Question

A mixture of N₂, H₂ and NH₃ is at equilibrium according to the equation

The volume is suddenly decreased (by increasing the external pressure), and a new equilibrium is established as depicted below.

a. If the volume of the final equilibrium mixture is 1.00 L, determine the value of the equilibrium constant K for the reaction. Assume temperature is constant.

b. Determine the volume of the initial equilibrium mixture assuming a final equilibrium volume of 1.00 L and assuming a constant temperature.

Short answer

(a) The equilibrium constant is 2.25

(b) The volume of the initial equilibrium mixture will be 33.9L

Step-by-step solution

Determine the equilibrium constant

(a)

From the 2^nd figure with decreased volume, it has been noted that

Moles of H₂=2

Moles of N₂=2

Moles of NH₃=6

As per the formula

$Concentration=\frac{mole}{volume}$

Given volume=1L

Therefore, for this case

Concentration=moles

So,

[H₂]=2

[N₂]=2

[NH₃]=6

The equilibrium constant will be

$$\begin{gathered} K=\frac{{\left[N{H}_3\right]}^2}{{\left[H_2\right]}^3\left[N_2\right]} \\ K=\frac{{\left[6\right]}^2}{{\left[2\right]}^3\left[2\right]} \\ K=2.25 \end{gathered}$$

Therefore, the equilibrium constant is 2.25

Determine the volume

(b)

Let the volume of the initial equilibrium mixture assuming a final equilibrium volume of 1.00 L and assuming a constant temperature be V

From the 1^st figure, it has been noted that

Moles of H₂=8

Moles of N₂=4

Moles of NH₃=2

As per the formula

$Concentration=\frac{mole}{volume}$

Given volume=V L

Therefore, for this case

$Concentration=\frac{mole}{V}$

So,

[H₂]=$\frac{8}{V}$

[N₂]=$\frac{4}{V}$

[NH₃]= $\frac{2}{V}$

$\begin{array}{rcl}K& =& \frac{{\left[N{H}_3\right]}^2}{{\left[H_2\right]}^3\left[N_2\right]}\\ K& =& \frac{{\left[{\displaystyle \frac{2}{V}}\right]}^2}{{\left[{\displaystyle \frac{8}{V}}\right]}^3\left[{\displaystyle \frac{4}{V}}\right]}\\ 2.25& =& \frac{{\left[{\displaystyle \frac{2}{V}}\right]}^2}{{\left[{\displaystyle \frac{8}{V}}\right]}^3\left[{\displaystyle \frac{4}{V}}\right]}\\ V& =& 33.9L\end{array}$

The volume of the initial equilibrium mixture will be 33.9L