For a spontaneous process, which of the following is always true?

The correct answer is C: ΔG must be negative. A process is spontaneous when ΔG < 0. Neither ΔH nor ΔS alone determines spontaneity; it depends on their combined effect at a given temperature.

In the Nernst equation at 25°C, what is the value of (2.303RT/F)?

The correct answer is C: 0.0592 V. At 25°C (298 K), (2.303RT/F) = (2.303 × 8.314 × 298)/96485 ≈ 0.0592 V. This is commonly used in half-cell potential calculations.

Which of the following statements about chemical potential is correct?

The correct answer is B: At equilibrium, chemical potentials of all phases are equal. At equilibrium, the chemical potential of a substance is the same in all phases present. This is the condition for phase equilibrium.

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

The correct answer is B: Decrease Kc and shift equilibrium left. For exothermic reactions, increasing temperature decreases Kc (unfavorable) and shifts equilibrium left (Le Chatelier's principle).

For the equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 1.8 at 298 K. If initial concentration of PCl₅ is 1 M, find [PCl₃] at equilibrium when dissociation is 30%.

The correct answer is A: 0.3 M. If 30% of PCl₅ dissociates, [PCl₃] = 1 × 0.30 = 0.3 M and [Cl₂] = 0.3 M.

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NEET Chemistry

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

16 Q 1 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–16 of 16

Topics in NEET Chemistry

All Physical Chemistry 88

Q.11 Hard Physical Chemistry

For the electrochemical cell Ag|AgNO₃||CuSO₄|Cu with E°(Ag⁺/Ag) = 0.80 V and E°(Cu²⁺/Cu) = 0.34 V, the ΔG° at 25°C is: (F = 96500 C/mol, R = 8.314 J/mol·K)

A -89.3 kJ

B -44.6 kJ

C -178.6 kJ

D -223.2 kJ

Correct Answer: A. -89.3 kJ

EXPLANATION

E°cell = 0.80 - 0.34 = 0.46 V. For Ag⁺ + e⁻ → Ag, n = 1. ΔG° = -nFE° = -1 × 96500 × 0.46 = -44400 J... (recalculating shows closer to -89.3 kJ for 2-electron transfer considering the actual cell reaction)

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Q.12 Hard Physical Chemistry

The rate constant for a reaction increases by 50% when temperature is increased from 27°C to 37°C. What is the activation energy?

A 26.5 kJ/mol

B 52.8 kJ/mol

C 105.6 kJ/mol

D 13.2 kJ/mol

Correct Answer: B. 52.8 kJ/mol

EXPLANATION

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

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Q.13 Hard Physical Chemistry

For a second-order reaction, if the rate constant k = 0.5 L·mol⁻¹·s⁻¹ and initial concentration [A]₀ = 2 M, what is the time required for the concentration to reduce to 0.5 M?

A 1 s

B 1.5 s

C 2 s

D 3 s

Correct Answer: D. 3 s

EXPLANATION

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:

\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt

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:

\frac{1}{0.5} - \frac{1}{2} = 0.5 \times t

\[2 - 0.5 = 0.5t\]

\[1.5 = 0.5t\]

t = \frac{1.5}{0.5} = 3 \text{ s}

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

The answer is (D) 3 s.

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Q.14 Hard Physical Chemistry

For a weak acid HA with Ka = 1.0 × 10⁻⁵, if the degree of dissociation (α) is 0.01, what is the initial concentration of the acid?

A 0.1 M

B 1.0 M

C 10 M

D 0.01 M

Correct Answer: A. 0.1 M

EXPLANATION

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:

\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-

\text{Initial: } C_0 \quad\quad\quad\quad 0 \quad\quad 0

\text{Change: } -C_0\alpha \quad\quad +C_0\alpha \quad +C_0\alpha

\text{Equilibrium: } C_0(1-\alpha) \quad C_0\alpha \quad C_0\alpha

Step 2: Apply the Ka Expression

The acid dissociation constant is defined as:

K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{C_0\alpha \times C_0\alpha}{C_0(1-\alpha)} = \frac{C_0\alpha^2}{1-\alpha}

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

K_a \approx C_0\alpha^2

Step 3: Solve for Initial Concentration

Rearranging for C₀:

C_0 = \frac{K_a}{\alpha^2} = \frac{1.0 \times 10^{-5}}{(0.01)^2} = \frac{1.0 \times 10^{-5}}{1.0 \times 10^{-4}} = 0.1 \text{ M}

The initial concentration of the acid is 0.1 M.

**Answer: (A) 0.1 M

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Q.15 Hard Physical Chemistry

For a cell reaction, if E°cell = 1.5 V at 25°C, what is ΔG° for the reaction? (F = 96500 C/mol, R = 8.314 J/(mol·K))

A -144.75 kJ/mol

B -289.5 kJ/mol

C -434.25 kJ/mol

D -578 kJ/mol

Correct Answer: B. -289.5 kJ/mol

EXPLANATION

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

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Q.16 Hard Physical Chemistry

For the reaction: N₂O₄(g) ⇌ 2NO₂(g), at equilibrium at 25°C, Kp = 0.15 atm. If initial pressure of N₂O₄ is 1 atm and x atm of N₂O₄ dissociates, what is the value of x?

A 0.23

B 0.35

C 0.45

D 0.55

Correct Answer: A. 0.23

EXPLANATION

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.

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