Question #b354b
1 Answer
Explanation:
So, you know that you're dealing with the decomposition of nitrosyl bromide,
Moreover, you know that this reaction is second order with respect to nitrosyl bromide nad scond-order overall. This tells you that the rate of the reaction will vary with the square of the concentration of the reactant.
This means that you have
#"rate" = k * ["NOBr"]^2#
In this case, the rate of the reaction is actually the rate of dissapereance of the reactant, which means that you have
#-(d["NOBr"])/(dt) = k * ["NOBr"]#
The negative sign tells you that the cocnentration is decreasing with time.
The integrated rate law for this reaction will take the form - I'll skip the derivation because I assume you're familiar with it
#1/( ["NOBr"]) - 1/(["NOBr"]_0) = k * t#
This tells you that the concentration of nitrosyl bromide at a time
#1/(["NOBr"]) = 1/(["NOBr"]_0) + kt#
#["NOBr"] = 1/(1/(["NOBr"]_0) + kt)#
Plug in your values to find
#1/(["NOBr"]_0) + kt = 1/"0.12 M" + 25"M"^(-1)color(red)(cancel(color(black)("min"^(-1)))) * 1.0color(red)(cancel(color(black)("min")))#
#1/(["NOBr"]_0) + kt = "8.33 M"^(-1) + "25 M"^(-1) = "33.33 M"^(-1)#
This means that
#["NOBr"] = 1/"33.33 M"^(-1) = color(green)("0.030 M")#