FIRST AND SECOND-ORDER REACTIONS

For any reaction that is known to be first order in a particular species, as in the N2O5 decomposition discussed earlier, the rate can be written as

For a first-order reaction, this means that doubling the concentration of A doubles the reaction rate. This equation may be integrated to give a more useful result:

This can also be written in a more user-friendly way as:

In similar fashion, the integrated rate equation for a second-order reaction is:

Note that [A]0 represents the concentration of species A at time zero, which may or may not be the same as the iinitial concentration, while [A] represents its concentration at any time t in the future.

HALF-LIFE OF A REACTION

A quantity defined as the half-life of the reaction can also be determined. The half-life is the time required for the concentration of the reactant to reach half of its initial value. For a first-order reaction, the half-life is independent of the initial concentration and is given as:

For a second-order reaction, the half-life depends on initial concentration and is given as:

Example 2.18

Consider the following reaction between peroxydisulfate ion, S2O82-, and iodide ion, I- and the experimental data about the concentrations and reaction rates at 25°C given in Table 2.3.

Table 2.3. Experimental data for the reaction kinetics of peroxydisulfate

Expt.

Number

[S2Os2- (aq)]

[I- (aq)]

Initial Rate

1

0.080 M

0.034 M

2.2 x 10-4 M/sec

2

0.080 M

0.017 M

1.1 x 10-4 M/sec

3

0.16 M

0.017 M

2.2 x 10-4 M/sec

A. Write the rate law expression and determine the overall order, that is, (x + y), for this reaction.

B. Determine the rate constant k for this reaction at 25°C.

Solution

A. Examine the data given in Table 2.3. Comparing Experiment #1 with #2, it can be seen that doubling the iodide ion concentration while holding the peroxydisulfate ion concentration constant, would double the reaction rate. Similarly, comparing Experiment #2 with #3 reveals that doubling the peroxydisulfate ion concentration while holding the iodide ion concentration constant, would also double the reaction rate.

Thus, this reaction is first order (or linear) in iodide ion concentration, and first order (or linear) in peroxydisulfate ion concentration. The rate law can be written as follows:

where x = 1 and y = 1.

Hence,

The overall order is (x + y) = (1 + 1) = 2, indicating second order overall.

B. The rate constant k can be determined using the data from any of the three experiments. Take the data from Experiment #1, and use the rate law expression determined in A. Thus,

Example 2.19

The thermal decomposition of phosphine, PH3, is known to be a first- order reaction:

The half-life, t1/2, is 35.0 sec at 680°C.

A. Compute the rate constant k for this reaction.

B. Find the time required for 75% of the initial concentration of PH3

to decompose.

Solution

A. For a first-order reaction, t1/2 = 0.693/k.

Hence,

B. To find the time, use the integrated, first-order rate law, equation

  • 2.18 or 2.19, and solve for At. If 75% of PH3 is to decompose,
  • 25% must remain! Thus,

Either equation 2.18 or 2.19 may be used. First, rearrange equation

2.19 and solve for At:

Now substitute these values into equation 2.19 and solve for At:

 
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