In a first-order decomposition reaction, 50.0% of a

Solution 33P

Here we have to calculate the rate constant and time required for 75.0% of the compound to decompose.

Step 1

(a) Calculation of  rate constant of the reaction for 1st order reaction.

In this question it has been given that, in a first-order decomposition reaction, 50.0% of a compound decomposes in 10.5 min.

The value of rate constant can be obtained by using the integrated rate law. For 1st order reaction, the rate law is,

ln ([A]0/ [A]t) = kt

Here it is given that 50% has decompose, hence the concentration of A has decreased by half, we can write it as, =([A]0/ [A]t) = (1/ (1/2)) = 2

Alternatively 50.0 % decomposition means, one half-life has passed. Hence the 1st order half life equation may be used as,

t1/2  = ln2/ k 

  k   = ln 2/ t = ln2/10.5 min = 0.066014 = 0.0660 min-1

Thus the rate constant for the 1st order reaction is found to be 0.0660 min-1.