lecture 6

lecture 6 - Thermochemistry example: diamond to graphite...

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Unformatted text preview: Thermochemistry example: diamond to graphite The webtext asks for G f (graphite) and G f, 298 (diamond) from the following information: H f (diamond) = 2032 . 9 J / mol S (diamond) = 2 . 899 J / mol K S (graphite) = 5 . 693 J / mol K The formation reaction for diamond is: C(graphite) C(diamond) G f (graphite) = ? From the definition G = H- TS , at constant T : G f (diamond) = H f (diamond)- T S f (diamond) where S f (diamond) = S (diamond)- S (graphite) =- 2 . 794 J / mol K Is this reaction possible at P = 1 atm? 1 Phase equilibria example Given that S vap , 100 C (H 2 O) = 109.0 J/mol K, estimate the vapour pressure of water that is in equilibrium with the liquid at T = 120 C. 2 Chemical equilibrium Consider a constant T gas-phase reaction involving ideal gases. Lets examine the pressure dependence of G . Recall that for one mole of a simple system d G = V dP- SdT which at constant T for an ideal gas is ( d G ) T = V dP = RT P dP Integrating we get ( G ) T = RT...
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lecture 6 - Thermochemistry example: diamond to graphite...

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