Lesson 5: Chemistry: Physical — Mole, Equilibrium, Electrochemistry, Kinetics
Physical Chemistry is the most calculation-heavy part of NEET Chemistry, and it rewards a single discipline: convert everything to moles (or to concentrations) before you use any ratio, constant or formula.
This lesson covers the mole concept and solutions, chemical equilibrium, electrochemistry and chemical kinetics. Thermodynamics appears only as background here — the ideas of enthalpy, entropy and Gibbs energy that make equilibrium and cell potentials intelligible.
Keep the standard values at hand: molar mass of NaOH is 40 g/mol, the Faraday constant is 96,500 C/mol, and at 298 K, 2.303RT/F is about 0.059 V for the Nernst equation in base-10 logarithms.
Mole, concentration and pH
Moles = mass / molar mass, and molarity = moles / volume in litres. For a strong acid, [H⁺] equals the acid concentration, and pH = −log₁₀[H⁺]; for a strong base, find pOH first and use pH + pOH = 14.
Moles in 8 g of NaOH: 8/40 = 0.2 mol.
Dissolved in 0.5 L, the molarity is 0.2/0.5 = 0.4 mol/L.
Diluting 100 mL of 0.5 mol/L solution to 250 mL: M₂ = M₁V₁/V₂ = 0.5 × 100/250 = 0.2 mol/L.
pH of 0.001 mol/L HCl: pH = −log₁₀(0.001) = 3. For 0.01 mol/L NaOH, pOH = −log₁₀(0.01) = 2, so pH = 14 − 2 = 12.
Equilibrium
For a reaction at equilibrium, Kc = [products]^coefficients / [reactants]^coefficients. Le Chatelier's principle predicts the direction of shift: increasing a reactant's concentration, changing pressure for a gaseous reaction with unequal moles, or changing temperature shifts equilibrium to counteract the change — but a catalyst changes only the speed, never the equilibrium position.
For A ⇌ 2B at equilibrium, [A] = 0.4 mol/L and [B] = 0.8 mol/L.
Kc = [B]²/[A] = 0.8²/0.4 = 0.64/0.4 = 1.6 (units mol/L for this stoichiometry).
For a first-order reaction, the rate constant and half-life are linked by k = 0.693/t½. A reaction with t½ = 20 minutes falls to 1/8 of its starting amount after 60 minutes (three half-lives), and k = 0.693/20 ≈ 0.0347 min⁻¹.
Electrochemistry and kinetics
A galvanic cell's standard potential is E°cell = E°cathode − E°anode. For the familiar zinc–copper cell, E°cathode (Cu²⁺/Cu) = +0.34 V and E°anode (Zn²⁺/Zn) = −0.76 V, so E°cell = 0.34 − (−0.76) = 1.10 V. Gibbs energy follows as ΔG° = −nFE°cell.
E°cell = 0.34 − (−0.76) = 1.10 V.
For n = 2 electrons: ΔG° = −nFE° = −2 × 96,500 × 1.10 = −212,300 J/mol ≈ −212.3 kJ/mol (negative, so the cell reaction is spontaneous).
A charge of 19,300 C passes: moles of electrons = 19,300/96,500 = 0.2 mol. Depositing copper (Cu²⁺ + 2e⁻ → Cu) needs 2 electrons per atom, so 0.1 mol of Cu forms: 0.1 × 63.5 = 6.35 g.
Kinetics: rate laws are determined experimentally, never read off the balanced equation. For a first-order reaction, doubling the concentration doubles the rate; for a second-order reaction in one reactant, doubling the concentration quadruples the rate. Raising temperature increases the fraction of molecules above the activation energy, which is why rate constants rise steeply with temperature.
Common pitfalls
Do not use masses or volumes directly in an equilibrium or rate expression: convert to moles or concentrations first, and square or cube concentrations exactly as the stoichiometry requires.
Do not forget the sign in E°cell = E°cathode − E°anode: subtracting a negative anode potential adds to the cell voltage.
Further study
- NCERT Class 11 and Class 12 Chemistry, chapters on mole concept and solutions, equilibrium, electrochemistry and chemical kinetics.
📝 Quiz
1. How many moles are in 8 g of NaOH (molar mass 40 g/mol)?
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a) Moles = 8/40 = 0.2 mol.
2. What is the pH of 0.001 mol/L HCl?
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b) pH = −log₁₀(0.001) = 3.
3. For the Zn–Cu cell, E°cell with E°cathode = +0.34 V and E°anode = −0.76 V is:
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c) E°cell = 0.34 − (−0.76) = 1.10 V.
4. A first-order reaction has half-life 20 minutes. What fraction remains after 60 minutes?
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c) 60 minutes is three half-lives: (1/2)³ = 1/8.