Transition state theory — activated complex (intro): Exercises

This page is a self-test ladder for Transition state theory — activated complex (intro). Work each problem before opening the collapsible solution. Every symbol used below is defined at first sight, so start from problem 1 even if the notation feels new.

Figure — Transition state theory — activated complex (intro)

Level 1 — Recognition

L1.1 — Maximum or minimum?

State whether each is a local energy maximum or a local energy minimum on a reaction coordinate diagram, and give its approximate lifetime: (a) an activated complex, (b) a reaction intermediate, (c) the reactants.

Recall Solution

(a) Activated complex → maximum (the hilltop). Lifetime ≈ $1 0^{- 13} s$ (one bond vibration). It is a point of passage, never a resting place.
(b) Intermediate → minimum (a small dip between two hills). Lifetime is long enough to potentially detect (nanoseconds up to seconds).
(c) Reactants → minimum (the left valley floor). They sit stably until given enough energy to climb.

Look at Figure s01: hilltops are maxima, valleys are minima. Only the hilltop is a transition state.

L1.2 — Read the diagram

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

Recall Solution

$E_{a}^{fwd} = E_{TS} - E_{reactants} = 220 - 50 = 170 kJ/mol$ . That is the climb from the left valley to the peak — trace it as the orange arrow in Figure s01.

L1.1 answer

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

L1.2 answer

170 kJ/mol

Level 2 — Application

L2.1 — Reverse barrier

Using the L1.2 numbers ( $E_{TS} = 220$ , $E_{products} = 60$ , all kJ/mol), find $E_{a}^{rev}$ and $Δ E$ . Is the forward reaction exo- or endothermic?

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 sit above reactants ⇒ endothermic. Check with the master identity: $E_{a}^{fwd} - E_{a}^{rev} = 170 - 160 = 10 = Δ E$ . ✓

L2.2 — Fill the missing height

A reaction has $E_{a}^{fwd} = 90 kJ/mol$ and is exothermic with $Δ E = - 40 kJ/mol$ . Find $E_{a}^{rev}$ .

Recall Solution

Rearrange the master identity for the reverse climb: $E_{a}^{rev} = E_{a}^{fwd} - Δ E = 90 - (- 40) = 130 kJ/mol .$ Why the reverse is bigger: the reaction went downhill by 40, so climbing back up to the same peak from the lower product valley is a taller climb. In Figure s01's language, a lower right base makes the reverse arrow longer.

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 — Why $E_{a} \neq =$ bond energy

For $H_{2} + I_{2} \to 2 HI$ : the H–H bond energy is $D_{H-H} = 436 kJ/mol$ , the I–I bond energy is $D_{I-I} = 151 kJ/mol$ , yet the measured $E_{a} = 170 kJ/mol$ . (a) What would $E_{a}$ be if both bonds broke fully before any new bond formed? (b) By how much does the real transition state undercut that value, and why?

Recall Solution

(a) Fully breaking both bonds first would cost $436 + 151 = 587 kJ/mol$ . (b) The real barrier is $170$ , so the transition state undercuts the "break-everything-first" cost by $587 - 170 = 417 kJ/mol$ . Why: at the activated complex the old bonds only partially stretch while the new H–I bonds partially form at the same time. That partial bonding pays back most of the cost, so you never have to fully break anything. This is why $E_{a}$ is not a bond dissociation energy.

L3.2 — Hammond reasoning

Hammond's postulate (stated so you can use it here): the transition state most resembles whichever species — reactants or products — it is closest to in energy. So an exothermic reaction (products far below reactants) has a peak that is reached early and looks like the reactants; an endothermic reaction (products above reactants) has a peak that is reached late and looks like the products.

Reaction X is strongly exothermic; reaction Y is strongly endothermic. State whether each transition state resembles reactants or products, and which reaction likely has the lower forward barrier (assume similar mechanisms).

Recall Solution

X (exothermic): transition state is reached early and resembles reactants — an "early barrier." Little geometric distortion needed, so it tends to have the lower $E_{a}^{fwd}$ .
Y (endothermic): transition state is late and resembles products — a "late barrier," typically the higher one. In Figure s02: the blue (exothermic) curve's peak sits to the left near the reactant side; the red (endothermic) curve's peak sits to the right near the product side. See Hammond's postulate.

L3.1a full-break cost

587 kJ/mol

L3.1b undercut

417 kJ/mol

Level 4 — Synthesis

L4.1 — Catalyst on the same diagram

A reaction has $E_{reactants} = 30$ , $E_{TS} = 180$ , $E_{products} = 70$ (kJ/mol). A catalyst provides a new path with $E_{TS}^{cat} = 120$ . Find: (a) uncatalyzed $E_{a}^{fwd}$ , (b) catalyzed $E_{a}^{fwd}$ , (c) uncatalyzed $E_{a}^{rev}$ , (d) catalyzed $E_{a}^{rev}$ , (e) does the catalyst change $Δ E$ ?

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. The catalyst lowers the peak only; the two valley floors (reactants, products) are untouched, so $Δ E$ is identical. Key insight: the catalyst lowered both barriers by the same $60 kJ/mol$ ( $150 \to 90$ and $110 \to 50$ ). See Catalysis.

L4.2 — Consistency puzzle

You are told $E_{a}^{fwd} = 120$ , $E_{a}^{rev} = 45$ , and $Δ E = + 90$ (kJ/mol). Show these numbers are inconsistent, and give the value of $E_{a}^{rev}$ that would make them consistent.

Recall Solution

The master identity demands $E_{a}^{fwd} - E_{a}^{rev} = Δ E$ . Check: $120 - 45 = 75 \neq = 90$ . Inconsistent. For consistency: $E_{a}^{rev} = E_{a}^{fwd} - Δ E = 120 - 90 = 30 kJ/mol$ . Why the identity must hold: both barriers are measured to the same peak. Their difference is (peak − left base) − (peak − right base) = right base − left base = $Δ E$ . The peak cancels, so no self-consistent diagram can violate it.

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

Two reactions share the same pre-exponential factor. Reaction P has $E_{a} = 50 kJ/mol$ ; reaction Q has $E_{a} = 75 kJ/mol$ . At $T = 300 K$ , by what factor is P faster than Q? Use $R = 8.314 J mol^{- 1} K^{- 1}$ .

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: $\frac{25000}{8.314 \times 300} = \frac{25000}{2494.2} = 10.023$ . $\frac{k _{P}}{k _{Q}} = e^{10.023} \approx 2.26 \times 1 0^{4} .$ A mere 25 kJ/mol difference makes P roughly twenty-two thousand times faster — this steep sensitivity is why small changes to the transition state (or a catalyst) matter enormously.

L5.2 — Design a diagram from clues

Sketch-and-solve. You must build a single reaction diagram obeying all: (i) exothermic by $60 kJ/mol$ ; (ii) forward barrier $= 40 kJ/mol$ ; (iii) reactants sit at $10 kJ/mol$ . Find $E_{TS}$ , $E_{products}$ , and $E_{a}^{rev}$ . Then state where Hammond places the peak.

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 resembles reactants (leans left). See Hammond's postulate and 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.