Chemistry · free practice

JEE Main Physical Chemistry: practice questions with solutions

21 questions on Atomic Structure, Chemical Equilibrium, Chemical Kinetics, Electrochemistry, Ionic Equilibrium, Mole Concept, Solutions, States of Matter, Thermodynamics. Try each one first, then open the solution.

Tip: solve on paper before opening a solution. Once you open it, that question is counted as practised and won't appear in your HeyGyan tests.
Mole ConceptEasy

Q1. What volume of O₂ at STP is needed for complete combustion of 4 g of methane? (Molar volume at STP = 22.4 L)

  1. 5.6 L
  2. 11.2 L
  3. 22.4 L
  4. 44.8 L
Show answer and solution

Answer: (B) 11.2 L

SolutionCH₄ + 2O₂ → CO₂ + 2H₂O. 4 g CH₄ = 0.25 mol, which needs 0.5 mol O₂ = 0.5 × 22.4 = 11.2 L.
Common trap5.6 L forgets the 2:1 ratio of O₂ to CH₄.
Atomic StructureMedium

Q2. What is the ratio of the radius of the 2nd Bohr orbit of He⁺ to that of the 3rd Bohr orbit of Li²⁺?

  1. 2 : 3
  2. 3 : 2
  3. 4 : 9
  4. 1 : 1
Show answer and solution

Answer: (A) 2 : 3

Solutionr ∝ n²/Z. He⁺ (n=2, Z=2): 4/2 = 2. Li²⁺ (n=3, Z=3): 9/3 = 3. Ratio = 2 : 3.
Common trap4 : 9 compares only n² and ignores the nuclear charge Z.
ThermodynamicsMediumNumerical

Q3. For a reaction, ΔH = −100 kJ mol⁻¹ and ΔS = −200 J K⁻¹ mol⁻¹. Above what temperature (in K) does the reaction become non-spontaneous? (Assume ΔH and ΔS do not change with temperature.)

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 500

SolutionΔG = ΔH − TΔS changes sign when T = ΔH/ΔS = (−100000)/(−200) = 500 K. Above this, −TΔS (positive) outweighs ΔH.
Common trapForgetting to convert kJ to J gives 0.5 K.
Ionic EquilibriumMedium

Q4. A buffer contains 0.02 M acetic acid and 0.2 M sodium acetate. What is its pH? (pKₐ of acetic acid = 4.74)

  1. 3.74
  2. 4.74
  3. 5.74
  4. 6.74
Show answer and solution

Answer: (C) 5.74

SolutionHenderson equation: pH = pKₐ + log([salt]/[acid]) = 4.74 + log(10) = 5.74.
Common trapInverting the ratio (acid/salt) gives 3.74.
ElectrochemistryMedium

Q5. E°cell for the Daniell cell (Zn | Zn²⁺ || Cu²⁺ | Cu) is 1.10 V. What is ΔG° for the cell reaction? (F = 96500 C mol⁻¹)

  1. −106.15 kJ
  2. −212.3 kJ
  3. +212.3 kJ
  4. −424.6 kJ
Show answer and solution

Answer: (B) −212.3 kJ

SolutionΔG° = −nFE° with n = 2 electrons: −2 × 96500 × 1.10 = −212300 J = −212.3 kJ.
Common trapTaking n = 1 gives −106.15 kJ.
Chemical KineticsEasy

Q6. A first-order reaction is 75% complete in 32 minutes. How long does it take to be 50% complete?

  1. 8 min
  2. 16 min
  3. 24 min
  4. 64 min
Show answer and solution

Answer: (B) 16 min

Solution75% completion means 25% remains = (1/2)², which takes two half-lives. So t½ = 32/2 = 16 min.
Common trapAssuming the reaction is linear (75% → 32 min, so 50% → 21.3 min).
SolutionsEasyNumerical

Q7. What is the van't Hoff factor (i) of K₄[Fe(CN)₆], assuming complete dissociation in water?

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 5

SolutionK₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻. The complex ion stays intact, so it gives 5 ions: i = 5.
Common trapBreaking the complex into Fe and CN ions. Coordination complexes stay intact.
States of MatterMedium

Q8. Equal masses of methane and oxygen are mixed in an empty container. What fraction of the total pressure is exerted by oxygen?

  1. 1/2
  2. 1/3
  3. 2/3
  4. 1/4
Show answer and solution

Answer: (B) 1/3

SolutionFor mass m: moles of CH₄ = m/16 and O₂ = m/32. Partial pressure ∝ moles, so O₂'s share = (1/32)/(1/16 + 1/32) = 1/3.
Common trapEqual masses do not mean equal moles.
Chemical EquilibriumEasy

Q9. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), how is Kₚ related to K꜀?

  1. Kₚ = K꜀(RT)²
  2. Kₚ = K꜀(RT)⁻²
  3. Kₚ = K꜀
  4. Kₚ = K꜀(RT)⁻¹
Show answer and solution

Answer: (B) Kₚ = K꜀(RT)⁻²

SolutionKₚ = K꜀(RT)^Δn, with Δn = moles of gaseous products − reactants = 2 − 4 = −2.
Common trapCalculating Δn as reactants − products flips the sign.
Chemical KineticsEasyNumerical

Q10. The rate of a reaction doubles for every 10 °C rise in temperature. By what factor does the rate increase when the temperature rises from 20 °C to 50 °C?

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 8

SolutionA 30 °C rise is three 10 °C steps, so the rate becomes 2³ = 8 times.
Common trapAnswering 6 by adding instead of multiplying the doublings.
SolutionsHard

Q11. A compound AB₂ dissociates 50% in water (AB₂ → A²⁺ + 2B⁻). What is the depression in freezing point of its 0.1 molal solution? (K_f = 1.86 K kg mol⁻¹)

  1. 0.186 K
  2. 0.279 K
  3. 0.372 K
  4. 0.558 K
Show answer and solution

Answer: (C) 0.372 K

SolutionEach AB₂ gives 3 ions, so i = 1 + (3 − 1)α = 1 + 2 × 0.5 = 2. ΔT_f = i K_f m = 2 × 1.86 × 0.1 = 0.372 K.
Common trap0.558 K assumes complete dissociation (i = 3).
ElectrochemistryMedium

Q12. For the cell Zn | Zn²⁺(0.1 M) || Cu²⁺(1 M) | Cu with E° = 1.10 V, what is the cell potential at 298 K? (2.303RT/F = 0.059 V)

  1. 1.07 V
  2. 1.10 V
  3. 1.13 V
  4. 1.16 V
Show answer and solution

Answer: (C) 1.13 V

SolutionE = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]) = 1.10 − 0.0295 × log(0.1) = 1.10 + 0.0295 ≈ 1.13 V.
Common trapPutting [Cu²⁺]/[Zn²⁺] in the log gives 1.07 V. Products go on top.
Mole ConceptEasy

Q13. 4 g of NaOH is dissolved in water to make 250 mL of solution. What is its molarity? (Na = 23, O = 16, H = 1)

  1. 0.1 M
  2. 0.2 M
  3. 0.4 M
  4. 1 M
Show answer and solution

Answer: (C) 0.4 M

SolutionMoles of NaOH = 4/40 = 0.1 mol. Molarity = 0.1 mol ÷ 0.25 L = 0.4 M.
Common trap0.1 M forgets to convert 250 mL to 0.25 L.
Atomic StructureMediumNumerical

Q14. What is the total number of nodes (radial + angular) in a 4f orbital?

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 3

SolutionTotal nodes = n − 1 = 3. For 4f (l = 3): angular nodes = l = 3, radial nodes = n − l − 1 = 0.
Common trapCounting only radial nodes gives 0.
ThermodynamicsEasy

Q15. Given: C + O₂ → CO₂, ΔH = −393.5 kJ and CO + ½O₂ → CO₂, ΔH = −283.0 kJ. What is the enthalpy of formation of CO?

  1. −676.5 kJ
  2. −110.5 kJ
  3. +110.5 kJ
  4. −283.0 kJ
Show answer and solution

Answer: (B) −110.5 kJ

SolutionHess's law: C + ½O₂ → CO is the first reaction minus the second. ΔH = −393.5 − (−283.0) = −110.5 kJ.
Common trapAdding the two values gives −676.5 kJ.
Ionic EquilibriumMedium

Q16. The solubility product of a salt AB₂ is 4 × 10⁻¹². What is its molar solubility?

  1. 2 × 10⁻⁶ M
  2. 1 × 10⁻⁴ M
  3. 2 × 10⁻⁴ M
  4. 1 × 10⁻⁶ M
Show answer and solution

Answer: (B) 1 × 10⁻⁴ M

SolutionAB₂ ⇌ A²⁺ + 2B⁻. With solubility s: Ksp = s(2s)² = 4s³. So 4s³ = 4 × 10⁻¹², s³ = 10⁻¹², s = 10⁻⁴ M.
Common trapWriting Ksp = s² (as for an AB salt) gives 2 × 10⁻⁶ M.
Chemical KineticsEasy

Q17. For a zero-order reaction, what happens to the half-life if the initial concentration is doubled?

  1. It halves
  2. It stays the same
  3. It doubles
  4. It becomes four times
Show answer and solution

Answer: (C) It doubles

SolutionFor zero order, t½ = [A]₀/2k, which is directly proportional to the initial concentration. Doubling [A]₀ doubles t½.
Common trap'Stays the same' is true for first-order reactions, not zero order.
ElectrochemistryEasy

Q18. How much charge is needed to deposit 1 mole of aluminium from Al³⁺ ions? (F = 96500 C mol⁻¹)

  1. 32167 C
  2. 96500 C
  3. 193000 C
  4. 289500 C
Show answer and solution

Answer: (D) 289500 C

SolutionAl³⁺ + 3e⁻ → Al, so 1 mol Al needs 3 mol of electrons = 3F = 3 × 96500 = 289500 C.
Common trapUsing 1F forgets that each Al³⁺ ion takes three electrons.
States of MatterMedium

Q19. Under the same conditions, what is the ratio of the time taken for equal volumes of O₂ and H₂ to effuse, t(O₂) : t(H₂)?

  1. 1 : 4
  2. 4 : 1
  3. 1 : 16
  4. 16 : 1
Show answer and solution

Answer: (B) 4 : 1

SolutionRate ∝ 1/√M, so time ∝ √M. t(O₂)/t(H₂) = √(32/2) = 4. The heavier gas takes 4 times longer.
Common trap1 : 4 is the ratio of rates, not times.
ThermodynamicsEasy

Q20. For the isothermal reversible expansion of an ideal gas, which statement is correct?

  1. ΔU = 0
  2. q = 0
  3. w = 0
  4. ΔH > 0
Show answer and solution

Answer: (A) ΔU = 0

SolutionThe internal energy of an ideal gas depends only on temperature. At constant temperature ΔU = 0 (and ΔH = 0), so the heat absorbed equals the work done by the gas.
Common trapq = 0 describes an adiabatic process, not an isothermal one.
Ionic EquilibriumEasyNumerical

Q21. What is the pH of a 0.01 M NaOH solution at 25 °C?

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 12

SolutionNaOH is a strong base: [OH⁻] = 10⁻² M, so pOH = 2. pH = 14 − 2 = 12.
Common trapAnswering 2, which is the pOH.

Frequently asked questions

Which Physical Chemistry chapters are covered here?

This page covers Atomic Structure, Chemical Equilibrium, Chemical Kinetics, Electrochemistry, Ionic Equilibrium, Mole Concept, Solutions, States of Matter, Thermodynamics. Questions follow the JEE Main pattern, with multiple-choice and numerical value types.

Are these previous year JEE questions?

No. These are original JEE Main-level practice questions written on the latest pattern, with solutions and the common trap for each.