Q17.87CP - Chemistry: Molecular Nature Of Matter And Change

Question

As an EPA scientist studying catalytic converters and urban smog, you want to find ${\mathbf{K}}_{\mathbf{c}}$ for the following reaction: 

$\mathbf{2}{\mathbf{NO}}_{\mathbf{2}}\left(g\right)\mathbf{\to }{\mathbf{N}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{2}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{K}}_{\mathbf{c}}\mathbf{=}\mathbf{?}$

Use the following data to find the unknown :

$$\begin{gathered} \frac{\mathbf{1}}{\mathbf{2}}{\mathbf{N}}_{\mathbf{2}}\left(g\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{NO}\left(g\right)\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{K}}_{\mathbf{c}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}} \\ \mathbf{2}{\mathbf{NO}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}\mathbf{NO}\left(g\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{K}}_{\mathbf{c}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}} \end{gathered}$$

Step-by-Step Solution

Answer

The value for the unknown equilibrium constant ${\mathrm{K}}_{\mathrm{c}}$ is ${\mathrm{K}}_{\mathrm{c}}=4.8\times {10}^{13}$.

Step 1: Concept Introduction

Chemical equilibrium is the state of a system in which the concentration of the reactant and the concentration of the products do not change over time and the system's attributes do not change.

Step 2: Manipulating the reactions

The given reaction is –

$2{\mathrm{NO}}_2\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right)$

Two reactions with known ${\mathrm{K}}_{\mathrm{c}}$ are given and it needs to be manipulated so that their combination will yield to the previous reaction.

$$\begin{gathered} \frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to \mathrm{NO}\left(\mathrm{g}\right) {\mathrm{K}}_{\mathrm{c}}=4.8\times {10}^{-10} \\ 2{\mathrm{NO}}_2\left(\mathrm{g}\right)\to {\mathrm{O}}_2\left(\mathrm{g}\right)+2\mathrm{NO}\left(\mathrm{g}\right) {\mathrm{K}}_{\mathrm{c}}=1.1\times {10}^{-5} \end{gathered}$$

Align the reactants and products, and identify the intermediates first. It can be observed that there is an $\mathrm{NO}$ in the step reaction but not on the overall reaction. Therefore, $\mathrm{NO}$ is our intermediate.

Next, ${\mathrm{N}}_2$ and ${\mathrm{O}}_2$ are part of our product side. Flip the equation (1) . The K_c will change inversely.

$$\begin{gathered} {\mathrm{NO}}_2\left(\mathrm{g}\right)\to \frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right) {\mathrm{K}}_{\mathrm{c}}=\frac{1}{4.8\times {10}^{-10}} \\ 2{\mathrm{NO}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) {\mathrm{K}}_{\mathrm{c}}=1.1\times {10}^{-5} \end{gathered}$$

Step 3: Finding the Equilibrium Constant

It can also be observed that we need 1N₂ and 2O₂. Therefore, we need to multiply our equation (1). This results to a squared K_c .

$$\begin{gathered} {\mathrm{NO}}_2\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) {\mathrm{K}}_{\mathrm{c}}={\left(\frac{1}{4.8\times {10}^{-10}}\right)}^2 \\ 2{\mathrm{NO}}_2\left(\mathrm{g}\right)\to 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) {\mathrm{K}}_{\mathrm{c}}=1.1\times {10}^{-5} \end{gathered}$$

If the overall reaction of the two is obtained, the original reaction equation is also obtained. Now multiply the two known K_c to obtain the K_c of the original reaction.

$$\begin{gathered} {\mathrm{NO}}_2\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) {\mathrm{K}}_{\mathrm{c}}={\left(\frac{1}{4.8\times {10}^{-10}}\right)}^2 \\ 2{\mathrm{NO}}_2\left(\mathrm{g}\right)\to 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) {\mathrm{K}}_{\mathrm{c}}=1.1\times {10}^{-5} \\ 2{\mathrm{NO}}_2\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) {\mathrm{K}}_{\mathrm{c}}={\left(\frac{1}{4.8\times {10}^{-10}}\right)}^2\times 1.1\times {10}^{-5} \\ {\mathrm{K}}_{\mathrm{c}}=4.8\times {10}^{13} \end{gathered}$$

Therefore, the value for Equilibrium constant is obtained as ${\mathrm{K}}_{\mathrm{c}}=4.8\times {10}^{13}$.