Order of reaction — how to determine from rate data, units, and half-life

Question

How do you determine the order of a reaction from experimental rate data? What are the units of the rate constant for zero, first, and second order reactions? How does half-life depend on concentration for each order?

(JEE Main 2024 asked half-life dependence; NEET tests order determination from data)

Solution — Step by Step

Compare two experiments where only one reactant concentration changes. If doubling $[A]$ doubles the rate, the reaction is first order in A. If doubling $[A]$ quadruples the rate, it is second order in A.

Mathematically: if $\text{Rate} = k[A]^n$ , then $\frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^n$ . Solve for $n$ .

The units of $k$ depend on the overall order:

Order	Rate law	Units of $k$
0	$\text{Rate} = k$	mol L⁻¹ s⁻¹
1	$\text{Rate} = k[A]$	s⁻¹
2	$\text{Rate} = k[A]^2$	L mol⁻¹ s⁻¹

If a question gives you the units of $k$ , you can immediately determine the order.

t_{1/2} \propto \frac{1}{[A_0]^{n-1}}

Zero order: $t_{1/2} = \frac{[A_0]}{2k}$ (half-life decreases as concentration decreases)
First order: $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$ (independent of concentration)
Second order: $t_{1/2} = \frac{1}{k[A_0]}$ (half-life increases as concentration decreases)

If half-life does not change with concentration → first order. This is the quickest test.

Plot concentration vs time data in different forms:

$[A]$ vs $t$ is linear → zero order
$\ln[A]$ vs $t$ is linear → first order
$1/[A]$ vs $t$ is linear → second order

Whichever plot gives a straight line tells you the order.

flowchart TD
    A[Determine reaction order] --> B{What data is given?}
    B -->|Rate at different concentrations| C["Initial rates method<br/>Rate₂/Rate₁ = (conc ratio)ⁿ"]
    B -->|Units of k| D["Zero: mol/L/s<br/>First: 1/s<br/>Second: L/mol/s"]
    B -->|Half-life data| E{"t₁/₂ changes with<br/>concentration?"}
    E -->|No change| F[First Order]
    E -->|Decreases| G[Zero Order]
    E -->|Increases| H[Second Order]
    B -->|Conc vs time data| I["Plot all three graphs<br/>Linear one gives order"]

Why This Works

The order of a reaction tells us how the rate changes with concentration. A first-order reaction’s rate is directly proportional to concentration — so as reactant is consumed, the rate slows down proportionally, giving the unique property of constant half-life.

For zero order, the rate is constant regardless of concentration — the reaction proceeds at the same speed until the reactant runs out. This makes the half-life proportional to initial concentration.

Alternative Method

The fastest way in an MCQ: check the units of k. If the question gives $k = 0.05 \text{ s}^{-1}$ , it is first order (no concentration unit). If $k = 0.05 \text{ mol L}^{-1}\text{s}^{-1}$ , it is zero order. The general formula for units: $k$ has units of $(\text{conc})^{1-n} \text{time}^{-1}$ .

Common Mistake

Students assume that the stoichiometric coefficient equals the order. For $2A \rightarrow \text{Products}$ , they write order = 2. This is WRONG. Order is determined experimentally, not from the balanced equation. The stoichiometric coefficient and order are the same only for elementary reactions (single-step). For complex (multi-step) reactions, order must be measured.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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