Κατέβασμα παρουσίασης
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1
2 CH3OH(l) + 3 O2(g) → CO2(g) + 2 H2O(g)
Calculate ∆G at 400 °C using the Gibbs Free Energy equation. State whether or not the reaction will be spontaneous. 2 CH3OH(l) + 3 O2(g) → CO2(g) + 2 H2O(g) ∆H (-238.7) (0) ( ) (-241.8) ∆S (0.1268) (0.205) (0.2136) (0.1887) (2(-241.8) + ( )) – (2(-238.7) + 3(0)) ΔH° = kJ (2(0.1887) + (0.2136)) – (2(0.1268) + 3(0.205)) ΔS° = kJ/K ∆G = ∆H° - T∆S° ∆G = ( ) – (673)( ) ΔG = kJ spontaneous
2
1) ∆G = ∆H° - T∆S° a) ∆G = (-638.4) – (298)(0.1569) ΔG = kJ spontaneous b) ∆G = (-57.2) – (298)( ) ΔG = -4.8 kJ spontaneous
3
C3H8 (l) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
2) C3H8 (l) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ∆H (-118.9) (0) (-393.5) (-285.8) ∆S (0.171) (0.2052) ( ) ( ) (4(-285.8) + 3(-393.5)) – ((-118.9) + 5(0)) ΔH° = kJ (4( ) + 3( )) – ((0.171) + 5(0.2052)) ΔS° = kJ/K ∆G = ∆H° - T∆S° ∆G = ( ) – (298)( ) ΔG = kJ spontaneous
4
Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
3) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ∆H (-822.0) ( ) (0) (-393.5) ∆S (0.090) (0.1976) (0.0273) ( ) (2(0) + 3(-393.5)) – ((-822.0) + 3( )) ΔH° = kJ (2(0.0273) + 3( )) – ((0.090) + 3(0.1976)) ΔS° = kJ/K ∆G = ∆H° - T∆S° -31.3 = (-26.91) – (T)( ) T = 293 K 20°C
5
N2(g) + 3 F2(g) → 2 NF3(g) 4) ∆G = ∆H° - T∆S°
(2(-131.2)) – ((0) + 3(0)) ΔH° = kJ (2(0.2607) ) – ((0.1915) + 3(0.2027)) ΔS° = kJ/K ∆G = ∆H° - T∆S° 0 = (-262.4) – (T)( ) T = 943 K 670°C
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