Equilibrium · Lesson 6 of 14
Changing an Equilibrium
“Explain how a system responds to added material, compression, heating, or a catalyst.”
• Use concentration changes and Q to predict a new equilibrium direction. • Explain pressure and inert-gas effects under specified constraints. • Distinguish temperature changes in K from other composition changes. • Explain how industrial processes balance equilibrium yield with reaction speed.
A disturbance changes what must balance
An equilibrium mixture can respond when its surroundings or composition change. Le Chatelier’s principle summarises this response: the net change tends to reduce the imposed disturbance. It is a qualitative guide, not a claim that the original state is restored completely. The quotient test provides a more precise way to justify many predictions.
When an equilibrium is disturbed, the system tends to change in a direction that counteracts the imposed change, subject to the constraints on the system.
Adding a reactant at fixed volume generally lowers the corresponding product-to-reactant quotient and promotes its consumption. Removing a product also promotes forward change. K remains the same at fixed temperature. After a reactant is added, some of it is consumed, but its new equilibrium amount need not fall all the way back to its old value. A shift is a net reaction during readjustment, not a permanent difference between the final opposing rates.
Colour changes make concentration effects visible
Mixing dilute iron(III) and thiocyanate solutions forms a deep-red complex, . At equilibrium its colour intensity becomes steady. Adding thiocyanate increases the red colour because more complex forms. Removing free iron(III) or thiocyanate encourages the complex to dissociate and its red colour fades.
Oxalate can bind iron(III), and mercury(II) can bind thiocyanate, removing free reactant ions from this equilibrium. These observations illustrate competing equilibria: a reagent can change the free concentration even when the element remains somewhere in solution. They are teacher-controlled demonstrations; the mercury reagent is not needed to reproduce the idea in independent practice. In industry, removing ammonia by condensation or removing CO2 from a lime kiln similarly promotes further product formation.
Problem
An equilibrium mixture for has Kc = 57. If hydrogen concentration doubles instantaneously while the other concentrations initially stay fixed, predict the response.
- 1.Before addition, Q equals K, or 57.
- 2.Doubling the hydrogen denominator halves Q to 28.5.
- 3.Since Q is now below K, the net reaction forms more HI. At the new equilibrium, the quotient returns to 57, not to a new K.
Compression depends on gas stoichiometry
Compressing an ideal gas mixture at constant temperature raises all its partial pressures and concentrations initially. If the two sides have different numbers of gaseous molecules, the quotient changes. The subsequent net reaction favours the side with fewer gas moles. Expansion favours the side with more gas moles. Pure solid and liquid coefficients do not enter this gas count.
Problem
Consider with all species gaseous. What happens when an equilibrium mixture is compressed to half its volume at constant temperature?
- 1.Each gas concentration initially doubles. The numerator contains two total powers and the denominator four.
- 2.Therefore .
- 3.Q is below K, so methane and water form overall. This reduces the total gas amount relative to the instant after compression, although the final pressure need not return to its original value.
For , compression promotes the reverse direction because one gaseous mole is favoured over two. If the gas-mole change is zero, uniform compression leaves the ideal-gas quotient unchanged. Adding an inert gas at constant volume raises total pressure but leaves reacting gases’ partial pressures unchanged, so their equilibrium is not shifted. The constant-volume condition is essential; an expansion performed to maintain constant total pressure is a different disturbance.
Temperature changes the equilibrium constant
Heating does more than change the current quotient: it changes K itself. For an exothermic forward reaction, higher temperature favours the reverse, heat-absorbing direction and K decreases. For an endothermic forward reaction, higher temperature favours forward change and K increases. Temperature can also alter the reaction rates, so its effect on speed must be considered separately.
| Disturbance | Idealised equilibrium response | Does K change at fixed written reaction? |
|---|---|---|
| Add reactant or remove product | Usually forward | No, if temperature is fixed |
| Compress a gas mixture | Towards fewer gaseous moles | No, if temperature is fixed |
| Add inert gas at fixed volume | No shift of reacting ideal gases | No |
| Heat an exothermic forward reaction | Reverse | Yes: K decreases |
| Heat an endothermic forward reaction | Forward | Yes: K increases |
| Add a catalyst | Equilibrium reached sooner | No |
For , the forward reaction is exothermic. Brown NO2 gives way to less-coloured N2O4 on cooling. In a demonstration, sealed samples placed in cold, room-temperature and warm baths show progressively different brown intensities. The gases and preparation reagents require appropriate laboratory controls; the useful evidence is the reversible colour response to temperature.
Another example is the pink hexaaquacobalt(II) complex exchanging ligands with chloride to form blue tetrachlorocobaltate(II): . The blue direction is endothermic under the described conditions, so warming favours blue and cooling favours pink. Both complexes contain cobalt in oxidation state +2.
A catalyst changes the route, not the destination
A catalyst provides a pathway with lower barriers and accelerates approach to equilibrium in both directions. It does not change the reaction’s overall standard Gibbs energy, K or equilibrium composition. A catalyst cannot make an equilibrium with negligible useful product become product-favoured. It can make a favourable but slow conversion practically accessible.
Problem
Ammonia synthesis is exothermic and changes four gas moles into two. Why does an industrial process use high pressure, a catalyst and a moderate elevated temperature?
- 1.Higher pressure favours the two-mole product side and increases the equilibrium ammonia yield.
- 2.Lower temperature favours the exothermic forward reaction, but very low temperature gives an impractically slow rate.
- 3.An iron-based catalyst makes useful rates possible at a compromise temperature. The chapter quotes conditions around 500 degrees Celsius and 200 atmospheres as illustrative operating conditions, not universal fixed settings.
- 4.Removing ammonia from the recycled mixture further encourages conversion. In the contact process, a catalyst similarly accelerates favourable but slow SO2 oxidation without changing its equilibrium constant.
A catalyst does not shift equilibrium, and increasing total pressure does not always shift a reaction. Identify how pressure changed, whether the volume changed, and which gases take part.
Quiz
Adding product at fixed temperature usually causes a net reaction that
Compressing N2 + 3H2 forming 2NH3 at fixed temperature favours
Adding argon at fixed volume to an ideal-gas equilibrium changes
Heating an exothermic forward reaction generally makes K
A catalyst added to an equilibrium mixture changes
Cooling the exothermic NO2 dimerisation equilibrium makes it
Practice Problems
- Predict the effects of adding hydrogen and removing methanol for 2H2(g) + CO(g) forming CH3OH(g).
- Explain why continuous CO2 removal favours limestone decomposition.
- Predict compression effects for PCl5 decomposition, CaO + CO2 forming CaCO3, and 3Fe + 4H2O(g) forming Fe3O4 + 4H2(g).
- For endothermic methane–steam conversion to CO and hydrogen, discuss heating, compression and a catalyst separately.
- Use the iron–thiocyanate equilibrium to explain the effect of binding free Fe3+ with oxalate.
- Explain why a catalyst can improve production per hour without raising the equilibrium yield at the same temperature and pressure.
Key Takeaways
A composition disturbance usually changes Q; temperature generally changes K. Pressure effects depend on gas stoichiometry and the way the pressure was changed. Catalysts improve rates, while industrial operating conditions also balance yield, energy and practical constraints.