Spring 2008 CH 302Worksheet 2 1. 100 g of ice at -25°C is heated to steam at 125°C. 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? Answer: Divide the process into several steps: Heat ice to 0°C: ∆H = mc∆T = (100 g)(2.093 J/g°C)(25°C) = 5.232 kJ Melt ice: ∆H = -m∆Hfusion = -(100 g)(-335.5 J/g) = 33.55 kJ Heat water to 100°C: ∆H = mc∆T = (100 g)(4.186 J/g°C)(100°C) = 41.86 kJ Boil water: ∆H = m∆Hvaporization = (100 g)(2.26 kJ/g) = 226 kJ Heat stem to 125°C : ∆H = mc∆T = (100 g)(2.009 J/g°C)(25°C) = 5.0225 kJ So the total ∆H = 5.232 kJ + 33.55 kJ + 41.86 kJ + 226 kJ + 5.0225 kJ = 312 kJ. 2. 1 MJ of heat is dumped into 2 kg of ice at -25°C. What is the final temperature and state (solid, liquid, or gas) of the water? Answer: First, see how much energy it takes to melt the ice: Heat ice to 0°C: ∆H = mc∆T = (2000 g)(2.093 J/g°C)(25°C) = 104.65 kJ Melt ice: ∆H = -m∆Hfusion = -(2000 g)(-335.5 J/g) = 671.00 kJ ∆H for this process = 104.65 + 671.00 kJ = 775.65 kJ. So we have 1000 kJ – 775.65 kJ = 224.35 kJ. So we solve for how much we can increase the temperature of the water with this amount of heat. Heat water to Tf: ∆H = mc(Tf - 0°C) Tf = ∆H/mc = 224350 kJ/(2000 g * 4.186 J/g°C) = 26.8°C 3. The phase diagram for water (shamelessly borrowed from last year’s quiz 2) is shown above. (a) What phase change(s) occur when going from 1 atm, 100 K to 1 atm, 400K? (b) From 0.1 atm, 100 K to 0.1 atm, 400K? Answer: (a) melting, then vaporization; (b) sublimation 4. For any temperature less than ~475 K, if you keep increasing the pressure on the system, what will be the eventual state of the system? Is this the same or different from most other substances? Answer: Liquid. Different; for most substance, the eventual state of the system will be solid.5. Describe the physical state of the system (a) at the point where the three lines meet, and (b) at temperatures above ~475K. Answer: (a) All three states are present in equilibrium. (b) The water is a supercritical fluid – it is no longer possible to distinguish between gas and liquid. 6. Give a basic explanation for the well-known rule “like dissolves like.” Answer: A solute molecule replaces solvent molecules. It is much less disruptive to the intermolecular forces at work if the solute molecule has similar intermolecular forces to the solvent molecules it replaces. 7. Describe the structure formed by soap molecules around a grease molecule. Answer: The nonpolar tails of the soap molecules orient themselves toward the nonpolar grease molecule and solvate it; the polar heads of the molecules then form a sphere around the micelle and are solvated by the polar water. 8. Which of the following will be most miscible in water: methanol (CH3OH), ethanol (CH3CH2OH), propanol (CH3CH2CH2OH), butanol (CH3CH2CH2CH2OH). Answer: Methanol; it has the smallest nonpolar alkyl tail. 9. Which of the above alcohols will be most miscible in hexane (C6H12)? Answer: Butanol; Hexane is nonpolar, so it is most miscible with other nonpolar molecules. Butanol has the largest nonpolar tail relative to the polar –OH group. 10. At 0°C, the vapor pressure of acetone is 0.095 atm, and at 20°C, the vapor pressure is 0.243 atm. What is the enthalpy of vaporization of acetone? Answer: ln(P2/P1) = ∆H/R(1/T1 – 1/T2) ∆H = R ln(P2/P1) (1/T1 – 1/T2)-1 = (8.314 J/molK) ln(0.243 atm/0.095 atm) (1/273 K – 1/293 K) = 31.3 kJ 11. Using the information from the above problem, what is the boiling point of acetone? Answer: Boiling occurs when the vapor pressure equals 1 atm. 1/T2 = 1/T1 - R/∆H ln(P2/P1) = 1/273 K – (8.314 J/molK)/(31300 kJ) ln(1 atm/0.095 atm) = 0.00304 K-1 T2 = 329 K = 56°C 12. Rank the following in terms of increasing ability to increase the boiling point of water: NaCl, sugar, CaCl2, BaS. Answer: BaS < sugar < NaCl < CaCl2. List in terms of increasing number of “things” dissolved for each molecule of the substance. BaS is insoluble in water. 13. Rank the osmotic pressure in increasing order when 1 mol of each of the compounds in #12 is dissolved in water. Answer: Same ranking as above. Osmotic pressure increases as the number of “things” dissolved increases.14. At room temperatures, he vapor pressures of ethylene glycol (HOCH2CH2OH) and water (H2O) are 0.06 torr and 17.54 torr, respectively. What is the vapor pressure of a mixture of 500 mL of ethylene glycol and 500 mL of water? Assume the densities of the two liquid are the same (1 g/mL). Answer: Find the number of moles and mole fractions: ne.g. = 500 mL * 1 g/mL * 1 mol/62 g = 8.065 mol nwater = 500 mL * 1 g/mL * 1 mol/18 g = 27.778 mol χe.g. = mol e.g./(mol e.g. + mol water) = (8.065 mol)/(8.065 mol + 27.778 mol) = 0.225 χwater = 1 – 0.225 = 0.775 Then the vapor pressure is Ptot = Pe.g. + Pwater = (0.06 torr)(0.225) + (17.54 torr)(0.775) = 13.6 torr 15. A 1 L mixture of ethylene glycol and water has a vapor pressure of 10 torr. What is the volume of ethylene glycol in the mixture? Answer: Ptot = pe.g.χe.g. + pwaterχwater = pe.g.χe.g. + pwater(1-χe.g.) 10 torr = (0.06 torr)χe.g. + (17.54 torr)(1-χe.g.) χe.g. = 0.43135 = mol e.g./(mol e.g. + mol water) = (Ve.g. 1 g/mL 1 mol/62 g)/[(Ve.g. 1 g/mL 1 mol/62 g)+ ((1000 mL - Ve.g.) 1 g/mL 1 mol/62 g) Ve.g. = 723 mL 16. Give a simple explanation for the depression of the vapor pressure of water by the addition of a solute. Answer: When a solute is added, there is less water at the surface of the solution than in pure water. 17. 5 g of table salt (NaCl) are dissolved in 100 mL of water. What is the vapor pressure of the solution, given P° = 17.54 torr? Answer: Find the number of moles of salt and water. nsalt = 5 g * (1 mol/58.5 g) = 0.08547 mol nwater = 100 mL * (1 g/mL) * (1 mol/18 g) = 5.556 mol The mole fraction is then χsalt = nsalt/(nsalt + nwater) = 0.08547 mol/(0.08547 mol + 5.556 mol) = 0.01515 So ∆P = i * χsaltP° = 2(0.01515)(17.54 torr) = 0.5316 torr P = 17.54 torr – 0.5316torr = 17.01 torr 18. What is the
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