Question

$\left[N_2{O}_4\right]$is soluble in the solventcyclohexane; however, dissolution does not prevent$\left[N_2{O}_4\right]$from breaking down to give $\left[N{O}_2\right]$according to the equation:

localid="1663589431242" ${\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}\mathbf{\left(}\mathbf{cyclohexane}\mathbf{\right)}}\underleftarrow{\mathbf{\to }}\mathbf{2}{\mathbf{NO}}_{\mathbf{2}\mathbf{\left(}\mathbf{cyclohexane}\mathbf{\right)}}$

An effort to compare this solution equilibrium with a similar equilibrium in the gas gave the following actualexperimentaldata at ${\mathbf{20}}^{\mathbf{°}}$:

localid="1663588653416" $\left[N_2{O}_4\right]$(mol/L)

$\mathit{N}{\mathit{O}}_{\mathbf{2}}$ (mol/L)

localid="1663589749124" $\begin{array}{l}\mathbf{0}\mathbf{.}\mathbf{192}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{0}\mathbf{.}\mathbf{721}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{1}\mathbf{.}\mathbf{61}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{2}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{3}\mathbf{.}\mathbf{95}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{7}\mathbf{.}\mathbf{90}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{11}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\end{array}$

$\begin{array}{l}\mathbf{2}\mathbf{.}\mathbf{80}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{5}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{7}\mathbf{.}\mathbf{26}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{10}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{11}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{17}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\\ \mathbf{21}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\end{array}$

Short answer

The equilibrium constant is 0.0374.

Step-by-step solution

Given information 

The known concentration is as follows:

$\left[{\mathrm{N}}_2{\mathrm{O}}_4\right]$(10-3mol/L)	$N{O}_2$(10-3mol/L)
$\begin{array}{l}0.190\\ 0.686\\ 1.54\\ 2.55\\ 3.75\\ 7.86\\ 11.9\\ \end{array}$	$\begin{array}{l}2.80\\ 5.20\\ 7.26\\ 10.4\\ 11.7\\ 17.3\\ 21.0\end{array}$

Concept of equilibrium and law of mass action  

The equilibrium expression is determined by the equilibrium concentrations of reactants and products, according to the Law of Mass Action. Even when all concentrations are varied, when a reactive system reaches equilibrium at a given temperature, the value of the equilibrium constant remains constant.

Calculate the equilibrium expression 

The equilibrium expression for the reaction is:

$K=\frac{{\displaystyle {\left[{\mathrm{NO}}_2\right]}^2}}{{\displaystyle \left[{\mathrm{N}}_2{\mathrm{O}}_4\right]}}$

Plot the graph

Thus, the plot of ${\left[{\mathrm{NO}}_2\right]}^2$vs $\left[{\mathrm{N}}_2{\mathrm{O}}_4\right]$must be a straight line with a slope of 1/K. And, the slope is K = 1/26.695 = 0.0374.