NEET Chemistry - MCQ Practice Questions

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

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.

Using Arrhenius equation: ln(k₂/k₁) = (Ea/R)[1/T₁ - 1/T₂]. ln(1.5) = (Ea/8.314)[$\frac{1}{300}$ - $\frac{1}{310}$]. Ea ≈ 52.8 kJ/mol

For first-order: [A] = [A₀]e^(-kt) = 0.1 × e^(-0.01×100) = 0.1 × e^(-1) = 0.1 × 0.368 = 0.0368 M

ICE table: At equilibrium, [A] = 1-2x, [B] = 1-x, [C] = x. Kc = x/[(1-2x)²(1-x)] = 0.5. Solving: x ≈ 0.414 M

Higher the reduction potential, stronger the oxidizing agent. Ag⁺ (E° = 0.80 V) > Cu²⁺ (E° = 0.34 V). Ag⁺ is strongest oxidizer.

α = √(Ka/C) = √(1.8 × 10⁻⁵/0.1) = √(1.8 × 10⁻⁴) = 0.0424 = 4.24%

ΔG° = -RT ln K. With ΔG° = -40 kJ/mol = -40000 J/mol, ln K = 40000/(8.314 × 298) ≈ 16.2, so K = e^16.2 >> 1 (very large).

Boiling points of hydrides depend on intermolecular forces, which vary based on molecular mass and hydrogen bonding strength.

Step 1: Identify Group 16 Hydrides and Their Boiling Points

Group 16 elements (chalcogens) form hydrides with the general formula H₂X. Let's list their boiling points:

Step 2: Analyze Intermolecular Forces

H₂O has exceptional boiling point due to strong hydrogen bonding between oxygen (highly electronegative) and hydrogen. However, S, Se, and Te are less electronegative, so H₂S, H₂Se, and H₂Te exhibit only weak dipole-dipole interactions and London dispersion forces:

Among H₂S, H₂Se, and H₂Te, molecular mass increases down the group, strengthening London dispersion forces. H₂S has the smallest molecular mass (34 g/mol) among these three, resulting in the weakest intermolecular forces.

The correct answer is (B) H₂S with a boiling point of -60°C, the lowest among the group.

For a first-order reaction, the relationship between time, concentration, and rate constant follows a logarithmic decay pattern that connects to the half-life.

Step 1: Apply the First-Order Rate Law

For a first-order reaction, we use the integrated rate equation that relates remaining concentration to time.

Since the reaction is 75% complete, 25% of the reactant remains, so $\frac{[A{]}_t}{[A{]}_0}=0.25$ or $\frac{[A{]}_0}{[A{]}_t}=4$

Step 2: Calculate Half-Life Using the Half-Life Formula

The half-life for a first-order reaction is independent of initial concentration and is given by:

The half-life of the reaction is 22.5 minutes.

The answer is (B) 22.5 minutes.

Initial: PA = 2, PB = 1. Change: -1.5 for A, -1.5 for B, +1.5 for C and D. Equilibrium: PA = 0.5, PB = -0.5 (invalid). Using Kp: 4 = (PC × PD)/(0.5 × PB). Since stoichiometry is 1:1:1:1, PC = PD = 1.5 atm.