Worked examples — Rate law — order vs molecularity

2.8.2 · D3 · Chemistry › Chemical Kinetics › Rate law — order vs molecularity

Yeh page parent topic ka drill hall hai. Yahan koi nayi ideas nahi seekhenge — hum har tarah ke questions ko hit karenge jo yeh topic throw kar sakta hai, ek worked example per cell. Agar "rate law" ya "molecularity" jaisi koi phrase blurry lagti hai, toh pehle parent note dobara padho; yahan hum assume karte hain ki tum already $\text{Rate}=k[\text{A}{]}^m[\text{B}{]}^n$ padh sakte ho aur jaante ho ki $m$ measure kiya jaata hai, guess nahi.

The scenario matrix

#	Case class	Kya cheez tricky banati hai	Example jo ise hit karta hai
C1	Initial-rates data se Integer order	ratios padhna, logs lena	Example 1
C2	Ek reactant mein Zero order	rate ek concentration ko ignore karta hai	Example 2
C3	Data se Fractional order	exponent whole number nahi hai	Example 3
C4	Negative order (inhibition)	zyada reactant → slower	Example 4
C5	Elementary step: order = molecularity	stoichiometry hi rate law hai	Example 5
C6	Multi-step: order ≠ stoichiometry	slow step sab decide karta hai	Example 6
C7	Degenerate / units of $k$	$k$ ki units order ke saath badlti hain	Example 7
C8	Real-world word problem	words → rate law mein translate karo	Example 8
C9	Exam twist: pseudo-order	ek reactant swamped, "hidden" order	Example 9

Hum inhe order mein karte hain. Har ek batata hai ki woh kaunsa cell cover karta hai.

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Example 1 — Integer order (Cell C1)

Forecast: A double karne se rate double hua — guess order 1 in A. B triple karne se rate 9 guna hua — guess order 2 in B. Overall 3.

Figure — ratio bars for A and B. Left pair of bars trials 1 aur 2 compare karti hai: concentration double hoti hai aur bar exactly double ho jaata hai → order 1. Right pair trials 1 aur 3 compare karti hai: concentration triple hoti hai lekin bar ×9 jump karta hai → order 2. Height jump versus width jump literally exponent hai.

Alt-text: two pairs of vertical bars on a white grid. Left pair: blue bars of height 2 and 4 under "[A] x2", labelled order 1. Right pair: orange bars of height 2 and 18 under "[B] x3", labelled order 2.

1. A ko isolate karo trials 1 aur 2 use karke ($[\text{B}]$ fixed): $\frac{4.0\times 10^{-3}}{2.0\times 10^{-3}}={\left(\frac{0.20}{0.10}\right)}^m⟹2=2^m⟹m=1$ Yeh step kyun? $[\text{B}]$ ko constant rakhne se $[\text{B}]$ wala factor cancel ho jaata hai, toh ratio sirf $[\text{A}]$ par depend karta hai akele — woh ratio hi order hai.

2. B ko isolate karo trials 1 aur 3 use karke ($[\text{A}]$ fixed): $\frac{18.0\times 10^{-3}}{2.0\times 10^{-3}}={\left(\frac{0.30}{0.10}\right)}^n⟹9=3^n⟹n=2$ Yeh step kyun? Same isolation trick, ab B ke liye. $9=3^2$ toh $n=2$.

3. Overall order $=m+n=1+2=3$. Yeh step kyun? Overall order defined hai individual exponents ke sum ke roop mein, kyunki yeh batata hai ki rate kaise respond karta hai jab tum saari concentrations ek saath scale karo — har concentration ko $c$ se multiply karne par rate $c^m\cdot c^n=c^{m+n}$ se multiply ho jaata hai. Toh exponents add hote hain.

4. $k$ solve karo trial 1 se: $k=\frac{\text{Rate}}{[\text{A}][\text{B}{]}^2}=\frac{2.0\times 10^{-3}}{(0.10)(0.10{)}^2}=2.0{\text{M}}^{-2}{\text{s}}^{-1}$ Yeh step kyun? Jab exponents pata ho jaate hain, rate law mein sirf ek unknown bachta hai ($k$); tab koi bhi ek trial simple division se usse fix kar deta hai.

Verify: Trial 3 ko $k=2.0$ se predict karo: $\text{Rate}=2.0\times 0.10\times 0.30^2=2.0\times 0.10\times 0.09=0.018=18\times 10^{-3}$ ✓. Diye gaye value se match karta hai.

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Example 2 — Zero order in one reactant (Cell C2)

Forecast: Trials 1→3 mein $[\text{Y}]$ 4 guna badha phir bhi rate unchanged hai — lagta hai order 0 in Y.

Figure — the flat bar. Picture mein do orange bars same height pe hain jabki $[\text{Y}]$ ×4 jump karta hai. Ek concentration jo quadruple ho jaaye aur rate bar flat rahe — exactly yahi "order 0" dikhta hai.

Alt-text: two equal-height orange bars on a white grid, both height 6, under labels "[Y]=0.10" and "[Y]=0.40", annotated "rate unchanged -> order 0".

1. X mein Order (trials 1,2, $[\text{Y}]$ fixed): $\frac{12.0}{6.0}={\left(\frac{0.10}{0.05}\right)}^m⟹2=2^m⟹m=1$ Yeh step kyun? X ka standard isolation — $[\text{Y}]$ cancel ho jaata hai, toh rate ratio hi X-order hai.

2. Y mein Order (trials 1,3, $[\text{X}]$ fixed): $\frac{6.0}{6.0}={\left(\frac{0.40}{0.10}\right)}^n⟹1=4^n⟹n=0$ Yeh step kyun? $4^n=1$ se force hota hai $n=0$: Y ko quadruple karna kuch nahi kiya, toh uska exponent zero hona chahiye (sirf $4^0=1$).

3. Rate law: $\text{Rate}=k[\text{X}{]}^1[\text{Y}{]}^0=k[\text{X}]$. Overall order $=m+n=1+0=1$. Yeh step kyun? $[\text{Y}{]}^0=1$ toh Y factor disappear ho jaata hai; exponents phir bhi add hote hain, overall order 1 dete hain.

4. $k$ trial 1 se: $k=\frac{6.0\times 10^{-4}}{0.05}=1.2\times 10^{-2}{\text{s}}^{-1}$. Yeh step kyun? Y ke gone hone ke baad, law hai $\text{Rate}=k[\text{X}]$, toh $k$ bas $\text{Rate}/[\text{X}]$ hai.

Verify: Ek first-order $k$ units ${\text{s}}^{-1}$ carry karta hai — yeh general units rule $[k]={\text{M}}^{1-n}{\text{s}}^{-1}$ se follow hota hai jo Example 7 mein derive kiya hai (order $n=1$ deta hai ${\text{M}}^0{\text{s}}^{-1}={\text{s}}^{-1}$) ✓. Trial 2 predict karo: $1.2\times 10^{-2}\times 0.10=1.2\times 10^{-3}=12\times 10^{-4}$ ✓.

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Example 3 — Fractional order (Cell C3)

Forecast: $[\text{A}]$ ×4 badha lekin rate sirf ×2. Kyunki $4^{1/2}=2$, guess order $\frac{1}{2}$.

Figure — a small-jump bar. Concentration bar quadruple ho jaata hai lekin rate bar sirf double hota hai. Jab height kam badhti hai width se, exponent 0 aur 1 ke beech hota hai — yahan exactly $\frac{1}{2}$.

Alt-text: two green bars of height 5 and 10 under labels "[A]=1.00" and "[A]=4.00", annotated "conc x4 but rate x2 -> order 0.5".

1. $m$ solve karo logs use karke — zaroori hai kyunki ratio 2 ka nice power nahi hai: $\frac{10.0}{5.0}={\left(\frac{4.00}{1.00}\right)}^m⟹2=4^m⟹m=\frac{\ln 2}{\ln 4}=\frac{\ln 2}{2\ln 2}=0.5$ Yeh step kyun? Jab $\text{Rate}$ ratio obvious power nahi hai, dono sides ka $\ln$ lo: unknown exponent power se neeche aa jaata hai (yahi pura reason hai ki logarithms exist karte hain), jo deta hai general order formula $m=\frac{\ln ({\text{Rate}}_2/{\text{Rate}}_1)}{\ln ([\text{A}{]}_2/[\text{A}{]}_1)}$. Is page ka har integer case bas yahi formula hai clean answer ke saath.

2. Rate law: $\text{Rate}=k[\text{A}{]}^{0.5}$. Yeh step kyun? Rate law bas general form $\text{Rate}=k[\text{A}{]}^m$ hai jisme solve kiya hua exponent ($m=0.5$) slot in kiya gaya hai — kuch aur nahi; ise likhne se single remaining unknown $k$ agले step ke liye isolate ho jaata hai.

3. $k$ trial 1 se: $k=\frac{5.0\times 10^{-3}}{1.00^{0.5}}=5.0\times 10^{-3}{\text{M}}^{0.5}{\text{s}}^{-1}$. Yeh step kyun? $\text{Rate}=k[\text{A}{]}^{0.5}$ ko $k=\text{Rate}/[\text{A}{]}^{0.5}$ ke liye rearrange karo aur trial 1 plug in karo.

Verify: Trial 2 predict karo: $5.0\times 10^{-3}\times 4^{0.5}=5.0\times 10^{-3}\times 2=10\times 10^{-3}$ ✓.

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Example 4 — Negative order / inhibition (Cell C4)

Forecast: Trials 1→3 mein $[\text{P}]$ ×4 badha aur rate drop hokar ek quarter reh gaya. Ek rate jo gir jaaye jab concentration badhe ka matlab hai ek negative exponent.

Figure — the shrinking bar. Pehle ki har picture ke ulta, yahan $[\text{P}]$ badhane se rate bar shrink hokar ek quarter ho jaata hai. Ek downward jump negative order ka visual signature hai.

Alt-text: two red bars, the left of height 8 and the right of height 2, under labels "[P]=0.10" and "[P]=0.40", annotated "conc x4 but rate to 1/4 -> order minus 1".

1. S mein Order (trials 1,2): $\frac{16.0}{8.0}={\left(\frac{0.40}{0.20}\right)}^m⟹2=2^m⟹m=1$ Yeh step kyun? S ko isolate karo: $[\text{P}]$ fixed hai aur cancel ho jaata hai, toh rate ratio S-order hai.

2. P mein Order (trials 1,3, $[\text{S}]$ fixed). Symbol $p$ ko P ke respect mein order denote karne do (rate law mein $[\text{P}]$ par exponent). Wahi general log method apply karo Example 3 se, ab zero se neeche answer ke saath: $\frac{2.0}{8.0}={\left(\frac{0.40}{0.10}\right)}^p⟹0.25=4^p⟹p=\frac{\ln 0.25}{\ln 4}=-1$ Yeh step kyun? Hum P ke liye ek named exponent $p$ introduce karte hain exactly jaisa ki $m,n$ ne A aur B ke liye exponents name kiye the; phir yeh log-isolation formula $p=\frac{\ln ({\text{Rate}}_3/{\text{Rate}}_1)}{\ln ([\text{P}{]}_3/[\text{P}{]}_1)}$ hai Example 3 se — sirf ek difference hai ki numerator $\ln 0.25<0$ hai, toh order negative aata hai. Negative order empirical rate law mein legal hai: yeh inhibition encode karta hai.

3. Rate law: $\text{Rate}=k[\text{S}][\text{P}{]}^{-1}=\frac{k[\text{S}]}{[\text{P}]}$. Overall order $=m+p=1+(-1)=0$. Yeh step kyun? Exponents pehle ki tarah add hote hain; yahan woh cancel hokar net overall order 0 dete hain, matlab dono S aur P ko ek saath scale karne se rate unchanged rehta hai.

4. $k$ trial 1 se: $k=\frac{\text{Rate}\cdot [\text{P}]}{[\text{S}]}=\frac{8.0\times 10^{-3}\times 0.10}{0.20}=4.0\times 10^{-3}{\text{s}}^{-1}$. Yeh step kyun? $k$ solve karne ke liye $\text{Rate}=k[\text{S}]/[\text{P}]$ rearrange karo, phir trial 1 plug in karo.

Verify: Trial 3 predict karo: $\frac{4.0\times 10^{-3}\times 0.20}{0.40}=2.0\times 10^{-3}$ ✓.

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Example 5 — Elementary step: order = molecularity (Cell C5)

Forecast: Ek NO plus ek O₃ collide karte hain, toh guess karo $\text{Rate}=k[\text{NO}][{\text{O}}_3]$, molecularity 2, order 2.

Figure — one collision. Do particles (ek NO, ek O₃) approach karte hain aur ek single point par touch karte hain. Kyunki exactly do particles milne chahiye, collision frequency $[\text{NO}]\times [{\text{O}}_3]$ ke proportional hai — picture hi rate law hai.

Alt-text: two circles labelled NO (blue) and O3 (orange) with arrows pointing toward a shared collision point marked with a green star, annotated "2 particles meet -> Rate proportional to [NO][O3]".

1. Elementary rule use karo: ek elementary step ke liye, exponents stoichiometric coefficients ke equal hote hain. $\text{Rate}=k[\text{NO}{]}^1[{\text{O}}_3{]}^1$ Yeh step kyun? Ek collision mein NO-meets-O₃ events ki frequency $[\text{NO}]\times [{\text{O}}_3]$ ke proportional hoti hai — stoichiometry literally hi collision requirement hai.

2. Molecularity = collide karne wali species ki sankhya = $1+1=2$ (bimolecular, yaani ek saath do particles milte hain). Yeh step kyun? Hum bas ek step mein reactant particles count karte hain: ek NO + ek O₃ = 2, jo upar diye gaye molecularity ki definition hai.

3. Order $=m+n=1+1=2$. Yahan order molecularity ke equal hai — lekin sirf isliye kyunki step elementary hai. Yeh step kyun? Overall order ke liye step 1 ke exponents add karo; ek elementary step ke liye woh exponents seedhe collision count se aaye hain, toh dono numbers coincide karte hain.

Verify: Agar hum dono concentrations double karein, rate ×4 hona chahiye: ${\left(\frac{2[\text{NO}]}{[\text{NO}]}\right)}^1{\left(\frac{2[{\text{O}}_3]}{[{\text{O}}_3]}\right)}^1=2\times 2=4$ ✓, overall order 2 ke consistent.

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Example 6 — Multi-step: order ≠ stoichiometry (Cell C6)

Figure — the bottleneck gate. Neeche diagram do-step mechanism ko ek pipe ke roop mein draw karta hai. Left par, reactants ${\text{NO}}_2+{\text{F}}_2$ ek narrow red gate (slow step 1) ki taraf badhte hain. Har second sirf ek trickle pass karta hai. Right par, jo bhi pass karta hai woh wide green box (fast step 2) se takraata hai, jo instantly clear ho jaata hai. Picture jo point visually banati hai: poori pipe ki rate sirf narrow gate se fix hoti hai — fast box ko wide karne se kuch nahi badlta. Isliye sirf step-1 reactants rate law mein appear karte hain.

Alt-text: a flow diagram; a blue "NO2 + F2" reactant box feeds through a narrow red "SLOW step 1" gate labelled Rate = k1[NO2][F2] into a wide green "FAST step 2" box, then to product P. Grey arrows show a few particles passing the gate and instant clearance afterward.

Forecast: Stoichiometry $2{\text{NO}}_2+{\text{F}}_2$ tumhe order 3 ki taraf tempt karti hai. Lekin sirf slow step count karta hai — guess order 2.

1. Rate = slow step ki rate (step 1 bottleneck hai): $\text{Rate}=k_1[{\text{NO}}_2][{\text{F}}_2]$ Yeh step kyun? Jaise figure dikhati hai, slow step ek narrow gate hai. Chahe step 2 kitna bhi fast chale, overall throughput gate se cap hoti hai, toh sirf slow step ke reactants rate set karte hain.

2. Orders padho: NO₂ mein order $=1$, F₂ mein order $=1$, overall $=1+1=2$. Yeh step kyun? Slow-step law elementary hai, toh iske exponents coefficients hain (dono 1); overall order 2 ke liye add karo.

3. Stoichiometry se compare karo: coefficients ka sum $=2+1=3\ne 2$. Extra NO₂ sirf fast step mein enter karta hai, toh woh rate law mein kabhi appear nahi karta. Yeh step kyun? Yahi cell ka poora lesson hai: overall order mechanism ke slow step se aata hai, balanced equation se nahi.

4. Molecularity check: step 1 bimolecular hai (2), step 2 bimolecular hai (2). Individual molecularities integers hain (2 each), phir bhi overall order (2) step-1 molecularity se match karna is mechanism ka coincidence hai, koi rule nahi. Yeh step kyun? Hum har elementary step mein colliding particles count karte hain (2 aur 2) yeh dikhane ke liye ki "molecularity" ek per-step count hai, overall order se alag — dono yahan sirf coincide karte hain.

Verify: Intermediate F step 1 mein produce hota hai aur step 2 mein steady state mein equal rates par consume hota hai, toh woh accumulate nahi hota aur observed law mein appear nahi kar sakta — $\text{Rate}=k_1[{\text{NO}}_2][{\text{F}}_2]$ ke consistent. (Numerically: sirf $[{\text{NO}}_2]$ double karne se rate double hota hai; sirf $[{\text{F}}_2]$ double karne se rate double hota hai — dono order 1.)

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Example 7 — Degenerate case: units of $k$ (Cell C7)

Forecast: Rate ki units hamesha ${\text{M s}}^{-1}$ hoti hain, toh $k$ ko woh sab absorb karna padega jo $[\text{A}{]}^n$ peeche chhod jaata hai.

1. General rule: $\text{Rate}=k[\text{A}{]}^n$ se, dono sides par units match karke $[k]$ solve karo: $[k]=\frac{[\text{Rate}]}{[\text{A}{]}^n}=\frac{{\text{M s}}^{-1}}{{\text{M}}^n}={\text{M}}^{1-n}{\text{s}}^{-1}$ Yeh step kyun? Kisi bhi physical equation ke dono sides par same units hone chahiye; left side ${\text{M s}}^{-1}$ hai, toh $k$ ko exactly ${\text{M}}^{1-n}{\text{s}}^{-1}$ supply karna hoga taaki $k[\text{A}{]}^n$ ${\text{M s}}^{-1}$ par land kare.

2. Har order plug in karo:

Order $n$	$[k]={\text{M}}^{1-n}{\text{s}}^{-1}$
$0$	${\text{M s}}^{-1}$
$1$	${\text{s}}^{-1}$
$2$	${\text{M}}^{-1}{\text{s}}^{-1}$
$0.5$	${\text{M}}^{0.5}{\text{s}}^{-1}$

Verify: Pehle ke examples se cross-check karo — Example 1 (order 3) ne ${\text{M}}^{-2}{\text{s}}^{-1}$ diya: formula deta hai ${\text{M}}^{1-3}{\text{s}}^{-1}={\text{M}}^{-2}{\text{s}}^{-1}$ ✓. Example 3 (order 0.5) ne ${\text{M}}^{0.5}{\text{s}}^{-1}$ diya ✓.

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Example 8 — Real-world word problem (Cell C8)

Forecast: "Time to fade" rate ka inverse hai. Faster rate ⇒ shorter time. Bleach ×2 se time aadha → rate double → order 1 in B. Dye ×2 se time nahi badla → order 0 in D.

1. Words ko rate mein translate karo: fade time $t\propto 1/\text{Rate}$.

   • Bleach ×2 → $t$ aadha → Rate ×2 → order in B: $2=2^b\Rightarrow b=1$.
   • Dye ×2 → $t$ same → Rate same → order in D: $1=2^d\Rightarrow d=0$. Yeh step kyun? Hum words se $k$ measure nahi kar sakte, lekin rate ke ratios exactly wahi hain jo orders ko chahiye — aur rate ratios time ratios ke inverse hain.

2. Rate law: $\text{Rate}=k[\text{B}{]}^1[\text{D}{]}^0=k[\text{B}]$. Overall order $=b+d=1+0=1$. Yeh step kyun? Dono exponents add karo; $[\text{D}{]}^0$ factor 1 hai aur drop out ho jaata hai, overall order 1 deta hai.

Verify: Sanity — reaction limit hota hai bleach kitna present hai, dye se nahi, jo "dye just along for the ride hai" se match karta hai. B double karna aur observed time aadha hona first order in B ke consistent hain ($2\times \text{rate}\Rightarrow \frac{1}{2}\times \text{time}$) ✓.

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Example 9 — Exam twist: pseudo-first order (Cell C9)

Forecast: Water bahut excess mein hai, toh $[{\text{H}}_2\text{O}]$ effectively constant hai. $k$ mein fold ek constant law ko sirf ester mein first order jaisa dikhata hai.

1. "Constant water" approximation justify karo: ester dilute hai, maano $\sim 0.1\text{M}$, jabki water $55.5\text{M}$ hai. Agar saara ester react kare, toh zyada se zyada $\sim 0.1\text{M}$ water consume hogi — $0.2\%$ se bhi kam ka change. Yeh step kyun? Yahi poore example ki key approximation hai. Ek quantity jo run ke dauran ek percent ke fraction se bhi kam change ho, use ek fixed number ki tarah treat kiya ja sakta hai, variable ki tarah nahi — yahi exactly woh hai jo hume agले step mein use rate constant mein fold karne deta hai.

2. Constant ko ek new constant $k^{\prime }$ mein absorb karo: kyunki $[{\text{H}}_2\text{O}]$ (approximately) fixed hai, define karo $k^{\prime }\equiv k[{\text{H}}_2\text{O}]$: $\text{Rate}=k[\text{Ester}]\underset{\approx \text{constant}}{\underbrace{[{\text{H}}_2\text{O}]}}=(k[{\text{H}}_2\text{O}])[\text{Ester}]=k^{\prime }[\text{Ester}]$ Yeh step kyun? Jab $[{\text{H}}_2\text{O}]$ ek number hai, toh $k[{\text{H}}_2\text{O}]$ bas ek aur number hai; law ki form collapse hokar $\text{Rate}=k^{\prime }[\text{Ester}]$ ho jaati hai, jo ek first-order law jaisi dikhti hai.

3. Naam do: yeh ek pseudo-first-order reaction hai — iska true order (aur elementary collision ki molecularity) 2 hai, lekin observed order 1 hai kyunki water ki dependence $k^{\prime }$ ke andar hidden hai. Yeh step kyun? Collapsed law $\text{Rate}=k^{\prime }[\text{Ester}]$ ko face value par padhne se observed order 1 milta hai; "pseudo" flag karta hai ki ek real second reactant $k^{\prime }$ mein lurk kar raha hai.

4. $k^{\prime }$ ko true $k$ se numerically relate karo. Maano true $k=2.0\times 10^{-4}{\text{M}}^{-1}{\text{s}}^{-1}$. Tab $k^{\prime }=k[{\text{H}}_2\text{O}]=2.0\times 10^{-4}{\text{M}}^{-1}{\text{s}}^{-1}\times 55.5\text{M}=1.11\times 10^{-2}{\text{s}}^{-1}$ Yeh step kyun? Fixed $[{\text{H}}_2\text{O}]$ ko $k^{\prime }=k[{\text{H}}_2\text{O}]$ mein substitute karna number dono deta hai aur units shift karta hai: $55.5\text{M}$ mein $\text{M}$, $k$ mein ek ${\text{M}}^{-1}$ cancel kar deta hai, second-order units ${\text{M}}^{-1}{\text{s}}^{-1}$ ko first-order units ${\text{s}}^{-1}$ mein turn karta hai.

Verify: Units: ${\text{M}}^{-1}{\text{s}}^{-1}\times \text{M}={\text{s}}^{-1}$ ✓ (Example 7 ke first-order units se match karta hai). Magnitude $2.0\times 10^{-4}\times 55.5=1.11\times 10^{-2}$ ✓.

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Recall Self-test — answers cover karo

A double karne se rate quadruple hota hai: A mein order? ::: 2 (kyunki $4=2^2$) B quadruple karne se rate unchanged: B mein order? ::: 0 (kyunki $1=4^0$) P quadruple hone par rate $1/4$ ho jaata hai: P mein order? ::: $-1$ (kyunki $\frac{1}{4}=4^{-1}$) $1\text{M}$ ka kya matlab hai? ::: $1{\text{mol L}}^{-1}$ — har litre solution mein ek mole solute Overall order $n$ ke liye $k$ ki units? ::: ${\text{M}}^{1-n}{\text{s}}^{-1}$ Ek multi-step reaction mein, kaunsa step rate law set karta hai? ::: slow (rate-determining) step Kya molecularity 0, fractional, ya negative ho sakti hai? ::: Nahi — sirf $1,2,3$; order in mein se koi bhi ho sakta hai Ester hydrolysis first order kyun lagti hai? ::: water vast excess mein hai, toh $[{\text{H}}_2\text{O}]$ $k^{\prime }$ mein fold ho jaata hai (pseudo-first order)