R2.3 - Equilibrium

1. Distinguish between reversible and irreversible reactions, giving one example of each.

A reversible reaction is a chemical change in which products can react to reform the original reactants; it is represented by a double arrow (⇌). In such systems both the forward and the reverse processes occur and, under suitable conditions, the two opposing rates can become equal so that the macroscopic composition no longer changes.

By contrast, an irreversible reaction proceeds essentially to completion under the given conditions: the forward reaction dominates and the reverse reaction is negligible. The difference is important because only reversible reactions can establish a dynamic equilibrium and be treated with equilibrium laws.

Example (reversible): the synthesis/decomposition of ammonia under closed conditions, N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g). Example (irreversible): combustion of methane in air (CH₄ + 2 O₂ → CO₂ + 2 H₂O) under normal conditions — products do not revert to reactants.

Memorise: Reversible ⇌ both directions occur; irreversible → proceeds to completion.

2. Define equilibrium in a chemical system and explain why it is important in industrial and biological processes.

Definition (WHAT): A chemical equilibrium is the state in a closed system in which the rate of the forward reaction equals the rate of the reverse reaction, so that the concentrations of reactants and products remain constant in time.

Why (WHY): This dynamic balance determines the composition of the reaction mixture and therefore the yield of desired products; controlling the position of equilibrium is central to maximizing industrial production (for example the Haber process to produce ammonia) and to maintaining steady conditions in biological systems (for example oxygen–haemoglobin binding curves).

How (HOW): By changing temperature, pressure or concentrations (Le Châtelier’s principle), chemists can shift the position of equilibrium to favour more product or more reactant, and engineers then choose the compromise of conditions that optimise both yield and rate.

Consequences / Example: In industry, choosing higher pressure and a moderate temperature in the Haber process increases ammonia yield while keeping reaction rates acceptable — a direct application of equilibrium control. In biology, metabolic pathways rely on equilibria to maintain homeostasis; small perturbations are often corrected by shifts of equilibria.

Memorise: Equilibrium: forward rate = reverse rate; composition constant but reactions continue.

3. Explain what is meant by dynamic equilibrium. How does it differ from static equilibrium?

Definition (WHAT): Dynamic equilibrium refers to the state in a closed chemical system where forward and reverse reactions continue to occur microscopically but at equal rates, so macroscopic properties (concentration, colour, density) remain constant.

Difference (WHY / HOW): The key point is ongoing microscopic activity — molecules are constantly reacting, but there is no net change in amounts. This differs from a static equilibrium (a mechanical or physical balance without microscopic change) because static equilibrium implies no motion or reaction at microscopic level. In chemistry, the equilibrium concept used in exam answers always means dynamic equilibrium.

Consequence / Example: For example, in a sealed container of ethanol the rate of evaporation equals the rate of condensation; the vapour pressure becomes constant even though molecules continually cross phases.

Memorise: Dynamic = microscopic change continues; static = no microscopic change.

4. Describe the conditions required for equilibrium to be established in terms of closed, open, and isolated systems.

WHAT: Equilibrium requires that the system is closed with respect to the species involved: no net loss or gain of reactant/product molecules to the surroundings. An open system allows matter exchange and so cannot necessarily sustain the concentrations required for equilibrium when volatile components escape; an isolated system exchanges neither matter nor energy with the surroundings.
WHY / HOW: For gas-phase equilibria, if a product gas escapes (open system) the reverse reaction cannot proceed, so equilibrium cannot be established. In contrast, solution equilibria can sometimes reach a steady state in a beaker open to air if none of the reactants or products are lost as gas. An isolated system will eventually reach equilibrium because neither energy nor matter is exchanged, but isolated systems are idealizations.
Consequence / Example: For the decomposition of CaCO₃(s) ⇌ CaO(s) + CO₂(g), equilibrium only exists if CO₂ is retained (closed vessel); if the system is open CO₂ escapes and the reaction proceeds to completion.

Memorise: Closed = equilibrium possible; open = matter loss prevents gas equilibria; isolated = no exchange (ideal).

1. State the equilibrium law and write the general expression for the equilibrium constant for the reaction.

WHAT (statement): The equilibrium law states that for a reversible reaction at equilibrium and fixed temperature, the ratio of the product of the concentrations of products (each raised to the power of its stoichiometric coefficient) to the corresponding product of reactant concentrations is constant. That constant is the equilibrium constant .
HOW (expression):

where square brackets denote equilibrium concentrations in mol dm⁻³ and the equation applies to the balanced chemical equation as written. Solids and pure liquids are omitted from the expression.

Memorise: K = (products)^(stoich) ÷ (reactants)^(stoich); omit pure solids & liquids.

2. Explain why the equilibrium constant is dimensionless and discuss what “units” seen in practice indicate.

WHAT: Formally the equilibrium constant is written as a dimensionless quantity because the concentrations used are ratios relative to a standard concentration (1 mol dm⁻³) or because activity (dimensionless) is the rigorous quantity in thermodynamics.
WHY / HOW: In practice, students may see K expressed with units if concentrations are substituted directly; these “units” are an artefact of treating molar concentrations numerically rather than as dimensionless activities. Thermodynamically, K is dimensionless and any apparent units cancel when activities (ratios) are used. For exam answers state that K is dimensionless, and if asked about units explain that apparent units come from the concentration values used numerically and are not fundamental.

Memorise: K is dimensionless — apparent units come from plugging in concentrations rather than using activities.

3. Calculate Kc for the esterification of ethanoic acid and ethanol, given equilibrium concentrations.

WHAT: To calculate use the equilibrium concentrations and substitute into the equilibrium expression.
HOW (worked steps you can reproduce):

1. Write the balanced equation: CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l).

2. Write the equilibrium expression:

(For this reaction all species are liquids; if treated by concentrations the expression still applies.)
3. Substitute the supplied equilibrium concentrations (or convert moles ÷ volume to concentrations) and evaluate.
Example (from your notes): Using the worked example where [CH₃COOH] = 0.470, [C₂H₅OH] = 0.070, [CH₃COOC₂H₅] = 0.364, [H₂O] = 0.364 (mol dm⁻³) gives

Show 3 s.f. as required.

Memorise: Always write K expression, convert to concentrations (moles ÷ volume if needed), substitute, calculate.

4. Explain what the magnitude of the equilibrium constant indicates about the relative concentrations of reactants and products.

WHAT: The numerical value of describes the relative amounts of products and reactants at equilibrium for the given balanced reaction at a fixed temperature.
WHY / HOW:

• If , the numerator (products) is much larger — equilibrium lies far to the right and products are predominant.

• If but not extremely large, substantial amounts of both exist but products are favoured.

• If , significant amounts of both are present; the reaction does not strongly favour either side.

• If , equilibrium lies to the left and reactants are predominant; the reaction hardly proceeds.
  CONSEQUENCE / EXAMPLE: This interpretation allows prediction of extent of reaction but not its rate — e.g. a large K for acid dissociation implies a strong acid (extent of dissociation large). Remember that K is temperature dependent: change the temperature and K may change.

Memorise: K≫1 → products favoured; K≈1 → neither favoured; K≪1 → reactants favoured.

5. Deduce the relationship between K for a reaction and for its reverse.

WHAT / HOW: If the reaction

has equilibrium constant as written, then the reverse reaction has equilibrium constant . This follows directly because reversing the fraction in the expression inverts its numerical value.

Memorise: K(reverse) = 1 / K(forward).

1. State Le Châtelier’s principle and explain how it can be used to predict changes in equilibrium position.

WHAT (statement): Le Châtelier’s principle states that if a stress (change in concentration, pressure, temperature, or other condition) is applied to a system at dynamic equilibrium, the system will respond in the direction that tends to minimise the effect of that change.
WHY / HOW: Use the principle qualitatively: if concentration of a reactant is increased the system shifts to consume some of that reactant (towards products); if pressure is increased for gas-phase reactions the system shifts to the side with fewer moles of gas; if temperature is increased the system shifts in the endothermic direction to absorb heat. This qualitative prediction helps choose conditions that favour product formation in industry.
Consequence / Example: For the Haber process N₂ + 3 H₂ ⇌ 2 NH₃ (ΔH negative, exothermic), increasing pressure favours ammonia (fewer gas moles) and lowering temperature favours product thermodynamically — but kinetics demands a compromise temperature.

Memorise: Apply change → predict shift that opposes change.

2. Explain the effect of increasing concentration of a reactant on both the position of equilibrium and the value of .

WHAT / HOW: Increasing the concentration of a reactant disturbs the equilibrium; to reduce the disturbance the system shifts to the right (toward products) so that some reactant is consumed and more product is formed.
Effect on K (WHY): Importantly, if only concentrations or partial pressures are changed and the temperature remains constant, the numerical value of the equilibrium constant does not change. depends only on temperature. The system adjusts concentrations so that the equilibrium expression returns to the same .

Memorise: Add reactant → equilibrium shifts to remove it; K unchanged (T fixed).

3. Describe the effect of increasing pressure on the position of equilibrium for reactions involving gases. Illustrate with the Haber process.

WHAT / HOW: For gaseous equilibria, increasing total pressure (by reducing volume) shifts the equilibrium toward the side with fewer moles of gas, because that shift reduces the pressure change. Conversely a decrease in pressure shifts the equilibrium toward the side with more gas moles.
Example (Haber): N₂(g)+3 H₂(g) ⇌ 2 NH₃(g). There are 4 moles of gas on the left and 2 on the right; increasing pressure therefore shifts the position of equilibrium toward ammonia formation (the right), increasing yield. Industrially, high pressures are used to exploit this, with the caveat that high pressures are costly and safety-limited.

Memorise: Increase pressure → shift to side with fewer gas moles (Haber → more NH₃).

4. (Table) — interpret pressure changes:

Model answer (short):

• Increase pressure: shifts to side with fewer gas molecules; unchanged at constant temperature.

• Decrease pressure: shifts to side with more gas molecules; unchanged.

5. Describe and explain the effect of increasing temperature on both (a) the position of equilibrium, and (b) the value of , for (i) exothermic and (ii) endothermic reactions.

WHAT / HOW: Temperature acts like addition or removal of heat. By Le Châtelier’s principle:

• Exothermic reaction (ΔH < 0): heat is a product. Increasing temperature adds heat → equilibrium shifts left (toward reactants) to absorb heat; decreasing temperature shifts right. Effect on K: increasing temperature decreases for exothermic reactions (numerator smaller at new eqm).

• Endothermic reaction (ΔH > 0): heat is a reactant. Increasing temperature shifts equilibrium right (toward products) to absorb heat; decreasing temperature shifts left. Effect on K: increasing temperature increases for endothermic reactions.

Consequence / Example: For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g) (exothermic), raising temperature reduces SO₃ yield and reduces K; for 2 HI ⇌ H₂ + I₂ (endothermic in the forward direction), increasing temperature raises K and shifts to products.

Memorise: Raise T: endothermic favoured (K↑); exothermic opposed (K↓).

6. Explain why catalysts do not change the position of equilibrium or the value of , but do affect the time taken to reach equilibrium.

WHAT / WHY / HOW: A catalyst provides an alternative reaction pathway with a lower activation energy, increasing the rate of both forward and reverse reactions by the same factor. Because it increases both rates equally, the system reaches the point where rates are equal faster, but the ratio of forward to reverse rates at equilibrium (and therefore ) is unchanged. Thus catalysts only affect kinetics (time to equilibrium), not thermodynamics (equilibrium composition).

Memorise: Catalyst: speeds both directions equally → equilibrium reached sooner but K unchanged.

1. Describe what is meant by a heterogeneous equilibrium. Why are the concentrations of solids and pure liquids omitted from the equilibrium constant expression?

WHAT: A heterogeneous equilibrium is one in which reactants and products are present in different phases (e.g. gas + solid, liquid + gas).
WHY / HOW: The equilibrium expression uses activities; for pure solids and liquids the activity is effectively constant (≈1) at a given temperature and so their concentration does not affect the position of equilibrium. Therefore these phases are omitted from the K expression to reflect that only the concentrations (or partial pressures/activities) of the variable species determine equilibrium. Consequence: For example in CaCO₃(s) ⇌ CaO(s) + CO₂(g), only the partial pressure of CO₂ appears in the equilibrium expression.

Memorise: Solids/pure liquids omitted because their activity is constant (≈1).

2. Define the reaction quotient and explain how its value predicts the direction of change towards equilibrium.

WHAT: The reaction quotient is calculated from the same expression as but using the instantaneous (not necessarily equilibrium) concentrations or partial pressures of reactants and products.
WHY / HOW: Compare with :

• If the system is at equilibrium.

• If the forward reaction is favoured and the system proceeds to the right to make more products.

• If the reverse reaction is favoured and the system proceeds to the left to make more reactants.
  This comparison allows prediction of the spontaneous direction of net change.

Memorise: Q < K → shift right; Q > K → shift left; Q = K → equilibrium.

1. Explain the purpose of an ICE table and outline the steps to construct one.

WHAT: An ICE (Initial–Change–Equilibrium) table organises known initial amounts, the algebraic change (in terms of a variable ) due to reaction stoichiometry, and resulting equilibrium amounts. It is the standard method to find unknown equilibrium concentrations when only partial data are given.
HOW (stepwise recipe to memorise):

1. Write the balanced equation and place initial concentrations (or moles) in the I row.

2. Assign change variables using stoichiometric ratios (e.g. if 2A ⇌ B + C then change for A = −2x, B = +x, C = +x).

3. Write equilibrium amounts as I + change (e.g. [A] = [A]₀ − 2x).

4. Substitute equilibrium expressions into the expression and solve for .

5. Check any approximations (e.g. that x ≪ [initial]) and compute final concentrations and answer.

Memorise: ICE: write Initial, assign Change (use stoich), form Equilibrium, substitute into K.

2. Calculate using an ICE table for a reaction with known initial concentrations and equilibrium yield. (Follow the worked example approach.)

1. Given: initial moles/volumes or concentrations. Convert moles → concentration if volume given (concentration = moles ÷ volume).

2. ICE table: put initial concentrations (I), express changes in terms of (C), write equilibrium concentrations (E).

3. Substitute E values into the expression for and solve the resulting equation for . If algebra gives a quadratic/cubic, solve appropriately; if K ≪ 1 an approximation simplifies the algebra.

4. Once is found, compute all equilibrium concentrations and evaluate , reporting to the correct number of significant figures.
   Example: the hydrolysis of ethyl ethanoate worked through in the notes gives using ICE and concentration conversion — follow steps exactly as on page 24–26 of your notes.

Memorise: Convert → ICE → substitute → solve → check approx → report.

3. Explain why, when is very small, the assumption simplifies calculations.

WHAT / WHY: If the reaction hardly proceeds to the right and the change from initial concentration to equilibrium concentration is negligibly small. Therefore in the ICE table the term can be approximated as , removing the variable from the denominator and simplifying algebra (usually to a solvable polynomial in of low order or directly to ). You must state and justify this approximation in an exam (e.g. show ).

Memorise: If K < 10⁻³, change ≈ 0 → [initial] ≈ [equilibrium] (state & check).

4. Solve a problem involving a very small K where approximation is required.

• Template: set up ICE, assume [R]eq ≈ [R]initial where K small, solve for x using the simplified equation, check % error.

• Short example (from notes): For 2 NH₃ ⇌ 3 H₂ + N₂ with K = 3.85 × 10⁻³ and initial [NH₃] = 0.20 mol dm⁻³, define change −2x etc., use approximation 0.20 − 2x ≈ 0.20, substitute into K expression, solve for x and hence [H₂]eq = 3x. Calculation gives [H₂] ≈ 3.6 × 10⁻⁴ mol dm⁻³ (see worked example). Be sure to check that 2x ≪ 0.20 to justify.

Memorise: ICE + small-K approximation → algebra → check percent change < 5%.

1. Deduce the relationship between Gibbs free energy change and the equilibrium constant .

WHAT (equation): The standard Gibbs free energy change is related to the equilibrium constant by:

where is the gas constant (8.314 J K⁻¹ mol⁻¹) and the absolute temperature in kelvin. This fundamental relationship links thermodynamics and equilibrium.
WHY / CONSEQUENCE: If then and is negative — the formation of products is thermodynamically favoured at standard conditions. Conversely implies and reactants are favoured. If , and neither side is favoured.

Memorise: ΔG° = −RT ln K; ΔG° negative → K>1 → products favoured.

2. Explain how changes in affect Gibbs free energy.

WHAT / HOW: Because , an increase in (more product-favouring at the given T) makes larger and thus more negative, indicating greater thermodynamic driving force for product formation. Conversely a smaller K gives a more positive . Temperature changes that alter K also change accordingly.

Memorise: ↑K → ΔG° becomes more negative (products more favoured).

3. Calculate for a reaction given at a specified temperature, and vice versa.

Model answer (procedure):

• To calculate from :

  1. Rearrange: .

  2. Compute then exponentiate: .

• To compute from : use .
  Notes: Units: convert to joules per mole if using in J K⁻¹ mol⁻¹. Check temperature in kelvin. Example: your notes calculate ΔG° = −4.38 kJ mol⁻¹ → K ≈ 5.86 at 298 K using the formula (worked example in notes).

Memorise: Use ΔG° = −RT ln K; convert kJ→J and °C→K before substituting.

Dynamic equilibrium — macroscopic & microscopic properties; why colour/density unchanged

At dynamic equilibrium microscopic events continue: molecules on the reactant side form products and molecules on the product side reform reactants at equal rates. Because the rates of forward and reverse processes are equal the macroscopic properties — such as concentration, colour and density — remain constant even though molecular change continues. Thus the absence of observable macroscopic change does not mean the reactions have stopped; rather the ongoing processes are balanced.

Closed vs open systems (gases vs solution example)

Equilibrium for gas-phase reactions requires a closed system because escaping gas removes species and prevents the reverse process; an open system therefore typically cannot establish gas equilibria. By contrast, some solution equilibria can be effectively treated as closed if none of the solutes volatilize; for example the esterification of ethanol and ethanoic acid carried out in a reflux apparatus can reach equilibrium even though the reaction mixture is accessible to the lab atmosphere (the volatile components are retained). If CO₂ produced by a reaction escapes (open vessel) the reaction will be driven to completion and no equilibrium is established.

Magnitude of K: K≫1, K≪1, K = 1 — relation to extent of reaction

If the numerator of the equilibrium expression is much larger than the denominator, so at equilibrium product concentrations are much greater than reactants and the reaction lies far to the right (large extent). If reactants predominate and the reaction hardly proceeds. If comparable amounts of reactants and products are present and neither side is strongly favoured. These magnitudes indicate extent (how far reaction proceeds) but give no information on rate.

Haber process: K expression & reverse relation

For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) the equilibrium constant (K) is

If the reaction is written in the reverse direction, the equilibrium constant becomes , because numerator and denominator are exchanged. This reciprocal relationship follows directly from the algebraic inversion of the equilibrium expression.

Effect of temperature on K (one exothermic, one endothermic)

For an exothermic equilibrium (heat as product), increasing temperature supplies heat and the equilibrium shifts left; the equilibrium constant decreases. For an endothermic equilibrium (heat as reactant), increasing temperature favours the forward (product) side and increases. Thus temperature changes alter both the position of equilibrium and the numerical value of . Use ΔH sign to decide direction and remember is temperature-dependent.

Catalysts — why affect rate but not K or position

A catalyst lowers activation energies and increases the rates of both forward and reverse reactions equally; therefore the system attains the balance of forward and reverse rates sooner but the equilibrium composition and the constant — which depend only on thermodynamic quantities — remain unchanged.

Heterogeneous equilibria — fizzy drink bottle example

In a sealed fizzy drink bottle an equilibrium exists between dissolved CO₂ and gaseous CO₂: CO₂(g) ⇌ CO₂(aq). Opening the bottle allows some gaseous CO₂ to escape (a stress) and by Le Châtelier’s principle the equilibrium shifts left, causing dissolved CO₂ to come out of solution as bubbles until a new equilibrium (at lower partial pressure of CO₂) is reached.

Q vs K — distinction and direction prediction (Q>K example)

The reaction quotient is the instantaneous ratio of product and reactant concentrations (same form as K) evaluated at any moment. If there are relatively too many products and the system will shift left to form reactants; if the system shifts right. Thus predicts the immediate direction of net change until equilibrium is restored.

ICE tables for esterification — why useful

ICE tables are used for esterification because often only initial amounts and one equilibrium amount are given. By expressing changes in terms of a variable and using stoichiometry, the unknown equilibrium concentrations can be determined algebraically and substituted into the expression to calculate the equilibrium constant. This systematic method avoids algebraic errors and makes approximations easy to justify.

Approximation when — justification & advantage

When the forward reaction proceeds only to a very small extent and the change in reactant concentration is negligible compared with its initial value. Therefore the approximation simplifies algebra and reduces the need to solve a quadratic; always check the assumption (2x/[initial] < 5%) after solving.

Equilibrium vs kinetics — same K different rates example

The equilibrium constant reflects only the final balance of concentrations and provides no information about how fast that balance is achieved. Two reactions can have identical at the same temperature but very different rate constants (activation energies). For example, a thermodynamically favourable reaction can be kinetically slow (large activation energy) and require a catalyst to proceed at a useful rate; remains unchanged by the catalyst.

ΔG and K relationship — sign interpretation

From ΔG° = −RT ln K: if ΔG° < 0 then K > 1 and products are favoured at standard conditions; if ΔG° > 0 then K < 1 and reactants are favoured; if ΔG° = 0 then K = 1 and the system is balanced. Thus the sign and magnitude of ΔG° directly indicate the thermodynamic favourability and extent of reaction.

Free energy & equilibrium calculations (worked outline)

Model answer (compact procedure you can memorise):

1. Convert ΔG° to J mol⁻¹ (if necessary) and T to K.

2. Use ΔG° = −RT ln K to find ln K; exponentiate to obtain K.

3. For the reverse, use K = e^(−ΔG°/RT). For the forward compute ΔG° = −RT ln K. (Remember units and significant figures.) Example in your notes: ΔG° = −4.38 kJ mol⁻¹ at 298 K gives K ≈ 5.86.