chap-19-2011W-b

# 5 atm nh3 n 2 g 3h 2 g 10atm q 30atm pnh 3 p

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Unformatted text preview: Most chemical reaction occur under nonstandard conditions: under Not all reactants and products at standard conditions ∆G: under any conditions: under • ∆G° : standard free energy change (kJ/mol) • R : gas constant (8.314×10-3 kJ/mol·K) kJ/mol· • T : absolute temp. (K) o ∆G = ∆G + RT ln Q Q: Q: reaction quotient ; varies according to the composition of reaction mixture ∆G = ∆G o + RT ln(1) = ∆G o (a) Under standard conditions: Q = 1 (b) At equilibrium: o o ∆G = ∆G + RT ln K = 0 ∆G = − RT ln K K =e −∆G o RT (a) (a) Calculate ∆G of Haber process at 25°C for a reaction mixture that consist of 10 atm N2, 30 atm H2 and 0.5 atm NH3 N 2 ( g ) + 3H 2 ( g ) 10atm Q= 30atm [ PNH 3 P ° ]2 [ PN 2 P ° ][ PH 2 2 NH 3 ( g ) ∆G ° = −33.32 KJ 0.5atm ∆G = ∆G o + RT ln Q [0.5atm /1atm]2 = = 9.26 × 10−7 P ° ]3 [10atm /1atm][30atm /1atm]3 ∆G = - 33.32 KJ/mol+(8.314J/mol ⋅ K × 10−3 KJ / J ) × (298.15K) × ln(9.26 × 10-7 ) = (- 33.32 KJ/mol)+(−34.44 KJ / mol ) = −67.76 KJ / mol < 0 (b) Calculate ∆G of Haber process at 25°C for a reaction mixture that consist of 0.1 atm N2, 0.3 atm H2 and 50 atm NH3 Q= [ PNH 3 P ° ]2 [ PN2 P ° ][ PH 2 [50atm /1atm]2 = = 9.26 × 105 P ° ]3 [0.1atm /1atm][0.3atm /1atm]3 ∆G = - 33.32 KJ/mol+(8.314J/mol ⋅ K × 10−3 KJ / J ) × (298.15K) × ln(9.3 × 103 ) = (- 33.32 KJ/mol)+(+34.06 KJ / mol ) = 0.735KJ / mol > 0 (a) (a) Calculate K for the reaction: 2C(diamond)+O2(g) →2CO(g) 2C(diamond)+O 2 (g) ∆G o = − RT ln K ∆G ° f 2.87 0 2CO(g) -137.3 (KJ/mol) ∆G o = 2 × ( −137.3) − [2 × 2.87 + 0] = −268.9kJ / mol ln K = −∆G RT = −(−268.9 KJ / mol ) /(8.314 J / mol ⋅ K ×10−3 KJ / J )(298.15K ) = 108.5 108.5 47 Favor the products! K =e = 1.27 ×10 (b) (b) Calculate ∆G° for the dissociation reaction of H2PO4-(aq), the acid dissociation constant of H2PO4-(aq) is 6.2×10-8 at 25°C. − H 2 PO4 (aq) − H 2 PO4 (aq) + H + (aq) K a = 6.2 ×10−8 ∆G o = − RT ln...
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