CH 302 Spring 2007 Worksheet 4 Practice Exam 1 1. Predict the signs of ∆H and ∆S for the sublimation of CO2. a. ∆H > 0, ∆S > 0 b. ∆H > 0, ∆S < 0 c. ∆H < 0, ∆S > 0 d. ∆H < 0, ∆S < 0 Answer: This reaction happens only at higher temperatures, which means ∆H > 0 and ∆S > 0. 2. Vapor pressure increases ________ with temperature. a. Linearly b. Exponentially c. Logarithmically d. Quadratically Answer: Recall the Clausius-Clapeyron equation. Pressure is related exponentially to temperature. 3. Which of the following salts will dissolve most easily in water? a. LiF b. MgO c. BN d. KBr Answer: The salt with the lowest charge density will dissolve the easiest. BN and MgO both have multiple charges, so they have high charge density. Li+ and F- are smaller than K+ and Br-, so LiF has a higher charge density than KBr. 4. For this question, refer to the phase diagram shown above. What is the phase of this substance at -56°C and 5.1 atm? a. Solid b. Liquid c. Gas d. Mixture of solid and gas e. Mixture of solid, liquid, and gas f. Supercritical fluid Answer: This is the triple point, which is the point at which solid, liquid, and gas are all at equilibrium with each other.5. For this question, refer to the phase diagram shown above question 4. The substance is originally held in a container at -60°C and 20 atm. It is then heated to room temperature, and next allowed to expand to atmospheric pressure. What happens to the substance? a. The liquid in the container boils. b. The liquid in the container becomes a supercritical fluid. c. The gas in the container becomes a supercritical fluid. d. The solid in the container sublimes. e. The solid in the container melts, then the resulting liquid boils. f. The solid in the container sublimes, and then the resulting gas condenses. Answer: Just trace the two steps on the phase diagram. 6. 1 kg of water starts at 200°C and is allowed to cool to room temperature. For water, the specific heats are cice = 2.093 J/g°C, cwater = 4.186 J/g°C, and csteam = 2.009 J/g°C. The enthalpy changes are ∆Hfusion = -335.5 J/g and ∆Hvaporization = 2.26 kJ/g. What is ∆Hsys for this process? a. -2775 J b. -2775 kJ c. +2775 kJ d. -1745 kJ e. +1745 kJ Answer: Divide the process into three stages. Cool steam 200°C → 100°C: ∆H = mcsteam∆T = (1000g)(2.009 J/g°C)(-100°C) = -201 kJ Condense steam: ∆H = m∆Hmelting = -m∆Hvaporization = (1000 g)(2.26 kJ/g) = -2260 kJ Cool water 100°C → 25°C: ∆H = mcwater∆T = (1000g)(4.186 J/g°C)(-75°C) = -314 kJ ∆Htot = -201 kJ + -2260 kJ + -314 kJ = -2775 kJ 7. Which of the following gases will be most soluble in water? a. CH4 b. O2 c. CCl4 d. He e. Cl2 Answer: Since all of these are nonpolar, solubility in water can be ranked based on size. Smaller molecules can more easily fit between water molecules. 8. Rank the following in terms of increasing miscibility with water: CH3OH, CH4, CH3CH2OH, CH3CH2CH2OH. a. CH3CH2CH2OH < CH3CH2OH < CH3OH < CH4 b. CH4 < CH3OH < CH3CH2OH < CH3CH2CH2OH c. CH4 < CH3CH2CH2OH < CH3CH2OH < CH3OH d. CH3OH < CH3CH2OH < CH3CH2CH2OH < CH4 Answer: CH4 is nonpolar, so it is not soluble in water. All of the alcohol (molecules with –OH) groups are polar, but the shorter the carbon backbone attached to the –OH, the soluble the molecule is in water.9. You’re cleaning your pet goldfish’s tank, and you put him in a bowl containing pure water. Because the fish has a certain electrolyte balance inside its body that doesn’t exist in the water, a concentration gradient is created. What is the name of the colligative property that explains why your fish blows up like a water balloon? a. Vapor pressure b. Freezing point depression c. Boiling point elevation d. Osmotic pressure e. Density depression f. Ion diffusion Answer: The concentration gradient causes osmotic pressure. 10. 25 g of acetic acid (CH3COOH) and 75 g of ethanol (CH3CH2OH) are mixed together. At 25°C, the vapor pressures of these compounds are 16 and 59 torr, respectively. What is the vapor pressure of the mixture? a. 37.50 torr b. 48.25 torr c. 26.75 torr d. 50.25 torr e. 24.75 torr Answer: nacetic = 25g /(60 g/mol) = 0.417 mol nethanol = 75 g/(46 g/mol) = 1.63 mol Pacetic = XaceticP°acetic = (0.417 mol)/(0.417 mol + 1.63 mol) x 16 torr = 3.26 torr Similarly, Pethanol = 46.99 torr. So Ptotal = 3.26 torr + 46.99 torr = 50.25 torr 11. Butanol boils at 118°C and has a ∆Hvap of 50 kJ/mol. What is butanol’s vapor pressure at room temperature, 25°C? Recall that 1 atm = 760 torr and R = 8.314 J/mol K. a. 6.28 torr b. 91965 torr c. 756.4 torr d. 763.7 torr Answer: Remember that the boiling point is the point at which the vapor pressure equals 1 atm. ln(P1/P2) = ∆H/R (1/T2 – 1/T1) P1 = P2exp[∆H/R(1/T2 – 1/T1)] P1 = (760 torr)exp[(50000 J/mol)/(8.314 J/molK)(1/391 K – 1/391 K)] P1 = 6.28 torr 12. 1 mol of each of the following is added to 1 L of water. Rank the solutions in terms of increasing freezing point. BaS, CaCl2, sugar, LiCl. a. BaS < sugar < LiCl < BaCl2 b. BaCl2 < LiCl < sugar < BaS c. sugar < BaS < LiCl < BaCl2 d. BaCl2 < LiCl < BaS < sugar Answer: Remember that dissolved substances decrease the freezing point of the solvent. So BaCl2 has the lowest (most depressed) freezing point, followed by LiCl, sugar (which doesn’t dissociate), and BaS (which doesn’t dissolve at all).13. Which of these is not an example of using a colligative property to your advantage? a. Adding salt to water so that your spaghetti cooks faster. b. Mixing ethylene glycol and water in your radiator so that the liquid remains liquid over a wide range of temperatures. c. Cooking your spaghetti in a pressure cooker so that it cooks faster. d. Salting the roads after it snows. Answer: Colligative properties are properties of solutions, where the solute affects a property of the solvent. A pressure cooker raises the boiling point, but is not a solution property. 14. 20 g of BaCl2 is added to 1 L of water (dwater = 1 g/mL). What is the boiling point of the water, given the boiling point of pure water is 100°C and Kb for water is 0.512 °C/m? a. 99.852 °C b. 100.148°C c. 99.951°C d. 100.0492°C e. 89.760°C f. 110.240°C Answer: nBaCl2 = 20 g / (208 mol/g) = 0.0962 mol m = nBaCl / mH2O = 0.0962 mol / (1 kg) = 0.0962 m ∆Tb = i m Kb =
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