Thermodynamic Quantities for Selected Substances at 298.15 K (25°C) Substance H
ID: 688604 • Letter: T
Question
Thermodynamic Quantities for Selected Substances at 298.15 K (25°C)
Substance H°f (kJ/mol) G°f (kJ/mol) S (J/K-mol)
Calcium
Ca (s) 0 0 41.4
CaCl2 (s) -795.8 -748.1 104.6
Ca2+ (aq) 226.7 209.2 200.8
Chlorine
Cl2 (g) 0 0 222.96
Cl- (aq) -167.2 -131.2 56.5
Oxygen
O2 (g) 0 0 205.0
H2O (l) -285.83 -237.13 69.91
Phosphorus
P2 (g) 144.3 103.7 218.1
PCl3 (g) -288.1 -269.6 311.7
POCl3 (g) &
Explanation / Answer
1) The given reaction is 2S (s,rhombic) + 3O2(g) 2SO3 (g) S°=Sfn(Products)-Sfn(Reactants) =2*S(SO3 (g)) -{ 2*S(S(rhombic)+3S(O2(g) } = 2*356.2 kJ/mol -{2*31.88 kJ/mol+3*205.0 kJ/mol} = 712.4 kJ/mol -{678.76 kJ/mol} = 33.64 kJ/mol ------------------------------------------------------------------- 2) The given reaction is 2SO3 (g) 2S (s,rhombic) + 3O2(g) S°=Sfn(Products)-Sfn(Reactants) =(2*31.88 kJ/mol+3*205.0 kJ/mol)-(2*256.2 kJ/mol) = 166.36 kJ/mol --------------------------------------------------------------------- 3) The givenreaction is P2(g) + O2 (g) + 3Cl2 (g) 2POCl3(g) S°=Sfn(Products)-Sfn(Reactants) =2*325 kJ/mol -(218.1 kJ/mol+205.0 kJ/mol) = 226.9 kJ/mol -------------------------------------------------------------------- 4) The given reaction is P2 (g) + 3Cl2 (g) 2PCl3(g) S°=Sfn(Products)-Sfn(Reactants) =2*311.7 kJ/mol -(218.1 kJ/mol +3*222.96 kJ/mol) = -263.58 kJ/mol -------------------------------------------------------------------- 5 ) The given reaction is CaCl2 (s) Ca(s) + Cl2 (g) H°=Hfn(Products)-Hfn(Reactants) = 0.00 kJ/mol +0.00 kJ/mol -(-795.8 kJ/mol) = + 795.8 kJ/mol ------------------------------------------------------------------- 6 ) The given reaxction is 2SO3 (g) 2S(s, rhombic) + 3O2(g) G°=Gfn(Products)-Gfn(Reactants) =2*0.00kJ/mol +3*0.00 kJ/mol -(2*-370.4 kJ/mol) = + 740.8 kJ/mol -------------------------------------------------------------- 7 ) The given reaction is P2 (g) + O2 (g) + 3Cl2 (g) 2POCl3 (g) G°=Gfn(Products)-Gfn(Reactants) =2* -502.5 kJ/mol -(103.7 kJ/mol +0.00 kJ/mol +3*0.00kJ/mol) = -1108.7 kJ/mol ------------------------------------------------------------ 8) The givn reaction is CaCl2(s) Ca (s) + Cl2 (g) G°=Gfn(Products)-Gfn(Reactants) = 0.00 kJ/mol +0.00 kJ/mol -(-748.1 kJ/mol) = +748.1 kJ/molRelated Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.