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1 quick approximate answer assume the volume is not

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Unformatted text preview: ction entropy change are independent of temperature, you calculate the reaction Gibbs free energy change at 278.15 K for three moles of iodine and three moles of hydrogen to be: a) - 6.8 kJ b) - 5.2 kJ c) - 1.3 kJ d) 0.0 kJ e) 0.8 kJ f) 4.4 kJ g) 9.7 kJ h) 20.2 kJ Show your calculations! Here we use ΔRG(278.15K, 1 bar) = ΔRH(278.15K, 1 bar) – 278.15 x ΔRS(278.15K, 1 bar) We approximate ΔRH(278.15K, 1 bar) by ΔRH(298.15K, 1 bar) and ΔRS(278.15K, 1 bar) by ΔRS(298.15K, 1 bar) We calculate ΔRH(298.15K, 1 bar) from ΔRH(298.15K, 1 bar) ΔRH(298.15K, 1 bar) = 2 x ΔFHo(298.15K)(HI(g)) = 2x 26.48 = 52.96 kJ/mol We calculate ΔRS(298.15K, 1 bar) from ΔRG(298.15K, 1 bar) and ΔRH(298.15K, 1 bar) ΔRG(298.15K, 1 bar) = ΔRH(298.15K, 1 bar) – 298.15 x ΔRS(298.15K, 1 bar) So, ΔRS(298.15K, 1 bar) = [ΔRH(298.15K, 1 bar) - ΔRG(298.15K, 1 bar)]/298.15K ΔRS(298.15K, 1 bar) = [ 52.96 kJ/mol – 3.4 kJ/mol] / 298.15 K = 0.16622 kJ/K.mol So, ΔRG(278.15K, 1 bar) = 52.96 kJ/mol – 278.15 K x 0.16622 kJ/K.mol = 6.72 kJ/mol For 3 mol H2(g) and 3 mol I2(s), we get ΔRG(278.15K, 1 bar) = 3 mol x 6.72 kJ/mol = 20.2 kJ 2 Question 7: (5 points) Consider the experimental values for molar mass, density, thermal expansion coefficient and compressibility coefficient for a substance X in the data section. Calculate as accurately as possible the o change in free energy for exactly one mole of that substance at 60.00 C as a result of the de...
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