the reaction will be non-spontaneous

273 T (K) = 298 K ΔGr ˚ = ΔHr ˚ - T ΔSr ˚ 104 KJ/mol = 86 KJ/mol – (298 K × ΔSr ˚) ΔSr ˚ = - 18 KJ/mol / 298 K = – 0.06 KJ/K.mol ΔS ° r (J/K.mol) = - 0.06 KJ/K.mol × 1000 ΔS ° r (J/K.mol) = - 60 J/K.mol Because of ΔSr ˚ =- the reaction will be more ordered

the reaction will be non- ° Because of ΔGr 300 K

Chemistry Chapter one عند درجة حرارة K 300 ΔGr ° ثانيا:- نجد ΔGr ° = ΔHr ° - TΔSr ° ΔGr ° = 6 KJ/mol – (300 K × 0.022 KJ/K.mol) ΔGr ° = 6 KJ/mol – (6.6 KJ/mol) = - 0.6 KJ/mol Because of ΔGr ° =- the reaction will be sponatanous at 300 K

273 T (K) = 338 K

273 T (K) = 298 K ΔGr ˚ = ΔHr ˚ - T ΔSr ˚ 195.7 KJ/mol = 235.8 KJ/mol – (298 K X ΔSr ˚) ΔSr ˚ = 40.1 KJ/mol 298 K ΔS ° r = 0.134 KJ/K.mol ΔGr ° االن نجد عند درجة حرارة 1000K ΔGr ° = ΔHr ° - TΔSr ° ΔGr ° = 235.8 – (1000 × 0.134) ΔGr ° = 235.8 KJ – (134 KJ) = 102 KJ

Chemistry Chapter one Lesson Seventy Examples ) Gibbs law ( Part 4 Example 47:(Q 1-15/A) For the reaction 3O2 → 2O3 At 25 °C and the pressure 1 atm S° for O2= 205 J/K.mol, O3 = 239 J/K.mol and ΔGf ° for O3 = 163 KJ/mol. Find ΔHf ° for O3? Solution: ΔSr ° = Ʃ n S° (products) - Ʃ n S° (reactants) ΔSr ° = [2 S˚ (O3 )] – [3 S˚ (O2 )] ΔSr ° = [(2 × 239)] – [(3 × 205)] ΔSr ° = 478 – 615 = - 137 J/K.mol ΔSr ° = −137 J/K.mol 1000 ΔSr ° = - 0.137 KJ/K.mol ΔGr ° = Ʃ n ΔGf ° (products) - Ʃ n ΔGf ° (reactants) ΔGr ° = [2 ΔGf ° (O3 )] - [3 ΔGf ° (O2 )] ΔGr ° = [2 × 163] – [0] ΔGr ° = 326 KJ/mol

273 T (K) = 298 K ΔGr ˚ = ΔHr ˚ - T ΔSr ˚ 326 = ΔHr ˚ - (298 ×( – 0.137)) ΔHr ˚ = 326 – 40.8 ΔHr ˚ = 285.2 KJ/mol ΔHr ° = Ʃ n ΔHf ° (products) - Ʃ n ΔHf ° (reactants) ΔHr ° = [ 2ΔHf ° O3 (g) ] – [3 ΔHf ° O2 (g) ] 285.8 = [ 2ΔHf ° O3 (g) ] – [ 0 ] 2ΔHf ° O3 (g) = 285.8 ΔHf ° O3 = 285.8 2 = 142.6 KJ/mol

Chemistry Chapter one Lesson Seventy- One Trouton’s Rule Calculation of Entropies of Physical Changes (Trouton’s Rule) Melting temperature (Tm): The temperature that is important to change substances from solid to liquid state. Boiling temperature (Tb ): The temperature that is important to change substances from liquid to gas state. Note: Trouton found out that the values of ΔS of most liquids is equal to a constant number (85 J/K.mol). Boiling Equation (Vaporizing) ΔSvap = ΔHvap Tb Melting Equation ΔSfus = ΔHfus Tm ∆S = ΔH T ∆Hvap = 85J/K. Mol × Tb

Chemistry Chapter one Why/ ΔSr ° of most liquids is equal to a constant number (85 J/K.mol)? Ans: Because of the presence of similarity in the movement of fluid particles, their composition and the movement of their vapors.

273 T(K) = 342 K ΔHvap = 85 J/K.mol × Tb K ΔHvap = 85 J/K.mol × 342K ΔHvap = 29070 J/mol 1000 = 29.07 KJ/mol

T(K) = t(C ΔSvap = ΔHvap Tb ΔSvap = 44 / 373 ΔHvap= 0.118 KJ/K.mol × 1000 J ΔHvap= 118 J/K.mol

292 KJ/mol H2O (s) ⟶ H2O (L) ΔHr ° = 6 KJ/mol

273 T (K) = 298 K ΔSfus = ΔHfus Tm ΔSfus = 6 273 ΔSfus = 0.022 KJ/K.mol ΔSfus = 0.022 KJ/K.mol × 1000 ΔSfus = 22 KJ/K.mol