Question

The rate law for the reaction $\mathbf{2}\mathit{N}\mathit{O}\mathit{B}\mathit{r}\left(g\right)\mathbf{}\mathbf{\to }\mathbf{2}\mathit{N}\mathit{O}\left(g\right)\mathbf{+}\mathit{B}{\mathit{r}}_{\mathbf{2}}\left(g\right)$at some temperature is$\mathit{R}\mathit{a}\mathit{t}\mathit{e}\mathbf{=}\mathbf{-}\mathit{d}\left[NOBr\right]\mathbf{/}\mathit{d}\mathit{t}\mathbf{=}\mathit{k}{\left[NOBr\right]}^{\mathbf{2}}$

a. If the half-life for this reaction is 2.00 s when ${\left[\mathrm{NOBr}\right]}_{\mathbf{0}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{900}\mathbf{M}$, calculate the value of k for this reaction.

b. How much time is required for the concentration of NOBr to decrease to 0.100 M?

Short answer

(a) The value of K is obtained as$0.556{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$.

(b)The time required for the [NOBr] to decrease is 15.9872 s.

Step-by-step solution

(a) Calculating the rate constant (k) for a second-order reaction

Given data

$\begin{array}{l}Rate=k{\left[NOBr\right]}^2\\ \left[NOBr\right]=0.900M\end{array}$

For half-life (t ½),

$$\begin{gathered} {\mathit{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{k}{\mathbf{A}}_{\mathbf{0}}} \\ \begin{array}{l}2=1/k\left(0.900\right)\\ k=0.556{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}\end{array} \end{gathered}$$

The value of K is obtained.

(b) Calculating the time required for concentration of NOBr decrease

Using the second order reaction concentration-time formula.

$\begin{array}{l}1/0.1=1/0.9+kt\\ 10=1.11+0.556t\\ 8.889=0.556t\\ t=8.889/0.556\\ t=15.9872s\end{array}$

The time required for the [NOBr] to decrease is 15.9872 s.