# Question fcc15

Mar 28, 2017

$\text{3.3 s}$

#### 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

$\text{A + A " -> " products}$

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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\frac{1}{{\left[\text{A"]) = 1/(["A}\right]}_{0}} + k t}}}$

Here

• $\left[\text{A}\right]$ is the concentration of the reactant after a time $t$
• ${\left[\text{A}\right]}_{0}$ is the initial concentration of the reactant
• $k$ is the rate constant
• $t$ is a given period of time

In your case, you have

$\frac{1}{{\left[\text{NOCl"]) = 1/(["NOCl}\right]}_{0}} + k \cdot t$

with

 ["NOCl"]_0 = "0.076 M"" " and ${\text{ "k = "3.2 M"^(-1)"s}}^{- 1}$

Rearrange the equation to solve for $t$

$t = \frac{\frac{1}{{\left[\text{NOCl"]) - 1/(["NOCl}\right]}_{0}}}{k}$

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.