## Midterm exam 2 answer key

Adding 65 g of Absolut to 1000 g of water gives you a solution of about 26 g of ethanol in 1039 g of water or about 0.54 m. Your mileage may vary slightly ...

1. (B) The melting is favorable (∆G < 0) above the melting point, unfavorable below (∆G > 0), and the system is at equilibrium at the melting point temperature (∆G = 0) 2. (C) The vapor pressure of the water at equilibrium does not depend on the volume of the container. It is an intrinsic property of water as a substance and is a constant at constant temperature. 3. (E). Here we use the Clausius-Clapeyron equation. ln(p2/p1) = ∆H/R (1/T1 – 1/T2) (T in Kelvin, pressure units do not matter) ∆H = R[ln(p2/p1)]/(1/T1 – 1/T2) = 8.31[ln(405/107)]/(1/298 – 1/331) = 33.1 kJ/mol 4. (D) The critical temperature is the temperature above which the liquid state is not possible. Here: 400 K. 5. (B) We are given standard ∆H and ∆S of this reaction. Assuming no change in these values with temperature, we can calculate ∆G of this reaction at 525 ºC (= 798 K). ∆G798 = ∆H – 798(∆S) ≈ 17 kJ From ∆G = -RT[ln(K)], we can now find the equilibrium constant K. K = exp(-∆G/RT) ≈ 0.08 6. They all stay the same. Introducing an inert gas does not affect the equilibrium. 7. Yes. Here, we need to calculate the reaction quotient Q and see how it compares with the equilibrium constant K. Recalling that activity of solvent (water) is 1, Q = [NO2-][H3O+]/[HNO2] = 1.0 × 10-10. Since Q < K, the reaction will tend to proceed in the forward direction, i.e., nitrous acid will tend to dissociate. 8. The Baltic Sea is seawater, whereas lake Mälaren is freshwater. Seawater contains dissolved salts and that depresses its freezing point. So, the temperature at which freshwater of the lake freezes should be close to 0 ºC, while the temperature at which the Baltic Sea would freeze is lower, probably on the order of a couple of degrees. Since Mälaren drains into the Baltic, at the point of drainage, the fresh water from the lake mixes with seawater, dilutes it, and raises its freezing point. 9. In order to get a solution with a freezing point of -1.0 ºC, we will need a solution with molality of 1.0/1.86 = 0.54. If the solute is ethanol, that means we will have to have 0.54 mol of ethanol (which is about 25 g) per kg of water. Absolut has 40% wt ethanol and 60% wt water. The desired ratio of methanol/mH2O = 0.025 (molality of 0.54).

methanol = 0.4mAbsolut; mH2O = 1000 + 0.6mAbsolut (all masses in g) Therefore: 0.4mAbsolut/(1000 + 0.6mAbsolut) = 0.025 Solving this for mAbsolut, we find that it equals ~65 g. Adding 65 g of Absolut to 1000 g of water gives you a solution of about 26 g of ethanol in 1039 g of water or about 0.54 m. Your mileage may vary slightly depending on how your round things up (or not). 10. Here, we need to convert the weight fractions into molar fractions in order to calculate the combined vapor pressure. 26 g of ethanol in 1039 g of water is 0.54 mol of ethanol in 1039/18 = 57.7 mol of water. So, the molar fraction of water is ~0.99 and the molar fraction of ethanol is ~0.01. Then, it is a simple matter of ptotal = 0.99×23.8 + 0.01×59.0 = 24.2 Torr Again, your result may differ slightly depending on how you rounded numbers up. I rounded them up fairly roughly, but then one can hardly be expected to weigh stuff precisely sitting in the middle of a frozen lake!