Second-Order Reaction Kinetics

A second-order reaction has a rate proportional to the square of one reactant concentration (or to the product of two concentrations), rate = k[A]2; the integrated law is 1/[A] = 1/[A]0 + kt, and the half-life t½ = 1/(k[A]0) increases as the reaction proceeds. In brief: for 2NO2 → 2NO + O2 at 300 °C with k = 0.54 M−1s−1 and [NO2]0 = 0.0100 M, the first half-life is 185 s, the next one 370 s and the next 740 s.

Second-order reactions are typical of collisions between two molecules: dimerizations, many gas-phase reactions and the reactions of radicals. Because the rate depends on the square of the concentration, the reaction slows down faster than a first-order one as the reactant is consumed. This article gives the integrated equation, the half-life relationship, a comparison with zero- and first-order kinetics and worked examples.

What is the rate law?

There are two forms of second-order kinetics:

The unit of the rate constant is M−1s−1 (L mol−1 s−1). When the rate is written for the disappearance of A, the second-order law is −d[A]/dt = k[A]2. Note that if the stoichiometry is 2A → products, some books define the rate constant for the reaction, and others for the loss of A, so the two values differ by a factor of 2.

What is the integrated rate law?

Integration of −d[A]/dt = k[A]2 gives:

1/[A] = 1/[A]0 + kt

A graph of 1/[A] against time is a straight line with a slope of +k and an intercept of 1/[A]0. This is the test for second order: a plot of [A] against t is curved, a plot of ln[A] against t is also curved, and a plot of 1/[A] is straight.

k 1/[A]₀ 1/[A] t

How does the half-life behave?

Setting [A] = [A]0/2 gives:

t½ = 1 / (k[A]0)

The half-life is inversely proportional to the initial concentration. Each successive half-life is twice as long as the one before, since the concentration of A is half as big and the collisions are less frequent. A reaction that appears to be half complete in a short time may take a very long time to finish.

Second-order decay with successive half-lives lengthening 0 10 30 70 0 0.25 0.5 0.75 1 Time (s) Concentration [A] (M)
Second-order decay for [A]0 = 1.00 M and k = 0.100 M−1 s−1. The concentration halves after 10 s, halves again 20 s later and again 40 s after that, so each half-life is twice the last.

Comparison of the three orders

Summary of integrated rate laws for the reaction A → products
Order Integrated law Straight-line plot Slope Half-life Units of k
0[A] = [A]0 − kt[A] vs t−k[A]0/2kM s−1
1ln[A] = ln[A]0 − ktln[A] vs t−k0.693/ks−1
21/[A] = 1/[A]0 + kt1/[A] vs t+k1/(k[A]0)M−1 s−1

Worked examples

Example 1: concentration after a given time

The decomposition 2NO2(g) → 2NO(g) + O2(g) at 300 °C is second order in NO2, with k = 0.54 M−1s−1. If [NO2]0 = 0.0100 M, what is the concentration after 60 s?

  1. 1/[A]0 = 1 / 0.0100 = 100 M−1.
  2. 1/[A] = 100 + 0.54 × 60 = 100 + 32.4 = 132.4 M−1.
  3. [A] = 1 / 132.4 = 0.00755 M.

Example 2: successive half-lives

The first half-life is t½ = 1 / (0.54 × 0.0100) = 185 s, when [A] = 0.00500 M. The second, with [A]0 = 0.00500 M, is 1 / (0.54 × 0.00500) = 370 s. The third, starting from 0.00250 M, is 740 s. After 185 + 370 + 740 = 1295 s, 87.5 % of the reactant has been used up.

0.0100 0.00500 0.00250 0.00125 ← 185 s → ← 370 s → ← 740 s →

Example 3: time to reach a concentration

How long does it take for [NO2] to fall to 0.00250 M? 1/[A] = 400 M−1, so t = (400 − 100) / 0.54 = 556 s, which is about 9.3 minutes. A first-order reaction with the same initial half-life of 185 s would reach 25 % of the original value after only 370 s, which shows that the second-order reaction slows down more.

Example 4: finding k from data

The concentration of a reactant falls from 0.400 M to 0.200 M in 250 s. Because 1/0.200 − 1/0.400 = 5.00 − 2.50 = 2.50 M−1, k = 2.50 / 250 = 0.0100 M−1s−1 (assuming a second-order reaction).

What is a pseudo-first-order reaction?

For a reaction with the rate law rate = k[A][B], if the concentration of B is much greater than that of A (for example 100 times), [B] hardly changes during the reaction, and the rate law simplifies to rate = k′[A], where k′ = k[B]. The kinetics then look like first order. This technique, called the isolation method, makes it easy to find the order with respect to each reactant. An example is the hydrolysis of an ester in a large excess of water.

Who studied second-order reactions?

The reaction of hydrogen with iodine to give hydrogen iodide was studied by Max Bodenstein in the 1890s, and was considered for decades as the classic example of a bimolecular, second-order reaction, until the mechanism was reexamined in the second half of the 20th century. The dimerization of 1,3-butadiene to 4-vinylcyclohexene is a standard second-order example in textbooks. Second-order kinetics is also relevant to the recombination of radicals.

Which mistakes are common?