Question 4 5 points for an ideal gas the expression

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Unformatted text preview: Vdp – TdS – SdT dG = Vdp - SdT Question 3: (5 points) Mathematically speaking, what are we saying when we state that a process is not spontaneous under conditions of constant volume and constant temperature? a) ΔG < 0 b) ΔG ≤ 0 c) ΔG ≥ 0 d) ΔG > 0 e) ΔA < 0 f) ΔA ≤ 0 g) Δ A ≥ 0 h) ΔA > 0 Explain your reasoning. A process occurring at constant temperature and volume is reversible if ΔG = 0. A process occurring at constant temperature and volume is spontaneous if ΔG < 0. Hence, a process occurring at constant temperature and volume is not spontaneous if ΔG = 0 or ΔG > 0. Question 4: (5 points) For an ideal gas, the expression dG – dA can be written as: a) + nR dT b) – pdV c) + Vdp d) + TdS e) – nRdT f) + pdV g) – Vdp h) – TdS i) – SdT j) + SdT Show how you arrive at the chosen result. G = H – TS and A = U – TS So, G – A = (H – TS) – (U – TS) = H – U = pV For an ideal gas, G – A = pV = nRT So, dG – dA = d(nRT) = nR dT 1 Question 5: (5 points) Consider the following chemical reaction at 298.15K and 10.6 bars: l2 (s) + H2 (g) à༎ 2HI (g) Using the data in the data section, you calculate the reaction Gibbs free energy change for three moles of iodine and three moles of hydrogen at 298.15K and 10.6 bars to be: a) - 12.0 kJ b) - 7.6 kJ c) - 7.4 kJ d) - 5.8 kJ e)...
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This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

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