NEET Chemistry - MCQ Practice Questions

The equilibrium constant expression follows the law of mass action where Kc = [products]^coefficients/[reactants]^coefficients. For this reaction, Kc = [NH₃]²/([N₂][H₂]³).

HCl is a strong acid and completely dissociates. [H⁺] = 0.01 M = 10⁻² M. Therefore, pH = -log[H⁺] = -log(10⁻²) = 2.

For a first-order reaction, t₁/₂ = 0.693/k = 0.$\frac{693}{0}$.693 = 1 s. This uses the relationship ln(2) ≈ 0.693.

The overall order of a reaction is the sum of all exponents in the rate law. Here, order = 2 + 1 = 3.

The Gibbs free energy equation is ΔG = ΔH - TΔS, where T is temperature in Kelvin. This is fundamental to predicting spontaneity.

For AgCl, Ksp = [Ag⁺][Cl⁻] = s². So s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/L

MgO has the smallest ionic radii and highest charge density among options, leading to maximum lattice energy (~3850 kJ/mol)

Given reaction releases 572 kJ for 2 moles of H₂. For 4 moles of H₂, heat released = ($\frac{572}{2}$) × 4 = 1144 kJ

π = nRT/V = (0.1 × 0.0821 × 300)/1 = 2.463 atm

For first-order reaction: t₁/₂ = 0.693/k, so k = 0.$\frac{693}{69}$.3 = 1.0 × 10⁻³ min⁻¹

E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 1.10 V

For l = 2 (d-subshell), ml values range from -2 to +2, giving 5 orbitals

Moles of sucrose = $\frac{342}{342}$ = 1 mol. Molarity = 1 mol/1 L = 1.0 M

The given reaction forms 2 moles of H₂O with ΔH° = -572 kJ/mol. Therefore, for 1 mole of H₂O, ΔH° = -$\frac{572}{2}$ = -286 kJ/mol.

At constant T and P, the Gibbs free energy change (ΔG = ΔH - TΔS) determines spontaneity. If ΔG < 0, the reaction is spontaneous.

E°cell = E°cathode - E°anode = 0.80 - (-0.40) = 1.20 V. A positive E°cell indicates a spontaneous galvanic cell.

For first-order reactions, t₁/₂ = 0.693/k = 0.$\frac{693}{0}$.693 = 1.0 s.

In endothermic reactions, ΔH is positive because heat is absorbed from surroundings. The system absorbs energy.

pH = -log[H⁺], so [H⁺] = 10⁻ᵖᴴ = 10⁻⁴ = 1 × 10⁻⁴ M.

Cu²⁺ + 2e⁻ → Cu. For deposition of 1 mole Cu, 2 moles of electrons are required. So 2 moles e⁻ deposits 1 mole Cu.