Chemistry — Physical Chemistry for the JEE Main Exam — Saudi candidates

Common failure modes

These are the patterns that cause most candidates to lose marks on this topic. Recognising them in advance is half the work.

• !Confusing ΔG with ΔG° — forgetting the Q (reaction quotient) term in ΔG = ΔG° + RT ln Q
• !Misapplying the Nernst equation when concentrations are non-standard
• !Using first-order half-life formula for a second-order reaction
• !Forgetting to convert temperature to Kelvin in Arrhenius and thermodynamic equations
• !Misidentifying buffer problems — Henderson-Hasselbalch requires a weak acid and its conjugate base, not any two acids

Study tips

• 1Build a formula card for physical chemistry: ΔG = ΔH − TΔS, Nernst equation, Arrhenius equation, Van't Hoff factor, and Raoult's law. JEE Main is formula-driven here.
• 2Practice electrochemistry cell-notation reading — JEE routinely asks cell EMF from given half-reactions.
• 3Drill rate-law problems by order determination from experimental data (method of initial rates).
• 4For solutions, practice colligative property calculations: depression of freezing point, elevation of boiling point, osmotic pressure.
• 5Do at least 5 past-paper thermodynamics questions per sitting — the multi-step Hess's law + entropy problems are high-value.
• 6Saudi candidates preparing for JEE Main can leverage the existing GAT (Qudurat) preparation infrastructure — many concepts (verbal reasoning, quantitative comparison) transfer directly.

Sample JEE Main Chemistry — Physical Chemistry questions

These sample items mirror the format and difficulty of real JEE Main questions. Practice with thousands more on the free Koydo question bank.

1. 1

   For a reaction at equilibrium at 298 K, if ΔG° = −5.7 kJ/mol, the equilibrium constant K is approximately (R = 8.314 J/mol·K):

   • A10Correct
   • B100
   • C1000
   • D0.1
   Why this answer?

   Illustrative JEE-style: ΔG° = −RT ln K, so ln K = −ΔG°/(RT) = 5700/(8.314 × 298) ≈ 2.30, giving K ≈ e^2.30 ≈ 10.

2. 2

   For a first-order reaction, the time required for 75% completion is:

   • At½
   • B2t½Correct
   • C3t½
   • D4t½
   Why this answer?

   Illustrative JEE-style: At 75% completion, 25% remains. Using N = N₀(1/2)^n, 0.25 = (1/2)^n gives n = 2. So time = 2t½.

3. 3

   Electrolysis of water produces hydrogen and oxygen in a volume ratio of:

   • A1 : 1
   • B2 : 1Correct
   • C1 : 2
   • D4 : 1
   Why this answer?

   Illustrative JEE-style: 2H₂O → 2H₂ + O₂. At the same temperature and pressure, volumes are proportional to moles, giving H₂ : O₂ = 2 : 1.

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