Daily Olympiad: Chemistry - Equilibrium [20260523]

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

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1. For the equilibrium reaction 2NO2(g) ⇌ N2O4(g), the partial pressures of NO2 and N2O4 at equilibrium are 0.4 atm and 0.2 atm, respectively. What is the value of Kp?

Solution
Correct: @
Kp is calculated as [N2O4]/[NO2]^2 = (0.2)/(0.4^2) = 0.2/0.16 = 1.25. However, since the correct answer is not listed, this indicates a need to recheck calculations or problem setup. Correct Kp = 1.25.

2. Which of the following conditions will shift the equilibrium of the reaction N2(g) + 3H2(g) ⇌ 2NH3(g); ΔH = -92 kJ/mol to the right?

Solution
Correct: D
The reaction is exothermic (ΔH negative). Increasing pressure favors the side with fewer moles of gas (2 NH3 vs. 4 moles total reactants). Le Chatelier's principle states pressure increase shifts equilibrium toward fewer gas moles.

3. The standard Gibbs free energy change (ΔG°) for a reaction is -249.5 kJ/mol. The corresponding standard cell potential (E°cell) at 25°C is:

Solution
Correct: A
Using ΔG° = -nFE°cell. Given ΔG° = -249,500 J/mol, F = 96,500 C/mol, solve for E°cell. Assume n=2 (common for two electrons). E°cell = -ΔG°/(nF) = 249,500/(2*96500) ≈ +1.3 V. Closest choice is +1.25 V.

4. The solubility product (Ksp) of Ag2CrO4 is 1.1×10⁻¹². What is the solubility of Ag2CrO4 in 0.1 M AgNO3?

Solution
Correct: A
Let s be solubility. Ksp = [Ag+]^2[CrO4^2-] = (0.1 + 2s)^2 * s ≈ (0.1)^2 * s = 1.1×10⁻¹². Solving gives s = 1.1×10⁻¹⁰ M.

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