Spontaneity Of Chemical Reactions — Practice Quiz

For ideal gases, ΔU = ΔH = 0 upon mixing at constant temperature because there are no intermolecular forces, so entropy increase is the only driving force.

Step 1: By definition, ideal gases have no intermolecular forces of attraction or repulsion. Step 2: Therefore, internal energy (U) and enthalpy (H) depend only on temperature for ideal gases, not on pressure or volume. Step 3: At constant temperature, mixing causes no change in U or H, so ΔU = ΔH = 0. Step 4: The driving force is purely the increase in entropy — the system moves to a state of greater disorder (maximum mixing). Step 5: This is a classic example showing that energy alone cannot explain all spontaneous processes; entropy is equally important.

ΔG = +79.0 kJ mol⁻¹, so the reaction is non-spontaneous at 298 K.

Step 1: ΔG = ΔH – TΔS. Step 2: TΔS = 298 × 41.5 × 10⁻³ = 298 × 0.0415 = 12.367 kJ mol⁻¹. Step 3: ΔG = 91.35 – 12.367 = +78.98 ≈ +79.0 kJ mol⁻¹. Step 4: Since ΔG > 0, the reaction is non-spontaneous at 298 K. Step 5: Here ΔH > 0 and ΔS > 0 (Case 3): at low temperatures, the positive ΔH dominates, making the reaction non-spontaneous. It could become spontaneous at very high temperatures where TΔS > ΔH.

ΔG(system) = –T × ΔS(total), so a negative ΔG corresponds to positive ΔS(universe).

Step 1: For a non-isolated system at constant T and P, ΔS(surroundings) = –ΔH(system)/T. Step 2: ΔS(total) = ΔS(system) + ΔS(surroundings) = ΔS(system) – ΔH(system)/T. Step 3: Multiplying both sides by –T: –TΔS(total) = ΔH(system) – TΔS(system) = ΔG(system). Step 4: Therefore, ΔG = –TΔS(total). Step 5: Since T is always positive, ΔG < 0 ↔ ΔS(total) > 0, which is the second law criterion for spontaneity. Gibbs energy beautifully combines both system and surroundings into one property of the system alone.

–818.0 kJ mol⁻¹

Step 1: ΔrG° = ΣΔfG°(products) – ΣΔfG°(reactants). Step 2: ΔfG°(O₂) = 0 (element in standard state). Step 3: ΔrG° = [ΔfG°(CO₂) + 2×ΔfG°(H₂O)] – [ΔfG°(CH₄) + 2×ΔfG°(O₂)]. Step 4: = [–394.4 + 2×(–237.2)] – [–50.8 + 2×0] = [–394.4 – 474.4] – [–50.8] = –868.8 + 50.8 = –818.0 kJ mol⁻¹. Step 5: The large negative ΔrG° confirms that methane combustion is highly spontaneous, which aligns with its use as a fuel in everyday life (LPG).