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Physical Chemistry 418-010 Fall 2010 TuTh 5:00-6:15 pm, 206 BrL Name __________KEY_______________ Problem 1 (7 points). a) Compare the pressure exerted by 30 mole of HCl enclosed in a 10 L cylinder at room temperature predicted by the ideal gas law with that predicted by the van der Waals description. Answer: Ideal gas law: J 30 mol × 8.3144 × 298.15 K nRT mol × K P(HCl )ideal = = = 7436815.1 Pa = 73.396 atm V 1 m3 10 L × 1000 L P(HCl )vdW Van der Waals law: Pa × m 6 J 0.3700 × 298.15 K 8.3144 nRT n2a RT a mol × K mol 2 = − 2 = − 2 = − = 3 3 3 2 V − nb V Vm − b Vm 10 L −3 m −6 m ⎛ ⎞ 10 L −3 m × 10 − 40.6092 × 10 ⎜ ⎟ 30 mol L mol ⎜ 30 mol × 10 L ⎟ ⎝ ⎠ = 8468513.8Pa − 3330000 Pa = 5138513.8 Pa = 50.713 atm b) Based on the result of this comparison, what type of interaction dominates at these conditions? Be concise in this answer; do not write more than a sentence or two. Because the answer predicted by the vdW law is over 30% lower than that predicted by the ideal gas law, attractive interactions between HCl molecules play the dominant role at these conditions reducing the pressure predicted by the ideal gas description that does not take into account intermolecular interactions. Problem 2 (3 points) From the possible statements in column B, select the best match for each phrase in column A and put its letter in the adjacent blank. There is only one best match for each phrase. Column A Column B 1. zeroth law of thermodynamics states that ___e___ a) molecules are hard spheres b) two systems in thermal equilibrium with one another always represent an open system c) gas can not be stable d) liquid can not be produced at any pressure e) two systems that are separately in thermal equilibrium with a third system are also in thermal equilibrium with one another f) collisions with a solid wall are inelastic g) molecules are point masses h) U = q + w i) is based on the activated complex theory j) solid can not be in one piece at any pressure 2. One of the ideal gas law approximations 3. Above critical temperature ___d___ ___g__ Physical Chemistry 418-010 Fall 2010 TuTh 5:00-6:15 pm, 206 BrL Name __________KEY_______________ Problem 1 (5 points). In class we calculated the heat required to raise the temperature of copper to 50°C at atmospheric pressure. This is indeed a practical exercise since copper wiring is used in cables and it is important to know the heat lost for cost estimates. The alternative material for cable manufacturing is aluminum. Assuming that the heat capacity of aluminum does not depend too much on the temperature within our range of interest, calculate the heat required to raise the temperature of a 53.96 g piece of aluminum from room temperature to 50°C at atmospheric pressure. Heat capacity of aluminum at room J temperature is Cp = 24.55 mol × K At constant pressure, the enthalpy change gives the heat required to change the state of the system: ⎛ ∂H ⎞ ⎛ ∂H ⎞ ⎛ ∂H ⎞ dH = ⎜ ⎟ dT = C P dT = dq ⎟ dP = ⎜ ⎟ dT + ⎜ ⎝ ∂T ⎠ P ⎝ ∂P ⎠ T ⎝ ∂T ⎠ P T2 q = ∫ C p dT = C P (T2 − T1 ) = 24.55 T1 J J × 25 K = 613.75 mol × K mol Since the molar mass of aluminum is 26.98 g/mol, the weight of the piece that we investigate corresponds to exactly 2 moles and the heat required to raise its temperature to 50°C at atmospheric pressure is twice what we have calculated for 1 mol. Thus q = 2 mol x 613.75 J/mol = 1227.5 J or 1.2275 kJ. Problem 2 (5 points). Starting with the slope formula for dU and using thermodynamic derivatives of the energy on page 5-2 of the Blue Book, prove that for any process involving ideal gas dU = CvdT. Show all your work clearly. ⎞ ⎛ ⎛ ∂P ⎞ ⎛ ∂U ⎞ ⎛ ∂U ⎞ dU = ⎜ ⎟ dT + ⎜ ⎟ dV = C v dT + ⎜⎜ T ⎜ ⎟ − P ⎟⎟dT = dq ⎝ ∂T ⎠V ⎝ ∂V ⎠ T ⎠ ⎝ ⎝ ∂T ⎠V Because for ideal gas: ⎛ ⎛ ⎛ nRT ⎞ ⎞ ⎞ ⎜ ⎜ ∂⎜ ⎟ ⎟⎟ ⎛ ⎛ ∂P ⎞ ⎞ ⎛ ⎛ nR ⎞ nRT ⎞ nRT ⎟ ⎜ ⎜ ⎝ V ⎠⎟ ⎜⎜ T ⎜ dT = ⎜⎜ T ⎜ − ⎟dT = 0 ⎟ − P ⎟⎟dT = ⎜ T ⎜ ⎟− ⎟ ∂T ⎟ V V ⎠ V ⎟⎠ ⎝ ⎝ ⎝ ⎝ ∂T ⎠V ⎠ ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ ⎠V ⎝ ⎝ ⎠ Physical Chemistry 418-010 Name __________KEY_______________ Fall 2010 TuTh 5:00-6:15 pm, 206 BrL Quiz#3 Problem 1 (5 points). It is possible that hydrogenation of coal was one of the important chemical processes that ultimately led to biochemical reactions on Earth. However, this is the type of reactions, for which determination of enthalpy (or heat at constant pressure) is difficult. Consider the reaction of ethane formation: 2C ( s ) + 3H 2 ( g ) → C 2 H 6 ( g ) (1) It is impossible to use simple calorimetry to do the measurement but the strategy similar to that discussed during the lecture can be employed. Use Hess’s law and the following set of reactions with Θ experimentally determined enthalpies to calculate ΔH rxn for reaction (1): 7 C 2 H 6 ( g ) + O2 ( g ) → 2CO2 ( g ) + 3H 2 O(l ) + ΔH I (2) 2 C ( s ) + O2 ( g ) → CO2 ( g ) + ΔH II (3) 1 H 2 ( g ) + O2 ( g ) → H 2 O(l ) + ΔH III (4), where 2 I ΔH = −1560.4 kJ ; ΔH II = −393.5 kJ ; ΔH III = −285.8 kJ ; Reaction (3) times 2 minus reaction (2) plus reaction (4) times 3 yields reaction (1) or: ΔH rxn (1) = 2ΔH II − ΔH I + 3ΔH III = −2 × (393.5 kJ ) + 1560.4 kJ − 3 × (285.8 kJ ) = −84.0 kJ , which, of course, also happens to be the standard enthalpy of ethane formation so you can easily check your answer based on Table 5.8. Problem 2 (5 points). Joule-Thompson experiment can be described as a study of an isenthalpic process. Based on the expression for Joule-Thompson process given on page 5-6 of the Blue Book, find the exact Joule-Thompson coefficient (μJT) for Argon at room temperature, assuming that it is an ideal gas. μ JT ⎛ ∂V ⎞ T⎜ ⎟ − V T nR − nRT ∂ T ∂ T ⎛ ⎞ ⎝ ⎠P P = 0 , which is correct for any ideal gas (argon or not…) =⎜ = P ⎟ = Cp Cp ⎝ ∂P ⎠ H Physical Chemistry 418-010 Name __________KEY_______________ Fall 2010 TuTh 5:00-6:15 pm, 206 BrL Quiz#4 Problem 1 (5 points). The standard enthalpy of formation of gaseous H2O 1 ⎛ ⎞ ⎜ H 2 ( g ) + O2 ( g ) → H 2 O( g ) ⎟ at 25ºC is -241.82 kJ/mol. Estimate its value at 100ºC given the 2 ⎝ ⎠ following values of the molar heat capacities at constant pressure: Cp(H2O, gas) = 33.58 J/mol/K, Cp(H2, gas) = 28.84 J/mol/K, Cp(O2, gas) = 29.37 J/mol/K. Assume that these values do not vary substantially with temperature in the temperature interval studied. T2 ΔH (T2 ) = ΔH (T1 ) + ∫ ΔC p ,reaction dT , where ΔC p , reaction = C p ( H 2 O, g ) − 0.5C p (O2 , g ) − C p ( H 2 , g ) T1 Since we assume that heat capacities are temperature-independent, it is simplified to: ΔH (T2 ) = ΔH (T1 ) + ΔC p ,reaction × ΔT = −241.82 kJ / mol + (33.58 − 0.5 × 29.37 − 28.84 )J / mol / K × 75 K = = −241.82 kJ / mol − 9.945 J / mol / K × 75 K = −241.82 kJ / mol − 745.845 J / mol = −242.57 kJ / mol Although ~1 kJ/mol difference does not seem substantial, it may translate into a real adjustment of the reaction temperature to save energy for a given reaction. Problem 2 (5 points). From the possible statements in column B, select the best match for each phrase in column A and put its letter in the adjacent blank. There is only one best match for each phrase. Column A 1. In a spontaneous process, entropy ___c___ 2. Perpetual motion machine of the second kind violates ___d__ 3. Carnot cycle describes a(n) ___g___ process 4. When considering an irreversible process __m__ 5. The expression of the second law stated as dS ≥ dq is referred to as ___j___ T Column B a) first law of thermodynamics b) entropy must be calculated along an irreversible path c) increases d) second law of thermodynamics e) Joule-Thompson coefficient f) entropic loss g) reversible h) decreases i) third law of thermodynamics j) Clausius inequality k) entropy can not be calculated l) does not change m) entropy must be calculated along a reversible path n) irreversible o) perpetual Physical Chemistry 418-010 Name __________KEY_______________ Fall 2010 TuTh 5:00-6:15 pm, 206 BrL Quiz#5 Problem 1 (6 points). Use the definition of Helmholtz free energy, first law of thermodynamics, and your knowledge of the Clausius inequality to prove that dA≤0 at constant V and T, and that the equality dA=0 is a characteristic of equilibrium at these conditions. Show all your work clearly. dA = dU − d (TS ) = dU − TdS − SdT = dq − Pext dV − TdS − SdT dA V ,T =Const = dq − TdS ≤ 0, since TdS ≥ dq, according to the Clausius inequality Thus dA V ,T =Const = dq − TdS < 0, for any irreversible (spontaneous) process and dA V ,T =Const = dq − TdS = 0, for a reversible process or for a condition of equilibrium Problem 2 (4 points total). Briefly answer the following questions: a) (2 points) State in a couple of sentences why Maxwell relations are useful The four Maxwell relationships described during the lecture are very useful in transforming seemingly obscure partial derivatives into the partial derivatives that can be directly measured. b) (2 points) Describe in a couple of sentences why dS≥0 may not be a sufficient criterion for the spontaneity of a process. The special case considered previously for understanding the change in entropy for an irreversible (spontaneous) transformation involved an isolated system, for which this is exactly the condition of spontaneity. However, most of the time we are interested in systems that actually interact with surroundings rather than in isolated systems, so a separate set of criteria had to be developed. Physical Chemistry 418-010 Name __________KEY_______________ Fall 2010 TuTh 5:00-6:15 pm, 206 BrL Quiz#6 Problem 1 (8 points). a) Calculate the Kp (as it is determined in the text) for the following reaction at room temperature: CH 4 ( g ) + Cl 2 ( g ) → CH 3 Cl ( g ) + HCl ( g ) 0 Answer: First we should calculate ΔGrxn : 0 ΔG rxn = 1 mol × ΔG 0f ,m (CH 3 Cl ( g )) + 1 mol × ΔG 0f , m ( HCl ( g )) − 1 mol × ΔG 0f ,m (CH 4 ( g )) − 1 mol × ΔG 0f ,m (Cl 2 ( g )) = = 1 mol × ( −57.4kJ / mol ) + 1 mol × (−95.3) − 1 mol × (−50.5) − 1 mol × 0kJ / mol = −102.2 kJ 0 0 ⎞ ⎛ ΔG rxn ΔGrxn 102200 J / mol ⎞ 17 ⎟ = exp⎛⎜ , K p = exp⎜⎜ − ⎟ = 8 × 10 ⎟ RT ⎝ 8.3144 J / mol / K × 298.15K ⎠ ⎝ RT ⎠ Thus, the reaction equilibrium is shifted towards the products. 0 0 b) For the reaction in part (a), calculate the ΔGrxn at 400 K assuming that ΔH rxn is independent of temperature. How does this temperature change affect Kp? Since ln K p = − 0 Answer: Since ΔH rxn is independent of temperature, we can use the Gibbs-Helmholtz equation but first 0 we need to calculate ΔH rxn : 0 ΔH rxn = 1 mol × ΔH 0f ,m (CH 3 Cl ( g )) + 1 mol × ΔH 0f ,m ( HCl ( g )) − 1 mol × ΔH 0f ,m (CH 4 ( g )) − 1 mol × ΔH 0f ,m (Cl 2 ( g )) = = 1 mol × (−81.9kJ / mol ) + 1 mol × (−92.3) − 1 mol × (−74.6) − 1 mol × 0kJ / mol = −99.6 kJ Now, using Gibbs-Helmholtz equation: 0 0 (at 400 K ) ΔG rxn (at 298.15 K ) ΔGrxn 1 ⎛ 1 ⎞ 0 (at 298.15 K )⎜ = + ΔH rxn − ⎟ 400 K 298.15 K ⎝ 400 K 298.15 K ⎠ (−102200 J / mol ) 1 ⎛ 1 ⎞ + 400 K × (−99600 J / mol )⎜ − ⎟= 298.15K ⎝ 400 K 298.15 K ⎠ = -137112.2 J/mol + 34024 J/mol = -103088.2 J/mol 0 ΔGrxn (at 400 K ) = 400 K 0 ⎞ ⎛ ΔG rxn 103088.2 J / mol ⎞ 13 ⎟ = exp⎛⎜ Then at 400 K: K p = exp⎜⎜ − ⎟ = 2.9 × 10 ⎟ ⎝ 8.3144 J / mol / K × 400 K ⎠ ⎝ RT ⎠ The equilibrium is still shifted towards the products formation but not as much as at room temperature. ⎛ ∂G ⎞ Problem 2 (2 points). If x is the extent of reaction parameter and ⎜ ⎟ > 0 , then (circle one): ⎝ dx ⎠ T , P a) the reaction proceeds spontaneously as written; b) the reaction proceeds spontaneously in the opposite direction; c) the reaction has reached equilibrium. Physical Chemistry 418-010 Name __________KEY_______________ Fall 2010 TuTh 5:00-6:15 pm, 206 BrL Quiz#7 Problem 1 (7 points). In class we discussed the problems associated with high altitude caused by lower boiling points of liquids at lower pressures. A different kind of problem may be caused by changing a boiling point of a liquid at altitudes below sea level. Northern Europe (especially Sweden and Norway) are known for their fjords (or lochs as in Loch Ness), valleys that are located below the sea level and may or may not be filled with water. Although the altitude change in this case is not really comparable with that for the height of the highest mountain ranges, the barometric pressure at the bottom a 300 meters deep fjord is about 4% higher than at the sea level. Estimate boiling point of water at this altitude assuming that the heat of vaporization for water is not affected substantially by this pressure change. Answer: we can use Clausius-Clapeyron equation Δ H P d (ln P) Δ v H = or ln 2 = − v 2 P1 R dT RT ⎛1 1⎞ ⎜⎜ − ⎟⎟ ⎝ T2 T1 ⎠ At 1 atm water boils at 100ºC or 373.15 K and ΔH vΘ = 40.657 kJ / mol (Table 6.1) P R 1 1 ln 2 + = ; − Thus: − Δ v H P1 T1 T2 J 1 1 mol × K ln 1.04 + = J 1 373.15 K T2 40657 mol 8.3144 T2 = 374.27 K. It only boils at a temperature that is about one degree higher than at a sea level. Probably not much of an effect on your cooking habits. Problem 2 (3 points). From the possible statements in column B, select the best match for each phrase in column A and put its letter in the adjacent blank. There is only one best match for each phrase. Column A 1. The point on a P-T phase diagram, where the liquidgas line terminates because both phases have the same density is called ___e___ 2. For a first order phase transition, the Cp at the phase transition temperature on a Cp vs. T diagram is ___i__ 3. ___c___ can be used as appropriate units for surface tension Column B a) b) c) d) e) f) g) h) i) zero moot point J/m2 undefined critical point N/m2 standard boiling point N2/m infinity by definition of Cp Physical Chemistry 418-010 Name __________KEY_______________ Fall 2010 TuTh 5:00-6:15 pm, 206 BrL Quiz#8 Problem 1 (7 points). According to Wikipedia (http://en.wikipedia.org/wiki/Mercury_(element)), pressure of mercury gas above liquid mercury is 1 Pa around room temperature and its density is 13.534 g·cm−3. What would be the pressure of mercury dispersed into uniform droplets with diameter of 10 nm? ⎛ P ⎞ 2γM and ln⎜ • ⎟ = ⎝ P ⎠ ρrRT ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ N kg 2 × 485.48 × 10 −3 × 200.59 × 10 −3 ⎜ ⎟ ⎛ 2γM ⎞ m mol ⎟ ⎟⎟ = 1 Pa × exp⎜ P = P • × exp⎜⎜ kg ⎜ ⎟ ⎝ ρrRT ⎠ 0.001 ⎜ ⎟ g J g −9 × × × 5 10 m 8 . 3144 298 . 15 K ⎜ 13.534 3 × ⎟ 3 mol × K cm −6 m ⎜ ⎟ 10 ⎜ ⎟ 3 cm ⎝ ⎠ P = 1 Pa × exp(1.161) = 3.2 Pa , more than three times higher than over a puddle Problem 2 (3 points). What would be the pressure inside the droplet of mercury described above in Problem 1? Pinside = Poutside 2γ + = 101325 Pa + r 2 × 485.48 × 10 −3 5 × 10 −9 m N m = 101325 Pa + 194192000 Pa = 19429333000 Pa = ~ 1917atm Physical Chemistry 418-010 Name __________KEY_______________ Fall 2010 TuTh 5:00-6:15 pm, 206 BrL Quiz#9 Problem 1 (5 points). Determine the molality and mole fraction of solvent and solute for a solution prepared by dissolving 119 g of potassium bromide (KBr) in a liter of pure water. M ( KBr ) = M ( K ) + M ( Br ) = (39.983 + 79.904) g / mol ≈ 119 g / mol 1000 g = 55.56 mol of water Thus we dissolve one mole of KBr in a 18 g / mol a) Molality is defined as a number of moles of a solute per 1 kg of solvent. In this case we dissolve 1 mole of KBr in 1 kg of water and the molality is 1 mol/kg. b) To determine the molar fraction we need to determine how many moles of either substantce is present in the solution. The total is 56.56 mol. To fine the molar fraction, we simply need to divide the number of moles of each component by the total number of moles in a solution: X water 55.56 mol = = 0.982; X water + X KBr 56.56 mol X KBr 1 mol = = = 1 − X water = 0.018 X water + X KBr 56.56 mol X water = X KBr Problem 2 (5 points). Apply your knowledge of Raoult’s law to determine the total pressure above a mixture of benzene and toluene at room temperature with the molar fraction of 0.3 for benzene in a liquid phase. Assume that the behavior of this mixture can be precisely described by Raoult’s law and that the pressures of toluene above a pure toluene is 28.9 torr and the pressure of benzene above pure benzene is 96.4 Torr. overs olution over solution vapor pure vapor pure liquid pure liquid pure Ptotal = Pbenzene + Ptoluene = y benzene Pbenzene + y toluene Ptoluene = xbenzene Pbenzene + xtoluene Ptoluene Ptotal = 0.3 × 96.4Torr + 0.7 × 28.9Torr = 49.15Torr = 6552.8 Pa Physical Chemistry 418-010 Name __________KEY_______________ Fall 2010 TuTh 5:00-6:15 pm, 206 BrL Quiz#10 Problem 1 (3 points). Estimate the molecular weight of a biopolymer if a solution of 20 mg of this compound in 10 ml of pure deionized water gives 0.1 torr of osmotic pressure at room temperature. 40 mg kg J × 10 −6 × 8.3144 × 298.15 K Pa nRT MW mg mol × K Π = CRT = = = 0.1 torr × 133.322 3 torr V m 10 ml × 10 −6 ml MW = 743.7 kg g = 743700 mol mol Problem 2 (7 points). Using the data on solubility of N2 (g) in water at 25°C from Table 7.2, find the ΔGfΘ(N2 (ao, m)) at 1 atm of N2 pressure and 25°C N 2 ( g , P) → N 2 (ao, m) mN2 ΔG Θ rxn Θ ΔGrzn γN 1mol / kg , but γ 2 m ≈ 1 and Φ ≈ 1 PN 2 Φ PΘ 1 mol g g ⎛ ⎞ ⎛ ⎞ × ⎜ 0.001751 ⎟ ⎜ 0.001751 ⎟ 100 g H 2 O ⎟ 0.1 kg H 2 O 2 × 14.0067 g ⎟ ⎜ ⎜ ⎜ ⎟ ⎜ ⎟ 1mol / kg 1mol / kg = − RT ln⎜ ⎟ ⎟ = − RT ln⎜ PN 2 101325 Pa ⎜ ⎟ ⎜ ⎟ Θ ⎜ ⎟ ⎜ ⎟ 100000 Pa P ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ J J Θ ΔGrzn = −8.3144 × 298.15 K × (− 7.39 ) = 18319 mol × K mol J J J Θ ΔG Θf ( N 2 , ao) = ΔGrzn + ΔG Θf ( N 2 , g ) = 18319 −0 = 18319 mol mol mol Θ Solubility is taken from Table 7.2 and the ΔG f ( N 2 , g ) = 0 Θ f Θ f = − RT ln K a = ΔG ( N 2 , ao) − ΔG ( N 2 , g ) = − RT ln 2