Daily Olympiad: Chemistry - Physical Chemistry [20260511]

Challenge yourself with today's KVPY practice! This test covers 'Physical Chemistry' for Chemistry (KVPY - 12). Level: Hard | Duration: 45 mins.

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1. A reaction with ΔH° = -50 kJ/mol and ΔS° = -100 J/mol·K is spontaneous at 298 K. What is the temperature range (in K) where the reaction becomes non-spontaneous?

Solution
Correct: B
Using ΔG° = ΔH° - TΔS°, for ΔG° ≥ 0 (non-spontaneous), T ≥ ΔH°/ΔS° = (-50 x 10³ J/mole)/(-100 J/mol·K) = 500 K. Thus, reaction is non-spontaneous above 500 K.

2. The van der Waals equation for a real gas is (P + a/n²V²)(V - nb) = nRT. Which term accounts for molecular volume?

Solution
Correct: B
The term 'b' corrects for the finite volume of gas molecules, reducing the available volume to (V - nb) in the equation.

3. A 0.1 M solution of acetic acid (Ka = 1.8 × 10⁻⁵) is mixed with 0.2 M HCl. Calculate the degree of dissociation of acetic acid.

Solution
Correct: D
HCl provides [H⁺] = 0.2 M, suppressing acetic acid dissociation. Using Ka = [H⁺][CH3COO⁻]/[CH3COOH], solve for α = (Ka × 0.1)/[(0.2)^2] = 4.5 × 10⁻⁵.

4. In a first-order reaction, the time required for 99% completion is

Solution
Correct: B
For first-order reactions, t = (1/k) ln([A₀]/[A]). t₉₉% = (1/k) ln(100), t₅₀% = (1/k) ln(2). Thus t₉₉% ≈ 4.60 × t₅₀%.

5. The molar heat capacity at constant pressure for CO₂ (gas) is 37.1 J/mol·K. Calculate q when 2 moles of CO₂ expand irreversibly from 1 L to 2 L at 1 atm.

Solution
Correct: A
Irreversible work is -PΔV. Heat q = nCpΔT from ΔT = (PΔV)/nR = (101.325 kPa)(2-1)L/(8.314 × 2). Calculations yield q ≈ 74.2 J.

6. The rate law for 2NO + O₂ → 2NO₂ is rate = k[NO]²[O₂]. If [NO] doubles and [O₂] halves, the rate

Solution
Correct: A
Rate depends on [NO]²×[O₂]. Doubling [NO] increases rate by 4×, halving [O₂] halves rate. Net factor: 4×0.5 = 2. Rate doubles.

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