Transition state theory — activated complex (intro): Exercises

2.8.10 · D4 · Chemistry › Chemical Kinetics › Transition state theory — activated complex (intro)

Yeh page Transition state theory — activated complex (intro) ke liye ek self-test ladder hai. Har problem ko collapsible solution kholne se pehle solve karo. Neeche use kiye gaye har symbol ko pehli baar dekhne par define kiya gaya hai, isliye problem 1 se hi shuru karo chahe notation naya lage.

Figure — Transition state theory — activated complex (intro)

Level 1 — Recognition

L1.1 — Maximum ya minimum?

Batao ki reaction coordinate diagram par har ek local energy maximum hai ya local energy minimum, aur uski approximate lifetime bhi do: (a) ek activated complex, (b) ek reaction intermediate, (c) reactants.

Recall Solution

(a) Activated complex → maximum (hilltop). Lifetime ≈ $1 0^{- 13} s$ (ek bond vibration). Yeh ek passage ka point hai, kabhi resting place nahi.
(b) Intermediate → minimum (do hills ke beech ek choti si dip). Lifetime itni lambi hoti hai ki potentially detect kiya ja sake (nanoseconds se lekar seconds tak).
(c) Reactants → minimum (left valley floor). Yeh stably baithe rehte hain jab tak inhe climb karne ke liye enough energy na mile.

Figure s01 dekho: hilltops maxima hain, valleys minima hain. Sirf hilltop transition state hai.

L1.2 — Diagram padhna

Figure s01 mein, $E_{reactants} = 50 kJ/mol$ , $E_{TS} = 220 kJ/mol$ , $E_{products} = 60 kJ/mol$ hai. $E_{a}^{fwd}$ nikalo.

Recall Solution

$E_{a}^{fwd} = E_{TS} - E_{reactants} = 220 - 50 = 170 kJ/mol$ . Yeh left valley se peak tak ki chadhai hai — ise Figure s01 mein orange arrow ke roop mein trace karo.

L1.1 answer

(a) maximum, (b) minimum, (c) minimum

L1.2 answer

170 kJ/mol

Level 2 — Application

L2.1 — Reverse barrier

L1.2 ke numbers use karte hue ( $E_{TS} = 220$ , $E_{products} = 60$ , sab kJ/mol mein), $E_{a}^{rev}$ aur $Δ E$ nikalo. Kya forward reaction exo- ya endothermic hai?

Recall Solution

$E_{a}^{rev} = E_{TS} - E_{products} = 220 - 60 = 160 kJ/mol$ . $Δ E = E_{products} - E_{reactants} = 60 - 50 = + 10 kJ/mol$ . Positive $Δ E$ ⇒ products reactants se upar hain ⇒ endothermic. Master identity se check karo: $E_{a}^{fwd} - E_{a}^{rev} = 170 - 160 = 10 = Δ E$ . ✓

L2.2 — Missing height nikalna

Ek reaction ka $E_{a}^{fwd} = 90 kJ/mol$ hai aur yeh $Δ E = - 40 kJ/mol$ ke saath exothermic hai. $E_{a}^{rev}$ nikalo.

Recall Solution

Reverse climb ke liye master identity ko rearrange karo: $E_{a}^{rev} = E_{a}^{fwd} - Δ E = 90 - (- 40) = 130 kJ/mol .$ Reverse zyada kyun hai: reaction downhill 40 gayi, isliye usi peak par neeche ki product valley se wapas chadhai karna zyada oonchi chadhai hai. Figure s01 ki bhaasha mein, ek neecha right base reverse arrow ko lamba banata hai.

L2.1 Ea_rev

160 kJ/mol

L2.1 dE

+10 kJ/mol (endothermic)

L2.2 Ea_rev

130 kJ/mol

Level 3 — Analysis

L3.1 — $E_{a} \neq =$ bond energy kyun

$H_{2} + I_{2} \to 2 HI$ ke liye: H–H bond energy $D_{H-H} = 436 kJ/mol$ hai, I–I bond energy $D_{I-I} = 151 kJ/mol$ hai, phir bhi measured $E_{a} = 170 kJ/mol$ hai. (a) $E_{a}$ kya hota agar dono bonds poori tarah toot jaate koi naya bond banne se pehle? (b) Real transition state us value se kitna kam hai, aur kyun?

Recall Solution

(a) Pehle dono bonds poori tarah todne mein $436 + 151 = 587 kJ/mol$ lagta. (b) Real barrier $170$ hai, isliye transition state "sab kuch pehle todo" cost se $587 - 170 = 417 kJ/mol$ kam hai. Kyun: activated complex par purane bonds sirf partially stretch hote hain jabki naaye H–I bonds partially usi waqt form ho rahe hote hain. Woh partial bonding zyaatartar cost wapas de deta hai, isliye tumhe kabhi bhi poori tarah kuch nahi todna padta. Yahi wajah hai ki $E_{a}$ ek bond dissociation energy nahi hai.

L3.2 — Hammond reasoning

Hammond's postulate (yahan use karne ke liye stated hai): transition state us species se sabse zyada milta-julta hota hai — reactants ya products — jo energy mein us se sabse kareeb hoti hai. Toh ek exothermic reaction (products reactants se kaafi neeche) ka peak early aata hai aur reactants jaisa lagta hai; ek endothermic reaction (products reactants se upar) ka peak late aata hai aur products jaisa lagta hai.

Reaction X strongly exothermic hai; reaction Y strongly endothermic hai. Batao ki har transition state reactants ya products se milta-julta hai, aur kis reaction ka likely lower forward barrier hoga (similar mechanisms assume karo).

Recall Solution

X (exothermic): transition state early reach hota hai aur reactants se milta-julta hai — ek "early barrier." Kam geometric distortion chahiye, isliye iska tend hota hai lower $E_{a}^{fwd}$ hone ka.
Y (endothermic): transition state late hai aur products se milta-julta hai — ek "late barrier," generally zyada ooncha. Figure s02 mein: blue (exothermic) curve ka peak left mein reactant side ke paas hai; red (endothermic) curve ka peak right mein product side ke paas hai. Dekho Hammond's postulate.

L3.1a full-break cost

587 kJ/mol

L3.1b undercut

417 kJ/mol

Level 4 — Synthesis

L4.1 — Usi diagram par catalyst

Ek reaction ka $E_{reactants} = 30$ , $E_{TS} = 180$ , $E_{products} = 70$ (kJ/mol) hai. Ek catalyst ek naya path provide karta hai jisme $E_{TS}^{cat} = 120$ hai. Nikalo: (a) uncatalyzed $E_{a}^{fwd}$ , (b) catalyzed $E_{a}^{fwd}$ , (c) uncatalyzed $E_{a}^{rev}$ , (d) catalyzed $E_{a}^{rev}$ , (e) kya catalyst $Δ E$ change karta hai?

Recall Solution

(a) $180 - 30 = 150 kJ/mol$ . (b) $120 - 30 = 90 kJ/mol$ . (c) $180 - 70 = 110 kJ/mol$ . (d) $120 - 70 = 50 kJ/mol$ . (e) $Δ E = 70 - 30 = + 40 kJ/mol$ , unchanged. Catalyst sirf peak ko lower karta hai; do valley floors (reactants, products) untouched rehte hain, isliye $Δ E$ same hai. Key insight: catalyst ne dono barriers ko usi $60 kJ/mol$ se lower kiya ( $150 \to 90$ aur $110 \to 50$ ). Dekho Catalysis.

L4.2 — Consistency puzzle

Tumhe bataya gaya hai ki $E_{a}^{fwd} = 120$ , $E_{a}^{rev} = 45$ , aur $Δ E = + 90$ (kJ/mol) hai. Dikhao ki yeh numbers inconsistent hain, aur $E_{a}^{rev}$ ki woh value do jo inhe consistent banaye.

Recall Solution

Master identity demand karti hai ki $E_{a}^{fwd} - E_{a}^{rev} = Δ E$ . Check karo: $120 - 45 = 75 \neq = 90$ . Inconsistent. Consistency ke liye: $E_{a}^{rev} = E_{a}^{fwd} - Δ E = 120 - 90 = 30 kJ/mol$ . Identity kyun hold karni chahiye: dono barriers usi peak tak measure kiye jaate hain. Unka difference hai (peak − left base) − (peak − right base) = right base − left base = $Δ E$ . Peak cancel ho jaata hai, isliye koi bhi self-consistent diagram ise violate nahi kar sakta.

L4.1 a

150 kJ/mol

L4.1 b

90 kJ/mol

L4.1 c

110 kJ/mol

L4.1 d

50 kJ/mol

L4.1 e

dE = +40 kJ/mol, unchanged

L4.2 consistent Ea_rev

30 kJ/mol

Level 5 — Mastery

L5.1 — Arrhenius meets TST

Do reactions ek hi pre-exponential factor share karte hain. Reaction P ka $E_{a} = 50 kJ/mol$ hai; reaction Q ka $E_{a} = 75 kJ/mol$ hai. $T = 300 K$ par, P, Q se kitne factor zyada fast hai? $R = 8.314 J mol^{- 1} K^{- 1}$ use karo.

Recall Solution

$\frac{k _{P}}{k _{Q}} = \frac{A e ^{- E_{a, P} / R T}}{A e ^{- E_{a, Q} / R T}} = e^{(E_{a, Q} - E_{a, P}) / R T} .$ Barrier gap: $E_{a, Q} - E_{a, P} = 75000 - 50000 = 25000 J/mol$ . Exponent: $\frac{25000}{8.314 \times 300} = \frac{25000}{2494.2} = 10.023$ . $\frac{k _{P}}{k _{Q}} = e^{10.023} \approx 2.26 \times 1 0^{4} .$ Sirf 25 kJ/mol ka difference P ko roughly baees hazaar guna faster banata hai — yeh steep sensitivity hi wajah hai ki transition state mein chhote changes (ya ek catalyst) enormously matter karte hain.

L5.2 — Diagram clues se design karo

Sketch-and-solve. Tumhe ek single reaction diagram banana hai jo sab satisfy kare: (i) $60 kJ/mol$ se exothermic; (ii) forward barrier $= 40 kJ/mol$ ; (iii) reactants $10 kJ/mol$ par hain. $E_{TS}$ , $E_{products}$ , aur $E_{a}^{rev}$ nikalo. Phir batao ki Hammond peak ko kahan rakhta hai.

Recall Solution

$E_{TS} = E_{reactants} + E_{a}^{fwd} = 10 + 40 = 50 kJ/mol$ .
$E_{products} = E_{reactants} + Δ E = 10 + (- 60) = - 50 kJ/mol$ .
$E_{a}^{rev} = E_{TS} - E_{products} = 50 - (- 50) = 100 kJ/mol$ .
Check: $E_{a}^{fwd} - E_{a}^{rev} = 40 - 100 = - 60 = Δ E$ . ✓
Hammond: strongly exothermic ⇒ early transition state, peak reactants se milta-julta hai (left ki taraf jhukta hai). Dekho Hammond's postulate aur Reaction coordinate diagrams.

L5.1 speed ratio

~2.26 × 10^4 (P faster than Q)

L5.2 E_TS

50 kJ/mol

L5.2 E_products

-50 kJ/mol

L5.2 Ea_rev

100 kJ/mol

Related vault notes: Eyring equation, Collision theory, Potential energy surfaces, SN1 vs SN2 mechanisms.