Chemistry · General chemistry II · Worked example

Find the order and k from concentration–time data

A reactant A decomposes at constant temperature. Its concentration is measured every 20 minutes. Use integrated rate law plots to find the order, k and the half-life, and predict [A] at 120 minutes.

A \to products, rate = k {[A]}^{m}

Transform the data three ways

Compute ln[A] and 1/[A] for each time so all three candidate plots can be compared.

Concentration–time data and transformations
t (min)	[A] (M)	ln[A]	1/[A] (M⁻¹)
0	0.500	−0.693	2.00
20	0.397	−0.924	2.52
40	0.316	−1.152	3.16
60	0.251	−1.382	3.98
80	0.199	−1.614	5.03

Look for the straight plot

ln[A] falls by 0.230 ± 0.002 every 20 minutes: equal steps, so ln[A] against t is a straight line (R² = 0.99999). [A] falls by 0.103, then 0.081, 0.065 and 0.052, and 1/[A] rises by 0.52, then 0.64, 0.82 and 1.05. Both bend, even though each still has R² ≈ 0.98. The reaction is first order.

ln {[A]}_{t} = - k t + ln {[A]}_{0}

Read k from the slope

A least-squares line through (t, ln[A]) has slope −0.01151 min⁻¹. For a first-order plot the slope is −k.

k = - slope = 1.15 \times 10^{- 2} \min^{- 1}

Half-life and a prediction

The half-life follows from k, and the integrated law predicts later concentrations. As a check, [A] falls from 0.500 M to 0.251 M in the first 60 minutes, which is almost exactly one half-life.

t_{1 / 2} = \frac{0.693}{0.01151 \min^{- 1}} = 60.2 \min, {[A]}_{120} = 0.500 e^{- (0.01151) (120)} = 0.126 M

Result

First order: k = 1.15 × 10⁻² min⁻¹, t½ = 60.2 min, and [A] ≈ 0.126 M at 120 minutes.

rate = 1.15 \times 10^{- 2} \min^{- 1} [A]

Your turn

A second-order reactant starts at 0.80 M and has k = 0.50 M⁻¹ min⁻¹. What is its first half-life, and how long does the next half-life take?

Show the answer and explanation

2.5 min, then 5.0 min.

For second order, t½ = 1/(k[A]₀). The first is 1/(0.50 × 0.80) = 2.5 min. After it, [A]₀ is effectively 0.40 M, so the next is 1/(0.50 × 0.40) = 5.0 min. Doubling half-lives are the second-order signature.

t_{1 / 2} = \frac{1}{k {[A]}_{0}}

Keep exploring

Fit the same data in the Kinetics studio. Switch the order hypothesis to 0 or 2 and compare the residuals.

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Tro, Chemistry: A Molecular Approach, 4th ed., §14.3 The Rate Law: The Effect of Concentration on Reaction Rate, pp. 629–633 (method of initial rates, p. 630)
Tro, Chemistry: A Molecular Approach, 4th ed., §14.4 The Integrated Rate Law: The Dependence of Concentration on Time, pp. 634–641
OpenStax Chemistry 2e — Rate laws
OpenStax Chemistry 2e — Integrated rate laws