Define activation energy — Biology notes

WHAT is activation energy?

WHY does a barrier even exist? For new bonds to form, old bonds must first be partially broken. Stretching/distorting existing bonds costs energy. At the top of the hill the molecule is in a strained, unstable arrangement called the transition state — half-old-bonds, half-new-bonds. Only molecules with enough energy to reach this point can react.

HOW the energy profile looks

Reading the diagram (left → right = reaction progress):

Reactants start at some energy level.
Energy rises to a peak = transition state. The height from reactants to this peak = $E_a$ .
Energy then falls to products.
The difference between reactants and products = $\Delta H$ (overall energy change), a separate quantity from $E_a$ .

Formula Two independent quantities — derive the relationship

Let energy of reactants $= E_R$ , transition state $= E_{TS}$ , products $= E_P$ .

Forward barrier: $E_{a,\text{forward}} = E_{TS} - E_R$ Reverse barrier: $E_{a,\text{reverse}} = E_{TS} - E_P$ Overall reaction energy change: $\Delta H = E_P - E_R$ Subtract the two barriers: $E_{a,\text{forward}} - E_{a,\text{reverse}} = (E_{TS}-E_R)-(E_{TS}-E_P) = E_P - E_R = \Delta H$ So: $\boxed{\Delta H = E_{a,\text{forward}} - E_{a,\text{reverse}}}$ Why this matters: it proves $E_a$ and $\Delta H$ are different things linked only through the shared transition state.

HOW enzymes & temperature affect $E_a$

The fraction of molecules with energy $\ge E_a$ rises steeply if you either lower $E_a$ (catalyst) or raise temperature (shifts the molecular speed distribution rightward). Conceptually: $\text{rate} \propto e^{-E_a / RT}$ Why this form? The exponential says only a small tail of molecules has the needed energy; making $E_a$ smaller or $T$ larger makes the negative exponent less negative, so the fraction (and rate) jumps.

Worked examples

Worked example Example 1 — finding

$E_a$ from an energy diagram Reactants sit at $30\ \text{kJ mol}^{-1}$ , transition state peak at $80\ \text{kJ mol}^{-1}$ , products at $10\ \text{kJ mol}^{-1}$ . Find $E_{a,\text{forward}}$ and $\Delta H$ .

$E_{a,\text{forward}} = E_{TS}-E_R = 80-30 = 50\ \text{kJ mol}^{-1}$ — Why? it's the climb from reactants to the peak.
$\Delta H = E_P-E_R = 10-30 = -20\ \text{kJ mol}^{-1}$ — Why? negative means products are lower ⇒ energy released ⇒ exergonic/exothermic.
Note both exist at once: a downhill (favourable) reaction that still needs a 50 kJ push.

Common mistakes (steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine a ball stuck in a little dip at the top of a slide. The bottom of the slide is lower, so the ball really wants to go down and will whoosh fast once it does. But first you have to push it over a tiny bump at the edge of the dip. That small push is the activation energy — the bit of effort needed to start. The big drop afterwards is the energy the reaction gives out. An enzyme is like a friend who makes the bump smaller, so the ball goes down much more easily — but the bottom of the slide stays in the same place.

Active recall flashcards

Activation energy is defined as

the minimum energy reactants must have to reach the transition state and form products.

What are the units of

E_a

kJ mol⁻¹ (energy per mole).

The unstable, highest-energy arrangement at the top of the energy hill is called the

transition state.

Forward activation energy formula in terms of energies

E_a = E_{TS} - E_{R}

(transition state minus reactants).

Relationship between forward/reverse barriers and ΔH

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

Does a catalyst change ΔH?

No — it only lowers

E_a

by offering a lower-energy pathway.

Why is sugar stable even though burning it releases energy?

The reaction is favourable (negative ΔH) but has a high

E_a

, so it is slow without a spark/heat/enzyme.

How does raising temperature speed a reaction in terms of

E_a

More molecules gain energy ≥

E_a

, so more can cross the barrier (rate

\propto e^{-E_a/RT}

Is activation energy the same as the energy released?

No —

E_a

is the barrier (reactants→peak); released energy is ΔH (reactants→products).

Connections

Enzymes lower activation energy
Transition state and enzyme-substrate complex
Exergonic vs endergonic reactions
Enthalpy change ΔH
Effect of temperature on reaction rate
Catalysts in metabolism

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho ek reaction tabhi hota hai jab molecules pehle ek "energy ki pahaadi" chadh lete hain. Is pahaadi ki height ko hi hum activation energy ( $E_a$ ) kehte hain — yeh wo minimum energy hai jo reactants ko chahiye taaki woh transition state (sabse upar wali unstable halat) tak pahunch saken aur phir product ban jaayen. Yaad rakhna: $E_a$ matlab reaction ko shuru karne ka kharcha hai, na ki reaction se nikalne wali energy.

Reaction se jo energy nikalti hai usko $\Delta H$ kehte hain, aur yeh bilkul alag cheez hai. Glucose jalne par bahut energy nikalti hai (ΔH negative), par cheeni almari me saalon tak waise hi padi rehti hai — kyunki uska $E_a$ bahut high hai, spark ya heat ke bina pahaadi paar nahi hoti. Isiliye "energy release hoti hai" ka matlab yeh nahi ki reaction fast hoga.

Enzymes ka kaam yahin aata hai. Enzyme pahaadi me ek tunnel khod deta hai — yaani $E_a$ ko kam kar deta hai by giving an alternative path. Lekin enzyme reactants aur products ki energy nahi badalta, isliye $\Delta H$ same rehta hai. Temperature badhane par bhi zyada molecules ke paas itni energy aa jaati hai ki woh barrier paar kar saken (rate $\propto e^{-E_a/RT}$ ), isliye reaction tez ho jaata hai.

Exam tip: agar diagram diya ho to $E_{a,forward} = E_{TS} - E_R$ , aur $\Delta H = E_P - E_R$ . Dono ko mat mix karo — yahi sabse common galti hai!

Test yourself — Enzymes & Bioenergetics Basics

Connections