Chapter 11, Problem 95QRT

Nitryl fluoride is an explosive compound that can be made by oxidizing nitrogen dioxide with fluorine:

2 NO₂(g) + F₂(g) → 2 NO₂F(g)

Several kinetics experiments, all done at the same temperature and involving formation of nitryl fluoride, are summarized in this table:

1. (a) Write the rate law for the reaction.
2. (b) Determine what the order of the reaction is with respect to each reactant and each product.
3. (c) Calculate the rate constant k and express it in appropriate units.

Interpretation Introduction

Interpretation:

The rate law for the reaction has to be written.

Explanation of Solution

  $2N{O}_2\left(g\right)+F_2\left(g\right)\to 2N{O}_{2}F\left(g\right)$

Suppose the rate law of the above reaction is as follows,

  $Rate=k{\left[N{O}_2\right]}^i{\left[F_2\right]}^j{\left[N{O}_{2}F\right]}^h$

Where, $k,i,jandh$ are unknown.  The values of $i,jandh$ can be calculated by taking given below assumptions.

Carefully examining the table, it can be said that $\left[N{O}_2\right]$ is constant in experiment number $2and3.$  Then, the rate law for experiment number $2and3$ can be written as follows,

  $\begin{array}{l}{\left(Rate\right)}_2=k{\left[N{O}_2\right]}^i{\left[F_2\right]}^j{\left[N{O}_{2}F\right]}^h\\ \\ \Rightarrow 4.0\times {10}^{-4}mol{L}^{-1}{s}^{-1}=k{\left(0.0020mol{L}^{-1}\right)}^i{\left(0.0050mol{L}^{-1}\right)}^j{\left(0.0020mol{L}^{-1}\right)}^h\\ \\ {\left(Rate\right)}_3=k{\left[N{O}_2\right]}^i{\left[F_2\right]}^j{\left[N{O}_{2}F\right]}^h\\ \\ \Rightarrow 1.6\times {10}^{-4}mol{L}^{-1}{s}^{-1}=k{\left(0.0020mol{L}^{-1}\right)}^i{\left(0.0020mol{L}^{-1}\right)}^j{\left(0.0020mol{L}^{-1}\right)}^h\end{array}$

Now, on dividing both the equations, the value of j can be found out.

  $\begin{array}{c}\frac{4.0\times {10}^{-4}mol{L}^{-1}{s}^{-1}}{1.6\times {10}^{-4}mol{L}^{-1}{s}^{-1}}=\frac{\cancel{k{\left(0.0020mol{L}^{-1}\right)}^i}{\left(0.0050mol{L}^{-1}\right)}^j\cancel{{\left(0.0020mol{L}^{-1}\right)}^h}}{\cancel{k{\left(0.0020mol{L}^{-1}\right)}^i}{\left(0.0020mol{L}^{-1}\right)}^j\cancel{{\left(0.0020mol{L}^{-1}\right)}^h}}\\ \\ {\left(\frac{5}{2}\right)}^1={\left(\frac{5}{2}\right)}^j\\ \\ \Rightarrow j=1.\end{array}$

Similarly, carefully examining the table, it can be said that $\left[F_2\right]$ is constant in experiment number $1and2.$  Then, the rate law for experiment number $1and2$ can be written as follows,

  $\begin{array}{l}{\left(Rate\right)}_1=k{\left[N{O}_2\right]}^i{\left[F_2\right]}^j{\left[N{O}_{2}F\right]}^h\\ \\ \Rightarrow 2.0\times {10}^{-4}mol{L}^{-1}{s}^{-1}=k{\left(0.0010mol{L}^{-1}\right)}^i{\left(0.0050mol{L}^{-1}\right)}^j{\left(0.0020mol{L}^{-1}\right)}^h\\ \\ {\left(Rate\right)}_2=k{\left[N{O}_2\right]}^i{\left[F_2\right]}^j{\left[N{O}_{2}F\right]}^h\\ \\ \Rightarrow 4.0\times {10}^{-4}mol{L}^{-1}{s}^{-1}=k{\left(0.0020mol{L}^{-1}\right)}^i{\left(0.0050mol{L}^{-1}\right)}^j{\left(0.0020mol{L}^{-1}\right)}^h\end{array}$

Now, on dividing both the equations, the value of i can be found out.

  $\begin{array}{c}\frac{2.0\times {10}^{-4}mol{L}^{-1}{s}^{-1}}{4.0\times {10}^{-4}mol{L}^{-1}{s}^{-1}}=\frac{{\left(0.0010mol{L}^{-1}\right)}^i\cancel{k{\left(0.0050mol{L}^{-1}\right)}^j{\left(0.0020mol{L}^{-1}\right)}^h}}{{\left(0.0020mol{L}^{-1}\right)}^i\cancel{k{\left(0.0050mol{L}^{-1}\right)}^j{\left(0.0020mol{L}^{-1}\right)}^h}}\\ \\ {\left(\frac{1}{2}\right)}^1={\left(\frac{1}{2}\right)}^i\\ \\ \Rightarrow i=1.\end{array}$

Similarly, carefully examining the table, it can be said that $\left[F_2\right]$ is constant in experiment number $3and4.$  Then, the rate law for experiment number $3and4$ can be written as follows,

  $\begin{array}{l}{\left(Rate\right)}_3=k{\left[N{O}_2\right]}^i{\left[F_2\right]}^j{\left[N{O}_{2}F\right]}^h\\ \\ \Rightarrow 1.6\times {10}^{-4}mol{L}^{-1}{s}^{-1}=k{\left(0.0020mol{L}^{-1}\right)}^i{\left(0.0020mol{L}^{-1}\right)}^j{\left(0.0020mol{L}^{-1}\right)}^h\\ \\ {\left(Rate\right)}_4=k{\left[N{O}_2\right]}^i{\left[F_2\right]}^j{\left[N{O}_{2}F\right]}^h\\ \\ \Rightarrow 1.6\times {10}^{-4}mol{L}^{-1}{s}^{-1}=k{\left(0.0020mol{L}^{-1}\right)}^i{\left(0.0020mol{L}^{-1}\right)}^j{\left(0.0010mol{L}^{-1}\right)}^h\end{array}$

Now, on dividing both the equations, the value of h can be found out.

  $\begin{array}{c}\frac{1.6\times {10}^{-4}mol{L}^{-1}{s}^{-1}}{1.6\times {10}^{-4}mol{L}^{-1}{s}^{-1}}=\frac{\cancel{k{\left(0.0020mol{L}^{-1}\right)}^i}\cancel{{\left(0.0020mol{L}^{-1}\right)}^j}{\left(0.0020mol{L}^{-1}\right)}^h}{\cancel{k{\left(0.0020mol{L}^{-1}\right)}^i}\cancel{{\left(0.0020mol{L}^{-1}\right)}^j}{\left(0.0010mol{L}^{-1}\right)}^h}\\ \\ 1={\left(2\right)}^h\\ \\ {\left(2\right)}^0={\left(2\right)}^h\\ \\ \Rightarrow h=0.\end{array}$

Therefore, the rate law for the reaction is $rate=k{\left[N{O}_2\right]}^1{\left[F_2\right]}^1.$

Interpretation Introduction

Interpretation:

The order of the reaction with respect to each reactant and each product has to be determined.

Explanation of Solution

The rate law for the reaction is $\text{rate=k}{\left[{\text{NO}}_{\text{2}}\right]}^{\text{1}}{\left[{\text{F}}_{\text{2}}\right]}^{\text{1}}{\left[N{O}_{2}F\right]}^0\text{.}$

Therefore, the order of the reaction with respect to $\left[N{O}_2\right]$, $\left[F_2\right]$ and $\left[N{O}_{2}F\right]$ is one, one and zero respectively.

Interpretation Introduction

Interpretation:

The rate constant $k$ for the reaction has to be calculated with appropriate units.

Answer to Problem 95QRT

The average of the rate constant is $40Lmo{l}^{-1}{s}^{-1}.$

Explanation of Solution

The rate constant can be expressed as shown below.

  $\begin{array}{c}Rate=k{\left[N{O}_2\right]}^1{\left[F_2\right]}^1\\ \\ k=\frac{Rate}{{\left[N{O}_2\right]}^1{\left[F_2\right]}^1}\end{array}$

For first experiment the rate constant can be calculated by plugging all the data given in the table.

In the first experiment, the initial concentration of $N{O}_2$ and $F_2$ are $0.0010mol/Land0.0050mol/L$ respectively.  The initial rate of the reaction in first experiment is $2.0\times {10}^{-4}molL{}^{-1}{s}^{-1}.$

Now, the rate constant for first experiment is given below.

  $k_1=\frac{Rate}{{\left[N{O}_2\right]}^1{\left[F_2\right]}^1}=\frac{2.0\times {10}^{-4}molL{}^{-1}{s}^{-1}}{\left(0.0010mol/L\right)\left(0.0050mol/L\right)}=40Lmo{l}^{-1}{s}^{-1}.$

Similarly, the rate constant for rest of the experiments can be calculated.  The calculated rate constant values are given in the table below.

$\begin{array}{c}\left[N{O}_2\right]\\ \left(mol/L\right)\end{array}$	$\begin{array}{c}\left[F_2\right]\\ \left(mol/L\right)\end{array}$	Initial rate $\left(molL{}^{-1}{s}^{-1}\right)$	Rate constant $\left({\text{Lmol}}^{\text{-1}}{\text{s}}^{\text{-1}}\right)$
$0.0010$	$0.0050$	$2.0\times {10}^{-4}$	$40$
$0.0020$	$0.0050$	$4.0\times {10}^{-4}$	$40$
$0.0020$	$0.0020$	$1.6\times {10}^{-4}$	$40$

The average of the rate constant is $40Lmo{l}^{-1}{s}^{-1}.$