Exam04answers pg4 - 1. He(g, 2 atm) → He(g, 10 atm) 2. O2...

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Unformatted text preview: 1. He(g, 2 atm) → He(g, 10 atm) 2. O2 (g) → O2 (aq) 3. H2 (g) + I2 (s) → 2 HI(g) correct 4. 2 Ag(s) + Br2 ( ) → 2 AgBr(s) 5. 2 NO2 (g) → N2 O4 (g) Explanation: 011 (part 1 of 1) 10 points In a system at equilibrium at constant temperature and pressure, ∆H is necessarily equal to 1. ∆U. 2. ∆G. 3. T ∆S. correct 4. P ∆V. 5. 0. Explanation: ∆G = 0 for a system at equilibrium, so ∆G = ∆H − T ∆S 0 = ∆H − T ∆S ∆H = T ∆S 012 (part 1 of 1) 8 points The process of water freezing to form ice at 0◦ C is 1. an endothermic phase change. 2. an exothermic chemical reaction. 3. an exothermic phase change. correct 4. neither exothermic nor endothermic. 5. an endothermic chemical reaction. Heat must be removed from the water for it to freeze. This is exothermic. 013 (part 1 of 1) 12 points In the isothermal compression of 66.0 mmol of a perfect gas at 327 K, the volume of the gas is reduced to 12.0% of its initial value. Calculate w for this process. Correct answer: 380.4 J. Explanation: w = −n R T ln Vf Vi w = −(66/1000)8.314(327) ln(12/100) w = 380.4 J 014 (part 1 of 1) 10 points If you have an endothermic process in which the change in entropy is positive, how can you make it spontaneous? 1. Increasing the temperature correct 2. Decreasing the volume 3. Increasing the pressure 4. Reducing the entropy change 5. Decreasing the temperature Explanation: ∆G = ∆H − T ∆S ∆H > 0 for endothermic processes. ∆G < 0 for spontaneous processes. T is always positive, so ∆G = ∆H − T ∆S = (+) − T ∆S ∆G is negative if T is very large, so increasing the temperature makes the process endothermic. 015 (part 1 of 1) 10 points Which one of the following thermodynamic ...
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