(10 points please help!) Calculate the free energy of the reaction?
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2NO2(g) <----> N2O4(g) ; deltaH= -56.8 kJ deltaS= -175 J/K In a container at 298 K, N2O4 and NO2 are mixed with initial partial pressures of 2.4 atm and 0.42 atm respectively. Calculate the free energy of the reaction.
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Answer:
Kp = 0.42atm / (2.4atm)^2 Kp = 0.073 ∆G = -RT x lnK ∆G = -8.314J/mole-K x 298K x ln(0.073) = +6484.54J/mole
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