Question #b354b

1 Answer
Sep 18, 2015

Answer:

#["NOBr"] = "0.030 M"#

Explanation:

So, you know that you're dealing with the decomposition of nitrosyl bromide, #"NOBr"#, so in essence your reaction has only one reactant.

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 #t#, #["NOBr"]#, which depends on the initial concentration of the reactant and on the rate constant #k#, will be equal to

#1/(["NOBr"]) = 1/(["NOBr"]_0) + kt#

#["NOBr"] = 1/(1/(["NOBr"]_0) + kt)#

Plug in your values to find #["NOBr"]# after 1.0 minutes

#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"]# will be

#["NOBr"] = 1/"33.33 M"^(-1) = color(green)("0.030 M")#