Orders of Reaction (0, 1, 2)

Learning Objectives (OCR Spec 5.1.1)

By the end of this lesson you should be able to:

• Explain the physical meaning of zero, first and second order kinetics
• Determine orders from rate-concentration data using ratios
• Use the initial rates method to compare experiments and deduce orders
• Apply orders to reactions with more than one reactant

Zero Order — Rate Independent of Concentration

If a reaction is zero order with respect to reactant A:

rate = k[A]^0 = k

Changing [A] has no effect on the rate. This seems counter-intuitive, but occurs when some other step — for example a catalytic surface reaching saturation, or enzyme active sites being fully occupied — limits the rate. Doubling [A] simply means more molecules sitting in a queue waiting to react.

Example: The decomposition of ammonia on a hot platinum wire is zero order in NH3 at high pressures — the platinum surface is saturated.

First Order — Rate Proportional to Concentration

If a reaction is first order with respect to A:

rate = k[A]^1 = k[A]

Doubling [A] doubles the rate. Tripling [A] triples the rate. The rate is directly proportional to concentration.

Example: Radioactive decay and simple SN1 nucleophilic substitutions such as (CH3)3CBr + OH- -> (CH3)3COH + Br- (rate = k[(CH3)3CBr]) — the C-Br bond breaks in the slow step which involves only one species.

Second Order — Rate Proportional to the Square of Concentration

If a reaction is second order with respect to A:

rate = k[A]^2

Doubling [A] quadruples the rate; tripling [A] increases it 9-fold. Or a reaction can be second order overall but first order in each of two different reactants:

rate = k[A][B]

Doubling either [A] or [B] alone doubles the rate; doubling both quadruples the rate.

Example: The SN2 reaction of OH- with CH3Br is first order in CH3Br and first order in OH-, giving rate = k[CH3Br][OH-] — both species are present in the slow step of the mechanism.

Summary Table

Order	Rate equation (simple)	Doubling [A] gives	Physical meaning
0	rate = k	no change	[A] does not appear in slow step
1	rate = k[A]	x2	one molecule of A in slow step
2	rate = k[A]^2	x4	two molecules of A in slow step

Determining Orders from Data — The Ratio Method

The simplest way to find orders is to compare experiments in which one concentration is changed at a time. This is the initial rates method (covered in more detail in Lesson 6).

Worked Example 1 — Two Reactants

Consider A + B -> products. Four experiments give the following initial rates:

Expt	[A] / mol dm^-3	[B] / mol dm^-3	initial rate / mol dm^-3 s^-1
1	0.10	0.10	2.0 x 10^-4
2	0.20	0.10	4.0 x 10^-4
3	0.30	0.10	6.0 x 10^-4
4	0.10	0.20	8.0 x 10^-4

Step 1 — Find order w.r.t. A