# Question #7b731

##### 1 Answer

#### Explanation:

All you have to do here is to use the **integrated rate law** for a second-order reaction that takes the form

#"XY + XY " -> " products"#

which will look like this

#1/(["XY"]_ t) - 1/(["XY"]_ 0) = kt#

Here

#["XY"]_ 0# is theinitial concentrationof the reactant#["XY"]_ t# is the concentration of the reactant after a period of time#t# #k# is therate constant

In your case, you have

#K = 6.82 * 10^(-3)# #"M"^(-1)"s"^(-1)#

and

#{(["XY"]_ 0 = "0.140 M"), (["XY"]_ t = 6.60 * 10^(-2)"M") :}#

Rearrange the integrated rate law to solve for

#t = (1/(["XY"]_ t) - 1/(["XY"]_ 0))/k#

Plug in your values to find

#t = ( 1/(6.60 * 10^(-2)color(red)(cancel(color(black)("M")))) - 1/(0.140color(red)(cancel(color(black)("M")))))/(6.82 * 10^(-3)color(red)(cancel(color(black)("M"^(-1))))"s"^(-1)) = color(darkgreen)(ul(color(black)("1170 s")))#

The answer is rounded to three **sig figs**

If you want, you can convert this to *minutes* or to *hours* by using

#"1 hour = 60 minutes = 3600 seconds"#