Question #7b731
1 Answer
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"XY + XY → products
which will look like this
1/(["XY"]_ t) - 1/(["XY"]_ 0) = kt1[XY]t−1[XY]0=kt
Here
["XY"]_ 0[XY]0 is the initial concentration of the reactant["XY"]_ t[XY]t is the concentration of the reactant after a period of timett kk is the rate constant
In your case, you have
K = 6.82 * 10^(-3)K=6.82⋅10−3 "M"^(-1)"s"^(-1)M−1s−1
and
{(["XY"]_ 0 = "0.140 M"), (["XY"]_ t = 6.60 * 10^(-2)"M") :}
Rearrange the integrated rate law to solve for
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"
