Chem 420/523 Chemical Thermodynamics Homework Assignment # 6 1. * Solid monoclinic sulfur (Sα ) spontaneously converts to solid rhombic sulfur (Sβ ) at 298.15 K and 0.101 MPa pressure. For the conversion process (we designate monoclinic as the “α” form and rhombic as the “β” form for convenience): Sα (298.15K, 0.101MPa) → Sβ (298.15K, 0.101MPa), calculate (a) ∆trans Sm , (b) ∆trans Hm ,and (c) ∆trans Gm , given ◦ =22.59 J K−1 mol−1 for rhombic sulfur Cp,m(β) ◦ Cp,m(α) =23.64 J K−1 mol−1 for monoclinic sulfur ∆trans H ◦ =397.9 J mol−1 for the transition of rhombic sulfur to monoclinic sulfur at the reversible transition temperature of 368.60 K and 0.101 MPa. Answer Note that the ∆trans H ◦ given is ∆βα H ◦ ,where β is the initial state and α is the final state: Sβ (298.15 K,0.101 MPa) → Sα (298.15 K,0.101 MPa). Therefore, for the transition of monoclinic (α) to rhombic (β), ∆αβ H ◦ = −397.9 J mol−1 . Also, we ◦ ◦ − Cp,m(α) , i.e., final state − initial state. denote ∆αβ Cp◦ = Cp,m(β) (a) At 368.60 K and the standard pressure of 0.101 MPa, the two forms of sulfur are in equilibrium (the ◦ = ∆ H ◦ /368.6 K = −1.0789 transition is reversible) and, therefore, ∆αβ G◦ = 0. Therefore, ∆αβ Sm αβ J K−1 mol−1 . At 298.15 K, ◦ ∆αβ Sm =∆αβ Sm + 298.15 Z K 368.60 K ∆αβ Cp◦ dT T =−1.0789 J K−1 mol−1 + (22.59 − 23.64) J K−1 mol−1 × ln =−0.8562 J K−1 mol−1 µ 298.15 368.60 ¶ (b) Similarly, ◦ ∆αβ Hm =∆αβ Hm + 298.15 Z K ∆αβ Cp◦ dT 368.60 K −1 =−397.9 J mol + (22.59 − 23.64) J K−1 mol−1 × (298.15 − 368.60) K =−323.9 J mol−1 (c) From the entropy and enthalpy for the transition, we get ∆αβ Gm =∆αβ Hm − T ∆αβ Sm =−68.62 J mol−1 The negative value for ∆αβ Gm is consistent with the fact that the transition is spontaneous at this temperature. 1 2. The vapor pressures of solid and liquid hydrogen cyanide in the appropriate temperature ranges are given, in Clausius-Clapeyron form, by 4252.4 T 3345.8 liquid :ln p/(kPa) = 15.818 − T solid : ln p(kPa) = 19.489 − (a) (b) (c) (d) (e) Find Find Find Find Find the the the the the molar enthalpy of sublimation for HCN. molar enthalpy of vaporization for HCN. molar enthalpy of fusion for HCN. triple point temperature and pressure. normal boiling point of HCN. Answer: ∆H + c, we get ∆sub H from the expression for the vapor (a) Since the Clausius-Clapeyron form is ln p = − RT pressure of the solid: ∆sub H = 4252.4R = 35, 357 J mol−1 (b) The enthalpy of vaporization is obtained from the expression for the vapor pressure of the liquid: ∆vap H = 3345.8R = 27, 819 J mol−1 (c) The enthalpy of fusion is calculated by recognizing that enthalpies are state functions and, therefore, ∆sub H = ∆f us H + ∆vap H. Therefore, ∆f us H = ∆sub H − ∆vap H = 7, 538.0 J mol−1 (d) At the triple point, the solid and liquid lines intersect. Therefore, 19.489 − 3345.8 4252.4 = 15.818 − . T T Solving for the triple point temperatrure, we get T = 246.96 K. Substituting the temperature back into one of the equations, we get 3345.8 246.96 = 2.2701 ln p/(kPa) = 15.818 − Therefore, p = 9.6804 kPa. (e) The normal boiling point corresponds to the temperature at which p = 101 kPa (or 0.101 MPa or 1.01 bar = 1 atm): T = −3345.8 ln(101) − 15.818 = 298.66 K. 2 3. Given below are the total vapor pressures for {(x1 or y1 )n-C6 H14 + (x2 or y2 )c-C6 H12 } at T = 308.15 K, where x is the mole fraction in the liquid phase and y is the mole fraction in the vapor phase. See p. 429 of text for data (a) Make a graph of x2 against p, p1 and p2 , where p1 and p2 are the vapor pressures of the individual components. For comparison, include on the graph, the ideal solution predictions for p, p1 and p2 . (b) Construct the (vapor + liquid) phase diagram at T = 308.15 K, by making a graph of p against x2 and against y2 . Label all regions of the phase diagram. Answer: (a) The figure is shown below. The solid lines represent actual data and the dashed lines are the ideal solution predictions. Vapor-pressure vs. mole fraction 35.0 30.0 p/(kPa) 25.0 20.0 15.0 10.0 5.0 0.0 0.00 0.20 0.40 0.60 0.80 1.00 x2 (b) The phase diagram is shown below. Liquid + Vapor Phase Diagram 32.0 Liquid 30.0 p(kPa) 28.0 Bubble point line 26.0 Liquid+Vapor Dew-point line 24.0 22.0 20.0 0.00 Vapor 0.20 0.40 0.60 x2 3 0.80 1.00 4. * Given below are the boiling temperatures of {(x1 or y1 )CHCl3 + (x2 or y2 )(CH3 )2 CO} at p = 0.101 MPa, where x is the mole fraction in the liquid phase and y is the mole fraction in the vapor phase. See p. 429 of text for data (a) Construct (vapor + liquid) phase diagram. The system has a maximum boiling azeotrope. Label all regions in the diagram with phases present. Estimate from the phase diagram, the temperature and composition of this azeotrope. (b) An x2 = 0.80 liquid mixture is heated at p = 101 MPa. Use the diagram to estimate (i) the temperature at which the liquid will start to boil, (ii) the temperature at which the last drop of liquid evaporates, (iii) the temperature where 50 mol% of liquid is left in the pot. Answer (a) The boiling point diagram is shown below. Boiling point diagram 337 Vapor 336 335 Liquid + Vapor T/K 334 333 332 331 330 Liquid 329 328 0.00 0.20 0.40 0.60 0.80 1.00 x2 , y 2 The azeotropic composition is estimated to be (x2 = y2 , T ) = (0.33, 336.4 K). (b) (i) The liquid starts to boil at 331.50 K. (ii) Since the vapor that forms at 331.5 K from a liquid mixture with x2 = 0.80 will be richer in component 2, the liquid composition will move towards the azeotrope. Therefore, the temperature as the last drop of liquid evaporates will be equal to that at the azeotropic composition, i.e., 336.4 K. (iii) The tempearture corresponding to x2 = 0.50 is 335.7 K. 4 5. The following cooling curve data for the Magnesium-Copper system are given: Wt % Mg First break (◦ C) Eutectic Halt (◦ C) 5 900 680 10 702 680 15 785 680 20 765 560 30 636 560 35 565 560 40 581 560 45 575 360 50 546 360 60 448 360 70 423 360 80 525 360 90 600 360 Pure copper melts at 1085 ◦ C while pure magnesium melts at 659 ◦ C. Two compounds are formed, one at 16.05 wt % Mg with a melting point of 800 ◦ C, and another at 43.44 wt % Mg with a melting point of 583 ◦ C, respectively. (a) Construct the phase diagram and identify the compositions of the eutectics. (b) What are the empirical formulae of the two compounds? Answer (a) The phase diagram is shown below. Mg-Cu Phase Diagram 1100 1000 t (°C) 900 800 700 600 500 400 300 0 20 40 60 80 100 Wt % Mg The eutectics, as best as can be determined from the given data, appear to be at 9.5 wt % Mg, 34 wt % Mg, and 65 wt % Mg. (b) Compound 1 corresponds to 16.05 wt % Mg. In other words, 100 g of compound will contain 16.05 g Mg and 83.95 g Cu. The Mg : Cu ratio is 83.95 g Cu 16.05 g Mg : = 0.660 mol Mg : 1.321 mol Cu −1 24.305 g mol 63.546 g mol−1 =1.000 mol Mg : 2.002 mol Cu Therefore, the formula of Compound 1 is MgCu2 . Compound 2 corresponds to 43.44 wt % Mg. In other words, 100 g of compound will contain 43.44 g Mg and 56.56 g Cu. The Mg : Cu ratio is 43.44 g Mg 56.56 g Cu : = 1.787 mol Mg : 0.890 mol Cu −1 24.305 g mol 63.546 g mol−1 =2.008 mol Mg : 1.000 mol Cu Therefore, the formula of Compound 2 is Mg2 Cu. 5 6. Construct the ternary phase diagram for the NiSO4 -H2 SO4 -H2 O system from the data given in problem 7.29 of Physical Chemistry by Alberty & Silbey. Answer: The phase diagram is shown below. The various hydrated forms of NiSO4 appear along the base of the triangle (the NiSO4 -H2 O side) since these solids do not contain any H2 SO4 . NiSO4-H2SO4-H2O System H2SO4 20 80 60 40 60 40 2O H 40 4 .7 4 .6 H 2O 20 Ni SO Ni SO NiSO4 20 Ni SO 4 .H 2O 80 6 60 80 H2O

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