Question

Question #fcc15

Answer 1

`"3.3 s"`

Explanation:

Your tool of choice here will be the integrated rate law for a second-order reaction, which for a general-form second-order reaction

`"A + A " -> " products"`

that is second order with respect to `A"`, takes the form

`color(blue)(ul(color(black)(1/(["A"]) = 1/(["A"]_0) + kt)))`

Here

• `["A"]` is the concentration of the reactant after a time `t`
• `["A"]_0` is the initial concentration of the reactant
• `k` is the rate constant
• `t` is a given period of time

In your case, you have

`1/(["NOCl"]) = 1/(["NOCl"]_0) + k * t`

with

`["NOCl"]_0 = "0.076 M"" "` and `" "k = "3.2 M"^(-1)"s"^(-1)`

Rearrange the equation to solve for `t`

`t = ( 1/(["NOCl"]) - 1/(["NOCl"]_0))/k`

and plug in your values to find

`t = (1/(0.042 color(red)(cancel(color(black)("M")))) - 1/(0.076color(red)(cancel(color(black)("M")))))/(3.2 color(red)(cancel(color(black)("M"^(-1))))"s"^(-1)) = color(darkgreen)(ul(color(black)("3.3 s")))`

The answer is rounded to three sig figs.