Worked examples — Transition state theory — activated complex (intro)

2.8.10 · D3 · Chemistry › Chemical Kinetics › Transition state theory — activated complex (intro)

Yeh page ek drill hall hai. Parent note mein tumne yeh idea seekha tha ki har reaction ek energy hill chadhhti hai aur ek knife-edge peak par rukti hai — woh hai activated complex. Yahan hum us hill ki arithmetic practice karte hain jab tak koi bhi case tumhe surprise na kar sake.

Koi bhi number likhne se pehle, hum vocabulary fix karte hain taaki koi bhi symbol bina reason ke na aaye.

Yahan har symbol sirf ek height hai. $E_a$ ("upar ka climb") hamesha top minus woh jagah hoti hai jahan se tumne shuru kiya. $\Delta H$ (Greek capital delta H, "change in enthalpy" padho) bas "finish height minus start height" hai — positive agar tum upar khatam karo, negative agar neeche. Hiking map padhne se zyada mysterious kuch nahi hai.

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The scenario matrix

Is topic ke har energy problem ke liye in mein se ek cell hogi. Aage aane waale worked examples mein har ek announce karta hai ki woh kaun si cell mein aata hai, aur mil ke woh sab ko cover karte hain.

#	Case class	Kya special hai	Example
A	Exothermic ($\Delta H<0$), reverse barrier find karo	product ground neeche hai	Ex 1
B	Endothermic ($\Delta H>0$), reverse barrier find karo	product ground upar hai	Ex 2
C	Thermoneutral ($\Delta H=0$) — degenerate	dono barriers equal hain	Ex 3
D	Zero barrier ($E_a=0$) — limiting case	koi hill nahi; barrierless	Ex 4
E	Sign/consistency trap	diya hua data ek sign force karta hai	Ex 5
F	$E_a\ne$ bond energy (partial bonding)	multi-bond concerted step	Ex 6
G	Real-world word problem	catalyst hill ko lower karta hai	Ex 7
H	Exam twist (Hammond, early/late TS)	TS ki position ke baare mein reason karo	Ex 8

Recall Which quantity is kinetics and which is thermodynamics?

$E_a$ (top minus start) ::: kinetics — rate control karta hai. $\Delta H$ (product minus reactant) ::: thermodynamics — equilibrium/heat control karta hai.

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Example 1 — Cell A: exothermic, reverse barrier find karo

Forecast: reverse climb $60$ se badi honi chahiye — tum lower ground se shuru kar rahe ho, toh peak tumse aur door hai. Aage padhne se pehle guess karo.

1. Seam identity likho. Kyun yeh step? Yeh woh ek equation hai jo forward, reverse, aur $\Delta H$ ko link karti hai — bilkul wahi teen quantities jo yahan kaam aa rahi hain. $E_a^{\text{fwd}}-E_a^{\text{rev}}=\Delta H$
2. Unknown ke liye solve karo. Kyun yeh step? Hum $E_a^{\text{rev}}$ akela chahte hain. $E_a^{\text{rev}}=E_a^{\text{fwd}}-\Delta H=60-(-25)=85\text{kJ/mol}$
3. Geometry padho. Kyun yeh step? Parent ki hill picture match karni chahiye: products reactants se $25$ neeche hain, toh peak products se $85$ upar hai.

Verify: upar chadhhna ($60$) phir doosri taraf slide karna ($85$) matlab net height change $=60-85=-25=\Delta H$. ✓ Reverse barrier forward barrier se zyada hai, jaise forecast tha. Units: sab kJ/mol, consistent.

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Example 2 — Cell B: endothermic, reverse barrier find karo

Forecast: endothermic matlab products upar hain, toh peak se wapas aana ek chhota drop hai — expect karo $E_a^{\text{rev}}<170$.

1. Phir seam identity. Kyun? Same teen quantities. $E_a^{\text{rev}}=E_a^{\text{fwd}}-\Delta H=170-10=160\text{kJ/mol}$
2. $\Delta H$ ke sign par sanity check. Kyun? Endothermic must give $E_a^{\text{rev}}<E_a^{\text{fwd}}$; $160<170$ confirm karta hai ki humne $+10$ sahi se plug kiya.

Verify: $E_a^{\text{fwd}}-E_a^{\text{rev}}=170-160=10=\Delta H$. ✓

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Example 3 — Cell C: the degenerate thermoneutral case

Forecast: agar hill ke dono taraf ground level same hai, toh dono climbs equal honi chahiye.

1. $\Delta H=0$ plug karo. Kyun? Degenerate ki definition — start aur finish heights ek hi hain. $E_a^{\text{rev}}=E_a^{\text{fwd}}-0=50\text{kJ/mol}$
2. Symmetry interpret karo. Kyun? Ek symmetric hill ka peak bilkul beech mein hota hai; activated complex do equal wells ke beech mein exactly aadhe energy par hota hai.

Verify: $E_a^{\text{fwd}}=E_a^{\text{rev}}$ aur $\Delta H=50-50=0$. ✓ Degenerate case matrix ka woh "diagonal" hai jahan dono barriers merge ho jaate hain.

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Example 4 — Cell D: barrierless limit ($E_a=0$)

Forecast: koi forward hill nahi, toh dono wells ke beech ki only height well depth hi hai, isliye reverse barrier $370$ hona chahiye.

1. Seam identity $E_a^{\text{fwd}}=0$ ke saath. Kyun? Barrierless matlab peak reactant ground ke saath coincide karta hai — koi extra climb nahi. $E_a^{\text{rev}}=E_a^{\text{fwd}}-\Delta H=0-(-370)=370\text{kJ/mol}$
2. Physics se match karo. Kyun? Koi forward hill na hone par naye bane C–C bond ko todna sirf uski full bond dissociation energy pay karna hai — woh limiting case jahan $E_a^{\text{rev}}=D$.

Verify: $E_a^{\text{fwd}}-E_a^{\text{rev}}=0-370=-370=\Delta H$. ✓ Yeh parent ke Mistake 2 ka degenerate limit hai: yahan, aur tabhi jab doosri taraf barrierless ho, $E_a$ ek bond energy ke equal hoti hai.

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Example 5 — Cell E: the sign/consistency trap

Forecast: seam identity ek hard constraint hai — teen mein se kisi par bhi trust karne se pehle ise check karo.

1. Identity test karo. Kyun? Teeno ek shared peak se fixed hain; woh independent nahi ho sakte. $E_a^{\text{fwd}}-E_a^{\text{rev}}=40-90=-50\ne +50=\Delta H?$
2. Sign error diagnose karo. Kyun? Computed value $-50$ hai, reported $\Delta H$ ka negative. Reaction actually exothermic hai: reverse barrier forward se bada matlab products lower hain.
3. Corrected value state karo. Kyun? Consistency demand karti hai $\Delta H=E_a^{\text{fwd}}-E_a^{\text{rev}}=-50\text{kJ/mol}.$

Verify: $\Delta H=-50$ ke saath: $40-90=-50$. ✓ Barriers theek the; reported $\Delta H$ ka sign flip tha. Bada reverse barrier ⇒ exothermic, hamesha.

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Example 6 — Cell F: $E_a$ broken bonds ka sum NAHI hai

Forecast: barrier $436+151=587$ se kaafi neeche hai, toh partial new bonds zyaatar difference pay kar rahe honge.

1. Cost agar bonds pehle fully break hote. Kyun? Upper-bound scenario — ek stepwise "pehle sab todo, phir banao" path. $436+151=587\text{kJ/mol}$
2. Real barrier se compare karo. Kyun? Antar woh hai jo concerted partial bonding bachata hai. $587-170=417\text{kJ/mol saved by partial H–I bond formation at the TS}$
3. Interpret karo. Kyun? Yeh parent ka Mistake 2 quantitative form mein hai: kyunki do H–I bonds tab form hone shuru hote hain jab H–H aur I–I sirf aadhe todti hain, peak naïve full-break estimate se $417\text{kJ/mol}$ neeche hai.

Verify: $587-170=417\ge 0$ (stabilisation positive honi chahiye) aur $170<587$ (barrier full break se neeche). ✓

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Example 7 — Cell G: real-world word problem, catalyst

Forecast: ek catalyst peak ko lower karta hai, toh dono climbs same $30$ se shrink hoti hain; $\Delta H$ (jo peak ko kabhi touch nahi karta) untouched rehta hai.

1. New forward barrier. Kyun? Peak $30$ girta hai; reactant ground wahi rehta hai. $E_a^{\text{fwd,new}}=120-30=90\text{kJ/mol}$
2. Seam se old reverse barrier. Kyun? Subtract karne ke liye ek baseline chahiye. $E_a^{\text{rev,old}}=E_a^{\text{fwd}}-\Delta H=120-(-40)=160\text{kJ/mol}$
3. New reverse barrier. Kyun? Same lowered peak, product ground unmoved ⇒ reverse climb bhi $30$ girti hai. $E_a^{\text{rev,new}}=160-30=130\text{kJ/mol}$
4. $\Delta H$? Kyun? Yeh sirf do ground levels par depend karta hai, dono unchanged. $\Delta H=-40\text{kJ/mol (unchanged)}$

Verify: catalysis ke baad bhi seam hold karta hai: $E_a^{\text{fwd,new}}-E_a^{\text{rev,new}}=90-130=-40=\Delta H$. ✓ Catalyst ne peak ko dono directions ke liye equally drop kiya — parent ke reaction coordinate diagram logic dekho.

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Example 8 — Cell H: exam twist, early vs late transition state

Forecast: exothermic + tiny forward barrier ⇒ peak reactant side ke paas hai (early, "reactant-like"); endothermic ⇒ peak products ke paas (late, "product-like").

1. Reaction P reverse barrier. Kyun? Seam identity. $E_a^{\text{rev}}=20-(-180)=200\text{kJ/mol}$ Chhota upar chadhhna ($20$), bada neeche aana ($200$): peak reactants ke bilkul baad hai ⇒ early, reactant-like TS.
2. Reaction Q reverse barrier. Kyun? Same identity. $E_a^{\text{rev}}=170-150=20\text{kJ/mol}$ Bada upar chadhhna ($170$), chhota neeche aana ($20$): peak products ke bilkul pehle hai ⇒ late, product-like TS.
3. Rule state karo. Kyun? Hammond: TS us species se milti hai jiske energy mein woh zyada close hai. Chhota $E_a^{\text{fwd}}$ vs bada $E_a^{\text{rev}}$ ⇒ reactant-like; aur vice versa.

Verify: P: $20-200=-180=\Delta H$ ✓; Q: $170-20=150=\Delta H$ ✓. Dono seams close hain, aur chhota barrier sahi se "resembled" side ki taraf point karta hai.

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Recall Reaction P above mein kaun sa transition state hai?

Early / reactant-like ::: kyunki woh exothermic hai aur forward barrier tiny hai — peak reactants ke paas hai (Hammond).

Recall Ek catalyst

$E_a^{\text{fwd}}$ ko $30$ se lower karta hai. $E_a^{\text{rev}}$ kitna change hoga? Also by $30$ (shared peak ko lower karta hai) ::: jabki $\Delta H$ bilkul same rehta hai.

In barriers ko actual rate constants mein convert karne ke liye Arrhenius equation bhi dekho, aur TST-native version ke liye Eyring equation dekho.