Chemical Kinetics (Class 12): Rate Law, Order and Half-Life Made Simple

Order and molecularity are not the same thing

This is the distinction examiners test most often, and the one most commonly answered wrongly.

Because order is experimental, a reaction written with a coefficient of 2 may still be first order. Never infer the rate law from stoichiometry.

The rate equations worth memorising

For a first-order reaction:

The first-order half-life is independent of the starting concentration — a favourite one-mark question. For a zero-order reaction:

And the temperature dependence, which links kinetics to the activation energy:

Where marks are actually lost

• Using the balanced equation to write the rate law instead of the experimental data.
• Mixing natural log and log base 10 — the 2.303 factor exists precisely to convert between them.
• Forgetting that the unit of k changes with order: s⁻¹ for first order, mol L⁻¹ s⁻¹ for zero order.
• Substituting temperature in Celsius into the Arrhenius equation instead of kelvin.

FAQs

Can the order of a reaction be zero?

Yes. Order is experimental, so it can be zero, fractional or negative. Molecularity cannot.

Why is first-order half-life independent of concentration?

Because t½ = 0.693/k contains only the rate constant. Halving the starting amount halves the time-zero rate too, so the time taken is unchanged.

What is the unit of the rate constant?

It depends on order. First order is s⁻¹; zero order is mol L⁻¹ s⁻¹; second order is L mol⁻¹ s⁻¹.

Does a catalyst change the activation energy?

It provides an alternative path with lower activation energy. It does not change ΔH or the position of equilibrium.