What you'll be able to do: Predict the direction an equilibrium shifts when concentration, volume, pressure or temperature is changed, and say whether K changes.
Industrial chemists do not accept the yield a reaction happens to give. They change the conditions to push it. Le Chatelier's principle is the reasoning behind those choices.
These are recommended, not required. You can start this lesson at any time.
If a system at equilibrium is disturbed, it shifts in the direction that partially counteracts the disturbance. The shift restores Q = K when equilibrium is attainable, but does not generally restore every original concentration.
In a heterogeneous system, a remaining pure phase can buffer an activity until that phase is exhausted.
Adding a reactant makes Q too small, so the system shifts right to consume it. Removing a product also lowers Q and shifts the system right. This is why continuously removing a product, for example condensing out water, drives a reaction to high yield.
“Shift” describes the subsequent reaction relative to the moment just after the disturbance. Adding NH₃ to a Haber equilibrium raises [NH₃] immediately; some is then consumed in the reverse reaction. The final amount is not generally lower than it was before the addition. Likewise a volume decrease raises all gas concentrations immediately, even for a species later consumed by the shift. Separate the instantaneous physical change from the slower chemical response.
Compressing a gaseous system raises every partial pressure, and the system responds by shifting toward whichever side has fewer moles of gas. Expanding does the opposite. If both sides have equal moles of gas, changing the volume causes no shift at all.
For N₂O₄ ⇌ 2 NO₂, Qp = PNO₂²/PN₂O₄. Halving volume at fixed T initially doubles each partial pressure, so Qnew = (2²/2)Qold = 2 K. Since Q > K, the reaction shifts toward N₂O₄, the side with fewer gas moles. More generally, multiplying all gas partial pressures by f multiplies Q by fΔn. If Δn = 0, the factors cancel. The fewer-moles rule concerns a pressure change caused by compression, not any increase in total pressure from any cause.
Compression shifts toward fewer moles of gas
Adding argon at constant volume raises the total pressure but leaves every partial pressure and every concentration unchanged. Q is untouched, so nothing shifts. This trips up more students than any other case.
Write heat as a product for an exothermic reaction and as a reactant for an endothermic one, then treat a temperature change like adding or removing that term. Heating an exothermic reaction shifts it left and lowers K; cooling it shifts right and raises K. This is the only stress that changes the value of K.
For an exothermic forward reaction, raising temperature lowers K; the mixture's old composition then has Q greater than the new K and reverse conversion is favored. Heat written beside an equation is a memory aid, not a chemical species to include as [heat] in K. A catalyst can accelerate both directions while leaving K unchanged. If temperature is changed while also compressing a vessel, consider both effects rather than attributing everything to pressure.
A + B ⇌ C + heat (exothermic)
N₂ + 3 H₂ ⇌ 2 NH₃ is exothermic with four moles of gas becoming two. High pressure and low temperature both favor ammonia, but low temperature makes the reaction unbearably slow, so industry compromises at about 450 C with a catalyst and very high pressure.
At constant volume and temperature, adding an inert gas changes total pressure but not the reacting gases' partial pressures, so their Q remains unchanged. At constant total pressure, adding inert gas expands the container and dilutes each reacting gas; equilibrium can shift toward the side with more gas particles. K changes only with temperature.
Adding more of an already present pure solid or pure liquid does not directly change its activity or shift equilibrium; removing the entire phase can change what equilibrium is possible. At fixed T and V, adding inert gas leaves reacting partial pressures unchanged. At fixed T and total P, adding inert gas expands the gas mixture and can favor the side with more gas moles. State the fixed conditions explicitly. In industrial systems, product removal and continuous feed also require material-flow accounting beyond a single sealed-flask shift.
A system at equilibrium shifts to partially relieve an applied stress.
Changing a concentration that enters Q can drive a shift; adding more of an existing pure solid or liquid does not directly change its activity.
Reducing the volume shifts the equilibrium toward the side with fewer moles of gas.
No partial pressure changes, so there is no shift.
The only stress that changes the value of K.
reactants ⇌ products + heat
reactants + heat ⇌ products
For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), ΔH = -92 kJ. Predict the effect of (a) adding H₂, (b) decreasing the volume, (c) raising the temperature.
Try it first: For each change, decide whether Q rises, falls, or stays the same.
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Argon is added to the equilibrium N₂O₄(g) ⇌ 2 NO₂(g) at constant volume. What happens?
Try it first: Ask whether any partial pressure in the Q expression changed.
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Cooling a flask of N₂O₄/NO₂ turns the brown color pale. Is the forward reaction N₂O₄ → 2 NO₂ endothermic or exothermic?
Try it first: Decide which species is brown and which way the system moved.
0 of 3 steps revealed.
For N₂O₄(g) ⇌ 2 NO₂(g), compare adding Ar at fixed volume versus fixed total pressure at constant T.
Try it first: Count one gas mole on the left and two on the right.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Saying an equilibrium shift changes the value of K. | Only temperature changes K; every other stress moves the system to a new position with the same K. | Ask whether the temperature changed. |
| Predicting a shift when an inert gas is added at constant volume. | No partial pressure in the expression changes, so Q is unchanged. | Test whether Q moved. |
| Thinking adding a solid to a heterogeneous equilibrium shifts it. | Pure solids do not appear in the expression. | Check whether the species is in the K expression at all. |
| Claiming a catalyst increases equilibrium yield. | It speeds both directions equally. | A catalyst changes only the time taken. |
| Applying the moles-of-gas rule while counting aqueous or solid species. | Volume changes act on gases. | Count only species labelled (g). |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
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Le Chatelier's principle states that a system at equilibrium responds to a stress by shifting in the direction that partly relieves it. Adding a species shifts the system away from that species and removing it shifts the system toward it. Compressing a gaseous system shifts it toward the side with fewer moles of gas, while adding an inert gas at constant volume changes nothing because no partial pressure changes. Temperature is the one stress that changes K itself: treat heat as a product for an exothermic reaction and as a reactant for an endothermic one. A catalyst changes neither K nor the position of equilibrium.
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