Equilibrium Constant (Kp) For Homogeneous Systems

## Identify a homogeneous dynamic equilibrium - A homogeneous equilibrium has all participating species in one phase. A mixture of several gases is homogeneous; one solid plus a gas is not. - In a closed system at equilibrium, forward and reverse rates are equal while molecular reactions continue. Species amounts need not be equal. - For a specified all-gas equation, Kp relates equilibrium partial pressures at a stated temperature. It is a composition relation, not a measure of how quickly equilibrium is reached. ## Relate partial pressure to mole fraction - For an ideal mixture, p(A)=y(A)Ptotal with y(A)=n(A)/ntotal. Partial pressure is what A alone would exert at the same volume and temperature. - The supplied mixture has y(A)=2/(2+3)=0.40 and p(A)=0.40×150=60 kPa. - Both species occupy the whole container. The qualitative mixture illustration shows intermingling, not separated volumes or a measured particle ratio. ## Include every gas in the mole-fraction denominator - Add equilibrium amounts of all gas species, including inert gases, when calculating total gas moles. An inert gas can contribute pressure without appearing in the reaction equation. - The supplied mixture has10.0 mol in total. Nitrogen has mole fraction0.10, giving p(N₂)=20 kPa at200 kPa total. - The mole fractions sum to one and partial pressures sum to total pressure. Use equilibrium amounts and retain unrounded intermediate values. ## Construct a pressure-power equilibrium expression - Raise each gas partial pressure to its stoichiometric coefficient. Multiply product terms and divide by reactant terms. - For N₂+3H₂⇌2NH₃, Kp=p(NH₃)²/[p(N₂)p(H₂)³]. These powers are coefficients, not equilibrium amounts in moles. - Use p(X) for partial pressure. Square brackets usually denote concentration; do not substitute concentrations into a pressure-based expression. ## Derive the units in the AQA pressure-power convention - Pressure-power Kp has units(pressure)^Δν, where Δν is total gaseous product coefficients minus total gaseous reactant coefficients. - Using kPa, ammonia formation has kPa⁻², N₂O₄ dissociation has kPa, and SO₃ formation has kPa⁻¹. Equal gas totals give cancellation. - A thermodynamic equilibrium constant instead uses dimensionless activities such as p/p°. Keep this distinction clear; use the convention and pressure unit requested in the question. ## Recognise when total pressure cancels - Substitute p(X)=y(X)P into Kp. A factor P^Δν remains, so mole fractions alone are generally insufficient. - For H₂+I₂⇌2HI, Δν=0 and P cancels. The supplied fractions give Kp=0.60²/(0.20×0.20)=9.0. - This cancellation is a special algebraic result, not permission to drop pressure in equations with unequal gas totals. Fixed-temperature ideal compression does not shift this equal-gas-coefficient equilibrium. ## Calculate Kp from equilibrium partial pressures - For2SO₂+O₂⇌2SO₃, Kp=p(SO₃)²/[p(SO₂)²p(O₂)]. All pressures must use the same unit. - The supplied values give40²/(20²×10)=0.400 kPa⁻¹. Square the complete numerical pressure values. - These are equilibrium pressures, not initial pressures or concentrations. Deduce units from the same expression used for the calculation. ## Use stoichiometry before calculating fractions - The supplied nitrogen loss is0.50 mol. The reaction consumes1.50 mol H₂ and forms1.00 mol NH₃. - Equilibrium amounts are0.50,1.50 and1.00 mol, so the total is3.00 mol rather than the initial4.00 mol. - At120 kPa the partial pressures are20,60 and40 kPa respectively. Thus Kp=40²/(20×60³)=3.70×10⁻⁴ kPa⁻² to three significant figures. ## Rearrange Kp to find total pressure - For the supplied ammonia fractions, substitution gives Kp=[.25²/(.25×.50³)]P⁻²=2/P² when P is in kPa. - Rearrange to P²=2/Kp. With Kp=2.00×10⁻⁴ kPa⁻², the positive pressure is100 kPa. - Check dimensions and physical roots. A negative pressure root is inadmissible, and a reaction with Δν=0 cannot yield total pressure from Kp and fractions alone. ## Solve for an unknown equilibrium fraction - Let x be the NO₂ mole fraction. With only N₂O₄ and NO₂ present, the other fraction is1−x. - Kp=x²P/(1−x). The supplied values give100x²=50(1−x), or2x²+x−1=0. - The roots are0.50 and−1. Reject−1 because mole fractions must lie from zero to one. The valid pressures are50 kPa each. ## Account for changing total moles during dissociation - Starting with1 mol N₂O₄ only, dissociation fraction α gives n(N₂O₄)=1−α, n(NO₂)=2α and ntotal=1+α. - The fractions are(1−α)/(1+α) and2α/(1+α). Thus Kp=4α²P/(1−α²) for the ideal mixture. - For α=.20 and P=120 kPa, equilibrium amounts are.80 and.40 mol, pressures80 and40 kPa, and Kp=20 kPa. ## Transform the constant when changing the equation - Reversing an equation inverts its pressure-power Kp and its units. Multiplying every coefficient by a factor raises Kp to that power. - If N₂O₄⇌2NO₂ has Kp=50 kPa, the reverse has0.020 kPa⁻¹ and the doubled equation has2500 kPa². - These are different written reactions, not changes in the equilibrium temperature. Always compare constants for the same equation and unit convention. ## Distinguish a pressure disturbance from changing Kp - At fixed temperature, halving the volume doubles every reacting gas partial pressure before further reaction occurs. - For N₂O₄⇌2NO₂ the current quotient doubles because Δν=+1. It exceeds the unchanged Kp, so net reaction proceeds left toward fewer gas moles. - The initial supplied Kp is40²/20=80 kPa. Immediately after compression Qp=80²/40=160 kPa; subsequent re-equilibration restores Qp=Kp. ## Treat inert addition at fixed volume correctly - For an ideal gas, p(X)=n(X)RT/V. Adding nonreacting argon at unchanged V and T leaves reacting partial pressures unchanged. - Total pressure rises and reacting mole fractions fall, but their product y(X)P remains the same. The pressure quotient therefore remains equal to Kp. - No ideal-gas equilibrium shift follows from this addition. A statement that any increase in total pressure favours fewer gas moles misses the constraint. ## Treat inert addition at fixed total pressure separately - Adding inert gas at fixed total pressure and temperature requires expansion. The reacting amounts initially stay constant but their partial pressures decrease. - For N₂O₄⇌2NO₂ the quotient scales with P-like dilution to the power Δν=+1. It falls below Kp and net reaction proceeds right. - This favours more gas moles and differs from adding inert at fixed volume. State the physical constraints before using a pressure rule. ## Link temperature trends to forward reaction enthalpy - For an exothermic forward reaction, increasing temperature decreases Kp and favours the reverse equilibrium direction under the usual comparison conditions. - For an endothermic forward reaction, increasing temperature increases Kp and favours products. Cooling reverses these trends. - The trend refers to the equation as written and the applicable temperature/phase range. A change in Kp does not describe how fast the new equilibrium is attained. ## Separate equilibrium composition from attainment time - A suitable catalyst supplies a faster reversible pathway while leaving Kp unchanged at the same temperature and reaction convention. - The same closed-system constraints and initial mixture give the same final equilibrium composition, reached more quickly if the catalyst is effective. - Forward and reverse rates are equal at equilibrium. Do not claim they are equal at every composition during the approach to equilibrium. ## Keep the pressure unit consistent - The supplied pressures give Kp=40²/20=80 kPa. In pascals, Kp=40000²/20000=80000 Pa. - The numerical change reflects the pressure unit, not a temperature change or a different chemical equilibrium. In general multiply the numerical value by1000^Δν when changing kPa to Pa. - For Δν=−2, this conversion factor is10⁻⁶. A dimensionless activity-based constant uses a specified reference pressure and must not be confused with the dimensional pressure-power value. ## Use Qp for a current non-equilibrium mixture - Qp has the same expression as Kp but uses current partial pressures. Compare them with the same equation and pressure-unit convention. - If Qp<Kp, net forward reaction is favoured; if Qp>Kp, net reverse reaction is favoured. At equilibrium Qp=Kp. - For the supplied mixture Qp=20²/20=20 kPa, below Kp=80 kPa, so net dissociation proceeds right. This does not require every individual collision to be forward. ## Evaluate equilibrium yield alongside rate and cost - Ammonia formation is exothermic and reduces gas moles. Lower temperature favours its equilibrium yield but can slow attainment. - Compression at fixed temperature favours ammonia without changing Kp. It also requires energy and pressure-rated equipment. - A catalyst helps reach equilibrium more quickly while leaving its position unchanged. Process choices combine equilibrium yield, kinetics, energy, separation and practical constraints.