Chemical Equilibrium
Chemical Equilibrium
Definition: Chemical equilibrium is the state of a reversible reaction in which the forward and reverse reaction rates are equal, so the concentrations of reactants and products remain constant over time even though both reactions keep occurring.
How It Works
- Most reactions don’t go to completion in one direction; they’re reversible:
A + B ⇌ C + D- As reactants form products, the reverse reaction, products reforming reactants, speeds up as product concentration builds.
- Equilibrium is reached when these two rates become equal, a dynamic balance, not a static one.
- Molecules are still reacting in both directions constantly.
- There’s just no further net change in composition.
Le Chatelier’s Principle predicts how a system at equilibrium responds to a disturbance: it shifts in the direction that partially counteracts the change.
- Adding reactant shifts equilibrium toward products, consuming some of the added reactant.
- Removing product shifts equilibrium toward products, replacing what was removed.
- Increasing pressure on a gas-phase equilibrium shifts toward the side with fewer moles of gas.
- Increasing temperature shifts an endothermic reaction forward.
- Increasing temperature shifts an exothermic reaction backward, since heat behaves like a reactant or product respectively.
- Adding a catalyst does not shift equilibrium at all.
- A catalyst speeds up both directions equally, only changing how fast equilibrium is reached.
Under the Hood
The equilibrium constant expresses the ratio of product to reactant concentrations, each raised to its stoichiometric coefficient, for:
aA + bB ⇌ cC + dD
Kc = [C]^c[D]^d / [A]^a[B]^b- Pure solids and liquids are omitted; their “concentration,” density over molar mass, is constant and folded into K.
- Kp uses partial pressures instead of concentrations:
Kp = Kc(RT)^Δn- Δn = (moles gas products) - (moles gas reactants).
The reaction quotient Q has the identical form to K but uses whatever concentrations exist at a given moment, not necessarily equilibrium.
- Q < K: the system shifts forward, toward products.
- Q > K: the system shifts backward, toward reactants.
- Q = K: the system is already at equilibrium.
Worked example (ICE table).
- Given:
H2(g) + I2(g) ⇌ 2HI(g), Kc = 54.3 at 425°C- Start: 1.00 M H2, 1.00 M I2, no HI initially
| H2 | I2 | HI | |
|---|---|---|---|
| Initial | 1.00 | 1.00 | 0 |
| Change | -x | -x | +2x |
| Equilibrium | 1.00-x | 1.00-x | 2x |
- Step 1: set up the equilibrium expression
Kc = (2x)² / [(1.00-x)(1.00-x)] = 54.3- Step 2: simplify, since both reactants start equal
(2x/(1.00-x))² = 54.3
2x/(1.00-x) = √54.3 = 7.37- Step 3: solve for x
2x = 7.37(1.00-x) = 7.37 - 7.37x
9.37x = 7.37
x = 0.786- Step 4: find equilibrium concentrations
[H2] = [I2] = 1.00 - 0.786 = 0.214 M
[HI] = 2(0.786) = 1.572 M- Answer: [H2] = [I2] = 0.214 M, [HI] = 1.572 M
- Check:
(1.572)²/(0.214 × 0.214) = 2.471/0.0458 = 54.0- This matches Kc = 54.3 within small rounding error.
Why It Matters
- Understanding equilibrium lets engineers maximize product yield in industrial processes.
N2 + 3H2 ⇌ 2NH3 (exothermic, Δn < 0)- In the Haber process, industrial plants run at high pressure, favoring fewer gas moles, shifting toward NH3.
- Plants use a moderate rather than low temperature, trading some equilibrium yield for a practical reaction rate.
- The reaction would be too slow at the low temperature that would maximize equilibrium conversion.
- Biologically, equilibrium principles explain oxygen loading and unloading by hemoglobin.
- They also explain the body’s tight regulation of blood pH via the bicarbonate buffer.
Common Pitfalls
- Including solids or pure liquids in the K expression; their concentration is constant and already absorbed into the constant.
- Confusing Q and K: forgetting that Q uses current, possibly non-equilibrium, concentrations.
- Using Q to determine shift direction rather than treating it as another form of K.
- Assuming a change in concentration or pressure changes the value of K itself.
- Only temperature changes K; concentration and pressure changes shift the position of equilibrium, not the constant.
- Forgetting that a large K (K >> 1) means products are favored, while a small K (K << 1) means reactants dominate.
- Applying the “x is small” approximation in an ICE table when K is not small enough relative to initial concentrations.
- As in the worked example above, the approximation can fail and the quadratic must be solved properly.
- Forgetting that adding an inert gas at constant volume doesn’t shift equilibrium at all.
- An inert gas doesn’t change any reacting species’ partial pressure or concentration.
Comparison
| Constant | Based On | Used For |
|---|---|---|
| Kc | Molar concentrations | General equilibria in solution |
| Kp | Partial pressures | Gas-phase equilibria |
| Ka / Kb | [H+] or [OH-] products | Acid/base dissociation equilibria |
| Ksp | Ion concentrations | Solubility of sparingly soluble salts |
| Condition | Interpretation |
|---|---|
| K >> 1 | Products strongly favored at equilibrium |
| K ≈ 1 | Comparable amounts of reactants and products |
| K << 1 | Reactants strongly favored at equilibrium |
| Q < K | Reaction shifts forward, toward products |
| Q > K | Reaction shifts backward, toward reactants |
| Q = K | System is already at equilibrium |
Real-World Application
Industrial esterification, producing ethyl acetate for use as a solvent, is an equilibrium reaction:
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O- Kc is around 4 at room temperature.
- A plain mixture of acid and alcohol only converts to roughly two-thirds ester at equilibrium.
- Manufacturers push the yield higher by continuously removing water, often by distillation, as it forms.
- Removing water lowers Q below K and forces the equilibrium to keep shifting toward products.
- This is a direct, industrial-scale application of Le Chatelier’s Principle.
Example
In the Haber process, N2 and H2 reach equilibrium with ammonia:
N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH° = -92 kJ/molBecause the product side has fewer gas moles (2 versus 4), increasing pressure shifts equilibrium toward NH3, which is why industrial reactors run at 150-300 atm to push conversion higher despite the added engineering cost.
FAQ
Does equilibrium mean the reaction has stopped?
- No. Both forward and reverse reactions continue at the molecular level.
- Equilibrium just means their rates are equal, so there’s no further net change in observable concentrations.
Can you speed up how quickly equilibrium is reached without changing K?
- Yes, that’s exactly what a catalyst does.
- It accelerates both the forward and reverse reactions equally, reaching the same equilibrium position faster.
Why does removing product keep shifting equilibrium instead of the system just re-establishing the original ratio?
- Removing product drops Q below K instantaneously.
- The system responds by continuously producing more product to try to restore Q = K.
- If product keeps being removed, that shift can continue indefinitely, pulling the reaction toward far higher conversion than a closed system would ever reach.
Related Terms
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