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e1p302key - CH302 Practice Exam 1 TA style Answer Key 1...

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CH302 Practice Exam 1 TA style Answer Key 1. Rearrange the Gibbs free energy equation ( G = H - T S) to solve for the temperature at a phase transition. a. T = ( H - G)/ S b. T = H/ S c. T = - G d. T = - G/ S At a phase transition delta G = 0. The temperature dependence at a phase transition can be determined from thermodynamics. 2. What will happen to vapor pressure when non-volatile solute A is added to a pure solution B? a. Vapor pressure of A increases b. Vapor pressure of B decreases c. Vapor pressure doesn't change d. Boiling point decreases of A inrcreases. e. Boiling point of B decreases Adding a non-volatile solute (its vapor pressure is essentially zero) to any solvent, will decrease the measured vapor pressure of the solvent, by a factor equal to the solvents new mole fraction. 3. Which of the following is a possible combination of values for ¨ H lattice , ¨ H hydration and ¨ H solution . All values are in kJ mol -1 . ¨ H lattice is always + and ¨ H hydration is always -. Their sum is ¨ H solution , which can be positive or negative. 4. Which of the following is always true for phase diagrams? Melting curves can have a negative slope (ie water). The condensation curve always has a positive slope. Many chemicals have numerous solid phases, and some even have more than one distinct liquid phase. The critical point can occur at a variety of temperatures. 5. At what temperature will water boil on top of a mountain if the normal boiling point is 100 °C at 1 atm. (You should not need to look at this graph to answer the question)
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Pressure is lower at higher altitudes. Following the evaporation curve on the phase diagram from 1 atm to a lower pressure you see that temperature decreases. 6. Consider a 27 g sample of ice at 1 atm. Initially, the sample is frozen at -20 °C. How much heat must be added to the sample for it to become a liquid at 78 °C. Heat capacity of ice: 2 J/gK. Heat capacity of water: 4 J/gK. ¨ H of melting: 330 J/g a. 9.5 kJ b. 18.4 kJ c. 5.3 kJ d. 10.6 kJ q = mc ¨ T (for the ice) + m ¨ H (for melting) + mc ¨ T (for the water) q = 27 g * 2 J/gK * 20 K + 27 g * 330 J/g + 27 g * 4 J/gK * 78 K q = 18.4 kJ
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