15.2 Le Chatelier's Principle
PRE-HEALTH PREP
Le Chatelier's Principle
Concentration, Temperature & Pressure
General Chemistry 15.2
Le Chatelier's Principle describes how a chemical system at equilibrium responds to external stresses by shifting to counteract the disturbance. Concentration changes alter the reaction quotient Q relative to the equilibrium constant K, while temperature is unique in that it changes the value of K itself. Pressure and volume effects depend on the moles of gas on each side of the equation, with important exceptions for inert gas addition and reactions with equal mole counts.
What Is Le Chatelier's Principle?
Le Chatelier's Principle is one of the most important conceptual topics in equilibrium chemistry. While there isn't much math involved, the reasoning can trip students up with what amounts to a chicken-and-egg scenario. Getting comfortable with this principle takes careful thought — but once it clicks, it becomes an incredibly powerful tool for predicting how a chemical equilibrium responds to change.
The principle states: if you place a stress on a system at chemical equilibrium, the system will respond in a way that counteracts that stress.
A stress is anything that takes a system out of equilibrium. Once displaced, the system will shift to re-establish equilibrium. This is the fundamental drive of every chemical reaction — it wants to reach equilibrium, it's always trying to get there, and once it arrives, it stays put.
Key Definition: A stress is any change that disrupts a system already at chemical equilibrium. The system then shifts — either toward products (right) or toward reactants (left) — to counteract that stress and restore equilibrium.
Throughout this lesson, we'll use the following reaction as our running example:
2C(s) + O₂(g) ⇌ 2 CO(g) ΔH < 0 (exothermic)
The equilibrium expression for this reaction is:
K = [CO]² / [O₂]
How Do Q and K Predict Equilibrium Shift Direction?
Before diving into specific stresses, we need to define the reaction quotient, represented by the letter Q. The reaction quotient equals the ratio of product concentrations to reactant concentrations — the same mathematical form as the equilibrium constant K. For our reaction:
Q = [CO]² / [O₂]
You might be thinking, "That looks identical to K." It is the same ratio — but there's a critical distinction. For this ratio to equal K, the concentrations must be measured at equilibrium. The ratio always equals Q regardless of whether the system has reached equilibrium. When the system is at equilibrium, Q = K. When it's not, Q ≠ K.
Using Q vs. K to Predict Shift Direction
Comparing Q and K tells you exactly which direction a reaction needs to shift to reach chemical equilibrium:
| Condition | What It Means | Shift Direction |
|---|---|---|
| Q < K | Too few products relative to reactants | Shift right (toward products) |
| Q > K | Too many products relative to reactants | Shift left (toward reactants) |
| Q = K | System is at equilibrium | No shift |
Example: Q < K
When Q is less than K, the ratio of products to reactants is too small. The system must convert reactants into products — a "shift to the right." As reactants are consumed (denominator decreases) and products form (numerator increases), Q grows until it equals K, and equilibrium is restored.
Example: Q > K
When Q is greater than K, the ratio is too large. The system must convert some products back into reactants — a "shift to the left." The numerator decreases, the denominator increases, and Q falls until it equals K.
How Does Concentration Affect Chemical Equilibrium?
Concentration changes are the most straightforward type of stress. There are two equivalent ways to analyze them: comparing Q to K, or applying Le Chatelier's Principle conceptually. Both approaches always lead to the same conclusion.
Increasing a Reactant: Adding O₂
Consider what happens when we increase the concentration of O₂ in our system at equilibrium. First, is this a stress? Absolutely — adding O₂ changes the ratio [CO]² / [O₂], which means the system is no longer at chemical equilibrium.
Since O₂ appears in the denominator, increasing its concentration makes the overall ratio smaller. The ratio was equal to K a moment ago, but now Q < K. Because Q is less than K, the equilibrium must shift to the right to restore balance.
Le Chatelier's Reasoning: The stress is that we now have too much O₂. The system has two options — shift right (which uses up O₂) or shift left (which makes even more O₂). Shifting right counteracts the stress, so that's the direction the equilibrium shifts.
Decreasing a Product: Removing CO
Now consider decreasing the concentration of CO. Does it change the ratio? Yes — CO appears in the numerator, so lowering its concentration lowers Q. Once again, Q < K, and the equilibrium shifts to the right.
Le Chatelier's Reasoning: The stress is that we don't have enough CO to be at equilibrium. Shifting right produces more CO, which counteracts the deficit. Shifting left would consume even more CO — the opposite of what's needed.
General Pattern: Adding a substance shifts the equilibrium away from that substance. Removing a substance shifts the equilibrium toward that substance. The system always works to counteract the concentration change you imposed.
3 Ways to Analyze Concentration Changes at Equilibrium
There's a third way to analyze concentration effects — through the lens of kinetics. Recall that at dynamic equilibrium, the forward rate equals the reverse rate. Rate laws tell us that reaction rates typically depend on the concentrations of reactants.
- The forward rate depends on [O₂] — the reactant of the forward reaction.
- The reverse rate depends on [CO] — the reactant of the reverse reaction.
Example: Adding O₂ (Kinetics View)
When we add O₂, we increase the concentration of the forward reaction's reactant, which speeds up the forward rate. For that instant, the reverse rate remains unchanged. Since the forward rate now exceeds the reverse rate, the net reaction proceeds to the right — exactly what Le Chatelier's Principle predicts.
Example: Removing CO (Kinetics View)
When we remove CO, we decrease the concentration of the reverse reaction's reactant, which slows down the reverse rate. The forward rate remains unchanged. Since the forward rate now exceeds the reverse rate, the net reaction again proceeds to the right.
3 Roads, 1 Destination: Whether you analyze a concentration change using Q vs. K, the conceptual definition of Le Chatelier's Principle, or kinetics, you should always arrive at the same conclusion. If you get different answers, re-check your reasoning.
Exception: Adding or Removing a Solid (or Liquid)
Here's an important exception that catches many students off guard. Recall that solids and liquids do not appear in equilibrium constant expressions. Since K = [CO]² / [O₂], the solid carbon does not factor into this ratio at all.
If you add more solid carbon to the container, the ratio doesn't change. If the ratio doesn't change, you're still at chemical equilibrium, Q still equals K, and there is no shift — not left, not right, nothing. Le Chatelier's Principle would not classify this as a stress because the system was never taken out of equilibrium.
Important Distinction: Solids can be affected by an equilibrium shift, but they cannot cause one. When the equilibrium shifts right (due to some other stress), it consumes solid carbon — you end up with less of it. But adding or removing the solid itself doesn't trigger a shift.
How Does Temperature Affect Chemical Equilibrium?
Temperature is the trickiest stress to analyze correctly. To determine which direction a temperature change shifts the equilibrium, you must know whether the reaction is endothermic or exothermic.
Our reaction has ΔH < 0, making it exothermic — it releases heat. Since heat is produced, we treat heat as a product:
2C(s) + O₂(g) ⇌ 2 CO(g) + heat
Determining Which Side Heat Goes On:
- Exothermic (ΔH < 0): Heat is a product — write it on the right side.
- Endothermic (ΔH > 0): Heat is a reactant — write it on the left side.
Raising the Temperature (Adding Heat)
Raising the temperature is equivalent to adding heat. Since our reaction is exothermic, heat is a product. Adding heat therefore means adding a product — and the system responds by shifting to the left to consume that excess heat.
Think of it this way: if we had added CO (another product), the equilibrium would shift left to use up the excess CO. Adding heat when heat is a product works exactly the same way.
Example: Lowering the Temperature
If instead we lowered the temperature, we would be removing heat. Removing a product means the system is deficient — it needs to produce more. The equilibrium would shift to the right to generate additional heat.
Why Temperature Is the Only Factor That Changes K
Here's what makes temperature truly unique among all the stresses: temperature is the only factor that changes the value of the equilibrium constant K itself.
When we added O₂ or removed CO earlier, those concentration changes altered Q while K remained constant. The system then shifted until Q returned to the original K value. But with temperature, we don't change Q at all — we change K. The system must then shift until Q equals the new equilibrium constant.
| Temperature Change | Equilibrium Shift | Effect on K |
|---|---|---|
| Raise temperature (exothermic rxn) | Shift left | K decreases |
| Lower temperature (exothermic rxn) | Shift right | K increases |
| Raise temperature (endothermic rxn) | Shift right | K increases |
| Lower temperature (endothermic rxn) | Shift left | K decreases |
Quick Rule: Shifting left decreases K. Shifting right increases K. All other stresses — concentration and pressure — change Q but never change K.
5-Step Checklist for Temperature Problems:
- Determine whether the reaction is endothermic or exothermic (sign of ΔH).
- Write heat on the correct side of the equation.
- Determine whether the temperature is increasing (adding heat) or decreasing (removing heat).
- Apply Le Chatelier's Principle to predict the shift direction.
- If asked, determine whether K increases or decreases based on the shift direction.
How Do Pressure and Volume Affect Equilibrium?
Pressure effects are best understood through the relationship between pressure and volume. The most common way to increase pressure is by decreasing the volume of the container — and these two changes go hand in hand.
How Changing Volume Affects Molarity and Q
You might notice that pressure doesn't appear directly in our equilibrium expression K = [CO]² / [O₂]. However, volume is embedded in every molarity term, since molarity = moles / volume (in liters). When you decrease the volume, all molarities increase.
Here's where it gets interesting. Suppose you cut the volume in half, which doubles all molarities. In the numerator, [CO] is squared — so doubling [CO] quadruples the numerator (2² = 4). In the denominator, [O₂] is first-power — so doubling [O₂] merely doubles the denominator. The numerator increased faster than the denominator, so Q increased.
Since Q > K, the equilibrium must shift to the left to bring Q back down to K.
The Shortcut: Shift Toward Fewer Moles of Gas
Rather than working through the math each time, there's a much faster approach: when volume decreases, the equilibrium shifts toward the side with fewer moles of gas.
Gases occupy vastly more volume than solids or liquids — roughly 1,000 to 10,000 times more per molecule under normal conditions. So when the container shrinks, the system relieves the pressure by shifting toward fewer gas molecules.
Example: Our Reaction
2C(s) + O₂(g) ⇌ 2 CO(g)
The reactant side has 1 mole of gas (O₂ — ignore the solid carbon). The product side has 2 moles of gas (2 CO). Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas — the left.
Tip: Correlate the shift with volume, not pressure, because volume and moles of gas move in the same direction. Less volume shifts the equilibrium toward fewer moles of gas. More volume shifts toward more moles of gas.
What If Both Sides Have Equal Moles of Gas?
If the total moles of gas are the same on both sides of the equation, changing the pressure or volume produces no shift. Both the numerator and denominator increase proportionally, Q remains equal to K, and the system stays at chemical equilibrium.
Exception: Adding an Inert Gas at Constant Volume
Here's a classic exam trap. Suppose you add argon gas to a rigid container holding our reaction mixture. The overall pressure increases — but does the equilibrium shift?
No. There is no shift.
The critical distinction is how the pressure increased. In the previous section, pressure increased because the volume decreased — which changed all molarities. Here, the pressure increases because we added more gas particles, but the container volume hasn't changed. Since the volume is the same and the moles of each reacting species are the same, all molarities remain unchanged. The ratio [CO]² / [O₂] is unaffected, Q still equals K, and the system remains at chemical equilibrium.
Common Mistake: Don't confuse these 2 scenarios:
- Increase pressure by decreasing volume — molarities change — likely a shift.
- Increase pressure by adding an inert gas at constant volume — molarities unchanged — no shift.
The inert gas (typically a noble gas like argon, or sometimes N₂) doesn't participate in the reaction, and at constant volume it doesn't alter the concentrations of any species that do.
Practice Problem: Predicting Equilibrium Shifts
Indicate the resulting shift in equilibrium (if any) for each of the following changes to the system at equilibrium.
2C(s) + O2(g) ⇌ 2CO(g) ΔH° = −221 kJ/mol
| LE CHATELIER'S PRINCIPLE | |
|---|---|
| Change | Shift |
| Increase [O2] | |
| Decrease [CO] | |
| Add C(s) | |
| Increase Temperature | |
| Increase Pressure (decrease volume) | |
| Add Ar(g) (at constant volume) | |
8 Key Takeaways: Le Chatelier's Principle
- Le Chatelier's Principle states that a system at chemical equilibrium will shift to counteract any stress that displaces it from equilibrium.
- The reaction quotient Q uses the same ratio as K but applies whether or not the system is at equilibrium. When Q < K, shift right; when Q > K, shift left; when Q = K, the system is at equilibrium.
- Concentration changes alter Q: adding a substance shifts equilibrium away from it, while removing a substance shifts equilibrium toward it.
- Solids and liquids do not appear in equilibrium expressions, so adding or removing them causes no shift.
- Temperature is the only stress that changes K itself. Write heat on the product side for exothermic reactions (ΔH < 0) and the reactant side for endothermic reactions (ΔH > 0), then treat heat like any other species.
- A shift to the left decreases K; a shift to the right increases K — but only temperature can cause these changes to K.
- Decreasing volume (increasing pressure) shifts equilibrium toward the side with fewer moles of gas. If both sides have equal moles of gas, there is no shift.
- Adding an inert gas at constant volume increases total pressure but does not change molarities, so there is no shift.