Le Chatelier's Principle: Concentration and Pressure
Core Concepts
Apply Le Chatelier's Principle to predict how a system at equilibrium responds to stresses. This stack focuses specifically on changes in the concentration of reactants/products and changes in overall pressure or volume.
Le Chatelier's Principle: Concentration and Pressure
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In simple terms, what is Le Chatelier's Principle?
When a stress (like a change in concentration, pressure, or temperature) is applied to a system at equilibrium, the system will shift in a way that counteracts or relieves that stress.
For the reaction $$ PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) $$, what happens if more \(Cl_2\) gas is added to the container?
The equilibrium will shift to the right. The system consumes the added \(Cl_2\) (and some \(PCl_3\)) to produce more \(PCl_5\), thus relieving the stress of increased reactant concentration.
For the reaction $$ PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) $$, what happens if \(PCl_5\) is removed from the system?
The equilibrium will shift to the right. The system works to replace the removed product by consuming reactants (\(PCl_3\) and \(Cl_2\)) to produce more \(PCl_5\).
Does changing the concentration of a reactant or product alter the value of the equilibrium constant, \(K\)?
No. The value of \(K\) is constant for a given reaction at a specific temperature. The system shifts to re-establish the concentrations that satisfy the \(K\) ratio.
How does increasing the total pressure (by decreasing the volume) affect a gaseous equilibrium?
The equilibrium will shift to the side of the reaction that has the fewer number of moles of gas. This shift helps to reduce the overall pressure.
For $$ N_2O_4(g) \rightleftharpoons 2NO_2(g) $$, which way does the equilibrium shift if the pressure is increased?
The equilibrium will shift to the left. The reactant side has \(1\) mole of gas, while the product side has \(2\) moles. Shifting left reduces the total number of gas molecules, counteracting the pressure increase.
For $$ C(s) + H_2O(g) \rightleftharpoons CO(g) + H_2(g) $$, which way does the equilibrium shift if the volume is increased?
Increasing volume decreases pressure. The equilibrium will shift to the side with more moles of gas. The reactant side has \(1\) mole of gas, and the product side has \(2\) moles. The equilibrium shifts to the right.
What is the effect of changing pressure or volume on an equilibrium like \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\)?
There is no effect. The number of moles of gas is the same on both sides of the equation (\(2\) moles). Therefore, a change in pressure or volume cannot be relieved by a shift.
What is the effect of adding an inert gas (e.g., Argon) to an equilibrium mixture at "constant volume"?
There is no effect on the equilibrium position. While the total pressure increases, the partial pressures and concentrations of the reacting gases do not change, so \(Q\) remains equal to \(K\).
If you add more of a solid reactant to a heterogeneous equilibrium like $$ CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) $$, will the equilibrium shift?
No, assuming some solid was already present. Adding more of a pure solid (or liquid) does not change its concentration or activity, so it does not stress the equilibrium.
