Equilibrium and Acid Reactions, complete assessment covering dynamic equilibrium, Le Chatelier predictions, Keq, Q, solubility equilibrium, Ksp and common ion effects from L01-L18. 15 MC questions (auto-marked) + 5 written questions (self-marked). Complete all questions before submitting.
A dynamic equilibrium exists when:
A closed system is required for equilibrium because it:
At equilibrium, collision theory implies:
Adding more reactant to an equilibrium mixture initially causes the system to:
For N2O4(g) ↔ 2NO2(g), decreasing volume favours:
The industrial Haber process uses a compromise temperature because lower temperature:
For CaCO3(s) ↔ CaO(s) + CO2(g), the equilibrium expression includes:
In an ICE table, the C row represents:
If Q < Keq, the reaction will proceed:
Changing temperature can change Keq because:
A larger Ka value indicates:
For a saturated ionic solution at equilibrium:
A precipitation reaction occurs when mixed ions form:
For AgCl(s) ↔ Ag+(aq) + Cl-(aq), Ksp equals:
Adding Cl- to a saturated AgCl solution causes precipitation because:
Explain why equilibrium is described as dynamic, and distinguish microscopic activity from macroscopic constancy.
Equilibrium is dynamic because the forward and reverse reactions continue. It is not a stopped reaction. At equilibrium the two rates are equal, so the concentrations of reactants and products remain constant at the macroscopic level. At the microscopic level particles still collide successfully and convert between reactants and products. A closed system is needed so material cannot escape and both directions can continue.
Marks: 1, forward reaction continues | 1, reverse reaction continues | 1, rates equal | 1, concentrations/macroscopic properties constant | 1, closed system explainedFor 2SO2(g) + O2(g) ↔ 2SO3(g), ΔH < 0, predict the effect on SO3 yield of increasing pressure, increasing temperature and adding a catalyst.
Increasing pressure favours the side with fewer gas moles. The product side has 2 mol gas compared with 3 mol gas on the reactant side, so SO3 yield increases. Increasing temperature favours the endothermic direction. Since the forward reaction is exothermic, the reverse direction is favoured and SO3 yield decreases. Adding a catalyst increases the rates of both forward and reverse reactions by lowering activation energy, so equilibrium is reached faster but the final yield and Keq do not change.
Marks: 1, gas mole comparison | 1, pressure prediction | 1, temperature/endothermic reasoning | 1, temperature prediction | 1, catalyst effectDescribe how to set up an ICE table for an equilibrium calculation and explain why stoichiometric coefficients matter in the change row.
An ICE table records Initial concentration, Change in concentration and Equilibrium concentration for each aqueous or gaseous species. The initial row uses the starting concentrations. The change row uses an unknown such as x, with signs showing whether each species is consumed or produced. Stoichiometric coefficients matter because a coefficient of 2 means that species changes by 2x when a coefficient of 1 changes by x. The equilibrium row adds the initial and change rows, and these equilibrium values are substituted into the Keq expression.
Marks: 1, I row | 1, C row signs | 1, E row | 1, stoichiometric coefficients | 1, substitution into KeqExplain how Q is used to predict reaction direction, then connect Ka magnitude to acid strength.
The reaction quotient Q has the same form as Keq but uses current concentrations rather than equilibrium concentrations. If Q < Keq, there are too few products relative to equilibrium, so the reaction proceeds forward. If Q > Keq, there are too many products, so it proceeds in reverse. If Q = Keq, the system is at equilibrium. Ka is an acid equilibrium constant; a larger Ka means the acid ionises to a greater extent, so the equilibrium lies further toward products and the acid is stronger.
Marks: 1, Q definition | 1, Q < Keq | 1, Q > Keq | 1, Q = Keq | 1, Ka magnitude and acid strengthUse Ksp and Qsp to explain when a precipitate forms, including the common ion effect.
Ksp is the equilibrium ion product for a saturated solution of a sparingly soluble salt. Qsp is calculated from the current ion concentrations. If Qsp < Ksp, more solid can dissolve. If Qsp = Ksp, the solution is saturated and at equilibrium. If Qsp > Ksp, the ion concentrations are too high, so precipitation occurs until Qsp returns to Ksp. Adding a common ion increases one ion concentration, raises Qsp and shifts the dissolution equilibrium toward the solid, reducing solubility.
Marks: 1, Ksp defined | 1, Qsp defined | 1, Qsp < Ksp | 1, Qsp > Ksp precipitation | 1, common ion effect