30 Chemical Equilibrium Practice Problems for DAT Prep

IX. Chemical Equilibrium

Chemical equilibrium practice problems

This chapter contains 30 practice questions on chemical equilibrium, including equilibrium constants, Le Châtelier’s principle, acids and bases, pH, buffers, and solubility equilibria. Allow about 1 minute per question. After completing a test, select “See Result” to view your score, answer choices, and explanations. Correct answers appear in green, and incorrect answers appear in red.

CHEMICAL EQUILIBRIUM TEST 1

1 / 20

Which reaction is commonly treated as a reversible equilibrium in a closed system, while the other reactions are strongly product-favored under ordinary introductory laboratory conditions?

A. 2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(l)

B. Cu(s) + 4 HNO₃(aq) → Cu(NO₃)₂(aq) + 2 NO₂(g) + 2 H₂O(l)

C. LiOH(aq) + HI(aq) → LiI(aq) + H₂O(l)

D. Pb(NO₃)₂(aq) + 2 KI(aq) → 2 KNO₃(aq) + PbI₂(s)

E. CH₄(g) + H₂O(g) ⇌ CO(g) + 3 H₂(g)

Explanation:

Steam reforming (E) is routinely treated as a reversible gas-phase equilibrium. The other reactions are strongly product-favored under the stated ordinary conditions: combustion, oxidation of copper by nitric acid, strong-acid/strong-base neutralization, and precipitation. “Irreversible” is a practical approximation; it does not mean that a reverse process is fundamentally impossible.

2 / 20

Which statement correctly describes a closed reacting system at equilibrium under constant conditions?

A. Equilibrium requires a reaction that can proceed only in the forward direction.

B. The forward reaction is faster than the reverse reaction.

C. Both the forward and reverse reactions stop.

D. Reactant and product concentrations remain constant over time.

E. Reactant and product concentrations must be equal to each other.

Explanation:

Chemical equilibrium is dynamic: the forward and reverse rates are equal and both processes continue. Concentrations remain constant over time, but reactant and product concentrations need not be equal to each other.

3 / 20

For a specified reaction and standard-state convention, which factor changes the equilibrium constant K?

A. Concentration of reactants or products

B. Temperature

C. The presence of a catalyst

D. Reaction quotient (Q)

E. B and C

Explanation:

For a specified reaction, K depends on temperature. Concentration and pressure changes may change Q and the equilibrium composition, but do not change K at fixed temperature. A catalyst accelerates equilibration without changing K.

4 / 20

Which statement about chemical equilibrium is INCORRECT?

A. K greater than 1 indicates a product-favored equilibrium.

B. Pure solids and pure liquids are omitted from the usual equilibrium expression because their activities are 1.

C. K less than 1 means that the forward reaction never occurs.

D. Q = K at equilibrium.

E. The reaction Gibbs energy change ΔG is zero at equilibrium.

Explanation:

K less than 1 indicates a reactant-favored equilibrium; it does not mean the forward reaction never occurs. Both directions proceed at equilibrium. K greater than 1 indicates product-favored equilibrium, but does not by itself guarantee that the total numerical concentration of products exceeds that of reactants for every stoichiometry. At equilibrium, Q = K and ΔG = 0.

5 / 20

Explanation:

Use stoichiometric coefficients as exponents: K_c = [CO₂][CO₃²⁻]/[HCO₃⁻]². Pure liquid water is omitted because its activity is approximately 1. Both gases and dissolved species can appear in a concentration-based equilibrium expression. C is correct.

6 / 20

A. 1/2 M

B. 1 M

C. 2 M

D. 4 M

E. 1/4 M

Explanation:

Pure liquid water is omitted. K_c = [H₂]²[O₂] = 8. With [O₂] = 2 M, [H₂]² = 4 and [H₂] = 2 M (C), using the introductory concentration convention.

7 / 20

A. 0.67

B. 1

C. 2

D. 4

E. 8

Explanation:

Cl₂ decreases from 12 M to 9 M, so 3 M is consumed. The reaction produces twice as much Cl: [Cl] = 6 M. Therefore K_c = [Cl]²/[Cl₂] = 6²/9 = 4 (D).

8 / 20

Explanation:

Only gaseous B and D appear in K_p; pure liquid A and pure solid C are omitted. The exponents are the gas-phase stoichiometric coefficients: K_p = (P_D)⁴/(P_B)², so E is correct.

9 / 20

A. 6

B. 12

C. 18

D. 24

E. 48

Explanation:

Use equilibrium partial pressures: K_p = (P_CO₂)²/[(P_CO)²P_O₂] = (0.6)²/[(0.2)²(0.5)] = 18 (C). These are partial pressures, not molar concentrations.

10 / 20

A. Removing H₂

B. Adding C

C. Adding H₂O

D. Removing C

E. Both removing H₂ and adding H₂O

Explanation:

At fixed temperature and volume, removing H₂ or adding H₂O(g) lowers Q = [CO][H₂]/[H₂O] below K. Equilibrium shifts right and [CO] increases. Adding or removing some pure solid carbon has no effect while solid remains. E identifies both effective changes.

11 / 20

A. Decreasing the volume

B. Decreasing the temperature

C. Increasing the pressure by compression

D. Adding H₂O(g)

E. Removing NH₃

Explanation:

Compression favors the reactant side, which has 9 mol of gas versus 10 mol on the product side. Cooling favors the exothermic forward reaction, so B is correct. Adding H₂O(g) or removing NH₃ shifts equilibrium left. Pressure comparisons assume compression at constant temperature.

12 / 20

A. Adding O₂

B. Removing O₃

C. Increasing pressure by compression

D. Decreasing temperature

E. Increasing temperature

Explanation:

Ozone formation, 3 O₂(g) ⇌ 2 O₃(g), is endothermic: ΔH° is about +285 kJ at 298 K. Cooling shifts equilibrium left, so D does not favor ozone. Adding O₂, removing O₃, compression (3 gas moles to 2), or heating favors the forward direction.

13 / 20

A. NH₃

B. NH₄⁺

C. F^–

D. NH₃ and F^–

E. HF and NH₄⁺

Explanation:

A conjugate base is formed when an acid donates one proton. HF donates H⁺ and becomes F⁻, so C is correct. NH₃ is the proton acceptor in this reaction, and NH₄⁺ is its conjugate acid.

14 / 20

A. Reaction I favors the products because HBr is weaker than HF.

B. In reaction II, F⁻ acts as a Brønsted–Lowry base by accepting a proton.

C. BF₃ is an electron-pair donor in reaction II.

D. Reaction II is an Arrhenius neutralization that produces water from H⁺ and OH⁻.

E. Both reactions can be described using Lewis acid–base concepts.

Explanation:

Reaction I is a proton-transfer reaction and favors the weaker acid/base pair HF + Br⁻, because HBr is a stronger acid than HF. Proton transfer also involves electron-pair donation to H⁺, so it fits the Lewis framework. In reaction II, F⁻ donates an electron pair to electron-deficient BF₃; no proton is transferred. Thus E is correct. Reaction II is not an Arrhenius neutralization.

15 / 20

In the standard introductory acid–base classification, which species is identified as an electron-pair-accepting Lewis acid rather than a Brønsted–Lowry proton acceptor?

A. H₂PO₄^–

B. NH₃

C. Cl^–

D. H₂O

E. BeCl₂

Explanation:

A Brønsted–Lowry base accepts a proton. H₂PO₄⁻, NH₃, Cl⁻, and H₂O can form H₃PO₄, NH₄⁺, HCl, and H₃O⁺, respectively. BeCl₂ is classified here as a Lewis acid because its electron-deficient Be center accepts an electron pair.

16 / 20

Which acid-strength comparison is correct for the usual aqueous acid-strength trends?

A. H₂O > HF

B. HCl > HBr

C. HOCl > HOBr

D. HOClO₂ > HOClO₃

E. CH₃CO₂H > H₂SO₄

Explanation:

HOCl is stronger than HOBr because the more electronegative Cl atom withdraws electron density more strongly in otherwise analogous hypohalous acids. Across a period, the binary hydride acid trend is CH₄ < NH₃ < H₂O < HF; down the hydrogen-halide group it is HF < HCl < HBr < HI. For chlorine oxyacids, added oxygen strengthens the acid: HClO < HClO₂ < HClO₃ < HClO₄. H₂SO₄ is much stronger than acetic acid. C is correct.

17 / 20

Which listed substance is classified as a strong base in water?

A. Ba(OH)₂

B. NH₃

C. CH₃NH₂

D. NO₃^–

E. HClO₄

Explanation:

Dissolved Ba(OH)₂ dissociates completely to Ba²⁺ + 2 OH⁻. NH₃ and CH₃NH₂ are weak bases. NO₃⁻ has negligible basicity in water because its conjugate acid HNO₃ is strong. HClO₄ is an acid, not a base. A is correct.

18 / 20

Which is the conventional dilute-solution expression for the acid ionization constant K_a of HCl in water?

Explanation:

For HCl + H₂O ⇌ Cl⁻ + H₃O⁺, the pure-liquid solvent activity is approximately 1 and is omitted. The conventional expression is K_a = [Cl⁻][H₃O⁺]/[HCl] (B). This does not mean that the molar concentration of liquid water is 1 M. HCl is strongly ionized in dilute aqueous solution.

19 / 20

For an aqueous solution initially containing 0.020 M NH₃, estimate [OH⁻] if K_b = 1.8 × 10⁻⁵. Use the small-ionization approximation.

A. 3.0×10^-4 M

B. 6.0×10^-4 M

C. 1.2×10^-3 M

D. 1.8×10^-3 M

E. 3.6×10^-3 M

Explanation:

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Let x = [OH⁻]. K_b = x²/(0.020 − x) ≈ x²/0.020, giving x ≈ 6.0 × 10⁻⁴ M (B). The estimated ionization is 3.0%, so the stated approximation passes the usual 5% check. The quadratic value is 5.91 × 10⁻⁴ M.

20 / 20

What is [OH⁻] in 0.050 M H₂SO₄ at 25 °C? Treat the first ionization as complete, use K_a2 = 1.2 × 10⁻² for HSO₄⁻, and K_w = 1.0 × 10⁻¹⁴. Use concentrations in place of activities.

A. 1.0 × 10⁻¹² M

B. 1.7 × 10⁻¹³ M

C. 1.0 × 10⁻¹⁴ M

D. 3.0 × 10⁻¹⁵ M

E. 5.0 × 10⁻¹⁶ M

Explanation:

After the first ionization, [H₃O⁺] = [HSO₄⁻] = 0.050 M. For the second step, K_a2 = x(0.050 + x)/(0.050 − x) = 0.012. The positive root is x = 0.00851 M, so [H₃O⁺] = 0.05851 M. Thus [OH⁻] = 1.0 × 10⁻¹⁴/0.05851 = 1.7 × 10⁻¹³ M (B). The second ionization cannot simply be omitted at this concentration.

Your score is

The average score is 66%

CHEMICAL EQUILIBRIUM TEST 2

1 / 10

What is the pH at 25 °C of a solution with [OH⁻] = 4.0 × 10⁻¹¹ M? Use K_w = 1.0 × 10⁻¹⁴.

A. 3.0

B. 3.60

C. 4.0

D. 4.5

E. 5.0

Explanation:

[H₃O⁺] = K_w/[OH⁻] = 1.0 × 10⁻¹⁴/(4.0 × 10⁻¹¹) = 2.5 × 10⁻⁴ M. Therefore pH = −log(2.5 × 10⁻⁴) = 3.602 ≈ 3.60 (B).

2 / 10

A 48 g sample of LiOH is dissolved to make 4.0 L of solution at 25 °C. Using a molar mass of 24 g/mol, complete dissociation, and K_w = 1.0 × 10⁻¹⁴, calculate the pH. Use concentrations in place of activities.

A. 0.5

B. 4.5

C. 9.0

D. 11.5

E. 13.70

Explanation:

The amount is 48 g/(24 g/mol) = 2.0 mol. Because each formula unit produces one OH⁻, [OH⁻] = 2.0 mol/4.0 L = 0.50 M. Then pOH = −log(0.50) = 0.301 and pH = 14.00 − 0.301 = 13.699 ≈ 13.70 (E).

3 / 10

A solution is prepared by diluting 1.0 × 10⁻⁵ mol of HNO₂ to a total volume of 5.0 L. If K_a = 4.5 × 10⁻⁴, calculate the pH. Neglect water autoionization and use concentrations in place of activities; do not assume that acid ionization is small.

A. 3.0

B. 3.5

C. 4.0

D. 5.70

E. 5.0

Explanation:

The total acid concentration is C = 1.0 × 10⁻⁵ mol/5.0 L = 2.0 × 10⁻⁶ M. With x = [H₃O⁺], x²/(C − x) = 4.5 × 10⁻⁴. Solving the quadratic gives x = 1.991 × 10⁻⁶ M and pH = 5.7009 ≈ 5.70 (D). The small-x estimate, 3.0 × 10⁻⁵ M, exceeds the total acid concentration and is invalid. Including water autoionization would still give 5.70 to two decimal places.

4 / 10

A monoprotic weak-acid/conjugate-base buffer has pH 6.5. At equilibrium, 99% of the acid/base component is A⁻ and 1% is HA. What is the approximate pK_a?

A. 4.0

B. 4.5

C. 5.0

D. 5.5

E. 6.0

Explanation:

Use pH = pK_a + log([A⁻]/[HA]). The ratio is 99/1, so pK_a = 6.5 − log(99) = 4.504 ≈ 4.5 (B).

5 / 10

A. [Ag⁺][2SO₄^2-]

B. [2Ag⁺][SO₄^2-]

C. [Ag⁺][SO₄^2-]²

D. [Ag⁺]²[SO₄^2-]

E. [Ag⁺]²[SO₄^2-]⁴

Explanation:

Ag₂SO₄(s) ⇌ 2 Ag⁺(aq) + SO₄²⁻(aq). The solid is omitted and the Ag⁺ coefficient becomes an exponent: K_sp = [Ag⁺]²[SO₄²⁻] (D). Coefficients do not belong inside a species concentration bracket.

6 / 10

Let X be the molar solubility of CaF₂ in pure water. Neglect side reactions and the ionization of water. Which expression gives K_sp in terms of X?

A. K_sp = X(X)²

B. K_sp = X(2X)²

C. K_sp = X²(2X)

D. K_sp = (2X)(X)

E. K_sp = (2X)²(2X)

Explanation:

CaF₂(s) ⇌ Ca²⁺ + 2 F⁻. For molar solubility X, [Ca²⁺] = X and [F⁻] = 2X. Therefore K_sp = [Ca²⁺][F⁻]² = X(2X)² = 4X³ (B). X is a numerical concentration variable, not a chemical species.

7 / 10

Estimate the molar solubility of chromium(III) hydroxide, Cr(OH)₃, if K_sp = 2.7 × 10⁻²⁷ at 20 °C. For this simplified dissolution model, neglect water autoionization, hydrolysis, complex formation, and activity corrections.

A. 1 × 10^-7 M

B. 3 × 10^-7 M

C. 6 × 10^-6 M

D. 1 × 10^-5 M

E. 3 × 10^-5 M

Explanation:

Under the stated simplified model, Cr(OH)₃(s) ⇌ Cr³⁺ + 3 OH⁻. Let s be the molar solubility: [Cr³⁺] = s and [OH⁻] = 3s. Thus K_sp = s(3s)³ = 27s⁴, giving s = (2.7 × 10⁻²⁷/27)^1/4 = 1 × 10⁻⁷ M (A). This estimate uses the explicitly stated neglect of other equilibria.

8 / 10

What is [Cu²⁺] in a saturated solution of Cu₃(PO₄)₂ if K_sp = 1.08 × 10⁻²⁸? Assume dissolution in pure water, neglect hydrolysis and complex formation, and use concentrations in place of activities.

A. 3×10^-5 M

B. 1×10^-6 M

C. 2×10^-6 M

D. 3×10^-6 M

E. 2×10^-7 M

Explanation:

Cu₃(PO₄)₂(s) ⇌ 3 Cu²⁺ + 2 PO₄³⁻. If the molar solubility is s, K_sp = (3s)³(2s)² = 108s⁵. Therefore s = 1.0 × 10⁻⁶ M and [Cu²⁺] = 3s = 3 × 10⁻⁶ M (D), under the stated simplified model.

9 / 10

A saturated Al(OH)₃ solution has [Al³⁺] = 2.0 × 10⁻⁵ M. What is K_sp if the only ion source is stoichiometric dissolution, and hydrolysis, complex formation, water autoionization, and activity corrections are neglected?

A. 4.0× 10^-10

B. 2.7 × 10^-18

C. 4.3×10^-18

D. 8.6 × 10^-16

E. 2.0 × 10^-20

Explanation:

Under the stated model, Al(OH)₃(s) ⇌ Al³⁺ + 3 OH⁻, so [OH⁻] = 3(2.0 × 10⁻⁵) = 6.0 × 10⁻⁵ M. K_sp = [Al³⁺][OH⁻]³ = (2.0 × 10⁻⁵)(6.0 × 10⁻⁵)³ = 4.32 × 10⁻¹⁸ ≈ 4.3 × 10⁻¹⁸ (C).

10 / 10

Which statement about equilibrium and precipitation is INCORRECT?

A. K greater than 1 indicates a product-favored equilibrium.

B. If Q is greater than K, the reaction shifts toward reactants.

C. Precipitation begins when the ion product Q_sp is less than K_sp.

D. If Q_sp is less than K_sp, the solution is unsaturated with respect to the solid.

E. The value of K_sp must be considered with dissolution stoichiometry when comparing molar solubilities.

Explanation:

C is incorrect: precipitation is thermodynamically favored when Q_sp is greater than K_sp. If Q_sp is less than K_sp, the solution is unsaturated. K alone does not determine freezing point or electrical conductivity across different salts; those also depend on stoichiometry, concentrations, ion properties, and other conditions.

Your score is

The average score is 64%

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