Question

At a particular temperature, 8.0 moles of NO₂ is placed into a 1.0-L container and the NO₂ dissociates by the reaction

$\mathbf{2}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\stackrel{}{\mathbf{\rightleftharpoons }}\mathbf{2}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

At equilibrium the concentration of NO(g) is 2.0 M. Calculate K for this reaction.

Short answer

The K value for this dissociation reaction at a particular temperature is 0.11.

Step-by-step solution

Initial concentration of NO2

The equilibrium constant, K expression in terms of concentrations is written below:

$\text{K=}\frac{{\left[\text{NO}\right]}^{\text{2}}\left[{\text{O}}_{\text{2}}\right]}{{\left[{\text{NO}}_{\text{2}}\right]}^{\text{2}}}$ …(1)

The given number of moles of${\text{NO}}_{\text{2}}$is 8.0 mol.

The given volume of the reaction container is 1.0 L.

Therefore, the concentration of${\text{NO}}_{\text{2}}$ present initially can be calculated as follows:

$\begin{array}{c}\left[{\text{NO}}_{\text{2}}\right]\text{=}\frac{{\text{Moles\hspace{0.17em}of\hspace{0.17em}NO}}_{\text{2}}}{\text{Volume}}\\ \text{=}\frac{\text{8.0\hspace{0.17em}mol}}{\text{1.0\hspace{0.17em}L}}\\ \text{=8.0\hspace{0.17em}M}\end{array}$

The initial concentration of${\text{NO}}_{\text{2}}$ is 8.0 M.

Change in concentration

Assume x be the concentration change.

Setup the ICE table:

$\begin{array}{cccccc}& {\text{2NO}}_{\text{2}}\left(\text{g}\right)& \stackrel{}{\rightleftharpoons }& \text{2NO}\left(\text{g}\right)& \text{+}& {\text{O}}_{\text{2}}\left(\text{g}\right)\\ \text{Iinitial}\left(\text{M}\right)& \text{8.0}& & \text{0}& & \text{0}\\ \text{Change}\left(\text{M}\right)& \text{\hspace{0.17em}–\hspace{0.17em}2x}& & \text{+2x}& & \text{+x}\\ \text{Equilibrium}\left(\text{M}\right)& \text{8.0\hspace{0.17em}–\hspace{0.17em}2x}& & \text{2x}& & \text{x}\end{array}$

Substitute these equilibrium concentrations in terms of x into equation (1) as shown below.

$\text{K=}\frac{{\left(\text{2x}\right)}^{\text{2}}\left(\text{x}\right)}{{\left(\text{8.0\hspace{0.17em}–\hspace{0.17em}}2\text{x}\right)}^{\text{2}}}$ …(2)

The equillibrium concentration of NO

The given concentration ofNO(g) at equilibrium is 2.0 M.

Using the ICE table, the equilibriumconcentration of NO(g) is found to be 2x. Hence, the x value is,

$\begin{array}{l}\text{2.0\hspace{0.17em}=2x}\\ \text{x=1.0}\end{array}$

Hence, the x value is calculated to be 1.0.

Equilibrium constant of dissociation reaction

Substitute 1.0 for x into Equation (2) to calculate the K as shown below:
$\begin{array}{c}\text{K=}\frac{{\left(\text{2×1.0}\right)}^{\text{2}}\left(\text{1.0}\right)}{{\left(\text{8.0\hspace{0.17em}–\hspace{0.17em}2×1.0}\right)}^{\text{2}}}\\ \text{=}\frac{{\left(\text{2.0}\right)}^{\text{2}}\left(\text{1.0}\right)}{{\left(\text{6.0}\right)}^{\text{2}}}\\ \text{=0.11}\end{array}$

Therefore, the K value for this dissociation reaction is 0.11.