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.

pH = pKa + log([A⁻]/[HA]). pKa = -log(1.8 × 10⁻⁵) ≈ 4.74. Since [A⁻] = [HA] (equal concentrations), log(1) = 0. Therefore, pH = 4.74.

At equilibrium, Kc = kf/kr. Given Kc = 4, this means kf = 4kr, indicating the forward reaction is faster than the reverse reaction.

This is an exothermic reaction (ΔH negative). By Le Chatelier's principle, decreasing temperature shifts equilibrium towards the products (right), favoring the exothermic forward reaction.

A process is spontaneous when ΔG < 0. Neither ΔH nor ΔS alone determines spontaneity; it depends on their combined effect at a given temperature.

ΔH = Ea(forward) - Ea(reverse) = 50 - 35 = 15 kJ/mol. The positive value indicates an endothermic reaction.

Using Raoult's law for partial pressures and ideal solutions: Total P = P₁X₁ + P₂X₂. Solving for mole fraction of ethanol: X_ethanol = (Total P - P_water)/(P_ethanol - P_water) ≈ 0.30

Colligative properties (boiling point elevation, freezing point depression, osmotic pressure, vapour pressure lowering) depend only on the number of solute particles, not on their chemical nature.

If pH = 2, then [H⁺] = 10⁻² = 0.01 M. For a 0.1 M HCl solution, α = [H⁺]/C₀ = 0.$\frac{01}{0}$.1 = 0.1. However, HCl is a strong acid with α ≈ 1.0 (essentially complete dissociation).

At equilibrium: P(N₂O₄) = (1-x) atm, P(NO₂) = 2x atm. Kp = (2x)²/(1-x) = 0.15. Solving: 4x² = 0.15(1-x), which gives x ≈ 0.23.

ΔG° = -nFE°cell. For a typical 2-electron transfer: ΔG° = -2 × 96500 × 1.5 = -289,500 J/mol ≈ -289.5 kJ/mol.

For a weak acid, the degree of dissociation relates the initial concentration to the equilibrium constant through the ionization expression.

Step 1: Set up the Ionization Expression

For a weak acid HA, the degree of dissociation (α) is the fraction of acid molecules that dissociate. At equilibrium, if initial concentration is C₀ and α fraction dissociates:

Step 2: Apply the Ka Expression

The acid dissociation constant is defined as:

Since α = 0.01 is very small, (1 - α) ≈ 1:

Step 3: Solve for Initial Concentration

Rearranging for C₀:

The initial concentration of the acid is 0.1 M.

**Answer: (A) 0.1 M

Using Dalton's law: Partial pressure = Mole fraction × Total pressure. P(O₂) = (2/(2+3)) × 5 = ($\frac{2}{5}$) × 5 = 2 atm.

For second-order reactions, the integrated rate law relates concentration changes to time through an inverse relationship.

Step 1: [Identify the Integrated Rate Law for Second-Order Reactions]

For a second-order reaction, the integrated rate law is given by:

Step 2: [Substitute Given Values]

We have k = 0.5 L·mol⁻¹·s⁻¹, [A]₀ = 2 M, and [A]_t = 0.5 M. Substituting into the equation:

The time required for the concentration to reduce to 0.5 M is 3 seconds.

The answer is (D) 3 s.

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)