Worked examples — Rate of reaction — average vs instantaneous

2.8.1 · D3 · Chemistry › Chemical Kinetics › Rate of reaction — average vs instantaneous

Yeh page ek drill hai. Hum parent note ke ideas lete hain — average rate, instantaneous rate, negative sign, aur stoichiometric coefficients se divide karna — aur unhe har us situation ke saath thokate hain jo ek problem mein aa sakti hai.

Shuru karne se pehle, ek reminder seedhe shabdon mein, taaki koi bhi symbol undefined na rahe:

Kyunki neeche har example ek master equation par tikaa hua hai, hum use yahan poora likhte hain taaki yeh page self-contained rahe.

Scenario matrix

Har rate problem inhi cells mein se ek mein aata hai. Neeche ke examples us cell ke saath labelled hain jo woh cover karte hain.

Cell	Case class	Kya cheez tricky banati hai
A	Reactant disappearing	negative sign apply karna padta hai
B	Product appearing	sign already positive hota hai
C	Coefficient $=1$ (degenerate)	1 se divide karne se kuch nahi badalta — check karo ki aap notice karo
D	Coefficient $>1$	reaction rate paane ke liye divide karna hoga
E	Do species ki rates link karna	A ki rate ko C ki rate mein convert karo
F	Data table se average nikalna	pure $\Delta /\Delta t$
G	Tangent slope se instantaneous	derivative = tangent ka slope
H	Average vs instantaneous compare karna	limiting behaviour: rate time ke saath girti hai
I	Zero / degenerate input	$\Delta t\to 0$, ya koi change hi nahi
J	Real-world word problem	words ko $\Delta []/\Delta t$ mein translate karo
K	Exam twist (mixed units / hidden coefficient)	unit conversion + stoichiometry saath mein

Neeche ke nau examples saare gyaarah cells ko cover karte hain.

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Worked examples

Forecast: Kya reaction rate us raw rate se badi hogi ya chhoti jis par ${\text{O}}_2$ gayab ho rahi hai? (Padhne se pehle guess karo.)

1. ${\text{O}}_2$ mein raw change. $\Delta [{\text{O}}_2]=0.180-0.240=-0.060\text{M}$. Yeh step kyun? $\Delta$ ka matlab hamesha final minus initial hota hai; minus batata hai ki ${\text{O}}_2$ use ho rahi hai.
2. ${\text{O}}_2$ disappear hone ki rate. $-\frac{\Delta [{\text{O}}_2]}{\Delta t}=-\frac{-0.060}{30.0}=2.0\times 10^{-3}\text{M/s}$. Yeh step kyun? Definition mein negative sign girti concentration ko positive speed mein badal deta hai.
3. Coefficient se normalise karo. $r=\frac{1}{3}\times 2.0\times 10^{-3}=6.67\times 10^{-4}\text{M/s}$. Yeh step kyun? Har reaction event mein teen ${\text{O}}_2$ molecules gayab hote hain, isliye reaction khud $3\times$ slow hai ${\text{O}}_2$ ke disappear hone se — coefficient $a=3$ se divide karo, bilkul general equation ka $-\frac{1}{a}\frac{d[\text{A}]}{dt}$ term.

Verify: Reaction rate $6.67\times 10^{-4}$, $2.0\times 10^{-3}$ se chhota hai, forecast se match karta hai (coefficient $>1$ reaction rate ko chhota karta hai). Units: $\text{M/s}$ ✓.

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Forecast: Reaction rate se tez ya slow? (${\text{O}}_3$ ka coefficient 2 hai.)

1. Link likho. $r=\frac{1}{c}\frac{d[{\text{O}}_3]}{dt}$, jahan $c=2$ product ${\text{O}}_3$ ka coefficient hai general rate equation se. Yeh step kyun? Product term $+\frac{1}{c}\frac{d[\text{C}]}{dt}=r$ kehta hai ki ek product ki rate, apne coefficient $c$ se divide karke, $r$ ke barabar hoti hai. Koi negative sign nahi — products badhte hain, isliye slope already positive hai.
2. Formation rate solve karo. $\frac{d[{\text{O}}_3]}{dt}=2r=2\times 6.67\times 10^{-4}=1.33\times 10^{-3}\text{M/s}$. Yeh step kyun? Har reaction event mein do ${\text{O}}_3$ bante hain, isliye ${\text{O}}_3$ reaction se do guna tez accumulate hota hai.

Verify: Ratio $\frac{d[{\text{O}}_3]/dt}{-d[{\text{O}}_2]/dt}=\frac{1.33\times 10^{-3}}{2.0\times 10^{-3}}=\frac{2}{3}$, bilkul stoichiometric ratio $\frac{2{\text{O}}_3}{3{\text{O}}_2}$ ✓.

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Forecast: Kya coefficient se divide karne ka yahan koi effect hai?

1. A loss ki average rate. $-\frac{0.320-0.500}{6.0}=-\frac{-0.180}{6.0}=0.030\text{M/s}$. Yeh step kyun? Sign flip ke saath standard definition.
2. Normalise karo. A ka coefficient $a=1$ hai, isliye $r=\frac{1}{1}\times 0.030=0.030\text{M/s}$. Yeh step kyun? Yahi trap hai: 1 se divide karne se number nahi badlta, lekin aapko phir bhi consciously coefficient check karna chahiye, step skip mat karo.

Verify: Kyunki $a=1$ hai, $r$ bilkul A-loss rate ke barabar hai ✓. Units $\text{M/s}$ ✓.

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$t$ (s)	0	20	40	60
$[\text{X}]$ (M)	0.800	0.560	0.400	0.288

Pehle interval ($0\to 20\text{s}$) aur aakhri interval ($40\to 60\text{s}$) par average rate nikalo. X ka coefficient 1 hai.

Forecast: Kis interval ki average rate zyada hogi?

1. Pehla interval. $-\frac{0.560-0.800}{20-0}=-\frac{-0.240}{20}=0.0120\text{M/s}$. Yeh step kyun? Average rate hamesha chosen interval par $\Delta /\Delta t$ hota hai.
2. Aakhri interval. $-\frac{0.288-0.400}{60-40}=-\frac{-0.112}{20}=0.0056\text{M/s}$. Yeh step kyun? Same definition, alag do rows.

Verify: Pehle interval ki rate $0.0120>$ aakhri interval ki rate $0.0056$: reaction slow hoti jaati hai jaise X consume hota hai — expected limiting behaviour ✓.

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Forecast: Tangent left-to-right girti hai, isliye uska slope negative hai — rate ka kya sign hoga?

1. Tangent line ka slope. $\frac{d[\text{A}]}{dt}=\frac{0.28-0.60}{60-20}=\frac{-0.32}{40}=-0.0080\text{M/s}$. Yeh step kyun? Instantaneous rate derivative hai, aur derivative bilkul us point par tangent line ka slope hota hai (orange tangent dekho jo magenta dot par curve ko touch kar rahi hai).
2. Sign convention apply karo. $r_{\text{inst}}=-\frac{d[\text{A}]}{dt}=-(-0.0080)=0.0080\text{M/s}$. Yeh step kyun? Rate positive report ki jaati hai; definition ka minus tangent ke negative slope ko cancel kar deta hai.

Verify: Slope negative $\Rightarrow$ rate positive $0.0080\text{M/s}$, forecast se match ✓. Units: $\frac{\text{M}}{\text{s}}$ ✓.

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Forecast: Kaunsa bada hai, aur kyun?

1. Average rate. $-\frac{0.25-1.00}{200-0}=-\frac{-0.75}{200}=3.75\times 10^{-3}\text{M/s}$. Yeh step kyun? Poore safar par average total $\Delta [\text{A}]$ divided by total time hai — figure mein straight chord ka slope.
2. Compare karo. $\frac{r_{\text{inst}}(0)}{r_{\text{avg}}}=\frac{8.0\times 10^{-3}}{3.75\times 10^{-3}}=2.13$. Yeh step kyun? Initial instantaneous rate $t=0$ par steep tangent hai; average shallower chord hai. Tangent zyada steep hai, isliye ratio 1 se zyada hai.

Verify: $r_{\text{inst}}(0)>r_{\text{avg}}$ kyunki reaction tab sabse tez hoti hai jab $[\text{A}]$ sabse zyada hoti hai aur baad mein slow hoti jaati hai — chord ka slope fast-early aur slow-late ka blend hai ✓.

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Forecast: Kya zero rate physically allowed hai? Part (b) mein "limit" kya object hai?

1. (a) Koi change nahi. $-\frac{0.50-0.50}{\Delta t}=-\frac{0}{\Delta t}=0\text{M/s}$. Yeh step kyun? $\Delta [\text{A}]=0$ numerator ko zero kar deta hai, isliye rate genuinely zero hai — reaction ruk gayi hai (e.g. equilibrium, ya us window mein kuch measurable nahi hua).
2. (b) Shrinking window. Jaise $\Delta t\to 0$, $-\frac{\Delta [\text{A}]}{\Delta t}⟶-\frac{d[\text{A}]}{dt}$ at $t=40\text{s}$. Yeh step kyun? Yahi instantaneous rate ki definition hai: vanishing interval par average derivative ban jaata hai. Aap $\Delta t=0$ seedha nahi rakh sakte (woh $0/0$ dega, undefined) — limit hi hamaari cheez hai.

Verify: (a) exactly $0$ deta hai ✓. (b) Average rate ka ek well-defined limit hai (derivative) even though fraction $\Delta t=0$ par undefined hai — isliye hi chemists ko instantaneous definition ki zaroorat padi ✓.

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Forecast: Kaunsa sign, aur roughly kitne $\text{mmol/L}$ har ghante gayab hote hain?

1. Change identify karo. $\Delta [\text{urea}]=7.0-25.0=-18.0\text{mmol/L}$. Yeh step kyun? Urea remove ho raha hai, isliye ise bilkul disappearing reactant ki tarah treat karo.
2. Average removal rate. $-\frac{\Delta [\text{urea}]}{\Delta t}=-\frac{-18.0}{4.0}=4.5\text{mmol/L per hour}$. Yeh step kyun? Same definition — yahan "reaction" physical removal hai, lekin $\Delta /\Delta t$ ka maths identical hai.

Verify: $4.5\text{mmol/L/hr}\times 4.0\text{hr}=18.0\text{mmol/L}$ removed, $25.0-7.0$ se match karta hai ✓. Disappearing quantity ke liye positive rate ✓.

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Forecast: Yahan do traps chhupe hain — unit conversion (min→s) aur do alag coefficients (2 aur 3). Predict karo ki kya ${\text{H}}_2$ ${\text{NH}}_3$ banne se tez ya slow gayab hogi.

1. Units convert karo. $1.2\times 10^{-2}\frac{\text{M}}{\text{min}}\times \frac{1\text{min}}{60\text{s}}=2.0\times 10^{-4}\text{M/s}$. Yeh step kyun? Combine karne se pehle saari rates ka time unit same hona chahiye. $1\text{min}=60\text{s}$.
2. ${\text{NH}}_3$ se reaction rate. $r=\frac{1}{c}\frac{d[{\text{NH}}_3]}{dt}=\frac{1}{2}\times 2.0\times 10^{-4}=1.0\times 10^{-4}\text{M/s}$, product coefficient $c=2$ use karke. Yeh step kyun? ${\text{NH}}_3$ ka coefficient 2 hai, isliye normalised reaction rate paane ke liye 2 se divide karo.
3. ${\text{H}}_2$ consume hone ki rate. $-\frac{d[{\text{H}}_2]}{dt}=3r=3\times 1.0\times 10^{-4}=3.0\times 10^{-4}\text{M/s}$. Yeh step kyun? Reactant term $-\frac{1}{a}\frac{d[{\text{H}}_2]}{dt}=r$ with $a=3$ rearrange karke $-\frac{d[{\text{H}}_2]}{dt}=3r$ milta hai; har reaction event mein teen ${\text{H}}_2$ gayab hote hain.

Verify: $\frac{-d[{\text{H}}_2]/dt}{d[{\text{NH}}_3]/dt}=\frac{3.0\times 10^{-4}}{2.0\times 10^{-4}}=\frac{3}{2}$, bilkul coefficient ratio $\frac{3{\text{H}}_2}{2{\text{NH}}_3}$ ✓. ${\text{H}}_2$ ${\text{NH}}_3$ banne se tez gayab hoti hai, jaise forecast kiya tha ✓.

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Recall Kaunsi cell kaunsi hai? (saare gyaarah)

Cell A — reactant disappearing ko negative sign chahiye. ::: A Cell B — product appearing ka already positive sign hota hai (koi minus nahi). ::: B Cell C — degenerate case jahan coefficient 1 ke barabar hai (divide karne se kuch nahi badlta lekin phir bhi check karo). ::: C Cell D — coefficient 1 se bada hone par reaction rate paane ke liye coefficient se divide karna padta hai. ::: D Cell E — do species ko link karne ka matlab hai ek species ki rate ko doosre mein unke coefficient ratio ke zariye convert karna. ::: E Cell F — data table se average rate pure Δ[ ]/Δt hai ek interval par. ::: F Cell G — instantaneous rate curve par tangent line ka slope ke barabar hoti hai. ::: G Cell H — average aur instantaneous compare karne se pata chalta hai ki rate time ke saath girti hai (tangent chord se steep hota hai). ::: H Cell I — $\Delta t\to 0$ hone par average rate derivative (instantaneous rate) ban jaati hai. ::: I Cell J — word problem ka matlab sirf words ko translate karna hai ek Δ[ ]/Δt calculation mein. ::: J Cell K — exam twist mein unit conversion aur hidden stoichiometric coefficient saath bundle hote hain. ::: K

Connections

• Jab aap rates measure kar lete ho, tab aap poochte ho ki woh concentration par kaise depend karti hain → Rate law and order of reaction.
• Rate law ko time par integrate karna → Integrated rate laws aur Half-life of reactions.
• Kyun rates girti hain jaise reactant khatam hota hai, molecular level par → Collision theory aur Reaction mechanisms.
• Kaise temperature rates ko turbo-charge karta hai → Arrhenius equation; aur kaise Catalysts unhe consume hue bina tez karte hain.