For the reaction ${\mathbf{\text{M}}}_{\mathbf{2}}\mathbf{+}{\mathbf{\text{N}}}_{\mathbf{2}}\mathbf{\rightleftharpoons }\mathbf{2}\mathbf{\text{MN}}$, scene A represents the mixture at equilibrium, with $\mathbf{M}$black and $\mathbf{N}$orange. If each molecule represents$\mathbf{0}\mathbf{.10}\mathbf{}\mathbf{\text{mol}}$ and the volume is localid="1656945725597" $\mathbf{1}\mathbf{.0}\mathbf{}\mathbf{\text{L}}$ , how many moles of each substance will be present in scene B when that mixture reaches equilibrium?

Chemical equilibrium is the state in which no net change in the amounts of reactants and products occurs during a reversible chemical reaction.

Consider the given reaction.

${\text{M}}_2+{\text{N}}_2\rightleftharpoons 2\text{MN}$

The equilibrium constant for the given reaction:

data-custom-editor="chemistry" ${\mathrm{K}}_{\text{c}}=\frac{[{\text{MN}}^2}{\left[{\text{M}}_2\right]\left[{\text{N}}_2\right]}$

$\star$The scene A:

${\text{M}}_2-2$molecules

${\text{N}}_2-2$molecules

$\mathrm{MN}-4$molecules

$\text{V}=1.0\text{L}$

$\text{n}=0.1\text{mol}$

Firstly we will calculate the equilibrium concentrations of these molecules:

$\begin{array}{l}\left[{\text{M}}_2\right]=\frac{\text{n}}{\text{V}}\\ \left[{\text{M}}_2\right]=\frac{2\cdot 0.1\text{mol}}{1.0\text{L}}\\ \left[{\text{M}}_2\right]=0.2\text{mol}/\text{L}\end{array}$

$\left[{\text{N}}_2\right]=\frac{2\cdot 0.1\text{mol}}{1\text{L}}$

$\left[{\text{N}}_2\right]=0.2\text{mol}/\text{L}$

$\begin{array}{l}\left[\text{MN}\right]=\frac{4\cdot 0.1\text{mol}}{1\text{L}}\\ \left[\text{MN}\right]=0.4\text{mol}/\text{L}\end{array}$

Therefore,$\left[\text{MN}\right]=0.4\text{mol}/\text{L}$ .

Now, we can calculate the equilibrium constant:

$\begin{array}{c}{\mathrm{K}}_{\text{c}}=\frac{{\left[\text{MN}\right]}^2}{\left[{\text{M}}_2\right]\left[{\text{N}}_2\right]}\\ {\mathrm{K}}_{\text{c}}=\frac{{(0.4\text{mol}/\text{L})}^2}{0.2\text{mol}/\text{L}\cdot 0.2\text{mol}/\text{L}}\\ {\mathrm{K}}_{\text{c}}=4\end{array}$

The scene B:

${\text{M}}_2$- 6 molecules

${\text{N}}_2-3$molecules

$\mathrm{MN}-0$molecules

$\text{V}=1.0\text{L}$

Therefore,

$\text{n}=0.10\text{mol}$.

Firstly we will calculate the equilibrium concentrations of these molecules:

$\left[{\text{M}}_2\right]=\frac{\text{n}}{\text{V}}$

$\begin{array}{l}\left[{\text{M}}_2\right]=\frac{6\cdot 0.1\text{mol}}{1\text{L}}\\ \left[{\text{M}}_2\right]=0.6\text{mol}/\text{L}\end{array}$

$\begin{array}{l}\left[{\text{N}}_2\right]=\frac{3\cdot 0.1\text{mol}}{1\text{L}}\\ [{\text{N}}_2]=0.3\text{mol}/\text{L}\end{array}$

$\mathrm{x}$- the change in the concentration ofdata-custom-editor="chemistry" ${\text{M}}_2$.

The equilibrium concentrations by using the$\mathrm{x}$ :

data-custom-editor="chemistry" ${\text{M}}_2=0.6-x$

data-custom-editor="chemistry" ${\text{N}}_2=0.3-x$

Therefore,

data-custom-editor="chemistry" $\text{MN}=2x.$

Now write the equilibrium constant by using the x to solve it:

$\begin{array}{c}{K}_{\text{c}}=\frac{{\left[\text{MN}\right]}^2}{\left[{\text{M}}_2\right]\left[{\text{N}}_2\right]}\\ 4=\frac{{\left(2x\right)}^2}{(0.6-x)(0.3-x)}\\ 4x^2=4\cdot (0.18-0.9x+x^2)\\ 4x^2=0.72-3.6x+4x^2\\ 0=0.72-3.6x\\ 3.6x=0.72\\ x=\frac{0.72}{3.6}\end{array}$

Therefore, $x=0.2$.

Now as we have the value of x, we can calculate the equilibrium concentrations in scene B:

$\begin{array}{c}\left[{\text{M}}_2\right]=0.6-x\\ \left[{\text{M}}_2\right]=0.6-0.2\\ \left[{\text{M}}_2\right]=0.4\text{mol}\\ [{\text{N}}_2]=0.3-x\\ [{\text{N}}_2]=0.3-0.2\\ [{\text{N}}_2]=0.1\text{mol}\\ \left[\text{MN}\right]=2x\\ \left[\text{MN}\right]=2\cdot 0.2\\ \left[\text{MN}\right]=0.4\text{mol}\end{array}$

Hence, the required value is:$\left[{\text{M}}_2\right]=0.4\text{M,}\left[{\text{N}}_2\right]=0.1\text{M,}\left[\text{MN}\right]=0.4\text{M}$