Practice Problem 9

Acetaldehyde, CH3CHO, decomposes by second-order kinetics with a rate constant of 0.334 M-1s-1 at 500C.  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:

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

Ex22_9b.gif (1319 bytes)

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.


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