E. Chemical Systems and Equilibrium · Chemical EquilibriumLe Châtelier's Principle II: Pressure, Volume, Temperature and Catalysts
E3.3 E2.2 E2.1 A1.12About 75 minutes
Learning goals- I can predict how a change in volume or pressure shifts a gas equilibrium, using moles of gas.
- I can explain why adding an inert gas at constant volume causes no shift.
- I can predict the effect of temperature by treating heat as a reactant or product, and explain why K changes.
- I can explain why a catalyst does not shift an equilibrium.
Words to know
inert gas
A gas that does not take part in the reaction, such as argon or helium in most systems.
exothermic
A reaction that releases heat to the surroundings (ΔH negative); heat can be written as a product.
endothermic
A reaction that absorbs heat from the surroundings (ΔH positive); heat can be written as a reactant.
catalyst
A substance that speeds up a reaction by providing a path with lower activation energy, without being used up; it speeds up the forward and reverse reactions equally.
thermochemical equation
A balanced equation that includes the enthalpy change, either as ΔH beside it or as a heat term in the equation.
Squeeze it, heat it, speed it up
In a factory you cannot easily add or remove one gas from a sealed reactor, but you can turn up the pressure, change the temperature or add a catalyst. Chemical engineers use these three "knobs" every day. Each one affects an equilibrium differently, and only one of them changes the value of K.
Changing volume (and pressure) of a gas system
Squeezing a gas mixture into a smaller volume raises its pressure. The system responds by reducing the pressure: it shifts toward the side with fewer moles of gas, because fewer gas particles exert less pressure.
- Decrease volume (increase pressure) → shift toward fewer moles of gas.
- Increase volume (decrease pressure) → shift toward more moles of gas.
- If both sides have the same number of moles of gas (Δn = 0), volume changes cause no shift.
- Count only gases. Solids, liquids and dissolved species do not count.
For N₂O₄(g) ⇌ 2 NO₂(g): 1 mol of gas on the left, 2 on the right. Compressing the mixture shifts it left, toward colourless N₂O₄.
Worked exampleExample 1: Halving the volume, with Q
An equilibrium mixture at 25 °C has [N₂O₄] = 0.0898 mol/L and [NO₂] = 0.0203 mol/L (K = 4.59 × 10⁻³). The volume is suddenly halved. (a) Use Q to predict the shift. (b) The new equilibrium has [N₂O₄] = 0.1853 mol/L. Find [NO₂] and describe what you would see.
- (a) Halving the volume doubles every concentration: [N₂O₄] = 0.1796, [NO₂] = 0.0406 mol/L.
- Q = (0.0406)² ÷ 0.1796 = 9.18 × 10⁻³, which is twice K. Q > K, so it shifts left (fewer moles of gas). ✓
- (b) N₂O₄ increased by 0.1853 − 0.1796 = 0.0057 mol/L, so NO₂ decreased by 2 × 0.0057 = 0.0114 mol/L.
- [NO₂] = 0.0406 − 0.0114 = 0.0292 mol/L. Check: (0.0292)² ÷ 0.1853 = 4.6 × 10⁻³ ✓.
Shift left; [NO₂] = 0.0292 mol/L. You would see the colour darken immediately (the brown gas is squeezed into half the space) and then fade a little as some NO₂ turns into N₂O₄. It stays darker than before (0.0292 > 0.0203): the change is only partly undone.
Adding an inert gas
If you pump argon into a rigid (constant-volume) reactor, the total pressure rises, but the concentrations and partial pressures of the reacting gases do not change. So Q still equals K: no shift.
(If the inert gas is added at constant total pressure, the container must expand. That lowers every reacting concentration, which is like increasing the volume: the system shifts toward more moles of gas.)
Changing temperature: the one that changes K
Temperature is different from every other stress, because it changes the value of K. To predict the direction, write heat into the equation as if it were a substance:
- Endothermic forward reaction (ΔH > 0): heat is a reactant. N₂O₄(g) + 57.2 kJ ⇌ 2 NO₂(g)
- Exothermic forward reaction (ΔH < 0): heat is a product. N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) + 92 kJ
Then apply Le Châtelier: adding heat (raising T) shifts away from the heat term; removing heat (lowering T) shifts toward it.
- Endothermic forward: raise T → shift right, K increases; lower T → shift left, K decreases.
- Exothermic forward: raise T → shift left, K decreases; lower T → shift right, K increases.
Because K itself changes, the new equilibrium has a genuinely different ratio of products to reactants.
Worked exampleExample 2: Predicting temperature effects
For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), ΔH = −198 kJ. Predict the shift and the change in K when (a) the temperature is raised; (b) the temperature is lowered.
- The forward reaction is exothermic: 2 SO₂ + O₂ ⇌ 2 SO₃ + 198 kJ. Heat is a product.
- (a) Raising T adds a "product": shift left. Less SO₃ at equilibrium, so K decreases.
- (b) Lowering T removes a "product": shift right. More SO₃, so K increases.
(a) shift left, K decreases; (b) shift right, K increases. Yet industry runs this reaction hot. Why? See the Industrial Equilibria lesson: low temperature gives a high K but a slow rate.
Catalysts: faster, not farther
A catalyst lowers the activation energy of both the forward and reverse reactions by the same amount, so it speeds both up equally. The result:
- equilibrium is reached sooner;
- the equilibrium concentrations are unchanged: no shift;
- K is unchanged.
Catalysts are still hugely important in industry, because getting to equilibrium quickly at a lower temperature saves time and energy (you study activation energy in Energy Changes and Rates of Reaction).
| Stress | Shift | Does K change? |
|---|
| add reactant / remove product | right | no |
| add product / remove reactant | left | no |
| decrease volume (increase pressure) | toward fewer moles of gas | no |
| increase volume (decrease pressure) | toward more moles of gas | no |
| add inert gas at constant volume | no shift | no |
| raise temperature | in the endothermic direction | yes |
| lower temperature | in the exothermic direction | yes |
| add a catalyst | no shift (equilibrium is reached faster) | no |
Just for funA catalyst at a chemical equilibrium is like a moving walkway at an airport that runs both ways at the same speed: everyone gets where they are going faster, but the number of people on each side ends up the same.
Hands-on activityA pressure and temperature test with pop bottles
You need: two unopened plastic bottles of carbonated water (one from the fridge, one at room temperature), a sink, a grown-up's permission.
- Look at the bubbles in an unopened bottle of carbonated water: very few. Squeeze the sealed bottle firmly. Do bubbles form or disappear?
- Open the cap briefly to release pressure, then recap. Watch the bubbles.
- Explain both observations using the equilibrium CO₂(g) ⇌ CO₂(aq): which side has more moles of gas?
- Think about temperature: dissolving CO₂ in water is exothermic. Predict whether the warm or the cold bottle fizzes more when opened, then test it by opening the second bottle over the sink and comparing.
What to look for: squeezing raises the pressure, which favours the side with fewer moles of gas (CO₂(aq)), so few new bubbles form. Releasing the pressure shifts the system toward CO₂(g): bubbles. Heat is a product of dissolving, so warming shifts it back toward CO₂(g): the warm bottle fizzes more. Safety: open bottles slowly, pointed away from faces, over a sink. Do not heat a sealed bottle.
Your turn
1For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), which way does the system shift when the volume is decreased? Explain.
Right. The left side has 3 mol of gas and the right side has 2; a smaller volume favours fewer moles of gas.
2For H₂(g) + I₂(g) ⇌ 2 HI(g), predict the effect of doubling the pressure by halving the volume.
No shift: both sides have 2 mol of gas. (All concentrations double, and the purple colour gets darker because I₂ is more concentrated, but Q stays equal to K.)
3For CaCO₃(s) ⇌ CaO(s) + CO₂(g), predict the effect of decreasing the volume.
Shift left (0 mol of gas on the left, 1 on the right). Some CO₂ reacts with CaO to form CaCO₃.
4For N₂(g) + O₂(g) ⇌ 2 NO(g), ΔH = +181 kJ, predict the effect of raising the temperature on [NO] and on K.
The forward reaction is endothermic (heat is a reactant). Raising T shifts right: [NO] increases and K increases. This is why NO forms in hot engines and lightning.
5Argon is added to an equilibrium mixture of N₂, H₂ and NH₃ in a rigid steel tank. What happens?
No shift. The partial pressures (and concentrations) of N₂, H₂ and NH₃ do not change, so Q = K.
6A catalyst is added to a system at equilibrium. What happens to the amounts of products and to K?
Nothing: no shift and no change in K. A catalyst only helps the system reach equilibrium faster.
More practice: warm-up
1Which stress changes the value of K?
2An increase in pressure (by decreasing volume) favours the side with more or fewer moles of gas?
3In an exothermic reaction, is heat written as a reactant or a product?
4Does a catalyst change where equilibrium lies?
No; it only speeds up reaching equilibrium.
5For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), which side has more moles of gas?
The right side (2 mol versus 1 mol).
More practice: on your own
1For CO(g) + 2 H₂(g) ⇌ CH₃OH(g), ΔH = −91 kJ, what combination of temperature and pressure gives the largest equilibrium amount of methanol?
Low temperature (exothermic, so cooling shifts right and increases K) and high pressure (3 mol of gas on the left, 1 on the right).
2For C(s) + H₂O(g) ⇌ CO(g) + H₂(g), ΔH = +131 kJ, predict the shift for: (a) adding C(s); (b) raising T; (c) decreasing the volume; (d) removing H₂.
(a) no shift (solid); (b) right (endothermic forward), K increases; (c) left (1 mol gas left versus 2 right); (d) right.
3A sealed tube of brown NO₂/N₂O₄ is moved from a hot-water bath into ice water. Describe the colour change and what happens to K.
The colour becomes paler. Cooling favours the exothermic direction (2 NO₂ → N₂O₄), so the system shifts left and K decreases.
4Explain, using Q, why decreasing the volume of N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) shifts it right.
Halving the volume doubles each concentration. Q = (2[NH₃])² ÷ [(2[N₂])(2[H₂])³] = 4 ÷ 16 × (old Q) = ¼K. Q < K, so it shifts right.
5Why is it incorrect to say "raising the temperature always increases the amount of product"?
It depends on the sign of ΔH. Raising T favours the endothermic direction: products for endothermic reactions, reactants for exothermic reactions.
Challenge
1For a gas reaction, show that when the volume is multiplied by a factor f (every concentration multiplied by 1/f), Q becomes K × f^(−Δn). Use it to predict the shift when the volume of 2 SO₂ + O₂ ⇌ 2 SO₃ is halved.
Each concentration term [X]^c becomes ([X]/f)^c, so Q = K × (1/f)^(sum of product coefficients − sum of reactant coefficients) = K × f^(−Δn). For SO₃: Δn = 2 − 3 = −1, f = ½: Q = K × (½)¹ = ½K < K, so it shifts right (fewer moles of gas).
2For N₂O₄ ⇌ 2 NO₂ at 25 °C, K = 4.6 × 10⁻³. At 100 °C, would you expect K to be larger or smaller? Explain, and describe how an experiment with sealed tubes could test it.
Larger. The forward reaction is endothermic, so higher T shifts it right and increases K. Put identical sealed tubes in a 25 °C water bath and a boiling-water bath (about 100 °C); the hotter tube should be clearly darker brown. A colorimeter could measure [NO₂] and allow K to be calculated at each temperature.
3Industrial methanol is made from CO and H₂ (ΔH = −91 kJ) at about 250 °C with a copper-based catalyst. Explain why engineers do not run it at room temperature, even though K is larger there.
At room temperature the reaction is far too slow to be useful, even with a catalyst. A moderate temperature gives a reasonable rate; high pressure and the catalyst help make up for the smaller K. This trade-off between yield and rate is studied in the lesson on industrial equilibria.
Quiz
1. For 2 NO(g) + O₂(g) ⇌ 2 NO₂(g), decreasing the volume shifts the equilibrium:
2. For H₂(g) + Cl₂(g) ⇌ 2 HCl(g), increasing the pressure by compression will:
3. For an exothermic reaction, raising the temperature will:
4. Adding helium to a rigid container holding an equilibrium gas mixture:
5. What does a catalyst do to a reaction at equilibrium?
6. A sealed tube of NO₂/N₂O₄ (forward reaction endothermic) is placed in ice water. The colour:
7. Which change increases the equilibrium amount of NH₃ in N₂ + 3 H₂ ⇌ 2 NH₃ + 92 kJ?
Big idea (E3.3, E2.2): compressing a gas mixture shifts it toward fewer moles of gas; an inert gas at constant volume and a catalyst cause no shift; temperature shifts the system toward the endothermic direction when heated and changes K.
How to help: have your teen rewrite each equation with "heat" as a reactant or product before predicting temperature effects, and count moles of gas only before predicting pressure effects. The pop-bottle activity models both.
Common difficulties:
- Counting solids or liquids when predicting pressure effects.
- Thinking a catalyst increases yield.
- Forgetting that only temperature changes K.
- Thinking "higher temperature always means more product" (it depends on ΔH).
Safety: never heat a sealed container. The hot/cold NO₂ tubes are a teacher demonstration (NO₂ is toxic).