Gaokao Chemistry: Chemical Equilibrium and Reactions Flashcards

For the equilibrium CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium constant expression is:

Answer: K = [CO₂]

Pure solids have activity = 1 and are excluded from K; only the gas-phase species CO₂ appears: K = [CO₂] (or Kp = P_CO₂).

The water autoionisation equilibrium is H₂O ⇌ H⁺ + OH⁻. At 25°C, Kw = 1×10⁻¹⁴. In a neutral solution, [H⁺] equals:

Answer: Both A and C are correct

In neutral water [H⁺] = [OH⁻]; Kw = [H⁺][OH⁻] = (10⁻⁷)² = 10⁻¹⁴; so [H⁺] = 10⁻⁷ M, which equals [OH⁻]. Both A and C are correct.

Which factor does NOT affect the value of the equilibrium constant K for a given reaction?

Answer: Adding a catalyst

A catalyst has no effect on K; it only affects the rate at which equilibrium is achieved. K is temperature-dependent.

In redox reactions, the species that undergoes an increase in oxidation number has been:

Answer: Oxidised

Oxidation is defined as an increase in oxidation number (loss of electrons); reduction is a decrease in oxidation number.

The rate of a chemical reaction generally increases with temperature because:

Answer: More molecules have energy ≥ activation energy (Maxwell-Boltzmann distribution)

Higher temperature shifts the Maxwell-Boltzmann distribution so a larger fraction of molecules exceed the activation energy threshold, increasing the reaction rate.

NH₃ is a weak base (Kb = 1.8×10⁻⁵). A 0.10 M NH₃ solution has [OH⁻] approximately equal to:

Answer: 1.34×10⁻³ M

Kb = x²/0.10; x = √(1.8×10⁻⁵ × 0.10) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³ M.

Increasing the concentration of a reactant in a reaction at equilibrium causes the reaction quotient Q to:

Answer: Decrease below K, causing the forward reaction to proceed

Adding a reactant increases Q denominator... actually Q = [products]/[reactants]; adding reactant decreases Q (increases denominator), so Q < K and equilibrium shifts forward.