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

This page is a drill hall. In the parent note you met the idea that every reaction climbs an energy hill and pauses at a knife-edge peak — the activated complex. Here we practice the arithmetic of that hill until no case can surprise you.

Before any numbers, we fix vocabulary so no symbol appears unearned.

Every symbol here is just a height. $E_a$ (the "climb up") is always the top minus where you started. $\Delta H$ (Greek capital delta H, read "change in enthalpy") is just "finish height minus start height" — positive if you end up higher, negative if lower. Nothing more mysterious than reading a hiking map.

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The scenario matrix

Every energy problem this topic can throw is one of these cells. The worked examples that follow each announce which cell they land on, and together they touch all of them.

#	Case class	What's special	Example
A	Exothermic ($\Delta H<0$), find reverse barrier	product ground is lower	Ex 1
B	Endothermic ($\Delta H>0$), find reverse barrier	product ground is higher	Ex 2
C	Thermoneutral ($\Delta H=0$) — degenerate	both barriers equal	Ex 3
D	Zero barrier ($E_a=0$) — limiting case	no hill; barrierless	Ex 4
E	Sign/consistency trap	given data forces a sign	Ex 5
F	$E_a\ne$ bond energy (partial bonding)	multi-bond concerted step	Ex 6
G	Real-world word problem	catalyst lowers the hill	Ex 7
H	Exam twist (Hammond, early/late TS)	reason about TS position	Ex 8

Recall Which quantity is kinetics and which is thermodynamics?

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

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

Forecast: the reverse climb should be bigger than 60 — you're starting from the lower ground, so the peak is farther above you. Guess before reading on.

1. Write the seam identity. Why this step? It is the one equation linking forward, reverse, and $\Delta H$ — exactly the three quantities in play. $E_a^{\text{fwd}}-E_a^{\text{rev}}=\Delta H$
2. Solve for the unknown. Why this step? We want $E_a^{\text{rev}}$ alone. $E_a^{\text{rev}}=E_a^{\text{fwd}}-\Delta H=60-(-25)=85\text{kJ/mol}$
3. Read the geometry. Why this step? The parent's hill picture must match: products sit $25$ below reactants, so the peak is $85$ above the products.

Verify: climb up ($60$) then slide down the far side ($85$) means net height change $=60-85=-25=\Delta H$. ✓ Reverse barrier exceeds forward barrier, as forecast. Units: all kJ/mol, consistent.

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

Forecast: endothermic means products sit higher, so coming back downhill from the peak is a shorter drop — expect $E_a^{\text{rev}}<170$.

1. Seam identity again. Why? Same three quantities. $E_a^{\text{rev}}=E_a^{\text{fwd}}-\Delta H=170-10=160\text{kJ/mol}$
2. Sanity on sign of $\Delta H$. Why? Endothermic must give $E_a^{\text{rev}}<E_a^{\text{fwd}}$; $160<170$ confirms we plugged the $+10$ correctly.

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: if both sides of the hill are the same ground level, the two climbs must be equal.

1. Plug $\Delta H=0$. Why? The definition of degenerate — start and finish heights coincide. $E_a^{\text{rev}}=E_a^{\text{fwd}}-0=50\text{kJ/mol}$
2. Interpret the symmetry. Why? A symmetric hill has its peak dead centre; the activated complex sits exactly halfway in energy between two equal wells.

Verify: $E_a^{\text{fwd}}=E_a^{\text{rev}}$ and $\Delta H=50-50=0$. ✓ The degenerate case is the "diagonal" of the matrix where the two barriers merge.

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

Forecast: with no forward hill, the only height between the two wells is the well depth itself, so the reverse barrier should just equal $370$.

1. Seam identity with $E_a^{\text{fwd}}=0$. Why? Barrierless means the peak coincides with the reactant ground — no extra climb. $E_a^{\text{rev}}=E_a^{\text{fwd}}-\Delta H=0-(-370)=370\text{kJ/mol}$
2. Match to physics. Why? Breaking the newly formed C–C bond with no forward hill is just paying its full bond dissociation energy — the limiting case where $E_a^{\text{rev}}=D$.

Verify: $E_a^{\text{fwd}}-E_a^{\text{rev}}=0-370=-370=\Delta H$. ✓ This is the degenerate limit of Mistake 2 in the parent: here, and only when the other side is barrierless, does $E_a$ equal a bond energy.

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Example 5 — Cell E: the sign/consistency trap

Forecast: the seam identity is a hard constraint — check it before trusting any of the three.

1. Test the identity. Why? All three are fixed by one shared peak; they cannot be independent. $E_a^{\text{fwd}}-E_a^{\text{rev}}=40-90=-50\ne +50=\Delta H?$
2. Diagnose the sign error. Why? The computed value is $-50$, the negative of the reported $\Delta H$. The reaction is actually exothermic: reverse barrier bigger than forward means products are lower.
3. State the corrected value. Why? Consistency demands $\Delta H=E_a^{\text{fwd}}-E_a^{\text{rev}}=-50\text{kJ/mol}.$

Verify: with $\Delta H=-50$: $40-90=-50$. ✓ The barriers were fine; the reported sign of $\Delta H$ was flipped. Larger reverse barrier ⇒ exothermic, always.

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Example 6 — Cell F: $E_a$ is NOT the sum of broken bonds

Forecast: the barrier is far below $436+151=587$, so partial new bonds must be paying most of the difference.

1. Cost if bonds broke fully first. Why? Upper-bound scenario — a stepwise "break everything, then build" path. $436+151=587\text{kJ/mol}$
2. Compare to the real barrier. Why? The gap is what concerted partial bonding buys back. $587-170=417\text{kJ/mol saved by partial H–I bond formation at the TS}$
3. Interpret. Why? This is the parent's Mistake 2 made quantitative: because two H–I bonds start forming while H–H and I–I only partly break, the peak is $417\text{kJ/mol}$ lower than the naive full-break estimate.

Verify: $587-170=417\ge 0$ (stabilisation must be positive) and $170<587$ (barrier below full break). ✓

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

Forecast: a catalyst lowers the peak, so both climbs shrink by the same $30$; $\Delta H$ (which never touches the peak) is untouched.

1. New forward barrier. Why? The peak drops $30$; the reactant ground stays put. $E_a^{\text{fwd,new}}=120-30=90\text{kJ/mol}$
2. Old reverse barrier from the seam. Why? Need a baseline to subtract from. $E_a^{\text{rev,old}}=E_a^{\text{fwd}}-\Delta H=120-(-40)=160\text{kJ/mol}$
3. New reverse barrier. Why? Same lowered peak, product ground unmoved ⇒ reverse climb also drops $30$. $E_a^{\text{rev,new}}=160-30=130\text{kJ/mol}$
4. $\Delta H$? Why? It depends only on the two ground levels, both unchanged. $\Delta H=-40\text{kJ/mol (unchanged)}$

Verify: seam still holds after catalysis: $E_a^{\text{fwd,new}}-E_a^{\text{rev,new}}=90-130=-40=\Delta H$. ✓ The catalyst dropped the peak for both directions equally — see the parent's reaction coordinate diagram logic.

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

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

1. Reaction P reverse barrier. Why? Seam identity. $E_a^{\text{rev}}=20-(-180)=200\text{kJ/mol}$ Tiny climb up ($20$), huge drop down ($200$): the peak sits just past the reactants ⇒ early, reactant-like TS.
2. Reaction Q reverse barrier. Why? Same identity. $E_a^{\text{rev}}=170-150=20\text{kJ/mol}$ Big climb up ($170$), tiny drop down ($20$): the peak sits just before the products ⇒ late, product-like TS.
3. State the rule. Why? Hammond: the TS resembles whichever species it is closer to in energy. Small $E_a^{\text{fwd}}$ vs large $E_a^{\text{rev}}$ ⇒ reactant-like; and vice versa.

Verify: P: $20-200=-180=\Delta H$ ✓; Q: $170-20=150=\Delta H$ ✓. Both seams close, and the smaller barrier correctly points to the "resembled" side.

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Recall Reaction P above has which kind of transition state?

Early / reactant-like ::: because it is exothermic with a tiny forward barrier — the peak sits close to the reactants (Hammond).

Recall A catalyst lowers

$E_a^{\text{fwd}}$ by $30$. By how much does $E_a^{\text{rev}}$ change? Also by $30$ (it lowers the shared peak) ::: while $\Delta H$ stays exactly the same.

See also the Arrhenius equation to convert these barriers into actual rate constants, and the Eyring equation for the TST-native version.