E. Chemical Systems and Equilibrium · Chemical EquilibriumLe Châtelier's Principle I: Changing Concentrations
E3.3 E2.2 E2.1About 70 minutes
Learning goals- I can state Le Châtelier's principle.
- I can predict how adding or removing a reactant or product shifts an equilibrium.
- I can explain each shift using Q and K.
- I can interpret concentration-time graphs that show a change in concentration.
Words to know
Le Châtelier's principle
When a system at equilibrium is disturbed by a change in conditions, it shifts in the direction that partly counteracts the change, until a new equilibrium is reached.
stress
Any change that disturbs an equilibrium: a change in concentration, pressure or volume (for gases), or temperature.
shift
The net forward (right) or reverse (left) reaction that happens after a stress, until equilibrium is restored.
complex ion
An ion made of a central metal ion bonded to surrounding molecules or ions, such as FeSCN²⁺ or [CoCl₄]²⁻; many are brightly coloured.
Push it, and it pushes back
Try to push a full beach ball under the water at a lake and it pushes back up. Chemical equilibria behave in a similar way. In 1884 the French chemist Henri Le Châtelier summed up what happens when you disturb a system at equilibrium: the system shifts to partly undo the disturbance. This idea, Le Châtelier's principle, lets chemists predict the effect of a change without doing any calculations, and it is how industry pushes reactions toward the products it wants.
Le Châtelier's principle
When a system at equilibrium is subjected to a change (a stress), it responds by shifting in the direction that partly counteracts the change, and a new equilibrium is established.
The stresses you will study are changes in concentration (this lesson) and changes in pressure/volume and temperature (next lesson).
Changing a concentration
- Add a reactant: the system uses some of it up → shifts right.
- Remove a reactant: the system makes more of it → shifts left.
- Add a product: shifts left.
- Remove a product: shifts right.
"Partly counteracts" matters: after you add H₂, the new equilibrium still has more H₂ than before, just less than right after you added it.
Why it works: Q and K
Concentration changes do not change K (only temperature does). They change Q.
- Adding a reactant makes the denominator bigger, so Q < K → shift right until Q = K again.
- Adding a product makes the numerator bigger, so Q > K → shift left.
Le Châtelier's principle is a quick way to predict; Q is the reason it works and the way to calculate the new concentrations.
Worked exampleExample 1: Adding hydrogen to the HI equilibrium
An equilibrium mixture contains [H₂] = [I₂] = 0.110 mol/L and [HI] = 0.780 mol/L (K = 50.3). Enough H₂ is added to raise [H₂] to 0.210 mol/L instantly. (a) Predict the shift. (b) Show it with Q. (c) Find the new equilibrium concentrations.
- (a) A reactant was added, so the system shifts right to use some of it.
- (b) Q = (0.780)² ÷ [(0.210)(0.110)] = 0.6084 ÷ 0.02310 = 26.3 < K = 50.3, so it shifts right. ✓
- (c) ICE: I: 0.210, 0.110, 0.780. C: −x, −x, +2x. E: 0.210 − x, 0.110 − x, 0.780 + 2x.
- (0.780 + 2x)² = 50.3(0.210 − x)(0.110 − x). Expanding gives 46.3x² − 19.22x + 0.5535 = 0, so x = 0.0311 (the other root, 0.384, would make [I₂] negative).
- [H₂] = 0.179 mol/L, [I₂] = 0.079 mol/L, [HI] = 0.842 mol/L.
Shift right. New equilibrium: [H₂] = 0.179, [I₂] = 0.079, [HI] = 0.842 mol/L. Notice that [H₂] (0.179) is higher than the original 0.110 but lower than the 0.210 just after adding it: the change was only partly counteracted. Check: (0.842)² ÷ [(0.179)(0.079)] ≈ 50 ✓.
Concentration-time graph for Example 1. When H₂ is added, its line jumps up. Then H₂ and I₂ decrease and HI increases until a new equilibrium is reached. Every concentration changes, but K stays the same.Reading concentration-time graphs
- A sudden vertical jump in one line means that substance was added (up) or removed (down) at that moment.
- The other lines change smoothly afterwards as the system shifts.
- The direction of those smooth changes tells you the shift: if products rise and reactants fall, it shifted right.
- The changes after the stress are in the ratio of the coefficients (here H₂ and I₂ each drop by x while HI rises by 2x).
Colour equilibria you can see
Two classic lab equilibria show concentration shifts as colour changes:
Iron(III) thiocyanate: Fe³⁺(aq) (pale yellow) + SCN⁻(aq) (colourless) ⇌ FeSCN²⁺(aq) (blood red)
- Adding Fe³⁺ (as iron(III) nitrate solution) or SCN⁻ (as potassium thiocyanate solution) → shifts right → darker red.
- Adding sodium hydrogen phosphate removes Fe³⁺ (it binds the iron ions) → shifts left → paler.
Chromate and dichromate: 2 CrO₄²⁻(aq) (yellow) + 2 H⁺(aq) ⇌ Cr₂O₇²⁻(aq) (orange) + H₂O(l)
- Adding acid (H⁺) → shifts right → orange.
- Adding a base, which removes H⁺ → shifts left → yellow.
Notice that the ions that do not take part (such as K⁺, Na⁺ or NO₃⁻ from the added solutions) do not appear in the equilibrium, so they do not cause a shift. Safety: these are teacher-supervised lab chemicals (chromates are toxic and must never be handled at home).
Worked exampleExample 2: Predicting several stresses
For CH₄(g) + H₂O(g) ⇌ CO(g) + 3 H₂(g) at constant temperature and volume, predict the shift and the effect on [H₂] for: (a) adding CH₄; (b) removing CO; (c) adding H₂; (d) adding H₂O.
- (a) Reactant added → shift right → [H₂] increases.
- (b) Product removed → shift right → [H₂] increases.
- (c) Product added → shift left → [H₂] ends higher than before the addition, but some of the added H₂ is used up.
- (d) Reactant added → shift right → [H₂] increases.
(a), (b), (d): shift right, more H₂. (c): shift left. This reaction (steam reforming of natural gas) is a major source of the hydrogen used to make ammonia.
Solids and pure liquids
Adding or removing a pure solid or pure liquid does not shift an equilibrium, because they are not in the K or Q expression. Adding more CaCO₃(s) to CaCO₃(s) ⇌ CaO(s) + CO₂(g) changes nothing (as long as some of each solid is present).
Just for funLe Châtelier's principle, summed up as a chemist's motto: "Whatever you do, I'll partly undo."
Hands-on activityLe Châtelier in a bottle of pop
You need: an unopened bottle of carbonated water or pop, a clear glass, a grown-up's permission. Safety: open the bottle slowly over a sink.
Carbonated drinks contain the equilibrium CO₂(g) ⇌ CO₂(aq) (gas in the space at the top of the bottle and dissolved gas in the liquid).
- Look at the sealed bottle. Are bubbles forming? (Very few: the system is at equilibrium.)
- Open the cap slowly. What happens to the bubbles?
- Pour some into a glass and watch over 10 minutes. Then close the bottle again and leave it for 30 minutes before looking again.
- Explain each observation with Le Châtelier's principle.
What to look for: opening the bottle lets CO₂(g) escape. Removing CO₂(g) shifts CO₂(g) ⇌ CO₂(aq) to the left, so dissolved CO₂ comes out of solution as gas: lots of bubbles. In the open glass the gas keeps escaping (open system), so the drink goes flat. After re-capping, CO₂(g) collects at the top until a new equilibrium is reached and bubbling slows, but with less dissolved CO₂ than before. This is a concentration (and pressure) stress; the next lesson looks at pressure in more detail.
Your turn
1State Le Châtelier's principle in your own words.
If you disturb a system at equilibrium, it shifts in the direction that partly undoes the disturbance, then settles at a new equilibrium.
2For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), predict the shift when (a) N₂ is added; (b) NH₃ is removed; (c) H₂ is removed.
(a) right; (b) right; (c) left.
3For Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺, what colour change do you expect when KSCN solution is added? Explain.
Darker red. Adding SCN⁻ (a reactant) shifts the equilibrium right, making more red FeSCN²⁺. The K⁺ ion does not take part.
4For 2 CrO₄²⁻ + 2 H⁺ ⇌ Cr₂O₇²⁻ + H₂O, a yellow solution turns orange. What was probably added?
An acid (H⁺). Adding a reactant shifted the system right to orange dichromate.
5Explain, using Q, why adding NH₃ to N₂ + 3 H₂ ⇌ 2 NH₃ at equilibrium causes a shift left.
Adding NH₃ increases the numerator of Q = [NH₃]² ÷ ([N₂][H₂]³), so Q > K. The system shifts left until Q falls back to K.
6In Example 1, by how much did [I₂] change, and why did it change even though no I₂ was added?
[I₂] fell from 0.110 to 0.079 mol/L (by x = 0.031). The system shifted right to use up some of the added H₂, and the forward reaction uses I₂ too (1 : 1 ratio).
More practice: warm-up
1Does adding a reactant change K?
No. It changes Q; only temperature changes K.
2Adding a product makes Q larger or smaller than K?
3Which way does removing a product shift an equilibrium?
4Does adding more solid carbon to C(s) + CO₂(g) ⇌ 2 CO(g) shift the equilibrium?
No. A pure solid is not in the K or Q expression.
5Who stated the principle about equilibrium shifts in 1884?
Henri Le Châtelier, a French chemist.
More practice: on your own
1For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), describe the concentration-time graph after extra O₂ is injected at constant volume.
[O₂] jumps up suddenly, then decreases gradually. [SO₂] decreases gradually. [SO₃] increases gradually. The changes after the jump are in a 2 : 1 : 2 ratio (SO₂ falls twice as much as O₂; SO₃ rises by the same amount SO₂ falls). New flat lines show the new equilibrium, with [O₂] higher than originally.
2For CoCl₄²⁻(aq) (blue) + 6 H₂O(l) ⇌ Co(H₂O)₆²⁺(aq) (pink) + 4 Cl⁻(aq), predict the colour change when (a) concentrated HCl (a source of Cl⁻) is added; (b) silver nitrate is added (Ag⁺ removes Cl⁻ as solid AgCl).
(a) Cl⁻ is a product: shift left, toward blue CoCl₄²⁻. (b) Removing Cl⁻ (a product) shifts right, toward pink Co(H₂O)₆²⁺ (and a white AgCl precipitate forms).
3A student says, "Adding sodium nitrate solution to the iron(III) thiocyanate equilibrium will shift it." Is the student right? (Ignore the small dilution.)
No. Na⁺ and NO₃⁻ are not part of Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺, so they do not change Q; there is no shift (apart from a slight dilution effect).
4For PCl₅ ⇌ PCl₃ + Cl₂ (K = 0.0150), an equilibrium mixture has [PCl₅] = 0.0681, [PCl₃] = [Cl₂] = 0.0320 mol/L. Enough Cl₂ is added to make [Cl₂] = 0.100 mol/L. Calculate Q and predict the shift.
Q = (0.0320)(0.100) ÷ 0.0681 = 0.0470 > K = 0.0150: shifts left, forming more PCl₅ and using up some PCl₃ and Cl₂.
5Explain why a shift after adding a reactant is said to "partly" counteract the change, not fully.
If the system used up all the added reactant, every other concentration would return to its old value too, but then Q would not equal K (the shift also made more product). So the new equilibrium has more of the added reactant than before, and only some of the addition is used up.
Challenge
1Continue the last question: find the new equilibrium concentrations.
Shift left. E: [PCl₅] = 0.0681 + x, [PCl₃] = 0.0320 − x, [Cl₂] = 0.100 − x. (0.0320 − x)(0.100 − x) = 0.0150(0.0681 + x) → x² − 0.1320x + 0.003200 = 0.0010215 + 0.0150x → x² − 0.1470x + 0.0021785 = 0 → x = [0.1470 − √(0.021609 − 0.008714)] ÷ 2 = (0.1470 − 0.1136) ÷ 2 = 0.0167 (the other root, 0.130, would make [PCl₃] negative). [PCl₅] = 0.0848, [PCl₃] = 0.0153, [Cl₂] = 0.0833 mol/L. Check: (0.0153)(0.0833) ÷ 0.0848 = 0.0150 ✓.
2In blood, hemoglobin (Hb) carries oxygen: Hb + 4 O₂ ⇌ Hb(O₂)₄ (simplified). Use Le Châtelier's principle to explain why oxygen is picked up in the lungs and released in muscles.
In the lungs, [O₂] is high, so the equilibrium shifts right and hemoglobin loads oxygen. In active muscles, cells use O₂ quickly, so [O₂] is low; the equilibrium shifts left and Hb(O₂)₄ releases oxygen where it is needed.
3Diluting an aqueous equilibrium by adding water lowers every dissolved concentration. Predict the shift for Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺ when the solution is diluted to twice its volume, using Q.
Each concentration halves. Q = (½[FeSCN²⁺]) ÷ (½[Fe³⁺] × ½[SCN⁻]) = 2 × K. Q > K, so the system shifts left (toward the side with more dissolved particles). The solution becomes paler both from dilution and from the shift.
Quiz
1. For A + B ⇌ C at equilibrium, more A is added. The system:
2. Adding product to a system at equilibrium makes:
3. Which change will NOT shift CaCO₃(s) ⇌ CaO(s) + CO₂(g)?
4. A pale FeSCN²⁺ solution turns dark red after something is added. Which was added?
5. After a reactant is added and a new equilibrium is reached, its concentration is:
6. On a concentration-time graph, a vertical jump in one line followed by smooth changes in all lines shows:
Big idea (E3.3, E2.2): a system at equilibrium shifts to partly counteract a change. Adding a reactant or removing a product shifts right; adding a product or removing a reactant shifts left. Q explains why: concentration changes alter Q, not K.
How to help: for any change, ask "what does the system want to undo?" and "what happens to Q?". The pop-bottle activity is quick and safe.
Common difficulties:
- Thinking the added substance returns to its original concentration (it does not).
- Thinking concentration changes alter K.
- Shifting for solids, pure liquids or spectator ions.
- Reading concentration-time graphs: the jump is the stress; the smooth changes are the response.
Safety: lab equilibria such as chromate/dichromate and cobalt chloride use toxic chemicals and are done only in a school lab under supervision.