# Question b354b

Sep 18, 2015

["NOBr"] = "0.030 M"

#### Explanation:

So, you know that you're dealing with the decomposition of nitrosyl bromide, $\text{NOBr}$, 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

$\frac{1}{{\left[\text{NOBr"]) - 1/(["NOBr}\right]}_{0}} = k \cdot t$

This tells you that the concentration of nitrosyl bromide at a time $t$, $\left[\text{NOBr}\right]$, which depends on the initial concentration of the reactant and on the rate constant $k$, will be equal to

$\frac{1}{{\left[\text{NOBr"]) = 1/(["NOBr}\right]}_{0}} + k t$

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

Plug in your values to find $\left[\text{NOBr}\right]$ 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 $\left[\text{NOBr}\right]$ will be

$\left[\text{NOBr"] = 1/"33.33 M"^(-1) = color(green)("0.030 M}\right)$