Chem. 412 – Phys. Chem. I
Free Energy Comparisons Helmholtz F.E. (A)Gibbs F.E. (G) A = U - TSG = H - A sys = U sys - T S G sys = H sys - T S sys If A sys < 0, rxn spontaneous. (Constant V & T) If G sys < 0, rxn spontaneous. (Constant P & T) If A sys = 0, equilibrium.If G sys = 0, equilibrium. dA = -PdV – SdTdG = VdP - SdT
Free Energy Comparisons - I
Free Energy Comparisons - II
Free Energy Comparisons - III
Free Energy Comparisons - IV Helmholtz F.E. (A)Gibbs F.E. (G) A = U - TSG = H - A sys = U sys - T S G sys = H sys - T S sys If A sys < 0, rxn spontaneous. (Constant V & T) If G sys < 0, rxn spontaneous. (Constant P & T) If A sys = 0, equilibrium.If G sys = 0, equilibrium. dA = -PdV – SdTdG = VdP - SdT
Free Energy Comparisons – I – F12
Free Energy Comparisons – II – F12
Free Energy Comparisons – III – F12
Free Energy Comparisons – III – F11 10
Phase Diagrams
The Phase Diagrams of H 2 O and CO 2 Phase Diagrams
Phase Transitions: Clapeyron Equation Over moderate temperature ranges:
Phase Transitions: Clapeyron Equation – I – F14
Phase Transitions: Clapeyron Equation – II – F14
Phase Transitions: Clapeyron Equation – III – F14
Phase Transitions: Clapeyron Equation – I – F13
Phase Transitions: Clapeyron Equation – II – F13
Phase Transitions: Clapeyron Equation – III – F13
Application of Clapeyron Equation Consider:Ice Water (ice, 101 kPa, 273 K) = 0.917x10 3 kg m -3 (liq, 101 kPa, 273 K) = 0.988x10 3 kg m -3 H f = 6.01 kJ mol -1 ( s liq ) Triple point at 0.6 kPa and K What is the melting point at 1.5x10 5 kPa ( 1500 atm ) ? Application: Blade in Ice-Skating. Mathcad Key
Clausius-Clapeyron Equation Applicable only to:s g & liq g equilibria Integrated form: Indefinite Integrated form: T-dep form:
Clausius-Clapeyron Equation - I
Clausius-Clapeyron Equation - II
Clausius-Clapeyron Equation – I – F11 26
Clausius-Clapeyron Equation – II – F11
Standard States & G o rxn P o for gas:ideal gas; P o = kPa non-ideal gas; (leave for now) for liquid:pure liquid at P o for solid:most stable crystalline structure at P o T o for all substances: K exactly S o o = 0 at 0 K for pure crystals H o f (T o ) = 0 for elements at reference state G convention must follow that of H & S G rxn from formation values
Substance H f (kJ/mol) G f (kJ/mol) S (J mol -1 K -1 ) C(s, diamond) C(s, graphite)005.69
P/T-Dependent Equations Variation of G with P for an ideal gas: Variation of G with T: Variation of K P with T:
P/T-Dependent Equations
A = U - TSG = H - TS If A sys < 0, rxn spontaneous. (Constant V & T) If G sys < 0, rxn spontaneous. (Constant P & T) dA = -PdV – SdTdG = VdP - SdT