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6 thermodynamics

# 6 thermodynamics - of the following reaction PCl 3(g 3HCl(g...

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Thermodynamics Three laws of thermodynamics 1) Conservation of energy (Energy can’t be created or destroyed). 2) For a spontaneous process, the entropy of the universe increases. 3) A perfect crystal at 0K has zero entropy. E = q + w w = -P V Gas in a piston 1) Transfer of heat (lock piston) 2) Expanding gases cool; compressing gases warm 3) Isobaric ( P = 0) 4) Isochoric ( V = 0 so w = 0) 5) Isothermal ( T = 0 so E = 0) 6) Adiabatic (q = 0) State Functions Entropy (S) Phases of matter S for rxns S = Σ nS products - Σ nS reactants Enthalpy (H) Exothermic ( H<0) / Endothermic ( H>0) Bond Enthalpies H = Σ D broken - Σ D formed Bond breaking is endothermic Bond making is exothermic Whatis H for the following rxn? Enthalpies of Formation H = Σ n H f,products - Σ n H f,reactants Given the following standard enthalpy of formation values, calculate the H ° rxn

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Unformatted text preview: of the following reaction: PCl 3 (g) + 3HCl(g) → 3Cl 2 (g) + PH 3 (g) Compound PCl 3 (g) HCl(g) PH 3 (g) ∆ H f (kJ/mol)-288.07-92.30 5.40 Bond Enthalpies (kJ/mol) C-H 413 C-Cl 328 Cl-Cl 242 H-Cl 431 Hess’s Law C 2 H 4 (g) + 6F 2 (g) → 2CF 4 (g) + 4HF(g) ∆ H = ? H 2 (g) + F 2 (g) → 2HF(g) ∆ H = -537kJ C(s) + 2F 2 (g) → CF 4 (g) ∆ H = -680kJ 2C(s) + 2H 2(g) → C 2 H 4(g) ∆ H = +52.3kJ Gibb’s Free Energy (G) ∆ G = Σ n ∆ G f,products- Σ n ∆ G f,reactants ∆ G = ∆ G ° + RTlnQ ∆ G ° = -RTlnK eq ∆ G = ∆ H - T ∆ S ∆ G< Spontaneous ∆ G> Nonspontaneous ∆ G= At equilibrium ∆ H ∆ S-+ Spontaneous at all temperatures +-Non-spontaneous at all temperatures--Spontaneous at low temperatures + + Spontaneous at high temperatures...
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6 thermodynamics - of the following reaction PCl 3(g 3HCl(g...

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