Q17.94CP - Chemistry: Molecular Nature Of Matter And Change

Question

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 $\mathbf{1}\mathbf{.0}\mathbf{ }\mathbf{\text{L}}$  , how many moles of each substance will be present in scene B when that mixture reaches equilibrium?

Step-by-Step Solution

Answer

Total  $\left[{\text{M}}_2\right]=0.4\text{M, }\left[{\text{N}}_2\right]=0.1\text{M, }\left[\text{MN}\right]=0.4\text{M}$ of moles of each substance will be present in scene B when that mixture reaches equilibrium.

Step 1: Definition of the reaction quotient

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

Step 2: Calculate the equilibrium concentrations

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

Step 3: Calculate the equilibrium constants

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, 

$\text{n}=0.10 \text{mol}$ .

Step 4: Calculate the equilibrium concentrations of molecules

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 of  data-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" $\text{MN}=2x.$

Step 5: Write the equilibrium constant by using the x

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,  $x=0.2$ .

Step 6: Calculate the equilibrium concentrations in scene B

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: $\left[{\text{M}}_2\right]=0.4\text{M, }\left[{\text{N}}_2\right]=0.1\text{M, }\left[\text{MN}\right]=0.4\text{M}$