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

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Thermochemistry example: diamond to graphite The webtext asks for Δ G f (graphite) and Δ G f, 298 (diamond) from the following information: Δ 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) Δ G f (graphite) = ? From the definition G = H - TS , at constant T : Δ G f (diamond) = Δ H f (diamond) - T Δ S f (diamond) where Δ S f (diamond) = S (diamond) - S (graphite) = - 2 . 794 J / mol K Is this reaction possible at P = 1 atm? 1
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Phase equilibria example Given that Δ 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
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Chemical equilibrium Consider a constant T gas-phase reaction involving ideal gases. Let’s 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 ln P f P i Define (as before) the standard or reference state as P
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