Science Questions and Answers

Start Your Free Trial

The reaction 2A->B is 2nd order.The initial concentration of A is 13.198M and half-life is 198 sec.  How long will it take [A] to drop to 1/7 of its initial value? 

Expert Answers info

ncchemist eNotes educator | Certified Educator

calendarEducator since 2010

write1,405 answers

starTop subjects are Science, Math, and Social Sciences

To solve this problem we need to know and use two different equations.  The first is the equation for the half life of a second order reaction.  It is:


where t is the half life, k is the second order rate constant, and [A'] is the initial concentration of A.  We can use this equation to calculate the rate constant k:


k = 3.827 L/mol*sec

We can now use the integrated rate law for a second order reaction to calculate the time required for [A] to drop to 1/7 of the initial value.


Solving for t:

`t=(([A'])/([A]) - 1)(1/(k[A']))`

Now we can input the values and solve for t.  [A'] is the initial concentration (13.198 M).  [A] is 1/7 of the initial value (1.885 M).  And k is the rate constant that we calculated above.  Inputting all of these values into the above equation gives t=1,188.31 seconds.  So the answer is that it takes 1,188.31 seconds for the concentration to drop to 1/7 of its initial value.

check Approved by eNotes Editorial