Equilibrium MCQs

Molecules of water keep leaving the liquid surface (evaporation) while vapour molecules keep returning to the liquid (condensation).

At dynamic equilibrium the amounts of liquid and vapour stay constant because the two opposite processes proceed at the same speed.

Therefore the rate of evaporation equals the rate of condensation.

Kc is defined as the ratio of product concentration to reactant concentration at equilibrium.

Substitute the given values: Kc = [B] / [A] = 1.5 mol L⁻¹ ÷ 0.5 mol L⁻¹.

The division gives 3.

Thus the equilibrium constant Kc equals 3.

Loss of a proton (H⁺) from an acid gives its conjugate base.

HCO₃⁻ can act as an acid by donating its hydrogen ion.

Removing H⁺ from HCO₃⁻ leaves the species CO₃²⁻.

Therefore the conjugate base of HCO₃⁻ is CO3 2-

pH is defined as the negative logarithm (base 10) of the hydrogen‑ion concentration.

Substitute the given concentration: pH = –log(1 × 10⁻⁵).

Since log(10⁻⁵) = –5, the negative sign makes pH = 5.

Hence the pH of the solution is 5

Ka × Kb = Kw (1.0 × 10⁻¹⁴) for a conjugate acid‑base pair.

Kb = Kw / Ka = (1.0 × 10⁻¹⁴) ÷ (1.8 × 10⁻⁵).

Divide the powers of ten: 10⁻¹⁴ ÷ 10⁻⁵ = 10⁻⁹; 1.0 ÷ 1.8 ≈

For the forward reaction A ⇌ B, Kc = [B]/[A] = 4.

The equilibrium constant for the reverse reaction is the reciprocal: Kc (reverse) = 1/Kc (forward).

Thus Kc = 1/4 = 0.25.

In the equilibrium NH3 + H2O ⇌ NH4⁺ + OH⁻, water supplies a H⁺ ion to NH3, producing NH4⁺.

Since it donates a proton, water is acting as a proton donor.

A proton donor is defined as a Brønsted acid.

Hence water acts as a Bronsted acid.

NaOH is a strong base, so it dissociates completely giving [OH⁻] = 0.01 M.

pOH = –log[OH⁻] = –log(0.01) = 2.

pH = 14 – pOH = 14 – 2 = 12.

For the acid HA dissociation, Ka = [H⁺][A⁻]/[HA] and for its conjugate base A⁻ reacting with water, Kb = [OH⁻][HA]/[A⁻].

Multiplying the two expressions cancels [HA] and [A⁻] leaving Ka·Kb = [H⁺][OH⁻].

At 25 °C, the product [H⁺][OH⁻] is the ion‑product of water, Kw.

Thus Ka x Kb = Kw

For a dissolution equilibrium the solid phase is omitted from the equilibrium expression.

Write the equilibrium: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).

Ksp is defined as the product of the molar concentrations of the ions at equilibrium.

Thus Ksp = [Ag⁺][Cl⁻]

At a fixed temperature the liquid and its vapor are in dynamic equilibrium, so the rate of evaporation equals the rate of condensation.

Because the two opposite rates are equal, the partial pressure of the vapor does not change with time.

Thus the quantity that stays unchanged is the vapour pressure. Vapour pressure

K is defined as the ratio of the concentrations of products to reactants at equilibrium.

When K is much greater than 1, the numerator (product concentrations) must be far larger than the denominator (reactant concentrations).

Thus at equilibrium the reaction mixture contains mainly products, i.e., products predominate.

NH₃ acts as a Brønsted‑Lowry base because it can accept a proton (H⁺).

When NH₃ gains one H⁺, its formula becomes NH₄⁺, which is the species formed after protonation.

Thus the conjugate acid of NH₃ is NH₄+.

pH is defined as the negative logarithm of the hydrogen ion concentration: pH = –log[H⁺]

Re‑arrange to find [H⁺]: [H⁺] = 10^(–pH)

Substitute pH = 4: [H⁺] = 10^(–4) = 1 × 10⁻⁴ mol/L

1 × 10⁻⁴ mol/L.

pKa is defined as the negative logarithm of the acid dissociation constant: pKa = –log Ka.

Substituting Ka = 1 × 10⁻⁵ gives pKa = –log(1 × 10⁻⁵).

Since –log(10⁻⁵) = 5, the pKa equals 5.

At equilibrium the forward and reverse reactions continue to occur, so concentrations stay constant while molecules interconvert.

The rates of the forward and reverse steps become equal, making the system dynamic rather than static.

Because the rates are equal, the same equilibrium state can be attained whether the mixture starts with mainly reactants or mainly products.

A catalyst lowers the activation energy for both the forward and reverse reactions equally.

Because it speeds up the rates of the forward and reverse processes proportionally, the ratio of product to reactant concentrations at equilibrium remains the same.

Thus the equilibrium constant Kc is unchanged, although the system reaches that equilibrium faster.

BF3 has an incomplete octet on boron, giving it an empty p‑orbital that can accept a lone‑pair of electrons.

NH₃, OH⁻ and H₂O each possess lone pairs and therefore act as Lewis bases, not acids.

A Lewis acid must be an electron‑pair acceptor, which BF₃ fulfills.

BF₃

pH = –log[H⁺]; at 25 °C neutral water has pH 7 because [H⁺] = 1 × 10⁻⁷ M.

A pH value greater than 7 indicates a lower hydrogen‑ion concentration than neutral water, i.e., the solution is alkaline.

Therefore a solution with pH = 9.5 is Basic.

Acid strength increases as the acid dissociation constant Ka increases.

Since pKa = –log Ka, a smaller pKa corresponds to a larger Ka.

Among the given values, 2 is the lowest pKa, indicating the highest Ka.

Therefore the strongest acid has pKa 2.