For gas-phase reactions, changes in pressure can have a significant effect on the equilibrium composition of a reaction mixture. However, it is important to distinguish between the equilibrium constant and the equilibrium composition.
The equilibrium constant is related to the standard Gibbs energy change through
\[ \Delta G^\circ = -RT\ln K \]Since \(\Delta G^\circ\) is defined for reactants and products in their standard states, the value of \(K\) depends only on temperature.
Consequently, changing the pressure does not change the value of the equilibrium constant.
What pressure changes do affect is the composition of the equilibrium mixture. When pressure changes alter the reaction quotient \(Q\), the system responds by shifting its composition until the condition
\[ Q = K \]is restored.
One way to change the pressure of a gas-phase equilibrium is to change the volume of the reaction vessel.
Decreasing the volume increases the partial pressures of all gases present, while increasing the volume decreases all partial pressures.
Because the reaction quotient depends on the partial pressures of the reacting species, changing the volume generally changes the value of \(Q\).
The system then responds by shifting its composition until
\[ Q = K \]once again.
A useful guideline is:
For example, consider the equilibrium
\[ \mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)} \]The reactant side contains no gaseous species, while the product side contains one mole of gas. Decreasing the volume increases the partial pressure of \(\mathrm{CO_2}\), causing \(Q > K\). The reaction therefore shifts toward the reactants until equilibrium is restored.
This behavior is consistent with Le Chatelier's principle because the shift reduces the pressure by consuming gaseous molecules.
The effect of an inert gas depends on how it is introduced.
Suppose an inert gas such as argon is added to a reaction vessel while the volume remains constant.
The total pressure increases because additional gas molecules are present. However, the partial pressures of the reactants and products do not change because their numbers of moles and the volume of the container remain unchanged.
Since the reaction quotient depends on the partial pressures of the reacting species,
\[ Q_p \text{ remains unchanged} \]and therefore the equilibrium composition does not change.
In this situation, the total pressure changes, but the equilibrium does not shift.
Consider the equilibrium
\[ A(g)+2B(g) \rightleftharpoons 2C(g) \]The reactant side contains three moles of gas, while the product side contains two moles of gas.
In contrast, for
\[ \mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)} \]decreasing the volume favors the reactants because the reactant side contains fewer moles of gas than the product side.
Big picture: Pressure changes do not alter the equilibrium constant, but they can alter the reaction quotient by changing the partial pressures of gaseous species. The resulting equilibrium shift always acts to restore the condition \[ Q = K \] and generally favors the side of the reaction with fewer moles of gas when pressure is increased, or more moles of gas when pressure is decreased.
At \(1000\ \text{K}\), suppose the equilibrium constant for
\[ \mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)} \]is
\[ K_p=0.100 \]A sample of \(\mathrm{CaCO_3(s)}\) is placed in a sealed \(1.00\ \text{L}\) container and allowed to reach equilibrium. What is the equilibrium partial pressure of \(\mathrm{CO_2}\)? Then, if the volume is decreased to \(0.250\ \text{L}\), how many grams of \(\mathrm{CaCO_3(s)}\) form as equilibrium is re-established?
The pure solids do not appear in the equilibrium expression, so
\[ K_p=p_{\mathrm{CO_2}} \]Therefore, in the original container,
\[ p_{\mathrm{CO_2}}=0.100\ \text{atm} \]When the volume is decreased from \(1.00\ \text{L}\) to \(0.250\ \text{L}\), the pressure of the gas initially increases by a factor of 4:
\[ p_{\mathrm{CO_2,initial\ after\ compression}} = 0.400\ \text{atm} \]But at equilibrium, the partial pressure of \(\mathrm{CO_2}\) must return to
\[ p_{\mathrm{CO_2}}=K_p=0.100\ \text{atm} \]Therefore, some \(\mathrm{CO_2}\) must be consumed to form additional \(\mathrm{CaCO_3(s)}\).
\[ \Delta n_{\mathrm{CO_2}} = \frac{(0.400-0.100)(0.250)} {(0.08206)(1000)} \] \[ \Delta n_{\mathrm{CO_2}} = 9.14\times10^{-4}\ \text{mol} \]The reaction consumes one mole of \(\mathrm{CO_2}\) for every mole of \(\mathrm{CaCO_3}\) formed, so
\[ n_{\mathrm{CaCO_3}} = 9.14\times10^{-4}\ \text{mol} \] \[ m_{\mathrm{CaCO_3}} = (9.14\times10^{-4}\ \text{mol}) (100.09\ \text{g mol}^{-1}) \] \[ \boxed{m_{\mathrm{CaCO_3}}=0.0915\ \text{g}} \]Physical interpretation: Decreasing the volume increases the partial pressure of \(\mathrm{CO_2}\), making \(Q > K\). The system responds by shifting toward reactants, consuming \(\mathrm{CO_2}\) and forming additional \(\mathrm{CaCO_3(s)}\), until \(Q = K\) again.
Consider again the equilibrium
\[ \mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)} \]Suppose the system is at equilibrium in a rigid container. Nitrogen gas, \(\mathrm{N_2(g)}\), is then added until the total pressure in the container is quadrupled. What effect does this have on the equilibrium?
Adding \(\mathrm{N_2(g)}\) increases the total pressure in the container. However, \(\mathrm{N_2(g)}\) is not a reactant or product in the equilibrium expression.
For this reaction,
\[ K_p = p_{\mathrm{CO_2}} \]because the pure solids \(\mathrm{CaCO_3(s)}\) and \(\mathrm{CaO(s)}\) have unit activity and do not appear in the equilibrium expression.
In a rigid container, adding an inert gas does not change the number of moles of \(\mathrm{CO_2}\) or the volume of the container. Therefore, the partial pressure of \(\mathrm{CO_2}\) does not change:
\[ p_{\mathrm{CO_2}} = \frac{n_{\mathrm{CO_2}}RT}{V} \]Since \(n_{\mathrm{CO_2}}\), \(T\), and \(V\) are unchanged,
\[ p_{\mathrm{CO_2}} \text{ is unchanged} \]Therefore,
\[ Q_p = K_p \]remains true, and the system is still at equilibrium.
Conclusion: Adding \(\mathrm{N_2(g)}\) at constant volume increases the total pressure, but it does not change the partial pressure of \(\mathrm{CO_2}\). Therefore, the equilibrium does not shift.
Physical interpretation: Le Chatelier's principle must be applied carefully. The system responds to changes that affect the reaction quotient, not simply to changes in total pressure. In this case, the total pressure increases, but \(Q_p\) does not change because the reacting gas, \(\mathrm{CO_2}\), has the same partial pressure as before.
Big picture: Pressure changes do not alter the equilibrium constant, but they can alter the equilibrium composition by changing the partial pressures of reacting gases. The resulting shifts can be understood quantitatively through the reaction quotient and qualitatively through Le Chatelier's principle, which predicts that systems respond to pressure changes by favoring the side of the reaction that best relieves the stress.