Transition state theory — activated complex (intro): Question bank

This is a rapid-fire trap-clearing deck for Transition state theory — activated complex. Each line hides a reveal. Say your answer out loud with a reason before you peek — a bare "true" or "false" earns zero. The goal is to burn out every misconception the reaction coordinate picture tends to plant.

True or false — justify

TST replaces collision theory entirely and says collisions no longer matter.

False — molecules must still collide to react; TST adds what happens after the collision, namely climbing to the transition state. It refines Collision theory, it does not delete it.

The activated complex can be bottled and stored like a normal reagent.

False — it sits at an energy maximum and survives only about one bond vibration (

\sim 1 0^{- 13}

s), so it is a point of passage, never an isolable substance.

An intermediate and a transition state are the same thing drawn twice.

False — an intermediate rests in a local minimum (measurable lifetime), while the transition state is a local maximum that lasts one vibration; a valley and a peak are opposites.

If a reaction is exothermic, its activation energy must be small.

False —

Δ H

(reactant-to-product drop) and

E_{a}

(reactant-to-peak climb) are independent;

H_{2} + O_{2}

is wildly exothermic yet needs a spark because its peak is still high.

For

H_{2} \to 2 H

, the activation energy roughly equals the H–H bond energy.

True — a pure dissociation has no new bond forming to lower the peak, so you must pay nearly the full bond-dissociation cost to reach the top.

For

H_{2} + I_{2} \to 2 HI

E_{a}

equals the sum of the H–H and I–I bond energies.

False — new H–I bonds begin forming at the transition state and pay back much of the cost, so

E_{a} \approx 170

kJ/mol, far below

436 + 151 = 587

kJ/mol.

The transition state always lies exactly halfway between reactants and products in structure.

False — Hammond's postulate says it resembles whichever side it is closer to in energy: early (reactant-like) for exothermic, late (product-like) for endothermic reactions.

E_{a}^{forward} - E_{a}^{reverse} = Δ E_{reaction}

for any single-barrier reaction.

True — both activation energies are measured from the same peak, so their difference is just the vertical gap between the two valleys, which is

Δ E

A catalyst lowers

Δ H

of the reaction.

False — a catalyst opens a new lower-energy path (a shorter hill) that changes

E_{a}

; the valleys (reactants and products) are untouched, so

Δ H

is unchanged.

The reverse activation energy can be negative.

False for an ordinary single barrier — the peak is above both valleys, so both

E_{a}^{fwd}

and

E_{a}^{rev}

are

\geq 0

(only complex multi-step apparent

E_{a}

can appear negative, which is a different quantity).

Spot the error

"The activated complex is the most stable arrangement of atoms during the reaction."

Error: it is the least stable point along the path — a maximum in potential energy — which is precisely why it cannot be isolated.

"In an

S_{N} 2

transition state the carbon is

s p^{3}

with four full bonds."

Error: the carbon is transiently pentacoordinate (trigonal bipyramidal), with a partial bond to the incoming nucleophile and a partial bond to the leaving group simultaneously — see SN1 vs SN2 mechanisms.

"Because

Δ H = + 10

kJ/mol for

H_{2} + I_{2}

, the barrier must be about 10 kJ/mol."

Error:

Δ H

measures only the valley-to-valley change; the peak sits

\sim 170

kJ/mol above reactants because bonds break before new ones fully form.

"The Arrhenius

E_{a}

and the transition-state barrier height are unrelated numbers."

Error: they describe the same hill — the Arrhenius equation

E_{a}

is the experimental read-out of the climb to the transition state (the Eyring equation refines it into

Δ H^{‡}

"An exothermic reaction has its transition state closer to the products."

Error: Hammond's postulate gives the opposite — an exothermic step has an early, reactant-like transition state; the product-like (late) one belongs to endothermic steps.

"Since the

S_{N} 2

transition state has a bond forming, its energy should be lower than the reactants'."

Error: forming a bond partially offsets the cost of stretching the C–Br bond and squeezing five groups together, but the net is still a maximum above the reactants — hence a real barrier (

\approx 75

kJ/mol).

"The reaction coordinate is just time."

Error: it is a geometric progress variable (positions of atoms along the minimum-energy path on the potential energy surface), not the clock.

Why questions

Why must a potential-energy maximum exist between reactants and products at all?

Because bonds must stretch and partly break before new ones fully form, and stretching costs energy (loss of bonding electron density, Pauli repulsion), so the system pays an unavoidable up-front energy toll.

Why is the activation energy usually less than the total bond-dissociation energy of the bonds being broken?

Because new bonds begin forming at the transition state, releasing energy that partly cancels the breaking cost — you never fully snap a bond in isolation during a concerted step.

Why does TST predict the

S_{N} 2

transition state is trigonal bipyramidal?

The nucleophile attacks opposite the leaving group (backside), so the three remaining substituents flatten into a plane while the entering and leaving groups sit axial — a five-centre, over-coordinated carbon.

Why can two different reactions have almost the same

E_{a}

but opposite signs of

Δ H

Because

E_{a}

(climb to the peak) and

Δ H

(valley-to-valley) are set by different features of the curve; the peak height and the net drop are geometrically independent.

Why does a catalyst speed a reaction without being consumed?

It provides an alternative pathway with a lower transition state, so more molecules clear the barrier per second; the catalyst is regenerated because it only shapes the hill, it isn't a reactant valley.

Why is the transition state a "saddle point" rather than a simple hilltop?

On the full potential energy surface it is a maximum along the reaction path but a minimum in every direction perpendicular to it — like a mountain pass, high along the trail, low across it.

Edge cases

What is

E_{a}

for a barrierless reaction, e.g. two radicals combining?

Effectively zero — there is no bond-breaking to pay for and the new bond forms as the atoms approach, so the "peak" flattens away and the rate is limited only by how fast they meet.

In a two-step mechanism with an intermediate, how many transition states are there?

Two — one peak on each side of the intermediate valley; the higher of the two peaks (the rate-determining step) controls the overall observed rate.

What happens to the transition-state lifetime as the barrier gets very thin and sharp?

It stays on the order of one vibration (

\sim 1 0^{- 13}

s) regardless of height — the transition state is a passage, not a well, so it never accumulates population no matter the barrier shape.

For a perfectly thermoneutral reaction (

Δ H = 0

), are the forward and reverse activation energies equal?

Yes — since both are measured from the same peak and the two valleys are level,

E_{a}^{fwd} = E_{a}^{rev}

, consistent with

E_{a}^{fwd} - E_{a}^{rev} = Δ E = 0

If a reaction is extremely endothermic, what does its transition state look like structurally?

Very product-like (a late barrier) by Hammond's postulate — the peak sits close to the products in both energy and geometry, so the transition state already resembles the products.

Can raising temperature change

E_{a}

Δ H

Neither is created by temperature — both are fixed features of the energy landscape; temperature only changes the fraction of molecules with enough energy to clear the fixed barrier (that's the Arrhenius equation at work).

Recall One-line summary of every trap

Peak = kinetics = $E_{a}$ = a maximum = uncatchable; Valley gap = thermodynamics = $Δ H$ ; they move independently, and catalysts reshape the peak but never the valleys.