Q4. Calculate the entropy change of the surroundings when 1.00 mol of H2O(l) is formed under standard conditions (take 298 K); ΔfH⊖ = −286 kJ mol−1 (NCERT Exercise 5.22, page 167).
The surroundings gain the heat released: ΔSsurr = −ΔH/T = 286 000/298, about +960 J K−1 mol−1.
Q5. Comment on the thermodynamic stability of NO(g), given ½N2(g) + ½O2(g) → NO(g), ΔrH⊖ = +90 kJ mol−1, and NO(g) + ½O2(g) → NO2(g), ΔrH⊖ = −74 kJ mol−1 (NCERT Exercise 5.21, page 167).
The given formation enthalpy places NO above its elements in enthalpy. Adding the two supplied equations gives ΔH = +16 kJ mol−1 for NO2 formation, so these data place NO2 below NO plus oxygen but above the elements in enthalpy. Enthalpy data alone do not establish thermodynamic stability: reaction Gibbs energy and composition are needed. Also, these exercise values are not internally consistent with NCERT Table 5.2, which gives NO2 formation near +33 kJ mol−1; use the supplied numbers for the exercise, not as reference data.
Q6. Ice melts with ΔH = +6.01 kJ mol−1 and ΔS = +22.0 J K−1 mol−1. Taking both as constant, find ΔG at 263 K and at 283 K and say what happens.
At 263 K: ΔG = 6010 − 263 × 22.0 = +224 J mol−1, so ice does not melt. At 283 K: ΔG = 6010 − 283 × 22.0 = −216 J mol−1, so ice melts. ΔG = 0 at 6010/22.0, about 273 K, the melting point.
Q7. Which of these have a positive ΔS: (a) H2O(l) → H2O(g); (b) 2SO2(g) + O2(g) → 2SO3(g); (c) NH4Cl(s) → NH3(g) + HCl(g); (d) N2(g) + 3H2(g) → 2NH3(g)?
(a) and (c): a liquid becomes a gas, and a solid gives two gases. (b) and (d) turn more moles of gas into fewer, so ΔS is negative.
Q8. For C2H4(g) + H2(g) → C2H6(g), ΔrH⊖ = −137.0 kJ mol−1 and ΔrS⊖ = −120.6 J K−1 mol−1. Is it spontaneous at 298 K? Taking both as constant, above what temperature does it stop being spontaneous?
ΔG = −137.0 − 298 × (−0.1206) = −137.0 + 35.9 = −101.1 kJ mol−1: spontaneous. Both ΔH and ΔS are negative, so ΔG turns positive above T = 137.0/0.1206, about 1140 K.