Physical Chemistry Third Edition

496 11 The Rates of Chemical Reactions

Determination of Reaction Order Using the Half-Life

You can carry out several experiments with different initial concentrations at the same

temperature. If the half-life of the reaction is independent of concentration the reaction

is first order. To test for other orders we take the natural logarithm of both sides of

Eq. (11.2-18) to obtain

ln (t 1 / 2 )ln

(

2 n−^1 − 1

(n−1)kf

)

−(n−1) ln([A] 0 )(n1,n0) (11.2-21)

To use Eq. (11.2-21), one could construct a plot of ln(t 1 / 2 ) versus ln([A] 0 ). A straight

line should result, with slope equal to−(n−1) and with intercept equal to the first term

on the right-hand side of Eq. (11.2-21). Instead of carrying out different experiments,

one can also take the data for a single experiment and regard different times during

the experiment as “initial” times. The reverse reaction must be negligible for the entire

experiment.

EXAMPLE11.6

For the gas-phase reaction at 300◦C:

C 2 F 4 −→

1

2

cyclo-C 4 F 8

the following data on the concentration of C 2 F 4 were taken. Determine the order of the reac-

tion and the rate constant at this temperature. Assume that the reverse reaction is negligible.

Time/min Concentration /mol L−^1

00. 0500

250 0. 0250

750 0. 0125

1750 0. 00625

3750 0. 00312

Solution

This example can be solved by inspection. Each concentration is half of the previous con-

centration, so that the time interval from a given data point to the next is the half-life for the

reaction using the given data point as an “initial state.” We have

Half-life/min Initial concentration/mol L−^1

250 0. 0500

500 0. 0250

1000 0. 0125

2000 0. 00625

The half-life doubles each time the “initial” concentration is reduced to half its previous

value. This behavior indicates a second-order reaction. A linear least-squares fit of ln(t 1 / 2 )

versus ln([A] 0 ) confirms this result with order2.00 andkf 0 .080 L mol−^1 min−^1 

1. 3 × 10 −^3 L mol−^1 s−^1.