Collision theory — frequency factor, steric factor: Exercises

2.8.9 · D4 · Chemistry › Chemical Kinetics › Collision theory — frequency factor, steric factor

Yeh page parent topic ka self-test ladder hai. Har problem L1 (sirf idea pehchano) se lekar L5 (mastery — sab kuch combine karo) tak graded hai. Har solution ek collapsible callout ke andar chhupi hui hai taaki tum pehle khud try kar sako, phir reveal karo.

Shuru karne se pehle, aao har symbol ko pin down karte hain taaki koi line one se hi na bhatak jaye.

Definition Woh symbols jo tum is page par use karoge

$k$ ::: the rate constant — woh number jo batata hai ki ek given temperature par reaction kitni tez hoti hai.
$A$ ::: the frequency factor (pre-exponential factor) — woh value jo $k$ reach karta agar har collision mein enough energy hoti.
$Z$ ::: the collision frequency — kitne collisions unit volume mein unit time mein hote hain.
$ρ$ ::: the steric factor — energetic collisions ka woh fraction jo sahi direction mein bhi pointed hain.
$E_{a}$ ::: the activation energy — react karne ke liye minimum collision energy (dekho Activation Energy).
$R = 8.314 J mol^{- 1} K^{- 1}$ ::: the gas constant.
$T$ ::: kelvin (K) mein absolute temperature.
$e^{- E_{a} / R T}$ ::: the Boltzmann factor — un collisions ka fraction jo energy bar cross kar lete hain (Maxwell-Boltzmann Distribution se).

Ek master equation jisse neche sab kuch flow karta hai:

Figure — Collision theory — frequency factor, steric factor

Upar ke teen filters independent gates hain — ek collision ko teeno pass karne honge. Woh picture har problem ki key hai.

L1 — Recognition

Problem 1.1

Inn mein se kaun sa hamesha $s^{- 1}$ ke units mein hoga (kabhi $M^{- 1} s^{- 1}$ nahi) no matter what: steric factor $ρ$ , collision frequency $Z$ , ya frequency factor $A$ ?

Recall Solution

Steric factor $ρ$ ek pure fraction hai — yeh "aimed-right collisions" aur "energetic collisions" ka ratio hai, isliye saare units cancel ho jaate hain. Yeh dimensionless hai (jise hum loosely $s^{0}$ likh sakte hain, yani bilkul koi units nahi). $A$ aur $Z$ dono $k$ ke saath same units carry karte hain (ek bimolecular step ke liye, $M^{- 1} s^{- 1}$ ). Toh answer hai $ρ$ — yeh unitless hai.

Problem 1.2

Ek-ek sentence mein batao ki teen factors $Z$ , $e^{- E_{a} / R T}$ , aur $ρ$ physically kis cheez ko filter karte hain.

Recall Solution

$Z$ count karta hai kitni baar molecules milte hain.
$e^{- E_{a} / R T}$ un collisions ka fraction hai jo kaafi hard hit karte hain (energy $\geq E_{a}$ ).
$ρ$ un energetic hits mein se sahi direction mein pointed wale ka fraction hai.

L2 — Application

Problem 2.1

Ek reaction mein $Z = 4.0 \times 1 0^{10} M^{- 1} s^{- 1}$ aur measured frequency factor $A = 8.0 \times 1 0^{8} M^{- 1} s^{- 1}$ hai. Steric factor $ρ$ nikalo.

Recall Solution

KYA karte hain: $A = ρZ$ ko invert karte hain. KYUN: $A$ aur $Z$ diye gaye hain, $ρ$ akela unknown hai. $ρ = \frac{A}{Z} = \frac{8.0 \times 1 0 ^{8}}{4.0 \times 1 0 ^{10}} = 0.020$ Matlab: sirf 2% energetic collisions itne sahi oriented hain ki react kar sakein.

Problem 2.2

$T = 300 K$ par, ek reaction mein $E_{a} = 50 kJ mol^{- 1}$ hai. Kitne fraction collisions energy bar cross kar lete hain?

Recall Solution

KYA/KYUN: hum Boltzmann factor chahte hain, toh $e^{- E_{a} / R T}$ compute karte hain — yeh akela tool hai jo answer deta hai "kitne fraction ki energy $\geq E_{a}$ hai". Units consistent rakho: $E_{a} = 50000 J mol^{- 1}$ , $R = 8.314$ , $T = 300$ . $\frac{E _{a}}{R T} = \frac{50000}{8.314 \times 300} = \frac{50000}{2494.2} = 20.05$ $e^{- 20.05} = 1.96 \times 1 0^{- 9}$ Matlab: lagbhag har billion mein 2 collisions itne energetic hain.

L3 — Analysis

Problem 3.1

$298 K$ par $NO + O_{3} \to NO_{2} + O_{2}$ ke liye: observed $k = 1.8 \times 1 0^{7} M^{- 1} s^{- 1}$ , $E_{a} = 10 kJ mol^{- 1}$ , collision frequency $Z = 5.0 \times 1 0^{10} M^{- 1} s^{- 1}$ . $ρ$ nikalo aur interpret karo.

Recall Solution

Step 1 — energy gate. Boltzmann factor compute karo: $\frac{E _{a}}{R T} = \frac{10000}{8.314 \times 298} = 4.036, e^{- 4.036} = 0.0177$ Step 2 — master equation invert karo. $k = ρZ e^{- E_{a} / R T}$ se: $ρ = \frac{k}{Z e ^{- E_{a} / R T}} = \frac{1.8 \times 1 0 ^{7}}{( 5.0 \times 1 0 ^{10} ) ( 0.0177 )} = \frac{1.8 \times 1 0 ^{7}}{8.85 \times 1 0 ^{8}} = 0.0203$ Interpretation: lagbhag $2%$ energetic collisions mein NO ka N-end $O_{3}$ ki taraf sahi aim kiya hua hota hai electron transfer ke liye. Baaki $98%$ energetic hain lekin mis-aimed glancing blows hain.

Problem 3.2

Do reactions mein same $Z$ aur same $E_{a}$ hai. Reaction P (chhote atoms) mein $ρ = 1$ hai; reaction Q (bulky molecules) mein $ρ = 1 0^{- 5}$ hai. P, Q se kitne factor zyada fast hai?

Recall Solution

Kyunki $Z$ , $E_{a}$ , $T$ shared hain, woh ratio mein cancel ho jaate hain: $\frac{k _{P}}{k _{Q}} = \frac{ρ _{P} Z e ^{- E_{a} / R T}}{ρ _{Q} Z e ^{- E_{a} / R T}} = \frac{ρ _{P}}{ρ _{Q}} = \frac{1}{1 0 ^{- 5}} = 1 0^{5}$ Matlab: orientation akele P ko ek lakh guna zyada fast banata hai. Isliye complex-molecule reactions (dekho Transition State Theory) atom recombinations se itni zyada slow hoti hain.

L4 — Synthesis

Problem 4.1

Ek bimolecular gas reaction $300 K$ par study ki jaati hai phir $600 K$ par ( $T$ double ho jaata hai). Iska $E_{a} = 100 kJ mol^{- 1}$ hai, aur $ρ$ temperature-independent hai. Kyunki $Z \propto T$ hai, kitne factor se (a) $A$ change hota hai, aur (b) Boltzmann factor change hota hai? Kaun sa $k$ par dominate karta hai?

Recall Solution

(a) Frequency factor. Kyunki $A = ρZ$ aur $Z \propto T$ : $\frac{A ( 600 )}{A ( 300 )} = \frac{600}{300} = 2 = 1.41$ (b) Boltzmann factor. Dono exponents compute karo. $\frac{E _{a}}{R ( 300 )} = \frac{100000}{8.314 \times 300} = 40.09, \frac{E _{a}}{R ( 600 )} = 20.05$ $\frac{e ^{- 20.05}}{e ^{- 40.09}} = e^{40.09 - 20.05} = e^{20.04} = 5.0 \times 1 0^{8}$ Kaun dominate karta hai? Boltzmann factor $\sim 5 \times 1 0^{8}$ se badhta hai; $A$ sirf $1.41$ se badhta hai. Exponential bilkul dominate karta hai — isliye hum Arrhenius plots mein $A$ ko constant treat karte hain.

Problem 4.2

Collision-frequency formula ko master equation ke saath combine karo. Diya gaya hai $σ = 2.83 \times 1 0^{- 19} m^{2}$ , $\overset{v}{ˉ}_{r e l} = 631 m s^{- 1}$ , number densities $[A] = [B] = 6.02 \times 1 0^{25} m^{- 3}$ , raw collision frequency $Z_{A B}$ nikalo (collisions per m³ per s).

Recall Solution

KYUN yeh formula: $Z_{A B} = σ \overset{v}{ˉ}_{r e l} [A] [B]$ cross-section (target kitna bada hai), approach speed (kitni tez milte hain), aur dono mein se kitne hain ko multiply karta hai. $Z_{A B} = (2.83 \times 1 0^{- 19}) (631) (6.02 \times 1 0^{25})^{2}$ Pehle density product: $(6.02 \times 1 0^{25})^{2} = 3.624 \times 1 0^{51}$ . Phir $2.83 \times 1 0^{- 19} \times 631 = 1.786 \times 1 0^{- 16}$ . $Z_{A B} = 1.786 \times 1 0^{- 16} \times 3.624 \times 1 0^{51} = 6.47 \times 1 0^{35} collisions m^{- 3} s^{- 1}$

L5 — Mastery

Problem 5.1

$500 K$ par ek reaction mein $E_{a} = 80 kJ mol^{- 1}$ , $Z = 1.0 \times 1 0^{11} M^{- 1} s^{- 1}$ , aur measured $k = 3.0 \times 1 0^{- 2} M^{- 1} s^{- 1}$ hai. (a) $ρ = 1$ (perfect orientation) maante hue $k$ predict karo. (b) Measured $k$ se compare karo aur true $ρ$ extract karo. (c) $ρ$ reactive site ki geometry ke baare mein kya batata hai?

Recall Solution

(a) Ideal prediction, $ρ = 1$ . $\frac{E _{a}}{R T} = \frac{80000}{8.314 \times 500} = 19.245, e^{- 19.245} = 4.39 \times 1 0^{- 9}$ $k_{ρ = 1} = Z e^{- E_{a} / R T} = (1.0 \times 1 0^{11}) (4.39 \times 1 0^{- 9}) = 4.39 \times 1 0^{2} M^{- 1} s^{- 1}$ (b) $ρ$ extract karo. $ρ = \frac{k _{meas}}{k _{ρ = 1}} = \frac{3.0 \times 1 0 ^{- 2}}{4.39 \times 1 0 ^{2}} = 6.83 \times 1 0^{- 5}$ (c) Geometry. $ρ \approx 7 \times 1 0^{- 5}$ matlab sirf lagbhag har 100 000 mein 7 energetic collisions sahi aim ki hui hain. Yeh ek bulky reactant ki signature hai jahan reactive site surface ka ek tiny patch cover karti hai — molecularity ke saath consistent jo large, orientation-sensitive species involve karta hai.

Problem 5.2

Algebraically dikhao ki do rate constants ka ratio same temperature par sirf do frequency factors par depend karta hai jab activation energies equal hoon — yani $k_{1} / k_{2} = A_{1} / A_{2}$ derive karo, aur batao yeh kab fail hota hai.

Recall Solution

KYA/KYUN: har $k$ ko Arrhenius form mein likho aur divide karo, taaki shared machinery cancel ho jaaye. $\frac{k _{1}}{k _{2}} = \frac{A _{1} e ^{- E_{a, 1} / R T}}{A _{2} e ^{- E_{a, 2} / R T}} = \frac{A _{1}}{A _{2}} e^{- (E_{a, 1} - E_{a, 2}) / R T}$ Agar $E_{a, 1} = E_{a, 2}$ ho, toh exponential $e^{0} = 1$ hai, jo bacha deta hai $\frac{k _{1}}{k _{2}} = \frac{A _{1}}{A _{2}}$ Kyunki $A = ρZ$ hai, aur $Z$ similar-sized molecules ke liye lagbhag same hota hai, yeh ratio actually steric factors ka ratio hai. Yeh fail hota hai jab bhi activation energies alag hoon: tab exponential term cancel nahi hota aur dominate karta hai, exactly jaise Problem 4.1 mein.

Recall Quick self-check clozes

The frequency factor equals == $A = ρZ$ . Teen independent gates jo ek reactive collision ko pass karne hote hain woh hain collision frequency $Z$ , energy $e^{- E_{a} / R T}$ , aur orientation $ρ$ ==. $T$ double karne par (300→600 K, $E_{a} = 100$ kJ) $A$ == $2 \approx 1.41$ se badhta hai lekin Boltzmann factor $\sim 5 \times 1 0^{8}$ == se. Ek computed $ρ > 1$ matlab the hard-sphere collision model has broken down; use transition-state theory.