**Practice Problem 9**

Acetaldehyde, CH_{3}CHO, decomposes by second-order kinetics with a rate
constant of 0.334 *M*^{-1}s^{-1} at 500°C. Calculate the
amount of time it would take for 80% of the acetaldehyde to decompose in a sample that has
an initial concentration of 0.00750 *M*.

**Solution**

We start with the rate law for the decomposition of acetaldehyde, which follows second-order kinetics:

Rate = *k*(CH_{3}CHO)^{2}

We then write the integrated form of this rate law:

The initial concentration of acetaldehyde is 0.00750 *M*. The
final concentration is only 20% as large, or 0.00150 *M*:

Substituting the rate constant for this reaction (*k*= 0.334) into
this equation and then solving for *t* gives the following result:

*t* = 1600 s

It therefore takes slightly less than one-half of an hour for 80% of the acetaldehyde to decompose at this temperature.