E) The degree of dissociation of Na2SO4 in solution is significantly greater than that of NaCl. 2008 #44 Cl2(g) + 2 I-(aq) 2 Cl-(aq) + I2(aq) Which of the following best accounts for the fact that a galvanic cell based on the reaction represented above will generate electricity? A) Cl2 can easily lose two electrons. (48%) B) Cl2 is a stronger oxidizing than I2 C) I have more electrons than do atoms of Cl D) I- is a more stable species than I2 E) I2(s) is more soluble than Cl2(g) 2008 #74 An electric current of 1.00 ampere is passed through an aqueous solution of Ni(NO3)2. How long will it take to plate out exactly 1.00 mol of nickel metal, assuming 100 percent current efficiency? (1 faraday = 96,500 coulombs = 6.02 x 1023 electrons) A) 386,000 sec (24%) B) 193,000 sec C) 96,500 sec D) 48,200 sec E) 24,100 sec AP Chemistry Concepts - THERMODYNAMICS 1. rxn = 0Products - Reactants Hf Hf0 ∆H0 ∑ ∆ ∑∆ = Bond Energy Reactants - Bond energy Products ∑ ∑ - exothermic + endothermic ∆Hrxn ∆Hrxn 2. rxn = Products - Reactants Sf0 Sf0 ∆S0 ∑ ∑ - ordered + disordered ∆S0rxn ∆S0rxn 3. rxn = - 0rxn T ∆G0 ∆S - spontaneous + nonspontaneous ∆G0rxn ∆G0rxn
E) The degree of dissociation of Na2SO4 in solution is significantly greater than that of NaCl. 2008 #44 Cl2(g) + 2 I-(aq) 2 Cl-(aq) + I2(aq) Which of the following best accounts for the fact that a galvanic cell based on the reaction represented above will generate electricity? A) Cl2 can easily lose two electrons. B) Cl2 is a stronger oxidizing than I2 (48%) C) I have more electrons than do atoms of Cl D) I- is a more stable species than I2 E) I2(s) is more soluble than Cl2(g) 2008 #74 An electric current of 1.00 ampere is passed through an aqueous solution of Ni(NO3)2. How long will it take to plate out exactly 1.00 mol of nickel metal, assuming 100 percent current efficiency? (1 faraday = 96,500 coulombs = 6.02 x 1023 electrons) A) 386,000 sec B) 193,000 sec (24%) C) 96,500 sec D) 48,200 sec E) 24,100 sec AP Chemistry Concepts - THERMODYNAMICS 1. ∆H0 rxn = ∑ ∆ Hf 0Products - ∑∆ Hf0 Reactants = ∑ Bond Energy Reactants - ∑ Bond energy Products ∆Hrxn - exothermic ∆Hrxn + endothermic 2. ∆S0 rxn = ∑ Sf0 Products - ∑ Sf0 Reactants ∆S0rxn - ordered ∆S0rxn + disordered 3. ∆G0 rxn = - T ∆S 0rxn ∆G0rxn - spontaneous ∆G0rxn + nonspontaneous
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