# Question #7fc24

Mar 24, 2017

Here's why that is the case.

#### Explanation:

Changing the volume of the system implies changing the pressure at which the reaction takes place.

Assuming that temperature is being held constant, increasing the volume of the reaction vessel will cause the pressure to decrease. Similarly, decreasing the volume of the reaction vessel will cause the pressure to increase.

Now, both equilibrium systems are governed by Le Chatelier's Principle, which states that a system at equilibrium will counteract any stress placed on its current position in a way that will reduce that stress.

Now, take the first equilibrium reaction

${\text{N"_ (2(g)) + 3"H"_ (2(g)) rightleftharpoons 2"NH}}_{3 \left(g\right)}$

Notice that you have a total of

$\text{1 mole N"_2 + "3 moles H"_2 = "4 moles gas}$

on the reactants' side and $2$ moles of gas on the products' side.

When you change the volume of the reaction vessel, you also change the total pressure. The system will counteract this change in pressure by shifting in a way that will reduce or increase the number of moles of gas present in the reaction vessel.

So if volume increases, pressure decreases and the equilibrium will favor the reaction that counteracts this change, i.e. that increases the pressure by increasing the number of moles of gas present in the reaction vessel.

If the volume decreases, pressure increases and the equilibrium will favor the reaction that counteracts this change, i.e. that decreases the pressure by decreasing the number of moles of gas present in the reaction vessel.

The second equilibrium looks like this

${\text{H"_ (2(g)) + "Cl"_ (2(g)) rightleftharpoons 2"HCl}}_{\left(g\right)}$

This time, you have

$\text{1 mole H"+2 + "1 mole Cl"_2 = "2 moles gas}$

on the reactants' side and $2$ moles of gas on the products' side.

This means that a change in volume, which will induce a change in pressure, cannot be counteracted by the system because both reactions will end up producing the same number of moles in the reaction vessel.

So regardless of how the equilibrium shifts, the total number of moles will neither decrease nor increase.

Therefore, the position of the equilibrium will not be affected by the change in volume.