Question

Question #b354b

Answer 1

`["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")`