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Class XI · Chemistry

Equilibrium: NEET MCQ practice

345 practice questions on Equilibrium, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

345

Practice questions

119

Easy

198

Medium

28

Hard

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What Equilibrium covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Meaning of equilibrium, concept of dynamic equilibrium
  • Equilibria involving physical processes- solid-liquid, liquid-gas, solid-gas equilibria, Henry's law
  • Equilibria involving chemical processes- law of chemical equilibrium, equilibrium constants (Kp, Kc)
  • Le Chatelier's principle
  • Ionic equilibrium- ionization of electrolytes, acid-base equilibria, ionization of water, pH scale
  • Common ion effect, hydrolysis of salts, buffer solutions, solubility product

Sample questions on Equilibrium

Try each one, then open the answer and explanation.

Question 1 · Practice question

0.70 g of mixture (NH₄)₂SO₄ was boiled with 100 mL of 0.2 N NaOH solution till all the NH₃ (g) evolved and get dissolved in solution itself. The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H₂SO₄ solution. The percentage purity of the (NH₄)₂SO sample is

  1. A.94.3
  2. B.50.8
  3. C.47.4
  4. D.79.8
Show answer

Answer: A. 94.3

This involves a back-titration where ammonium sulphate reacts with NaOH to release NH₃ gas, which is boiled off, leaving behind excess NaOH. The leftover NaOH is titrated with H₂SO₄. First, calculate initial NaOH equivalents (20 meq). Let x moles of (NH₄)₂SO₄ react, consuming 2x moles NaOH. NaOH left = (0.02 − 2x) moles. After dilution and aliquot titration, NaOH left corresponds to 1 meq acid, giving x = 0.005 mol and mass = 0.66 g. Purity = (0.66/0.70)×100 ≈ 94.3%. The tricky part is recognizing that NH₃ is completely removed, so only excess NaOH is titrated; misreading the phrase about NH₃ dissolving might wrongly make you include NH₃ in the titration.

Question 2 · Practice question

Conversion factor for converting partial pressures (in K p ) to active masses (in K c ) is

  1. A.nRT
  2. B.1/RT
  3. C.(RT) 2
  4. D.1/RT 2
Show answer

Answer: B. 1/RT

For a gas, partial pressure is related to concentration by P = CRT, so converting pressure to concentration requires dividing by RT, i.e., factor = 1/RT. Option (a) nRT is wrong because it multiplies instead of dividing, and 'n' here is confusing—should not appear in the factor.

Question 3 · Practice question

Solubility of s block halides in water depends upon:

  1. A.Ionization energy of metal
  2. B.Lattice energy of compound
  3. C.Hydration energy of ions
  4. D.Both 2 and 3
Show answer

Answer: D. Both 2 and 3

Solubility of ionic compounds like s-block halides depends on the balance between lattice energy (energy needed to break the solid) and hydration energy (energy released when ions are surrounded by water). If hydration energy outweighs lattice energy, the compound is more soluble. Ionization energy is not a direct factor here because we are dealing with pre-formed ions in the solid.

Question 4 · Practice question

Once the equilibrium is reached under given condition:

  1. A.Cone, remains the same in spite of the change in temperature
  2. B.Cone, of all the substances presents do not change
  3. C.Cone, of reactants remairfs same
  4. D.Cone, of products remains same
Show answer

Answer: B. Cone, of all the substances presents do not change

At equilibrium, the concentrations of reactants and products no longer change with time — all species remain constant. Option (b) correctly states this definition. Option (a) is wrong because temperature changes will shift equilibrium and change concentrations; options (c) and (d) are incomplete because both reactants and products remain constant, not just one of them.

Question 5 · Practice question

Which of the following will produce a buffer solution when mixed in equal volumes?

  1. A.0.1 mol dm –3 NH₄OH and 0.1 mol dm –3 HCl
  2. B.0.05 mol dm –3 NH₄OH and 0.1 mol dm –3 HCl
  3. C.0.1 mol dm –3 NH₄OH and 0.05 mol dm –3 HCl
  4. D.0.1 mol dm –3 CH₄COONa and 0.1 mol dm –3 NaOH
Show answer

Answer: C. 0.1 mol dm –3 NH₄OH and 0.05 mol dm –3 HCl

NH₄OH (weak base) reacts with HCl (strong acid) to form NH₄Cl. In option C, after mixing equal volumes, the amounts are such that some NH₄OH remains unreacted along with the NH₄Cl formed, creating a basic buffer. Option A results in complete neutralization leaving only salt, not a buffer. Option D combines NaOH with CH₃COONa, which lacks the weak acid needed for a buffer.

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