Thermodynamics filled in class notes_Part_103

# Thermodynamics filled in class notes_Part_103 - 6.2...

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Unformatted text preview: 6.2. EQUILIBRIUM CONDITIONS 213 Take the reaction to be isothermal and isobaric with T = 6000 K and P = 100 kPa . Initially one has 1 kmole of N 2 and 0 kmole of N . Use the extremization of Gibbs free energy to find the equilibrium composition. First find the chemical potentials at the reference pressure of each of the possible constituents. μ o T,i = g o i = h o i − T s o i = h o 298 ,i + Δ h o i − T s o i . (6.62) For each species, one then finds μ o N 2 = 0 + 205848 − (6000)(292 . 984) = − 1552056 kJ kmole , (6.63) μ o N = 472680 + 124590 − (6000)(216 . 926) = − 704286 kJ kmole . (6.64) To each of these one must add RT ln parenleftbigg n i P nP o parenrightbigg , to get the full chemical potential. Now P = P o = 100 kPa for this problem, so one only must consider RT = 8 . 314(6000) = 49884 kJ/kmole . So, the chemical potentials are μ N 2 = − 1552056 + 49884 ln parenleftbigg n N 2 n N + n N 2 parenrightbigg , (6.65) μ N = − 704286 + 49884 ln parenleftbigg n N n...
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Thermodynamics filled in class notes_Part_103 - 6.2...

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