7.1 Chemical equilibria and dynamic equilibrium

A reversible reaction can form products from reactants and also regenerate reactants from products. It is represented with a double arrow when both directions are relevant.

The forward and reverse reactions may have different rates. In a closed system, the composition can settle at a dynamic equilibrium where both continue.

N₂O₄(g) ⇌ 2NO₂(g) allows colourless N₂O₄ to form brown NO₂ and brown NO₂ to recombine. The observed colour depends on the equilibrium composition.

Reversible does not mean the reaction stops or that the amounts become equal. It means both directions occur under the conditions.

When a system at equilibrium is disturbed, it shifts in the direction that opposes the disturbance, partially restoring the balance.

For concentration changes, the system consumes an added reactant or replaces a removed product. For pressure changes in gases, it favours the side with fewer gas molecules; temperature must be treated as heat in the equation.

For N₂ + 3H₂ ⇌ 2NH₃, adding H₂ shifts right, while increasing pressure also shifts right because four gas moles become two.

A catalyst changes how quickly equilibrium is reached, not the equilibrium position. “Opposes” does not mean the disturbance is completely cancelled.

A change in concentration, pressure or temperature disturbs an equilibrium. The system responds in the direction that reduces the effect of that change.

Write the balanced equation first. For gases, compare total gas moles; for temperature, treat heat as a reactant in an endothermic direction or a product in an exothermic direction.

For 2SO₂ + O₂ ⇌ 2SO₃, adding oxygen shifts right and increasing pressure shifts right because three gas moles become two.

The equilibrium shift is not a complete cancellation and a catalyst does not change the final composition. It only speeds both directions.

For aA + bB ⇌ cC + dD, Kc = [C]^c[D]^d / ([A]^a[B]^b), using equilibrium concentrations in mol dm⁻³.

Products appear in the numerator, reactants in the denominator, and equation coefficients become powers. Pure solids and liquids are omitted because their effective concentration is constant.

For N₂ + 3H₂ ⇌ 2NH₃, Kc = [NH₃]²/([N₂][H₂]³). Do not use initial concentrations unless the question states they are equilibrium values.

Kc is not the same as a simple product-minus-reactant ratio. Coefficients become indices, and the expression depends on the balanced equation.

Mole fraction is the moles of one gas divided by total moles. For an ideal gas mixture, partial pressure pᵢ = xᵢP_total, where xᵢ is the mole fraction.

The partial pressures add to the total pressure. Use mole fraction rather than mass fraction because gas behaviour in the ideal model is tied to particle amount.

A mixture with 2 mol N₂ and 1 mol O₂ has x(N₂)=2/3. At total pressure 300 kPa, p(N₂)=200 kPa.

Partial pressure is not the pressure of a separated sample at the same volume. It is the component’s contribution within the mixture.

For gaseous reactions, Kp is written like Kc but uses equilibrium partial pressures. Coefficients in the balanced equation become powers.

Omit solids and liquids, and use a consistent pressure convention. The numerical value of Kp depends on the chosen pressure units unless the standard-state convention is fixed.

For N₂(g)+3H₂(g)⇌2NH₃(g), Kp = p(NH₃)²/(p(N₂)p(H₂)³).

Do not insert total pressure for every species. Each p value is that gas’s equilibrium partial pressure, and the reaction equation controls the powers.

To calculate Kc or Kp, first write the correct expression, then substitute equilibrium concentrations or partial pressures with the required powers and units.

Keep brackets grouped, evaluate powers before division, and check whether omitted phases are pure solids or liquids. A large or small value describes composition, not reaction speed.

If Kc=[C]²/([A][B]) and [A]=0.20, [B]=0.10, [C]=0.30 mol dm⁻³, then Kc=0.09/(0.02)=4.5.

Do not use concentrations at the start of the reaction unless they are explicitly equilibrium values. K is defined at equilibrium.

At equilibrium, the concentrations or partial pressures of all species are related by the balanced equation and the value of K. They are not required to be equal.

Use an ICE-style change table or stoichiometric ratios to express unknown equilibrium amounts in terms of one variable, then substitute into K and solve.

For A ⇌ 2B, if A decreases by x, B increases by 2x. The equilibrium values are [A]₀−x and [B]₀+2x, not independent guesses.

Dynamic equilibrium means forward and reverse rates are equal, not concentrations. The composition depends on K and the starting conditions.

For a given reaction, changing temperature changes the equilibrium constant because it changes the energy balance. Changing concentration or pressure shifts the equilibrium position but leaves K unchanged at that temperature.

Use Le Chatelier for the direction of a shift. For temperature, treat heat as a reactant or product and identify whether the forward reaction is endothermic or exothermic.

For N₂ + 3H₂ ⇌ 2NH₃ + heat, raising temperature shifts left and lowers K; adding nitrogen shifts right but does not change K.

A catalyst changes the rates of both directions, not K or the final equilibrium composition. Pressure changes K only indirectly if temperature also changes.

The Haber process makes ammonia: N₂(g)+3H₂(g)⇌2NH₃(g), an exothermic equilibrium with fewer gas moles on the product side.

High pressure favours ammonia, lower temperature favours yield but slows the reaction, and an iron catalyst increases rate without changing equilibrium position. Industry chooses a compromise condition and recycles unreacted gases.

Raising pressure improves equilibrium yield but increases equipment cost. A moderate temperature with an iron catalyst gives an economically useful throughput rather than the maximum possible yield.

The catalyst does not move equilibrium right. “Compromise temperature” means balancing rate and yield, not ignoring the exothermic nature of the reaction.