# Question #cef19

##### 1 Answer

Here's what I got.

#### Explanation:

The **integrated rate law** for a first-order reaction

#"A " -> " products"#

looks like this

#color(blue)(ul(color(black)( ln( (["A"]_t)/(["A"]_0)) = - k * t)))#

Here

#[A]_t# is the concentration of the reactant after a period of time#t# #[A]_0# is the initial concentration of the reactant#k# is the rate constant

In your case, you know that

#k = "200 s"^(-1)#

Now, in order for **decrease** by

In other words, you need the concentration of the reactant to be down to *initial value*.

Therefore, you can say that

#[A]_t = 25/100 * [A]_0#

Plug this into the integrated rate law and solve for

#ln( (25/100 * color(red)(cancel(color(black)([A]_0))))/color(red)(cancel(color(black)([A]_0)))) = - "200 s"^(-1)#

This will be equivalent to

#ln(25/100) = - "200 s"^(-1)#

which will get you

#t = (ln(25/100))/(-"200 s"^(-1)) = color(darkgreen)(ul(color(black)("0.007 s")))#

The answer is rounded to one **significant figure**, the number of sig figs you have for the rate constant.

If you want, you can convert this to *milliseconds* by using the fact that

#"1 s" = 10^3# #"ms"#

You should end up with

#color(darkgreen)(ul(color(black)("time needed = 7 ms")))#