For the reaction: 2A + B → C, if the rate law is Rate = k[A]²[B], what is the overall order of the reaction?

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

For the reaction: 2A + B ⇌ C, Kc = 0.5 at 298 K. If initial concentrations are [A] = 1 M, [B] = 1 M, [C] = 0, what is the equilibrium concentration of C?

The correct answer is A: 0.414 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

The rate constant of a first-order reaction is 0.693 s⁻¹. What is the half-life of this reaction?

The correct answer is B: 1 s. For a first-order reaction, t₁/₂ = 0.693/k = 0.693/0.693 = 1 s. This uses the relationship ln(2) ≈ 0.693.

A gas mixture contains 2 moles of O₂ and 3 moles of N₂. At total pressure of 5 atm, what is the partial pressure of O₂?

The correct answer is B: 2 atm. Using Dalton's law: Partial pressure = Mole fraction × Total pressure. P(O₂) = (2/(2+3)) × 5 = (2/5) × 5 = 2 atm.

In electrochemistry, the reduction potential of Cu²⁺/Cu is +0.34 V and Zn²⁺/Zn is -0.76 V at 25°C. What is E°cell for Zn-Cu cell?

The correct answer is B: 1.10 V. E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 1.10 V

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NEET Chemistry
Physical Chemistry

Chemistry questions for NEET UG — Physical, Organic, Inorganic Chemistry.

41 Q 1 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–20 of 41

Topics in NEET Chemistry

All Physical Chemistry 88

Q.11 Medium Physical Chemistry

At 25°C, for the reaction: Cu²⁺(aq) + Zn(s) → Cu(s) + Zn²⁺(aq), E° = 1.1 V. What is ΔG°? (F = 96500 C/mol)

A -212.3 kJ/mol

B -106.15 kJ/mol

C -424.6 kJ/mol

D -53.075 kJ/mol

Correct Answer: A. -212.3 kJ/mol

EXPLANATION

ΔG° = -nFE° = -2 × 96500 × 1.1 = -212300 J/mol = -212.3 kJ/mol

Test

Q.12 Medium Physical Chemistry

In a galvanic cell, at the cathode:

A Oxidation occurs and it's the positive electrode

B Reduction occurs and it's the negative electrode

C Reduction occurs and it's the positive electrode

D Oxidation occurs and it's the negative electrode

Correct Answer: C. Reduction occurs and it's the positive electrode

EXPLANATION

Cathode is where reduction occurs. It's the positive electrode in a galvanic cell as it attracts cations.

Test

Q.13 Medium Physical Chemistry

A buffer solution is prepared by mixing 100 mL of 0.1 M CH₃COOH and 100 mL of 0.1 M CH₃COONa. The pKa of acetic acid is 4.74. What is the pH?

A 3.74

B 4.74

C 5.74

D 6.74

Correct Answer: B. 4.74

EXPLANATION

Using Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.1/0.1) = 4.74

Test

Q.14 Medium Physical Chemistry

For the exothermic reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), increasing temperature will:

A Increase Kc and shift equilibrium right

B Decrease Kc and shift equilibrium left

C Increase Kc and shift equilibrium left

D No effect on equilibrium

Correct Answer: B. Decrease Kc and shift equilibrium left

EXPLANATION

For exothermic reactions, increasing temperature decreases Kc (unfavorable) and shifts equilibrium left (Le Chatelier's principle).

Test

Q.15 Medium Physical Chemistry

The osmotic pressure of a solution is given by π = iMRT. If 0.1 M glucose solution has osmotic pressure 2.46 atm at 273 K, what is the van't Hoff factor?

A 0.5

B 1

C 2

D 4

Correct Answer: B. 1

EXPLANATION

i = π/(MRT) = 2.46/(0.1 × 0.0821 × 273) ≈ 1. Glucose is non-electrolyte with i = 1

Test

Q.16 Medium Physical Chemistry

A buffer solution contains 0.1 M acetic acid and 0.1 M sodium acetate. If Ka = 1.8 × 10⁻⁵, calculate the pH:

A 4.74

B 3.74

C 5.74

D 2.74

Correct Answer: A. 4.74

EXPLANATION

Using Henderson-Hasselbalch equation: pH = pKa + log([salt]/[acid]) = 4.74 + log(0.1/0.1) = 4.74 + 0 = 4.74.

Test

Q.17 Medium Physical Chemistry

What is the entropy change when 18 g of ice at 273 K is converted to water at 273 K? (ΔfusH = 6 kJ/mol)

A 22.0 J/K

B 11.0 J/K

C 44.0 J/K

D 5.5 J/K

Correct Answer: A. 22.0 J/K

EXPLANATION

ΔS = ΔH/T = (6000 J/mol) / 273 K = 22.0 J/(mol·K). For 18 g (1 mole) of ice, ΔS = 22.0 J/K.

Test

Q.18 Medium Physical Chemistry

For the reaction: A → B, if the order is 2 and rate constant k = 0.5 L·mol⁻¹·s⁻¹, find the half-life when initial concentration is 2 M.

A 1.0 s

B 2.0 s

C 0.5 s

D 4.0 s

Correct Answer: A. 1.0 s

EXPLANATION

For a second-order reaction, the half-life depends inversely on both the rate constant and the initial concentration.

Step 1: [Identify the Reaction Order and Given Data]

We are dealing with a second-order reaction (order = 2) with rate constant k = 0.5 L·mol⁻¹·s⁻¹ and initial concentration [A]₀ = 2 M. We need to find the time required for the concentration to reduce to half its initial value.

\text{Order} = 2, \quad k = 0.5 \text{ L·mol}^{-1}\text{·s}^{-1}, \quad [A]_0 = 2 \text{ M}

Step 2: [Apply the Half-Life Formula for Second-Order Reaction]

For a second-order reaction, the half-life formula is derived from the integrated rate law and is expressed as:

t_{1/2} = \frac{1}{k[A]_0}

Substituting the values:

t_{1/2} = \frac{1}{0.5 \times 2} = \frac{1}{1.0} = 1.0 \text{ s}

The half-life is 1.0 s, so the answer is (A).

Test

Q.19 Medium Physical Chemistry

The standard cell potential for a cell is E° = 1.10 V. Calculate ΔG° (F = 96500 C/mol, n = 2):

A -212.3 kJ/mol

B +212.3 kJ/mol

C -106.15 kJ/mol

D +106.15 kJ/mol

Correct Answer: A. -212.3 kJ/mol

EXPLANATION

ΔG° = -nFE° = -2 × 96500 × 1.10 = -212,300 J/mol = -212.3 kJ/mol.

Test

Q.20 Medium Physical Chemistry

For a reversible adiabatic process of an ideal gas, which statement is correct?

A ΔS = 0 (entropy is constant)

B ΔU = 0

C q = 0 and ΔS > 0

D ΔH = 0

Correct Answer: A. ΔS = 0 (entropy is constant)

EXPLANATION

For a reversible adiabatic process, q = 0 and ΔS = 0 because there is no heat exchange and the process is reversible, making it isentropic.

Test

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