Question

Question #7b731

Answer 1

`"1170 s"`

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 the initial concentration of the reactant
• `["XY"]_ t` is the concentration of the reactant after a period of time `t`
• `k` is the rate 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`

`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"`