Exam II CHM 3410, Dr. Mebel, Fall 2005 1. Prior to the discovery that freon-12 (CF2Cl2) was harmful to Earth’s ozone layer, it was frequently used as the dispersing agent in spray cans for hair spray, etc. Its enthalpy of vaporization at its normal boiling point of –29.2°C is 20.25 kJ mol-1. Estimate the pressure that a can of hair spray using freon-12 had to withstand at 40°C (the temperature of a can that has been standing in sunlight). Assume that ΔvapH is a constant over the temperature range involved. 3. At high altitudes, where the atmospheric pressure is lower, mountain climbers are unable to absorb a sufficient amount of O2 into their blood streams to maintain a high activity level. Assuming that the Henry’s law constant for blood is the same as for water, calculate the amount of O2 dissolved in 1.00 L of blood for atmospheric pressures of 1 bar and 0.500 bar. The mole fraction of O2 in air is 20.99%. Assume that the density of blood is 998 kg m-3 and take the molar mass of blood equal to the molar mass of water. 4. (a) Calculate the cryoscopic and ebullioscopic constants of NH3. (b) Calculate the amount of salt (NaCl), which needs to be added to 2.0 L of water to make it boil at 100.75°C. 5. A non-perfect gas has the equation of state pVm = RT + αp where α is a function of temperature only and does not depend on pressure. Obtain an expression for the fugacity and the fugacity coefficient as a function of T, p, and α. 1 Exam II CHM 3410, Dr. Mebel, Fall 2006 (Each problem is 20 pts.) 1. The normal melting point of H2O is 273.15 K, and ΔfusH = 6010 J mol-1. Calculate the decrease in the freezing temperature at 100 and 500 bar assuming that the density of the liquid and solid phases remains constant at 997 and 917 kg m-3, respectively. 2. At 100°C benzene has a vapor pressure of 180.9 kPa, and toluene has a vapor pressure of 74.4 kPa. Assuming that these substances form ideal binary solutions with each other, calculate the composition of the solution that will boil at 1 bar at 100°C and the vapor composition. 3. On the surface of Mars, the mean atmospheric pressure is 600 Pa, and the air mostly consists of carbon dioxide. The mole fraction of CO2 in Martian air is 0.9532, whereas the mole fraction of O2 is only 0.0013. If a Martian sea were to exist, what would be the molalities of CO2 and O2 solutions in Martian water at 298 K? 4. Calculate ΔrG∅(500 K) for the reaction of combustion of heptane, C7H16(l), from the values of ΔrG∅(298 K) and ΔrH∅(298 K) assuming that ΔrH∅ does not change in this temperature range. 5. (a) Derive the thermodynamic equation of state ∂H ∂V = V − T ∂T p ∂ p T using the expression for dH and a Maxwell relation. € ∂H (b) Use the above thermodynamic equation of state to determine for a perfect ∂ p T gas. € 1 Exam II CHM 3410, Dr. Mebel, Fall 2013 (Each problem is 20 pts.) 1. Carbon tetrachloride melts at 250 K. The vapor pressure of the liquid is 10539 Pa at 290 K and 74518 Pa at 340 K. The vapor pressure of the solid is 270 Pa at 232 K and 1092 Pa at 250 K. (a) Calculate ΔHvaporization and ΔHsublimation. (b) Calculate ΔHfusion. (c) Calculate the normal boiling point and ΔSvaporization at the boiling point. 2. The average human with a body weight of 70 kg has a blood volume of 5.00 L. The Henry’s law constant for the solubility of N2 in water is 9.04×104 bar at 298 K. Assume that this is also the value of the Henry’s law constant for blood and that the density of blood is 1.00 kg L-1. (a) Calculate the number of moles of N2 absorbed in this amount of blood in air of composition of 80% N2 at sea level, where the pressure is 1 bar, and at a pressure of 50 bar. (b) Assume that a diver was breathing compressed air at a pressure of 50 bar and then was suddenly brought to the sea level. What volume of N2 gas was released as bubbles in the diver’s bloodstream? 3. A sample of glucose (C6H12O6) of mass 4.55 g is placed in a test tube of radius r = 2.50 cm. The bottom of the test tube is a membrane that is semipermeable to water. The tube is partially immersed in a beaker of water at 298 K so that the bottom is only slightly below the level of the water in the beaker. The density of water at this temperature is 997 kg m-3. After equilibrium is reached, how high is the water level of the water in the tube above that in the beaker. What is the value of the osmotic pressure? Assume that the test tube is cylindrical and the volume of the solution in the tube can be calculated as V = πr2h and that the density of the solution in the tube is equal (approximately) to the density of water. 4. The phase diagram for two metals A and B contains one eutectic and no compound between A and B forms. The freezing points of solutions of A and B are found to vary linearly with the mole fraction of B in the solution. The melting points of pure A and B are 650 K and 600 K, respectively. The cooling curve for a mixture containing 40% mole fraction of B exhibits a single flat region (eutectic halt) at 400 K. Using this information, sketch the A-B phase diagram and indicate the phases present in each region. 5. Describe how Le Chatelier’s principle is manifested in changes of equilibrium constant (or equilibrium composition of a reaction mixture) with temperature and pressure. Support your explanation with formal equations showing the described changes. 1 Exam II CHM 3410, Dr. Mebel, Fall 2005 1. Prior to the discovery that freon-12 (CF2Cl2) was harmful to Earth’s ozone layer, it was frequently used as the dispersing agent in spray cans for hair spray, etc. Its enthalpy of vaporization at its normal boiling point of –29.2°C is 20.25 kJ mol-1. Estimate the pressure that a can of hair spray using freon-12 had to withstand at 40°C (the temperature of a can that has been standing in sunlight). Assume that ΔvapH is a constant over the temperature range involved. We use the Clausius-Clapeyron equation: p Δ H 1 1 ln * = − vap − * R T T p p* = 1 bar (normal boiling point) T* = 273.15 – 29.2 = 243.95 K T = 273.15 + 40 = 313.15 K ln(p/p*) = {-(20.25×103 J mol-1)/(8.3145 J K-1 mol-1)}×{1/(313.15 K) – 1/(243.95 K)} = 2.20618 p = p* exp(2.20618) = € 9.081 bar 3. At high altitudes, where the atmospheric pressure is lower, mountain climbers are unable to absorb a sufficient amount of O2 into their blood streams to maintain a high activity level. Assuming that the Henry’s law constant for blood is the same as for water, calculate the amount of O2 dissolved in 1.00 L of blood for atmospheric pressures of 1 bar and 0.500 bar. The mole fraction of O2 in air is 20.99%. Assume that the density of blood is 998 kg m-3 and take the molar mass of blood equal to the molar mass of water. First, we need to use Dalton’s law to determine the pressure of O2 in the atmosphere: p(O2) = y(O2)p, where y(O2) = 0.2099 – the mole fraction of O2 in air. Next, we use Henry’s law to find the mole fraction of O2 in blood in equilibrium with O2 in the vapor phase (in air): p(O2) = x(O2)K(O2) K(O2) – Henry’s law constant x(O2) = p(O2)/K(O2) = y(O2)p/K(O2) Finally, we determine the amount of O2: x(O2) = n(O2)/{n(O2) + n(solvent)} ≈ n(O2)/n(solvent) = n(O2)/{m(solvent)/M(solvent)} = n(O2)/{ρ(solvent)×V(solvent)/M(solvent)} nO2 = a) xO2 ρ solventVsolvent M solvent = yO2 pρ solventVsolvent M solvent KO2 5 p = 1 bar = (10 Pa /133.32 Pa) × 1 Torr = 750 Torr n(O2) = (0.2099) × (750 Torr) × (998 kg m-3) × (1.0×10-3 m3) /{(18.02×10-3 kg mol-1) × (3.30×107 Torr)} = 2.64×10-4 mol m(O2) = n(O2)M(O2) = (2.64×10-4 mol) × (32 g mol-1) = 8.455 mg € b) p = 0.5 bar = 375 Torr 1 n(O2) = (0.2099) × (375 Torr) × (998 kg m-3) × (1.0×10-3 m3) /{(18.02×10-3 kg mol-1) × (3.30×107 Torr)} = 1.32×10-4 mol m(O2) = n(O2)M(O2) = (1.32×10-4 mol) × (32 g mol-1) = 4.228 mg 4. (a) Calculate the cryoscopic and ebullioscopic constants of NH3. (b) Calculate the amount of salt (NaCl), which needs to be added to 2.0 L of water to make it boil at 100.75°C. (a) Ebbulioscopic constant Kb = RT *2 M NH 3 Δ vap H Kb = (8.3145 J K-1 mol-1)×(239.7 K)2×(17.03×10-3 kg mol-1)/(23.35×103 J mol-1) = 0.348 K kg mol-1 Kf = Cryoscopic constant: € -1 -1 2 RT *2 M NH 3 Δ fus H Kf = (8.3145 J K mol )×(195.4 K) ×(17.03×10-3 kg mol-1)/(5.652×103 J mol-1) = 0.9565 K kg mol-1 (b) ΔT = Kbb b = ΔT/Kb Kb = 0.51 K kg mol-1 for water ΔT = 0.75 K b = (0.75 K)/(0.51 K kg mol-1) = 1.47 mol kg-1 n(NaCl) in 2 kg of H2O = 2.94 mol -1 m(NaCl) = (2.94 €mol)×(58.44 g mol ) = 171.88 g 5. A non-perfect gas has the equation of state pVm = RT + αp where α is a function of temperature only and does not depend on pressure. Obtain an expression for the fugacity and the fugacity coefficient as a function of T, p, and α. ln φ = We use p ∫0 Z −1 dp p pVm pVm = RT + αp RT Z −1 α pV αp = Therefore, Z = € m = 1+ p RT RT RT p α α p αp € p Z −1 ln φ = ∫ dp = ∫ dp = dp = ∫ 0 0 RT p RT 0 RT Z= € € φ = eαp RT € f = φp = peαp € € € 2 RT