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Chem33 1st Midterm SOLUTION SHEET

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1. (1 point) Calculate the equilibrium constant of the reaction 2CO (g) + O2 (g) ↔ 2CO2 (g) The Gibbs Free Energy of reaction at std. conditions and 25oC is ΔGro = - 514 kJ / mol. Calculate the equilibrium constant K. ANSWER: ΔG r 0 = − RT ln( K ) ⇓ 0 −− 514 kJ / mol K e − ΔGr RT 8.3 J K mol x 298.15 K 1.33 x10 90 = = e =2. (1 point) For the reaction SF4 (g) + F2 (g) ↔ SF 6 (g) the Gibbs Free Energy of reaction at std. conditions and 25oC is ΔGro = - 374 kJ / mol. ΔGfo(SF6)= -1105 kJ / mol Calculate the Gibbs Free Energy of formation for SF4 (g). ANSWER: ΔGfo(F2)= 0 kJ / mol Therefore, ΔGfo(SF4 ) = -1105 kJ / mol - (-374 kJ / mol) = -731 kJ / mol
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