Chemical Thermodynamics — Practice Quiz

-94.8 J K⁻¹ mol⁻¹

Step 1: ΔS° = ΣS°(products) - ΣS°(reactants). Step 2: S°(products) = S°[SO₃(g)] = 256.2. Step 3: S°(reactants) = S°[SO₂(g)] + ½ × S°[O₂(g)] = 248.5 + ½×205.0 = 248.5 + 102.5 = 351.0. Step 4: ΔS° = 256.2 - 351.0 = -94.8 J K⁻¹ mol⁻¹. The negative sign makes sense because gas moles decrease (fewer gaseous molecules = less disorder). Option B has wrong sign. Option C uses full O₂ (not ½).

509 K

Step 1: At equilibrium, ΔG = 0, so ΔH = TΔS. Step 2: T = ΔH/ΔS. Step 3: T = 30.54 kJ mol⁻¹ ÷ 0.06 kJ K⁻¹ mol⁻¹ = 509 K. Step 4: Below 509 K, TΔS < ΔH, so ΔG > 0 (non-spontaneous forward). Above 509 K, TΔS > ΔH, so ΔG < 0 (spontaneous forward — Ag₂O decomposes). Option B incorrectly uses ΔS in J units without converting. Option C uses an incorrect formula.

1.8 × 10⁻⁷

Step 1: Use ΔG° = -2.303RT log K. Step 2: log K = -ΔG°/(2.303RT) = -(38740)/(2.303 × 8.314 × 300) = -38740/5748 = -6.74. Step 3: K = 10⁻⁶·⁷⁴ = 1.8 × 10⁻⁷. Step 4: A positive ΔG° means the products are not favoured at standard conditions (K < 1). Option B is the reciprocal (K > 1, which would require ΔG° < 0). Option C uses wrong sign in the exponent.

Zero

Step 1: Internal energy (U) is a state function — it depends only on the current state of the system, not the path. Step 2: In a cyclic process, the initial state and final state are identical. Step 3: Therefore, ΔU = U_final - U_initial = 0. Step 4: Even though q = +400 J and w = -150 J in one step (giving ΔU = 250 J for that step), returning to the initial state means the sum of all ΔU over the entire cycle is zero. The values of q and w mentioned are distractors — the key concept is that ΔU_cycle = 0.