Worked examples — Collision theory — frequency factor, steric factor

2.8.9 · D3 · Chemistry › Chemical Kinetics › Collision theory — frequency factor, steric factor

Yeh page parent topic ki practice arena hai. Parent ne ideas build kiye the: collision frequency $Z$, Boltzmann energy filter $e^{-E_a/RT}$, orientation filter $\rho$, aur yeh sab milke rate constant $k=\rho Z{e}^{-E_a/RT}$ kaise banate hain. Yahan hum unhe use karte hain — aur deliberately har tarah ke sawaal dhundh ke laate hain jo yeh topic throw kar sakta hai.

Kisi bhi symbol ke solution mein aane se pehle, yaad karo har ek kya hai, seedhe alfazon mein:

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The scenario matrix

Collision theory ka har problem actually inhi cells mein se ek hota hai. Hum ensure karenge ki har cell ko ek worked example mile.

Cell	Kya vary karta hai / kya degenerate hai	Example
C1 Forward count	$Z$ compute karo radii, $T$, masses se	Ex 1
C2 Identical partners	$A+A$ case with the $\frac{1}{2}$ double-count fix	Ex 2
C3 Solve backward	Given $k$, $Z$, $E_a$ → find $\rho$	Ex 3
C4 Limiting $\rho$	$\rho \to 1$ (noble-gas / atomic) vs $\rho \ll 1$ (big molecule)	Ex 4
C5 Degenerate speed	Same velocity → relative speed $\to 0$; aur $T\to 0$ / $T\to \infty$	Ex 5
C6 Temperature scaling	$A$ aur $e^{-E_a/RT}$ dono kaise move karte hain jab $T$ change hota hai	Ex 6
C7 Real-world word problem	Atmospheric ozone destruction, full chain	Ex 7
C8 Exam twist	"Measured $A$ bigger kyun hai $Z$ se?" trap	Ex 8

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[!example] Ex 1 (C1) — Collisions count karo: H₂ + I₂

Statement. $T=500\text{K}$ par, lo $r_{{\text{H}}_2}=1.0\text{A˚}$, $r_{{\text{I}}_2}=2.0\text{A˚}$, $[{\text{H}}_2]=[{\text{I}}_2]=0.1\text{M}$. Collision frequency $Z$ nikalo.

Forecast: order of magnitude guess karo — $Z$ nearer hai $10^5$ ya $10^{36}$ collisions per m³ per second? Aage padhne se pehle apna guess likho.

Step 1 — Collision cross-section. $\sigma =\pi (r_{{\text{H}}_2}+r_{{\text{I}}_2}{)}^2=\pi (3.0\times 10^{-10}\text{m}{)}^2=2.83\times 10^{-19}{\text{m}}^2$ Yeh step kyun? Ek hit tab hoti hai jab do centres $r_A+r_B$ ke andar aate hain; woh distance "target disc" ka radius hai jo ek molecule sweep out karta hai. Neeche figure mein, molecule A (blue) centre par baitha hai aur yellow dashed circle woh target disc hai — koi bhi B-centre (red) usme enter kare toh woh collision count hoti hai. Dekho kaise disc radius hai $r_A+r_B$, dono radii ka sum, sirf koi ek nahi.

Step 2 — Reduced mass. $\mu =\frac{m_{{\text{H}}_2}{m}_{{\text{I}}_2}}{m_{{\text{H}}_2}+m_{{\text{I}}_2}}=\frac{(2)(254)}{256}\frac{\text{g/mol}}{N_A}=3.30\times 10^{-27}\text{kg}$ Yeh step kyun? Jo matter karta hai woh hai pair kitni tezi se approach karta hai, aur woh approach ek single body ki tarah behave karta hai jis mass $\mu$ ho. (Grams ko kg mein convert karo: $\text{g/mol}$ ko $N_A$ se divide karo grams per molecule ke liye, phir $10^3$ se kg ke liye.)

Step 3 — Mean relative speed. ${\bar{v}}_{\text{rel}}=\sqrt{\frac{8k_{B}T}{\pi \mu }}=\sqrt{\frac{8(1.38\times 10^{-23})(500)}{\pi (3.30\times 10^{-27})}}\approx 2.31\times 10^3\text{m/s}$ Yeh step kyun? Collisions per second scale hoti hain is se ki molecules kitni jaldi gap close karte hain; $\sqrt{T}$ seedha the speed distribution se aata hai. (Halka, low-mass reduced pair fast move karta hai — few km/s exactly wahi hai jo kinetic theory predict karta hai.)

Step 4 — Concentration convert karo, phir multiply karo. $0.1\text{M}=0.1\frac{\text{mol}}{\text{L}}\times 10^3\frac{\text{L}}{{\text{m}}^3}\times N_A=6.02\times 10^{25}{\text{molecules/m}}^3$ $Z=\sigma {\bar{v}}_{\text{rel}}[{\text{H}}_2][{\text{I}}_2]=(2.83\times 10^{-19})(2.31\times 10^3)(6.02\times 10^{25}{)}^2\approx 2.37\times 10^{36}\frac{\text{coll}}{{\text{m}}^3\text{s}}$ Yeh step kyun? Har factor ek ingredient hai: target kitna bada, kitni tezi, kitne molecules.

Verify. Units: ${\text{m}}^2\cdot \frac{\text{m}}{\text{s}}\cdot \frac{1}{{\text{m}}^3}\cdot \frac{1}{{\text{m}}^3}=\frac{1}{{\text{m}}^3\text{s}}$ ✓. Huge number ($\sim 10^{36}$) expected hai — isliye hi reactions ko energy aur orientation filters chahiye hote hain unhe slow karne ke liye.

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[!example] Ex 2 (C2) — Identical partners: the ½ trap

Statement. ${\text{NO}}_2+{\text{NO}}_2\to$ products at $500\text{K}$ ke liye, $\sigma =4.0\times 10^{-19}{\text{m}}^2$, $m=46\text{g/mol}$, $[{\text{NO}}_2]=6.02\times 10^{24}{\text{m}}^{-3}$. $Z_{AA}$ nikalo.

Forecast: kya $Z_{AA}$ "naive" $\sigma \bar{v}[A{]}^2$ se bada hoga ya chhota? Factor guess karo.

Step 1 — Identical partners ki reduced mass. $\mu =\frac{m\cdot m}{m+m}=\frac{m}{2}=\frac{46}{2}\frac{\text{g/mol}}{N_A}=3.82\times 10^{-26}\text{kg}$ Yeh step kyun? Do equal masses dete hain $\mu =m/2$ — effective inertia half ek molecule ki hoti hai.

Step 2 — Relative speed. ${\bar{v}}_{\text{rel}}=\sqrt{\frac{8k_{B}T}{\pi \mu }}=\sqrt{\frac{8(1.38\times 10^{-23})(500)}{\pi (3.82\times 10^{-26})}}\approx 6.78\times 10^2\text{m/s}$ Yeh step kyun? ${\text{NO}}_2$ H₂/I₂ reduced pair se bahut bhaari hai, isliye uski thermal speed lower hai — few hundred m/s.

Step 3 — ½ apply karo. $Z_{AA}=\frac{1}{2}\sigma {\bar{v}}_{\text{rel}}[A{]}^2=\frac{1}{2}(4.0\times 10^{-19})(678)(6.02\times 10^{24}{)}^2\approx 4.92\times 10^{33}\frac{\text{coll}}{{\text{m}}^3\text{s}}$ Yeh step kyun? Molecule 1 ka molecule 2 se takrana same event hai jaise 2 ka 1 se takrana. Pair ko ek baar count karne ka matlab hai do se divide karna.

Verify. $\frac{1}{2}$ drop karo toh $9.83\times 10^{33}$ milega — exactly double, yaani har collision do baar count ki gayi. Physical (correct) answer chhota wala hai, $4.92\times 10^{33}$.

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[!example] Ex 3 (C3) — ρ ke liye backward solve karo: NO + O₃

Statement. $\text{NO}+{\text{O}}_3\to {\text{NO}}_2+{\text{O}}_2$ at $298\text{K}$ ke liye: measured $k=1.8\times 10^7{\text{M}}^{-1}{\text{s}}^{-1}$, $E_a=10\text{kJ/mol}$, computed $Z=5.0\times 10^{10}{\text{M}}^{-1}{\text{s}}^{-1}$. $\rho$ nikalo.

Forecast: kya $\rho$ $1$ ke kareeb hai ya $0.01$ ke? Ek teen-atom molecule ko carefully aim karna padta hai...

Step 1 — Boltzmann fraction. $e^{-E_a/RT}=\exp \left(\frac{-10000}{(8.314)(298)}\right)=e^{-4.036}=0.0177$ Yeh step kyun? Sirf yahi fraction collisions energy hill clear karta hai; orientation effect akele dekhne se pehle hume yeh strip out karna hoga. Joules use karo, isliye $E_a=10000\text{J/mol}$ match karta hai $R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1}$ se.

Step 2 — $k=\rho Z{e}^{-E_a/RT}$ rearrange karo. $\rho =\frac{k}{Z{e}^{-E_a/RT}}=\frac{1.8\times 10^7}{(5.0\times 10^{10})(0.0177)}=\frac{1.8\times 10^7}{8.85\times 10^8}\approx 0.020$ Yeh step kyun? Orientation ke alawa sab kuch ab right side par hai; jo bachta hai, $\rho$, woh pure geometry fraction hai. $k$ aur $Z$ dono ${\text{M}}^{-1}{\text{s}}^{-1}$ mein hain, isliye $\rho$ dimensionless aata hai — jaisa ek fraction hona chahiye.

Verify. $\rho =0.02<1$ ✓ — ek directional molecule ke liye sensible. $k$ rebuild karo: $(0.02)(5.0\times 10^{10})(0.0177)=1.77\times 10^7\approx k$ ✓. Interpretation: sirf 2% energetic hits sahi aim kiye gaye the (N-end ${\text{O}}_3$ ki taraf).

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[!example] Ex 4 (C4) — ρ ki do limits

Statement. Compare karo (a) $\text{Ar}+\text{Ar}$ (spherical atoms, koi preferred direction nahi) aur (b) ek protein-substrate collision jahan active site surface ka $10^{-4}$ cover karta hai. Collision theory har ek ke liye kya $\rho$ predict karta hai?

Forecast: ek sphere ke liye, kya orientation ka koi matlab bhi hai?

Step 1 — Spherical atom (limit $\rho \to 1$). Ek single atom har direction se identical dikhta hai, isliye har energetic hit "correctly oriented" hai. ${\rho }_{\text{Ar}}\approx 1$ Yeh step kyun? Orientation tabhi penalty hoti hai jab molecule ka koi special reactive spot ho. Sphere ka koi nahi hota.

Step 2 — Big molecule (limit $\rho \ll 1$). Agar reactive patch surface ka fraction $f=10^{-4}$ hai, toh sahi approach ka geometric chance roughly wahi fraction hai: ${\rho }_{\text{protein}}\approx 10^{-4}$ Yeh step kyun? Reader random aim se aata hai; sirf woh aim jo tiny patch par land kare react karta hai.

Verify. Yeh poori real range bracket karte hain: ${\text{H}}_2+{\text{I}}_2$ jaisi chhoti molecules beech mein hain $\rho \sim 0.1$–$0.5$ ke saath. Isliye $10^{-6}\le \rho \le 1$ sab kuch cover karta hai — koi fourth case nahi. (Deeper reason ke liye Transition State Theory dekho ki $\rho$ $1$ se bhi zyada kyun ho sakta hai — Ex 8.)

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[!example] Ex 5 (C5) — Degenerate & limiting speeds

Statement. (a) Do molecules side-by-side same velocity se travel karte hain. $Z$ kya hai? (b) ${\bar{v}}_{\text{rel}}$ ka kya hota hai jab $T\to 0$ aur jab $T\to \infty$?

Forecast: kya do tez chalte molecules kabhi nahi takra sakte?

Step 1 — Equal velocities → zero relative speed. ${\vec{v}}_{\text{rel}}={\vec{v}}_A-{\vec{v}}_B=\vec{0}\Rightarrow Z\propto {\bar{v}}_{\text{rel}}=0$ Yeh step kyun? Collisions depend karti hain gap close karne par. Agar dono identically move karein, gap kabhi nahi shrinkta — chahe kitna bhi tez chalein. Figure mein, top pair (green arrows, equal length aur direction) ek fixed separation forever maintain karta hai; bottom pair (blue aur red arrows ek doosre ki taraf point karte hue) gap close karta hai aur collide karta hai. Arrow directions dekho, unki lengths nahi — yahi decide karta hai collision.

Step 2 — Cold limit $T\to 0$. ${\bar{v}}_{\text{rel}}=\sqrt{\frac{8k_{B}T}{\pi \mu }}\to 0$ Yeh step kyun? Koi thermal energy nahi matlab koi motion nahi, isliye molecules ek doosre ko kabhi nahi miltay — $Z\to 0$ aur reactions freeze ho jaate hain.

Step 3 — Hot limit $T\to \infty$. ${\bar{v}}_{\text{rel}}\to \infty ,e^{-E_a/RT}\to e^0=1$ isliye $k\to \rho Z\to A$ (times ek slow $\sqrt{T}$ growth). Yeh step kyun? Infinite temperature par har collision mein enough energy hoti hai; energy filter disappear ho jaata hai aur rate constant apni ceiling $A$ tak pahunch jaata hai — exactly isliye $A$ ko "maximum" rate constant kehte hain.

Verify. Teeno ${\bar{v}}_{\text{rel}}\propto \sqrt{T}$ ke saath consistent hain: yeh $T=0$ par $0$ hai, grows without bound, aur ceiling $k\to A$ match karta hai ${\lim }_{T\to \infty }{e}^{-E_a/RT}=1$ se. ✓

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[!example] Ex 6 (C6) — T ke saath kaun sa factor zyada move karta hai?

Statement. Ek reaction mein $E_a=100\text{kJ/mol}$ hai. Compare karo ki frequency factor $A$ aur Boltzmann term $e^{-E_a/RT}$ dono kaise change karte hain jab $T$ $300\to 600\text{K}$ jaata hai.

Forecast: parent kehta hai "$A$ ko constant treat karo." Yeh kitne powers of ten tak ek achha approximation hai?

Step 1 — $A$ mein change ($Z\propto \sqrt{T}$ ke zariye). $\frac{A(600)}{A(300)}=\sqrt{\frac{600}{300}}=\sqrt{2}=1.41$ Yeh step kyun? $A$ ke andar sirf ${\bar{v}}_{\text{rel}}\propto \sqrt{T}$ $T$ par depend karta hai; $\rho$ aur $\sigma$ nahi karte.

Step 2 — Boltzmann term mein change. $\frac{e^{-E_a/(R\cdot 600)}}{e^{-E_a/(R\cdot 300)}}=\exp \left[\frac{E_a}{R}\left(\frac{1}{300}-\frac{1}{600}\right)\right]=\exp \left[\frac{100000}{8.314}\cdot \frac{1}{600}\right]=e^{20.05}\approx 5.1\times 10^8$ Yeh step kyun? Yeh exponential ka response isolate karta hai; yeh eight orders of magnitude se move karta hai.

Verify. Dono effects ka ratio: $\frac{5.1\times 10^8}{1.41}\approx 3.6\times 10^8$. Exponential $A$ ko hundreds of millions to one se outrun karta hai — yahi hai rigorous justification Arrhenius plot mein $A$ ko fixed rakhne ka.

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[!example] Ex 7 (C7) — Word problem: stratosphere mein ozone

Statement. Stratospheric ozone $\text{NO}+{\text{O}}_3\to {\text{NO}}_2+{\text{O}}_2$ se destroy hoti hai. Ex 3 ke numbers use karte hue ($k=1.8\times 10^7{\text{M}}^{-1}{\text{s}}^{-1}$), $[\text{NO}]=1.0\times 10^{-9}\text{M}$ aur $[{\text{O}}_3]=5.0\times 10^{-8}\text{M}$ ke saath, ozone ki destruction rate nikalo.

Forecast: ek fast $k$ lekin tiny concentrations ke saath, kya rate large hogi ya vanishing?

Step 1 — Rate law likho. $\text{Rate}=k[\text{NO}][{\text{O}}_3]$ Yeh step kyun? Collision theory ne ek bimolecular (one-step, two-body) mechanism build kiya, isliye rate har reactant mein first order hai.

Step 2 — Numbers plug karo. $\text{Rate}=(1.8\times 10^7)(1.0\times 10^{-9})(5.0\times 10^{-8})=9.0\times 10^{-10}\text{M/s}$ Yeh step kyun? Yeh molecular collision counting ko ek lab-measurable rate mein turn karta hai.

Verify. Units: ${\text{M}}^{-1}{\text{s}}^{-1}\cdot \text{M}\cdot \text{M}=\text{M/s}$ ✓. Small concentrations fast $k$ par dominate karti hain, ek slow lekin relentless rate deti hain — physically isliye thoda sa NO catalytically bahut saari ozone ko time ke saath damage karta hai.

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[!example] Ex 8 (C8) — Exam twist: measured A, Z se bada

Statement. Ek student measure karta hai $A=3\times 10^{13}{\text{M}}^{-1}{\text{s}}^{-1}$ lekin compute karta hai $Z=1\times 10^{13}{\text{M}}^{-1}{\text{s}}^{-1}$. Isse $\rho =A/Z=3>1$ milta hai. "Ek fraction 1 se exceed nahi kar sakta — kya data galat hai?"

Forecast: kya $\rho$ hamesha $1$ se kam hona chahiye?

Step 1 — Apparent $\rho$ compute karo. $\rho =\frac{A}{Z}=\frac{3\times 10^{13}}{1\times 10^{13}}=3$ Yeh step kyun? Yeh paradox ko resolve karne se pehle honestly expose karta hai.

Step 2 — Diagnose karo. Simple collision theory molecules ko hard spheres maanta hai. Real molecules mein long-range attractions hote hain (dipoles, ion pairs) jo partners ko funnel karte hain ek doosre ki taraf, isliye hard-sphere $Z$ predict kare se zyada reactive encounters hoti hain. Yeh step kyun? Yeh identify karta hai kaun sa assumption break hua — data nahi, model.

Step 3 — Fix. $\rho >1$ ko ek signal maano Transition State Theory par upgrade karne ka, jo attraction aur loose transition states naturally handle karta hai. Yeh step kyun? $\rho$ ek correction factor hai, literal probability nahi; $1$ se thoda upar ke values physically meaningful hain.

Verify. $\rho =3$ koi error nahi hai: ions ya strongly polar species ke beech reactions routinely $\rho >1$ dikhate hain. Lesson — "fraction over one" sphere model ki limits flag karta hai, koi mistake nahi. ✓

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Recall Self-test

Which matrix cell forces the $\frac{1}{2}$ factor? ::: C2 — identical collision partners ($A+A$), double counting avoid karne ke liye. Jab $T\to \infty$, $k$ kya approach karta hai? ::: $k\to A=\rho Z$, kyunki $e^{-E_a/RT}\to 1$. Agar side-by-side molecules ek velocity share karein, toh $Z$ kya hai? ::: Zero — relative speed zero hai, isliye gap kabhi close nahi hota. Measured $\rho$ of $3$ matlab data galat hai. True/False? ::: False — yeh long-range attraction flag karta hai; Transition State Theory par upgrade karo. Ex 3 mein, energetic collisions ka kitna fraction correctly aimed tha? ::: Lagbhag $2\%$ ($\rho \approx 0.02$). Ex 1 mein, raw collision frequency $Z$ roughly kis order of magnitude ka hai? ::: Lagbhag $10^{36}$ collisions per m³ per second.