Question #fcc15
1 Answer
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
#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 = ( 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.
