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PRACTICE QUIZ KEY THERMODYNAMICS (20 points)

PRACTICE QUIZ KEY THERMODYNAMICS (20 points). 1a. (2 pts) Calculate D G f for SO 3 (g). D G rxn = D G f (products) - D G f (reactants) = [2(-300.0) - 2( D G f (SO 3 ))] = 139.6 kJ (139.6 + 600.2)kJ = -2 D G f (SO 3 ) D G f (SO 3 ) = -369.kJ/mole

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PRACTICE QUIZ KEY THERMODYNAMICS (20 points)

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  1. PRACTICE QUIZ KEY THERMODYNAMICS(20 points)

  2. 1a. (2 pts) Calculate DGf for SO3(g). • DGrxn = DGf(products) - DGf(reactants) = [2(-300.0) - 2(DGf(SO3))] = 139.6 kJ(139.6 + 600.2)kJ = -2 DGf(SO3)DGf(SO3) = -369.kJ/mole • b. (1 pt) Formation of SO3 spontaneous? • yes, DGrxn = -139.6 kJ (for the reverse reaction) • c. (1 pt) What is the maximum work? • From reaction in b, DGrxn = -Wmax This is the work performed by the spontaneous reaction. Maximum work is +139.6 kJ

  3. 2. (3 pts) Give the expected sign on DS: • Process DS sign • a. CaSO4(s)  CaO(s) + SO3(g) + • b. Ag+(aq) + Cl-(aq)  AgCl(s) - • c. I2(g)  I2(s) -

  4. 3. (2 pts.) Circle T (true) or F (false): • T F a. Endothermic reactions which become more ordered are never spontaneous. • T F b. A system at equilibrium cannot perform any work. • 4. (2 pts.) What is the sign of DG? • +

  5. 5. (2 pts) How does free energy change? • The reaction going toward equilibrium is spontaneous. Therefore, DG is negative for the reaction, meaning DG(products) is less than DG(reactants), or free energy must be released. • The free energy decreases as the system goes to equilibrium.

  6. 6a. (1 pt) Why direction <-> temperature. • When a solid melts, it absorbs heat (DH is positive) and DS is also positive (more disorder); therefore, DG depends on the temperature (DG = DH - TDS). At hi T, the reaction is spontaneous (DG is negative since TDS > DH). At low T the reaction is non-spontaneous. The reverse reaction is spontaneous. The liquid freezes spontaneously below 5 ºC because DH > TDS. • 6b. (1pt) What does 5 ºC represent? • Melting point or freezing point. • 6c. (1 pt) What is DG at 5 ºC? • DG = 0 at equilibrium

  7. 7. (4 pts) Calculate DS. • DGrxn = DHrxn - TDSrxn • Since you know T and DGrxn; to calculate DSrxn, find DHrxn from DHrxn = DHf(products) - DHf(reactants) • DHrxn = [2(-173.9)+ 1(+90.7)] - [3(+34.0)+ 1(-286.9)] • = [-257.1] - [-184.9] • DHrxn = -72.2 kJ • DSrxn = (DHrxn - DGrxn) / T • = (-72.2 - (+8.4)) / 298 • = -80.6/298 = -0.27 kJ/K • (System goes to more order!!)

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