# Question 7b731

Apr 25, 2017

$\text{1170 s}$

#### Explanation:

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

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

which will look like this

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

Here

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

$K = 6.82 \cdot {10}^{- 3}$ ${\text{M"^(-1)"s}}^{- 1}$

and

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

Rearrange the integrated rate law to solve for $t$

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

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

$\text{1 hour = 60 minutes = 3600 seconds}$