Chemical Equilibria, Le Chatelier’S Principle & Kc

## Explain dynamic equilibrium in a closed system - A reversible reaction can proceed in both directions. At dynamic equilibrium the forward and reverse reactions continue at equal rates, so there is no net composition change under fixed conditions. - A closed system exchanges no matter with its surroundings, although heat can pass through its boundary. A sealed liquid–vapour system provides a physical analogy: evaporation and condensation continue. - Constant amounts alone do not prove equilibrium; a reaction might be stopped or an open flow might maintain a steady state. Use evidence for ongoing opposite processes and fixed conditions. ## Interpret forward and reverse rate curves - For the supplied A ⇌ B model, F is the forward species rate and R the reverse species rate. Starting with A only gives a nonzero forward rate and initially zero reverse rate. - As B forms, the reverse rate rises and the forward rate falls towards their common nonzero limiting value. Equilibrium involves continued turnover, not two zero rates. - The exact exponential curves belong to this first-order model. Their asymptotic approach does not imply that every real reaction follows the same time course or reaches an exact plateau at a finite plotted time. ## Distinguish constant concentrations from equal concentrations - The supplied concentration model describes A ⇌ B in a fixed volume, with A initially 1.00 mol dm⁻³ and B initially zero. - The long-time values are [A] = 0.600 and [B] = 0.400 mol dm⁻³. Both become approximately constant on the plotted scale while remaining unequal. - For this one-to-one model [A] + [B] remains 1.00 mol dm⁻³. A different balanced equation or changing volume would require the appropriate material balance. ## Separate an imposed addition from the later response - Le Chatelier’s principle predicts that a disturbed equilibrium tends to respond in a direction that partially opposes the disturbance, subject to the specified constraints. - Adding one reactant at fixed volume immediately raises that reactant concentration. Subsequent forward reaction can consume some of the added reactant and form more products. - The response does not generally restore every original concentration. State the initial intervention and the subsequent chemical change separately. ## Predict a response to removing product - Removing some product from an equilibrium mixture changes its composition immediately. At fixed temperature, the equilibrium constant for the same equation remains unchanged. - The subsequent net reaction can form more product, opposing some of the removal. This is a change in equilibrium composition, not a new value of the constant. - Industrial product separation and recycle apply this idea across an open process. They are not the same as retaining all matter in one sealed vessel. ## Predict the effect of isothermal gas compression - Compression at fixed temperature raises reacting-gas concentrations and partial pressures immediately. Compare the total gaseous stoichiometric coefficients on the two sides. - For N₂O₄(g) ⇌ 2NO₂(g), compression favours net formation of N₂O₄, the side with fewer gaseous molecules. The reaction response partially opposes the pressure increase. - This ideal-gas prediction refers to changing volume. It is not a universal rule that any increase in total pressure has the same effect. ## Recognise equal gaseous stoichiometric totals - For H₂(g) + I₂(g) ⇌ 2HI(g), each side has two gaseous stoichiometric units. - At fixed temperature, changing the ideal-gas volume scales all concentrations together. The squared concentration factor in the numerator cancels the two factors in the denominator. - There is therefore no shift in composition by subsequent reaction in this model, although the immediate concentrations and total pressure change. ## Distinguish an inert-gas pressure increase - Adding a chemically inert ideal gas at fixed volume and temperature raises total pressure but leaves reacting-gas concentrations and partial pressures unchanged. - The reacting mixture’s concentration quotient is unchanged, so an initially equilibrated ideal mixture has no composition shift from this intervention alone. - At fixed total pressure, inert-gas addition can expand the volume instead. Always specify the constraint before making a pressure prediction. ## Analyse dilution using the balanced equation - Dilution lowers solution concentrations, but whether an equilibrium shifts depends on how those concentrations enter its expression. - For A(aq) ⇌ 2B(aq), the quotient [B]²/[A] falls when all concentrations are halved. Subsequent forward reaction raises it back towards the same Kc. - Equal overall powers on each side would instead cancel the common dilution factor in the ideal concentration model. Do not infer no shift merely because every species was diluted equally. ## Predict temperature effects from reaction enthalpy - Increasing temperature favours the endothermic direction; decreasing temperature favours the exothermic direction, within the stated equilibrium model. - The supplied P forward reaction is endothermic, so heating favours its products. The supplied Q forward reaction is exothermic, so heating favours its reactants. - Temperature affects both rate and equilibrium composition. A faster approach does not by itself establish a greater final product yield. ## Explain catalysis without changing equilibrium yield - A catalyst provides an alternative pathway and facilitates approach to equilibrium in both directions. It does not change the equilibrium constant at fixed temperature. - The matched closed systems therefore reach the same final composition, though the catalysed one reaches it sooner. - Do not claim that forward and reverse rates must be equal at every stage after adding a catalyst. Their equality characterises equilibrium, not every earlier mixture. ## Evaluate an ammonia-synthesis compromise - The Haber reaction is exothermic and reduces the gaseous stoichiometric total from four to two. Lower temperature and higher pressure favour ammonia at equilibrium in the usual model. - A lower temperature also slows reaction. High compression requires energy and strong equipment, so operating choices balance rate, conversion, cost and safety rather than maximising one variable alone. - Iron-based catalysis supports a useful rate. Downstream ammonia separation and recycle improve overall utilisation; representative textbook temperatures and pressures are not universal plant settings. ## Evaluate sulfur-trioxide production and separation - The Contact-process equilibrium stage converts sulfur dioxide and oxygen to sulfur trioxide. The forward step is exothermic and reduces the gaseous stoichiometric total. - A suitable catalyst and a compromise temperature support useful production. Compression can favour SO₃, but the benefit must justify its equipment and energy costs. - Downstream SO₃ is absorbed into concentrated sulfuric acid to make oleum, followed by controlled dilution. Directly mixing SO₃ with water is difficult to control and produces acid mist; this is an industrial description, not a practical instruction. ## Construct a homogeneous equilibrium expression - Write product equilibrium concentrations in the numerator and reactant equilibrium concentrations in the denominator. Raise each concentration to its balanced-equation coefficient. - For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = [NH₃]²/([N₂][H₂]³). Brackets refer to equilibrium concentrations, not initial amounts. - The expression belongs to the particular equation and direction. It cannot be transferred unchanged to a reversed or rescaled equation. ## Apply phase conventions carefully - The main AQA Kc treatment here is for homogeneous systems. In a single liquid mixture, varying component concentrations can belong in the expression, including water when treated as a mixture component. - As an extension, the activity of a pure solid or pure liquid phase is conventionally one and is omitted from an activity expression while that phase remains present. - Do not remove every species marked (l), or treat adding more pure solid as changing its activity. A mixed-liquid component is different from a separate pure phase. ## Deduce the units of concentration-form Kc - Substitute mol dm⁻³ for each included concentration and combine the powers algebraically. The net product exponent minus reactant exponent determines the remaining unit power. - For A ⇌ 2B, concentration-form Kc has units mol dm⁻³. For A + B ⇌ 2C the units cancel; for ammonia synthesis they are mol⁻² dm⁶. - This A-level concentration-form convention differs from a dimensionless thermodynamic constant written with standard-state activities. Keep the convention and equation explicit. ## Convert equilibrium amounts to concentrations - Concentration equals amount divided by mixture volume in the required unit. Convert cm³ to dm³ by dividing by 1000. - The supplied 0.0600 mol in 200.0 cm³ occupies 0.2000 dm³, giving 0.300 mol dm⁻³. - Use equilibrium amounts for a direct Kc substitution. Initial quantities require a material-balance calculation before they become equilibrium concentrations. ## Calculate Kc from equilibrium concentrations - For the supplied H₂ + I₂ ⇌ 2HI dataset, write Kc = [HI]²/([H₂][I₂]) before substituting. - The numerator is 0.400² and the denominator is 0.100×0.200, giving Kc = 8.00 with no remaining concentration units. - Keep intermediate precision and report to the stated data precision. The number does not tell you how quickly equilibrium was reached. ## Build an initial–change–equilibrium balance - An ICE table separates initial concentration, stoichiometric change and equilibrium concentration. Define a change variable with a clear basis. - For the supplied A ⇌ B example, A falls by 0.300 mol dm⁻³. The one-to-one equation therefore forms the same concentration of B at constant volume. - At equilibrium [A] = 0.200 and [B] = 0.300, so Kc = 1.50. Verify nonnegative values and the conserved total rather than guessing the product concentration. ## Apply stoichiometric coefficients in an ICE calculation - For N₂ + 3H₂ ⇌ 2NH₃, forming two concentration units of ammonia consumes one of nitrogen and three of hydrogen in a fixed volume. - The supplied 0.400 mol dm⁻³ ammonia corresponds to a change basis of 0.200. Equilibrium nitrogen is 0.300 and hydrogen is 0.900 mol dm⁻³. - Substitute those equilibrium concentrations into the coefficient-powered expression. Equal numerical concentration changes would violate the balanced equation. ## Identify when volume factors do not cancel - For A ⇌ 2B, replacing each concentration by n/V gives K_{c} = n_{B}^{2}/(n_{A} V). One inverse volume remains because the net concentration exponent is one. - The supplied equilibrium amounts in 2.00 dm³ give [A] = 0.100 and [B] = 0.200 mol dm⁻³, hence Kc = 0.400 mol dm⁻³. - For equations with zero net exponent, common volume factors cancel. Do not extend that simplification to every equation merely because all species share the same vessel. ## Transform Kc when the equation changes - Reversing an equation replaces its equilibrium expression by the reciprocal, so the new constant is 1/Kc at the same temperature. - Multiplying every coefficient by a factor raises the original constant to that factor. Doubling all coefficients squares Kc. - Numerical constant magnitudes therefore depend on how the equation and units are written. Compare like expressions rather than using one universal large-number yield rule. ## Distinguish a current quotient from equilibrium Kc - The reaction quotient Qc uses the same concentration expression as Kc but can be calculated for a mixture that is not at equilibrium. - For the supplied A ⇌ B mixture, Qc = 0.400/0.200 = 2.00, below Kc = 4.00. Net forward reaction increases the quotient towards its equilibrium value. - This quotient method makes the direction test explicit. Kc is unchanged by the composition adjustment at fixed temperature; Qc changes as the reaction proceeds. ## Infer temperature effects on Kc - At fixed equation and phase model, temperature changes Kc. For an endothermic forward reaction, heating increases Kc; for an exothermic forward reaction, heating decreases it. - The supplied constant increases from 2.00 at 300 K to 5.00 at 350 K. That qualitative pattern supports an endothermic forward reaction over the stated interval. - Concentration changes, ideal-gas compression and catalysts do not change Kc at fixed temperature in this treatment. They can affect composition, reaction rate or approach time in different ways.