Chemistry - What would be the effect of the addition of an inert gas to a reaction at equilibrium?

Solution 1:

Dissociation obviously increases the number of moles.
The addition of an inert gas can affect the equilbrium, but only if the volume is allowed to change.

There are two cases on which equilibrium depends. These are:

• Addition of an inert gas at constant volume:
  When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change.

Hence, when an inert gas is added to the system in equilibrium at constant volume there will be no effect on the equilibrium.

• Addition of an inert gas at constant pressure:
  When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.

Consider the following reaction in equilibrium: $$\ce{2 NH_3(g) ⇌ N_2 (g) + 3 H2 (g)}$$ The addition of an inert gas at constant pressure to the above reaction will shift the equilibrium towards the forward direction(shift to the right) because the number of moles of products is more than the number of moles of the reactants.

Please read about the Effect of adding an Inert Gas

Solution 2:

Nick's answer is good. Let's add a little maths.

Let's take an example dissociation reaction $\ce{A<=>B + C}$ for which $K_p=1$. Since

$$K_p = \frac{P_B P_C}{P_A}$$

one equilibrium scenario is $P_A =P_B = P_C = 1 \text{ atm}$. The total pressure $P_T = 3 \text{ atm}$.

If we add an inert gas at constant pressure, the total pressure cannot change. Thus, the partial pressures of the three components $\ce{A}$, $\ce{B}$, and $\ce{C}$ must decrease. If we add some inert gas $\ce{D}$ so that $P_A=P_B=P_C=P_D=\frac{3}{4} \text{ atm}$, the total pressure is still $P_T=3\text{ atm}$. The reaction quotient $Q_p$ is now:

$$Q_p = \frac{P_B P_C}{P_A}=\frac{(\frac{3}{4}\text{ atm})(\frac{3}{4}\text{ atm})}{\frac{3}{4}\text{ atm}}=\frac{3}{4}\text{ atm}<K_p$$

Since $Q_p<K_p$, the equilibrium will shift toward products.

At constant volume, the partial pressures of the three components $\ce{A}$, $\ce{B}$, and $\ce{C}$ can remain constant and equilibrium will be unchanged. Adding inert gas $\ce{D}$ will increase the pressure of the system.