Practice Problem 9
Acetaldehyde, CH3CHO, decomposes by second-order kinetics with a rate constant of 0.334 M-1s-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.
We start with the rate law for the decomposition of acetaldehyde, which follows second-order kinetics:
Rate = k(CH3CHO)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.