Q.Consider the potential energy diagrams of reactions (I) and (II) given below, predict which reaction will go faster and why ? [figure: two potential-energy-vs-reaction-coordinate diagrams labelled (I) and (II). Each shows a reactant-to-product energy curve with a hump marked "Threshold energy" and the activation energy $E_a$ marked as the gap between the reactant energy level and the threshold. Diagram (II)'s hump/activation energy $E_a$ is drawn visibly lower/smaller than diagram (I)'s.]

Reaction (II) is faster than reaction (I) because its potential-energy diagram shows a smaller activation energy, and rate increases as activation energy decreases.

The activation energy $E_a$ is the minimum extra energy that reactant molecules must acquire (over and above their average energy) in order to cross the energy barrier (the 'threshold energy' hump) and be converted into products. The Arrhenius equation relates the rate constant $k$ of a reaction to its activation energy:

$k = A\,e^{-E_a/RT}$

where $A$ is the frequency (pre-exponential) factor, $R$ the gas constant, and $T$ the absolute temperature.

From this equation, $k$ depends exponentially and inversely on $E_a$: the smaller the activation energy, the larger the value of $e^{-E_a/RT}$, and hence the larger the rate constant $k$. Physically, this is because, at a given temperature, the fraction of molecules in a Maxwell-Boltzmann energy distribution that possess energy equal to or greater than $E_a$ increases sharply as $E_a$ decreases - so more collisions become 'effective' (successful) collisions leading to product formation per unit time.

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