 10 K badhne par lagbhag double, kyunki collisions ke bade fraction ke paas kam se kam activation energy hoti hai. Arrhenius equation, $k = A e^{- E_{a} / R T}$ , ise describe karti hai: jaise $T$ badhta hai, $E_{a} / R T$ girta hai aur $e^{- E_{a} / R T}$ exponentially badhta hai.
NumericalMediumNCERT Intext Question 3.8
Temperature $300 K$ se $310 K$ badhne par reaction ka rate teen guna ho jaata hai. $E_{a}$ calculate karo.
kJ mol⁻¹
Solution dikhao
$\log 3 = \frac{E_{a}}{2.303 \times 8.314} \cdot \frac{10}{300 \times 310}$
$\log 3 = 0.4771$ aur $2.303 \times 8.314 = 19.15$ , isliye $E_{a} = 0.4771 \times 19.15 \times 9300 \approx 8.5 \times 10^{4} J m o l^{- 1}$ , lagbhag $85 k J m o l^{- 1}$ .
NumericalMediumNCERT Intext Question 3.9
$600 K$ par $E_{a} = 150 k J m o l^{- 1}$ wali reaction ke liye, collisions ke kitne fraction ke paas kam se kam activation energy hai?
Solution dikhao
$E_{a} / R T = 150 000 / (8.314 \times 600) = 30.07$ .
$x = e^{- 30.07} = 8.7 \times 10^{- 14}$
Das trillion mein ek se bhi kam collision mein kaafi energy hoti hai.

Quick check

3 mein se 1

Collision theory ko steric factor

P

ki zaroorat kyun hai?

Kuch energetic collisions ki orientation galat hoti hai.Kuch molecules kabhi takraate hi nahi.Temperature badhne par collision frequency girti hai.Reaction ke dauraan activation energy badal jaati hai.

Answer dekho

Answer (A) Kuch energetic collisions ki orientation galat hoti hai.

Kaafi energy wale collisions bhi fail ho jaate hain agar molecules galat direction mein milein, jaise jab Cl, NOCl ke O end se takraata hai.

P

sahi nishaane wale collisions ka fraction hai.

Recap

Summary aur key points

Chemical kinetics padhta hai ki reactions kitni tezi se chalti hain aur kyun; thermodynamics sirf batata hai ki woh chal sakti hain ya nahi (Introduction).
Reaction ka rate per unit time concentration ka change hai. Average rate ek interval ke liye hota hai; instantaneous rate tangent ka slope hai, $- d [R] / d t$ (Section 3.1).
Poori reaction ka ek rate dene ke liye har species ke rate of change ko balanced equation mein uske coefficient se divide karo (Section 3.1).
Concentration, temperature, catalysts aur surface area, sab rate par asar daalte hain (Section 3.2).
Rate law, $Rate = k [A]^{x} [B]^{y}$ , experiment se milta hai; iski powers ka coefficients ke barabar hona zaroori nahi (Section 3.2.2).
Order $x + y$ hai; yeh zero, fractional ya negative ho sakta hai, aur yahi $k$ ki unit tay karta hai, $(m o l L^{- 1})^{1 - n} s^{- 1}$ (Section 3.2.3).
Molecularity, 1, 2 ya kabhi-kabhi 3, sirf elementary steps par laagu hoti hai; mechanism ka sabse slow step rate tay karta hai (Section 3.2.4).
Zero order: $[R] = [R]_{0} - k t$ , slope $- k$ wali straight line, aur $t_{1 / 2} = [R]_{0} / 2 k$ (Section 3.3.1).
First order: $k = (2.303 / t) \log ([R]_{0} / [R])$ ; $\ln [R]$ vs $t$ straight line hai (Section 3.3.2).
First-order half-life, $0.693 / k$ , starting concentration par depend nahi karti; ek reactant ka bahut zyada excess reaction ko pseudo first order bana deta hai (Section 3.3.3).
Arrhenius equation: $k = A e^{- E_{a} / R T}$ . $\ln k$ vs $1 / T$ ke plot ka slope $- E_{a} / R$ hai; room temperature ke paas 10 K se $k$ lagbhag double hota hai (Section 3.4).
Catalyst kam activation energy wala raasta deta hai; yeh $Δ H$ , $Δ G$ aur equilibrium constant ko nahi badalta (Section 3.4.1).
Collision theory: $Rate = P Z_{A B} e^{- E_{a} / R T}$ . Effective collisions ke liye kaafi energy aur sahi orientation chahiye (Section 3.5).

Flash cards · 10

Formula cards

1 / 10

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Intext Question 3.1	Practice: reaction rates (Intext Questions 3.1–3.2) · Q1
NCERT Intext Question 3.2	Practice: reaction rates (Intext Questions 3.1–3.2) · Q2
NCERT Intext Question 3.3	Practice: order (Intext Questions 3.3–3.4) · Q1
NCERT Intext Question 3.4	Practice: order (Intext Questions 3.3–3.4) · Q2
NCERT Intext Question 3.5	Practice: first-order reactions (Intext Questions 3.5–3.6) · Q1
NCERT Intext Question 3.6	Practice: first-order reactions (Intext Questions 3.5–3.6) · Q2
NCERT Intext Question 3.7	Practice: temperature aur energy (Intext Questions 3.7–3.9) · Q1
NCERT Intext Question 3.8	Practice: temperature aur energy (Intext Questions 3.7–3.9) · Q2
NCERT Intext Question 3.9	Practice: temperature aur energy (Intext Questions 3.7–3.9) · Q3
NCERT Exercises 3.1	Exercises · Q1
NCERT Exercises 3.2	Exercises · Q2
NCERT Exercises 3.3	Exercises · Q3
NCERT Exercises 3.4	Exercises · Q4
NCERT Exercises 3.5	Exercises · Q5
NCERT Exercises 3.6	Exercises · Q6
NCERT Exercises 3.7	Exercises · Q7
NCERT Exercises 3.8	Exercises · Q8
NCERT Exercises 3.9	Exercises · Q9
NCERT Exercises 3.10	Exercises · Q10
NCERT Exercises 3.11	Exercises · Q11
NCERT Exercises 3.12	Exercises · Q12
NCERT Exercises 3.13	Exercises · Q13
NCERT Exercises 3.14	Exercises · Q14
NCERT Exercises 3.15	Exercises · Q15
NCERT Exercises 3.16	Exercises · Q16
NCERT Exercises 3.17	Exercises · Q17
NCERT Exercises 3.18	Exercises · Q18
NCERT Exercises 3.19	Exercises · Q19
NCERT Exercises 3.20	Exercises · Q20
NCERT Exercises 3.21	Exercises · Q21
NCERT Exercises 3.22	Exercises · Q22
NCERT Exercises 3.23	Exercises · Q23
NCERT Exercises 3.24	Exercises · Q24
NCERT Exercises 3.25	Exercises · Q25
NCERT Exercises 3.26	Exercises · Q26
NCERT Exercises 3.27	Exercises · Q27
NCERT Exercises 3.28	Exercises · Q28
NCERT Exercises 3.29	Exercises · Q29
NCERT Exercises 3.30	Exercises · Q30

Quick check: MCQs

3.1 Rate of a Chemical Reaction4 questions
3.2.4 Molecularity of a Reaction4 questions
3.3.3 Half-Life of a Reaction4 questions
3.4.1 Effect of Catalyst4 questions
3.5 Collision Theory of Chemical Reactions3 questions

Glossary · 15

Key terms (English + Hindi)

activated complexसक्रियित संकुलsakriyit sankul: Reactants se products ke raaste mein atoms ka kam der tikne wala, sabse zyada energy wala arrangement.Chapter mein pehli baar yahan aaya →
activation energyसक्रियण ऊर्जाsakriyan urja: Reactants aur activated complex ke beech energy barrier $E_{a}$ : takraate molecules ko react karne ke liye chahiye extra energy.Chapter mein pehli baar yahan aaya →

catalystउत्प्रेरकutprerak: Aisa substance jo kam activation energy wala raasta khol kar reaction tez karta hai aur end mein chemically bina badle rehta hai.Chapter mein pehli baar yahan aaya →
chemical kineticsरासायनिक बलगतिकीrasayanik balgatiki: Chemistry ki woh branch jo padhti hai ki reactions kitni tezi se hoti hain, unki speed ko kya control karta hai aur woh kin steps se hoti hain.Chapter mein pehli baar yahan aaya →

effective collisionप्रभावी संघट्टprabhavi sanghatt: Kam se kam threshold energy aur sahi orientation wala collision, jisse products bante hain.Chapter mein pehli baar yahan aaya →
elementary reactionप्राथमिक अभिक्रियाprathmik abhikriya: Aisi reaction jo ek hi step mein hoti hai, bilkul waise jaise uski equation likhi hai.Chapter mein pehli baar yahan aaya →

half-lifeअर्धायुardhayu: Reactant ke concentration ko aadha hone mein lagne wala time; first-order reaction ke liye yeh $0.693 / k$ hai.Chapter mein pehli baar yahan aaya →

molecularityआण्विकताaanvikta: Elementary step mein ek saath takraane wali species ki sankhya: ek, do ya, kabhi-kabhi, teen.Chapter mein pehli baar yahan aaya →

order of a reactionअभिक्रिया की कोटिabhikriya ki koti: Rate law mein concentration terms ki powers ka sum. Yeh experiment se milta hai aur zero ya fraction bhi ho sakta hai.Chapter mein pehli baar yahan aaya →

pre-exponential factorआवृत्ति गुणांकaavritti gunank: $k = A e^{- E_{a} / R T}$ mein constant $A$ ; yeh batata hai ki reactant molecules kitni baar takraate hain.Chapter mein pehli baar yahan aaya →
pseudo first-order reactionछद्म प्रथम कोटि अभिक्रियाchhadm pratham koti abhikriya: Zyada order wali reaction jo first order ki tarah behave karti hai kyunki baaki reactants bahut zyada excess mein hain.Chapter mein pehli baar yahan aaya →

rate constantवेग स्थिरांकveg sthirank: Rate law mein constant $k$ ; diye gaye temperature par reaction ke liye fixed hota hai aur unit concentrations par rate ke barabar hota hai.Chapter mein pehli baar yahan aaya →
rate lawवेग नियमveg niyam: Experiment se mili equation jo rate ko reactant concentrations ke terms mein deti hai, jaise $Rate = k [A]^{x} [B]^{y}$ .Chapter mein pehli baar yahan aaya →
rate of reactionअभिक्रिया का वेगabhikriya ka veg: Per unit time reactant ka concentration kitni tezi se girta hai ya product ka kitni tezi se badhta hai, balanced equation mein us species ke coefficient se divide karke.
rate-determining stepवेग निर्धारक पदveg nirdharak pad: Mechanism ka sabse slow step, jo overall reaction ke rate ko limit karta hai.Chapter mein pehli baar yahan aaya →

Chemical Kinetics: Class 12 Chemistry Chapter 3 Hinglish mein

Is chapter mein aap kya seekhenge

3.1 Rate of a Chemical Reaction

Practice: reaction rates (Intext Questions 3.1–3.2): answers ke saath

3.2 Factors Influencing Rate of a Reaction

3.2.1 Dependence of Rate on Concentration

3.2.2 Rate Expression and Rate Constant

3.2.3 Order of a Reaction

3.2.4 Molecularity of a Reaction

Practice: order (Intext Questions 3.3–3.4): answers ke saath

3.3 Integrated Rate Equations

3.3.1 Zero Order Reactions

3.3.2 First Order Reactions

3.3.3 Half-Life of a Reaction

Practice: first-order reactions (Intext Questions 3.5–3.6): answers ke saath

3.4 Temperature Dependence of the Rate of a Reaction

3.4.1 Effect of Catalyst

3.5 Collision Theory of Chemical Reactions

Practice: temperature aur energy (Intext Questions 3.7–3.9): answers ke saath

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye