chap-19-2011W-b

# 88jmol k1605 jk rxn 0 0 at 25 o c gr0xn h rxn t s

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Unformatted text preview: K = −(8.314 × 10−3 kJ / mol ⋅ K ) × 298.15K × ln(6.2 ×10−8 ) = +41KJ / mol Favor Favor the reactants! Solve K for the following reaction at 25oC, when given: 2 NO( g ) N 2O4 ( g ) ∆H o f 33.84 9.66 ( KJ / mol ) So 240.45 304.3 ( J / mol ⋅ K ) o ∆ r H = 9.66 - 2 × 33.84 = -58.02 KJ ∆ r So = 304.3 - 2 × 240.45 = -176.6 J/K= - 0.1766KJ/K ∆ r G o = ∆ r H o - T∆ r So = -58.02 KJ - 298K × (- 0.1766) KJ/K = - 5.39 KJ ∆G o = - RTlnK - 5.39 KJ/mol = - (8.314 × 10-3 KJ/mol ⋅ K)(298.15K) × lnK lnK = 2.17, K = 8.8 Gibbs Free Energy and Temperature Predict whether the following reaction is spontaneous → at room temperature CaCO3 ( s) ← CaO( s) + CO2 ( g ) ∆H o (kJ/mol ⋅ K) -1207.1 f S o (J/mol ⋅ K) f 92.88 -635.5 -393.5 39.75 213.6 ∆H 0 = [(-635.5)+(-393.5)-(-1207.1)]=177.8 KJ rxn ∆S0 = [(39.75)+(213.6)-(92.88)]J/mol ⋅ K=160.5 J/K rxn 0 0 At 25 o C, ∆Gr0xn = ∆H rxn − T ∆S rxn ∆G ° RXN -3 = 177.8 KJ-298.15K ×160.5 ×10 KJ / K = 130kJ > 0 Non-spontaneous Gibbs Free Energy and Temperature At what temperature (if any) would the reaction become → spontaneous? CaCO3 ( s ) ← CaO( s ) + CO2 ( g ) 0 At 25 o C, ∆Grxn = 130 KJ ∆H 0 = 177.8 kJ rxn ∆S0 = 160.5 J/K rxn Assuming H° and S° do not change with temp.(T2>T1) Under standard conditions, Gibbs free energy energy change at different T: ° o o ∆Grxn (T ) ≈ ∆H rxn − T ∆S rxn ° o o ∆Grxn (T ) ≈ ∆H rxn − T ∆S rxn = 0 177.8 kJ-T × 160.5 ×10-3 kJ / K = 0 0 ∆...
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