Collision Theory, Activation Energy, Energy Profile

CHETemperature & Collision Theory

Collision theory explains rate in terms of effective collisions, requiring both sufficient energy and correct orientation. This subtopic centres on the definitions of threshold and activation energy and on sketching how a catalyst reshapes the energy profile.

Activation energy

E_a = E_{\text{threshold}} - E_{\text{reactants}}

extra energy reactants need to reach the threshold

Effective-collision rate

\text{rate} = P\,Z\,e^{-E_a/RT}

Z

collision frequency,

P

steric (orientation) factor

Enthalpy from forward and reverse barriers

\Delta H = E_{a(\text{forward})} - E_{a(\text{reverse})}

\Delta H<0

(exothermic) when forward barrier is smaller

Threshold energy is the minimum energy colliding molecules must possess to react; activation energy $E_a$ is the extra energy above the average reactant energy needed to reach it: $E_a = E_{\text{threshold}} - E_{\text{reactants}}$ .
For a collision to be effective it must (i) have energy $\ge E_a$ and (ii) occur with the proper orientation of the molecules.
Proper orientation is needed so that the reacting atoms/bonds are positioned to form the activated complex; wrongly oriented energetic collisions are ineffective.
A catalyst lowers $E_a$ by providing an alternative path through a lower-energy activated complex, increasing the fraction of effective collisions.
On an energy profile (potential energy vs reaction coordinate), the activated complex sits at the peak; the catalysed path shows a lower peak.
A catalyst changes neither $\Delta H$ nor the energies of reactants and products — only the height of the energy barrier ( $E_a$ ) is reduced.
For an exothermic reaction, products lie lower than reactants, so $\Delta H$ is negative; $E_a$ (forward) is smaller than $E_a$ (reverse).
The activated complex (transition state) is an unstable, high-energy arrangement at the barrier top that can collapse to either reactants or products.
Lowering $E_a$ shifts more of the Maxwell-Boltzmann energy distribution beyond the barrier, so a far larger fraction of collisions become effective — hence the faster rate.
The steric (orientation) factor $P$ corrects collision theory for the fact that only suitably oriented collisions react; $P\le 1$ for most reactions.
For an endothermic reaction, products lie above reactants ( $\Delta H>0$ ) and $E_a$ (forward) exceeds $E_a$ (reverse).
On the catalysed energy profile, draw a lower hump for the alternative mechanism; reactant and product levels stay exactly where they were.

Where the marks go

Equating threshold energy with activation energy — $E_a$ is the difference between threshold energy and the average reactant energy.
Drawing the catalysed profile with shifted reactant or product levels — only the barrier height changes, not the end levels.
Saying a catalyst changes $\Delta H$ — it lowers $E_a$ for both forward and reverse directions equally, leaving $\Delta H$ fixed.
Ignoring the orientation requirement and assuming every collision with energy $\ge E_a$ reacts — the steric factor $P$ accounts for misoriented collisions.
Confusing the activated complex (a transient transition state at the peak) with a stable reaction intermediate sitting in an energy well.

How the board asks it

Define / statethreshold energy, activation energy and effective-collision conditions
Define activation energy. State the two conditions a molecular collision must satisfy to be an effective collision.
Structure / namingenergy profile with catalysed and uncatalysed paths
Draw a labelled potential-energy versus reaction-coordinate diagram for an exothermic reaction, marking the energies of reactants and products, the activated complex, $E_a$ (forward), $E_a$ (reverse) and $\Delta H$ , and show how the curve changes in the presence of a catalyst.
Give reasonscatalyst lowers $E_a$ and shifts the maxwell-boltzmann fraction
Account for the fact that a small amount of catalyst greatly increases the rate of a reaction, even though it does not change the $\Delta H$ of the reaction.
Distinguishthreshold energy versus activation energy
Distinguish between threshold energy and activation energy, giving the relation between them.
Numerical $\Delta H = E_a(\text{forward}) - E_a(\text{reverse})$
For a reaction the activation energy of the forward reaction is $80\ \text{kJ mol}^{-1}$ and that of the reverse reaction is $120\ \text{kJ mol}^{-1}$ . Calculate $\Delta H$ and state whether the reaction is exothermic or endothermic.

Practise this topic

PreviousArrhenius Equation & Temperature Dependence NextMixed, Reasoning & Examiner-Favourite Long Answers

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.