Chem 420/523 Chemical Thermodynamics Homework Assignment # 6

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1 Chem 420/523 Chemical hermodynamics Homework Assignment # 6 1. * Solid monoclinic sulfur (S α ) spontaneously converts to solid rhombic sulfur (S β ) at K and MPa pressure. For the conversion process (we designate monoclinic as the α form and rhombic as the β form for convenience): S α (298.15K, 0.101MPa) S β (298.15K, 0.101MPa), calculate (a) trans S m, (b) trans H m,and (c) trans G m, given Cp,m(β) =22.59 JK 1 mol 1 for rhombic sulfur Cp,m(α) =23.64 JK 1 mol 1 for monoclinic sulfur trans H =397.9 Jmol 1 for the transition of rhombic sulfur to monoclinic sulfur at the reversible transition temperature of K and MPa. Answer Note that the trans H given is βα H,where β is the initial state and α is the final state: S β ( K,0.101 MPa) S α ( K,0.101 MPa). herefore, for the transition of monoclinic (α) to rhombic (β), αβ H = Jmol 1. Also, we denote αβ Cp = Cp,m(β) C p,m(α), i.e., final state initial state. (a) At K and the standard pressure of MPa, the two forms of sulfur are in equilibrium (the transition is reversible) and, therefore, αβ G =0. herefore, αβ Sm = αβ H /368.6 K = JK 1 mol 1. At K, (b) Similarly, αβ S m = αβ S m Z K K αβ Cp d = JK 1 mol 1 +( ) JK 1 mol 1 ln = JK 1 mol 1 αβ H m = αβ H m Z K K αβ C pd µ = Jmol 1 +( ) JK 1 mol 1 ( ) K = Jmol 1 (c) From the entropy and enthalpy for the transition, we get αβ G m = αβ H m αβ S m = Jmol 1 henegativevaluefor αβ G m is consistent with the fact that the transition is spontaneous at this temperature. 1

2 2. he vapor pressures of solid and liquid hydrogen cyanide in the appropriate temperature ranges are given, in Clausius-Clapeyron form, by solid :lnp(kpa) = liquid :ln p/(kpa) = (a) Find the molar enthalpy of sublimation for HCN. (b) Find the molar enthalpy of vaporization for HCN. (c) Find the molar enthalpy of fusion for HCN. (d) Find the triple point temperature and pressure. (e) Find the normal boiling point of HCN. Answer: (a) Since the Clausius-Clapeyron form is ln p = H R +c, we get subh from the expression for the vapor pressure of the solid: sub H = R =35, 357 Jmol 1 (b) he enthalpy of vaporization is obtained from the expression for the vapor pressure of the liquid: vap H = R =27, 819 Jmol 1 (c) he enthalpy of fusion is calculated by recognizing that enthalpies are state functions and, therefore, sub H = fus H + vap H. herefore, fus H = sub H vap H =7, Jmol 1 (d) At the triple point, the solid and liquid lines intersect. herefore, = Solving for the triple point temperatrure, we get = K. Substituting the temperature back into one of the equations, we get ln p/(kpa) = = herefore, p = kpa. (e) he normal boiling point corresponds to the temperature at which p = 101 kpa (or MPa or 1.01 bar = 1 atm): = ln(101) = K. 2

3 3. Given below are the total vapor pressures for {(x 1 or y 1 )n-c 6 H 14 +(x 2 or y 2 )c-c 6 H 12 } at = 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 x 2 against p, p 1 and p 2,wherep 1 and p 2 are the vapor pressures of the individual components. For comparison, include on the graph, the ideal solution predictions for p, p 1 and p 2. (b) Construct the (vapor + liquid) phase diagram at = K, by making a graph of p against x 2 and against y 2. Label all regions of the phase diagram. Answer: (a) he figure is shown below. he solid lines represent actual data and the dashed lines are the ideal solution predictions. Vapor-pressure vs. mole fraction p/(kpa) x 2 (b) he phase diagram is shown below Liquid + Vapor Phase Diagram 30.0 Liquid 28.0 p(kpa) Dew-point line Bubble point line Liquid+Vapor 22.0 Vapor x 2 3

4 4. * Given below are the boiling temperatures of {(x 1 or y 1 )CHCl 3 +(x 2 or y 2 )(CH 3 ) 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. he 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 x 2 =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) he boiling point diagram is shown below. Boiling point diagram Vapor Liquid + Vapor /K Liquid x 2, y 2 he azeotropic composition is estimated to be (x 2 = y 2,) = (0.33, K). (b) (i) he liquid starts to boil at K. (ii) Since the vapor that forms at K from a liquid mixture with x 2 =0.80 will be richer in component 2, the liquid composition will move towards the azeotrope. herefore, the temperature as the last drop of liquid evaporates will be equal to that at the azeotropic composition, i.e., K. (iii) he tempearture corresponding to x 2 =0.50 is K. 4

5 5. he following cooling curve data for the Magnesium-Copper system are given: Wt % Mg First break ( C) Eutectic Halt ( C) Pure copper melts at 1085 C while pure magnesium melts at 659 C. wo compounds are formed, one at wt % Mg with a melting point of 800 C, and another at 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) he phase diagram is shown below Mg-Cu Phase Diagram 900 t ( C) Wt % Mg he 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 wt % Mg. In other words, 100 g of compound will contain g Mgand83.95gCu.heMg:Curatiois gmg gmol 1 : gcu =0.660 mol Mg : mol Cu gmol 1 =1.000 mol Mg :2.002 mol Cu herefore, the formula of Compound 1 is MgCu 2. Compound 2 corresponds to wt % Mg. In other words, 100 g of compound will contain g Mgand56.56gCu.heMg:Curatiois gmg gmol 1 : gcu = mol Mg :0.890 mol Cu gmol 1 =2.008 mol Mg : mol Cu herefore, the formula of Compound 2 is Mg 2 Cu. 5

6 6. Construct the ternary phase diagram for the NiSO 4 -H 2 SO 4 -H 2 O system from the data given in problem 7.29 of Physical Chemistry by Alberty & Silbey. Answer: he phase diagram is shown below. he various hydrated forms of NiSO 4 appear along the base of the triangle (the NiSO 4 -H 2 O side) since these solids do not contain any H 2 SO 4. NiSO 4 -H 2 SO 4 -H 2 O System H 2 SO NiSO 4 NiSO 4.H 2 O NiSO 4.6H 2 O NiSO 4.7H 2 O H 2 O 6