Does Adding an Inert Gas at Constant Volume Affect Equilibrium?

Adding an inert gas raises the total pressure of a gas mixture, which can make it seem as though equilibrium must shift. For an ideal-gas reaction at fixed temperature and volume, however, the reacting gases’ partial pressures do not change. Since the reaction quotient depends on those partial pressures, the equilibrium composition remains unchanged.

The fixed-temperature assumption matters: the result describes adding the gas and allowing the system to return to its original temperature. The main contrast is adding an inert gas at constant pressure, where the volume changes and equilibrium can shift.

Why the equilibrium is unchanged

For an ideal gas, the partial pressure of species $i$ is

$$p_i = \frac{n_iRT}{V},$$

where $n_i$ is the amount of that species, $T$ is temperature, and $V$ is volume. Adding an inert gas increases its own amount and partial pressure. It does not change $n_i$ for any reacting gas. If both $T$ and $V$ stay fixed, each reacting species therefore retains its original partial pressure.

For a general gas reaction,

$$aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g),$$

the pressure-based reaction quotient is

$$Q_p = \frac{p_C^c p_D^d}{p_A^a p_B^b}.$$

Because none of the reacting gases’ partial pressures change when the inert gas is added, $Q_p$ does not change. The system was initially at equilibrium, so $Q_p=K_p$. Since the equilibrium constant $K_p$ is determined by temperature and the temperature is fixed, the system is still at equilibrium; no net reaction occurs.

This does not mean the total pressure stays constant. The total pressure is the sum of all the gases’ partial pressures, including the added inert gas, so it increases even though the reacting gases’ partial pressures remain the same.

Worked example: nitrogen dioxide equilibrium

Consider the equilibrium

$$\mathrm{N_2O_4(g)}\rightleftharpoons 2\mathrm{NO_2(g)}.$$

Suppose an equilibrium mixture at a fixed temperature and volume has partial pressures $p_{\mathrm{N_2O_4}}=0.40\,\mathrm{atm}$ and $p_{\mathrm{NO_2}}=0.20\,\mathrm{atm}$. Its pressure quotient, using the usual classroom pressure convention, is

$$Q_p = \frac{(0.20)^2}{0.40}=0.10.$$

Now add argon until its partial pressure increases from $0.20\,\mathrm{atm}$ to $0.80\,\mathrm{atm}$. At constant volume and temperature, the nitrogen tetroxide and nitrogen dioxide partial pressures remain $0.40\,\mathrm{atm}$ and $0.20\,\mathrm{atm}$. Thus $Q_p$ is still $0.10$, equal to its original equilibrium value. The total pressure rises from $0.80\,\mathrm{atm}$ to $1.40\,\mathrm{atm}$, but the reaction does not shift.

Why constant pressure is different

At constant pressure, adding an inert gas generally makes the container expand. The reacting gases then occupy a larger volume, so their partial pressures decrease. For an ideal-gas reaction, this can change $Q_p$ and cause the reaction to shift.

The direction depends on the change in the number of gas molecules, calculated from the balanced equation:

$$\Delta n_{\mathrm{gas}} = \text{sum of gaseous product coefficients} - \text{sum of gaseous reactant coefficients}.$$

When the volume expands at constant pressure, the reaction tends to shift toward the side with more gas molecules. If the two sides have equal total gas coefficients, dilution changes the quotient by the same factor in numerator and denominator, so there is no shift under the ideal-gas model. This constant-pressure result is different from the constant-volume case, where the reacting gases’ partial pressures do not change in the first place.

Common mistakes

• Confusing total pressure with partial pressure. Total pressure rises when inert gas is added, but the partial pressure of a reacting species is $n_iRT/V$; it does not rise if its amount, temperature, and volume are unchanged.
• Assuming every pressure increase shifts equilibrium. A change in total pressure does not by itself determine the response. The relevant issue is whether the partial pressures or activities of the reacting species change.
• Forgetting temperature. If adding gas or operating the apparatus changes the temperature, $K_p$ may change. The no-shift result assumes the system remains at the same temperature.
• Applying the ideal-gas result to every mixture. At very high pressures or in strongly interacting mixtures, real-gas effects can matter; fugacities rather than simple partial pressures are then used to describe equilibrium.

The standard result is therefore specific but useful: adding a genuinely inert gas to an ideal-gas equilibrium at constant temperature and volume increases total pressure without changing the reaction quotient or equilibrium composition.