Problem Set 10 Solutions

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1 Chemistry 360 Dr Jean M Standard Problem Set 10 Solutions 1 Sketch (roughly to scale) a phase diagram for molecular oxygen given the following information: the triple point occurs at 543 K and 114 torr; the critical point occurs at 1546 K and torr; the normal melting point is 5475 K; and the normal boiling point is 9025 K Does solid molecular oxygen melt under applied pressure? he data points that are given for the phase diagram are plotted below with pressure on the y-axis and temperature on the x-axis Note that critical point occurs at such a high pressure compared to the other points that it is difficult to see the other regions (these regions will be focused on later) We know that the triple point, normal boiling point, and critical point must lie along the liquid-vapor coexistence curve, with the triple point and critical point corresponding to the terminators of the curve his is indicated in the plot above by dashed lines connecting these three points On the lower temperature side of the L-V coexistence curve lies the liquid phase, and on the higher temperature side lies the vapor phase We also know that the triple point and the normal melting point lie along the solid-liquid coexistence curve his curve most likely extends to pressures above that of the normal melting point, but we have no data to indicate the pressure and temperature values above 1 atm760 torr hus, the solid phase must lie somewhere on the left side of the plot, at temperatures lower than that of the normal melting point, as indicated he other coexistence curve, for the solid-vapor phase equilibrium, must also be present on the phase diagram However, we are given no data to characterize this portion of the phase diagram

2 2 1 continued o get a better view of the coexistence curves at lower pressures, the scale of the y-axis is adjusted to show pressures of 800 torr and lower; therefore, the position of the critical point is not shown in the figure below In this zoomed-in version of the O 2 phase diagram, the liquid-vapor coexistence curve connecting the triple point and the normal boiling point (and eventually the critical point) is clearly seen and the liquid and vapor regions noted Similarly, the solid-liquid coexistence curve connecting the triple point and the normal melting point is more easily seen and the solid (and liquid) regions defined Once again, however, no data is given to locate the solid-vapor coexistence curve It must occur at very low pressure, however, since it must lie below the triple point at 114 torr o answer the question of whether or not solid O 2 melts when the pressure is increased, we need to estimate the slope of the coexistence curve For a graph of pressure vs temperature, the slope is slope ΔP Δ P 2 P Using the two known points along the solid-liquid coexistence curve, the normal melting point ( K, P torr ) and the triple point ( K, P torr ), the slope is slope ΔP Δ P 2 P torr 760 torr 543 K 5475 K slope 1686 torr/k Since the slope of the solid-liquid coexistence curve is positive, the solid will not melt as the pressure increases

3 3 2 Calculate the melting point of ice under a pressure of 50 bar given that the melting point at 1 bar is 0 C Assume that the density of ice under these conditions is 092 g/ml and the density of liquid water is 100 g/ml he molar enthalpy of fusion of water is 601 kj/mol he Clapeyron equation for solid-liquid phase equilibrium is dp d ΔH fus,m fus ΔV m We can approximate the left side of the equation as dp d ΔP Δ Substituting, ΔP Δ ΔH fus,m fus ΔV m o get the melting point of ice at a pressure of 50 bar, we can solve for Δ and use the fact that the melting point of ice at 1 bar pressure is 0 C (27315 K) Solving for Δ yields, Solving for 2, Δ fus ΔV m ΔP ΔH fus,m, or 2 1 fus ΔV m ( P 2 P 1 ) ΔH fus,m fus ΔV m ( P 2 P 1 ) ΔH fus,m he molar volumes of the liquid and solid phases can be calculated from the molecular weight and the densities For ice, For liquid water, the molar volume is V s,m M D 1L 1000 ml g mol g ml 1 1L 1000 ml V s,m L/mol V l,m M D 1L 1000 ml g mol g ml 1 1L 1000 ml V l,m L/mol

4 4 2 Continued Substituting, fus ΔV m ( P 2 P 1) ΔH fus,m ( K) ( L/mol) ( 50 1bar) 27315K + ( 6010 J/mol) 27315K K L bar J J 1L bar 27315K 035K K or 035! C 3 If it takes an increase of 1334 megabars of pressure to change the melting point of a substance from 222 C to 122 C for a change in molar volume of 322 cm 3 /mol, what is the molar enthalpy of fusion of the substance in J/mol? he Clapeyron equation for solid-liquid phase equilibrium is dp d ΔH fus,m fus ΔV m Using an approximation for the left side in terms of finite changes leads to the relation ΔP Δ ΔH fus,m fus ΔV m Solving for the molar entropy of fusion, we have ΔH fus,m ΔP fus ΔV m Δ Substituting ΔP bar, Δ 100 K, fus K, and ΔV m L/mol yields ΔP ΔH fus,m fus ΔV m Δ bar 100 K ( 49515K )( L/mol) 100 J L bar/mol 1L bar ΔH fus,m J/mol or 2127 kj/mol

5 5 4 he vapor pressure of liquid benzene obeys the relation ln P , where the pressure is in torr and temperature is in Kelvin Calculate the molar enthalpy of vaporization of benzene and the normal boiling point he liquid-vapor equilibrium coexistence curve is given by the Clausius-Clapeyron equation as ln P ΔH vap,m R + C Since the equation above is in the same form, we have that ΔH vap,m 58935K, R or ΔH vap,m R 58935K 58935K 8314 J mol 1 K 1 ΔH vap,m J/mol or 490 kj/mol he normal boiling point can be determined by solving the equation given for and substitution of P 760 torr, hen, for P 760 torr, ln P ln P ln P ln K or 1445! C

6 6 5 At what pressure does the boiling point of water become 300 C? If oceanic pressure increases by 1 atm for every 10 m, to what ocean depth does this pressure correspond? Since we are dealing with a liquid/vapor equilibrium, the Clausius-Clapeyron equation applies he integrated form is ln P 2 ΔH vap,m 1 R 1 P In order to use this equation to determine P 2, we need the enthalpy of vaporization of water his can be obtained from values given in the appendix ( ΔH vap,m 4403 kj/mol ) Substituting, ln P 2 ln P 1 + ΔH vap,m R ln ( 1atm) + ln P aking the exponential gives the pressure, ( J/mol) J/molK 37315K K P 2 e P atm his would correspond to an ocean depth of about 1420 meters 6 he sublimation pressures of solid Cl 2 are 352 Pa at 112 C and 35 Pa at 1265 C he vapor pressures of liquid Cl 2 are 1590 Pa at 100 C and 7830 Pa at 80 C Calculate the molar enthalpies of sublimation, vaporization, and fusion Since the problem deals with S-V and L-V equilibria, the Clausius-Clapeyron equation applies he equation is ln P 2 ΔH m 1 R 1 P Solving for the enthalpy change, ΔH m R ln P 2 /P

7 7 6 continued For the S-V equilibrium, the equation is ΔH sub,m R ln P 2 /P For the L-V equilibrium, the equation becomes 2 ( 8314 J mol 1 K 1 ) 35Pa ln 352 Pa K K ΔH sub,m J/mol or 3128 kj/mol ΔH vap,m R ln P 2 /P ( 8314 J mol 1 K 1 ) 7830 Pa ln 1590 Pa K K ΔH vap,m J/mol or 2216 kj/mol Since enthalpy is a state function, the enthalpy of fusion can be calculated from the following relation Solving for the enthalpy of fusion, ΔH sub,m ΔH fus,m + ΔH vap,m ΔH fus,m ΔH sub,m ΔH vap,m 3128 kj/mol 2216 kj/mol ΔH fus,m 912 kj/mol

8 8 7 Freon-12 (CF 2 Cl 2 ) was commonly used in spray cans prior to the discovery that it was harmful to the ozone layer Its enthalpy of vaporization is 2025 kj/mol and its normal boiling point is 292 C Determine the vapor pressure of Freon-12 at 40 C he Clausius-Clapeyron equation for the liquid-vapor equilibrium is ln P 2 ΔH vap,m 1 R 1 P Solving for ln P 2 yields he pressure P 2 is then ln P 2 ln P 1 + ΔH vap,m R ln ( 10 atm) + ln P ( J/mol) J mol 1 K K K P 2 e P atm 8 he vapor pressure of solid uranium hexafluoride, UF 6, follows the equation ln P , where the pressure is in Pa and temperature is in Kelvin he vapor pressure of liquid uranium hexafluoride follows the equation ln P Determine the temperature and pressure of the triple point he triple point occurs at the intersection of the S-V and L-V coexistence curves Equating these from above, K hus, K is the triple point temperature Substituting this temperature back into either of the two coexistence equations and solving for pressure give the triple point pressure

9 9 8 continued Using the S-V equation, ln P ln P K ln P P Pa 9 he vapor pressure of liquid mercury is 0133 bar at 260 C and 0533 bar at 330 C Assume that the mercury vapor can be treated as an ideal gas and that the enthalpy of vaporization is independent of temperature Calculate the molar enthalpy and the molar Gibbs free energy of vaporization at 25 C he Clausius-Clapeyron equation for the liquid-vapor equilibrium is ln P 2 ΔH vap,m 1 R 1 P Solving for the enthalpy of vaporization, ΔH vap,m R ln P 2 /P Substituting, ΔH vap,m R ln P 2 /P ( 8314 J mol 1 K 1 ) 0533bar ln 0133 bar K K ΔH vap,m J/mol or 5302 kj/mol herefore, assuming that ΔH vap,m is independent of temperature, the molar enthalpy of vaporization at 25 C is 5302 kj/mol o determine the Gibbs free energy, we can use the equation ΔG vap,m ΔH vap,m ΔS vap,m

10 10 9 continued he entropy of vaporization can be determined from the alternate form of the Clausius-Clapeyron equation, ln P ΔH vap,m R + ΔS vap,m R Solving for the entropy of vaporization, ΔS vap,m R ln P + ΔH vap,m Substituting 0133 bar and K for the pressure and temperature, ΔS vap,m ( 8314 J mol -1 K 1 ) ln ( 0133bar) J/mol 53315K ΔS vap,m 8270 J mol -1 K 1 Finally, the Gibbs free energy at 25 C is ΔG vap,m ΔH vap,m ΔS vap,m J/mol ( 29815K) 8270 J mol 1 K 1 ΔG vap,m J/mol or 2836 kj/mol

11 10 Consider the phase diagram of sulfur given below (rhombic and monoclinic are two different solid forms of sulfur) Starting at 298 K and 1 atm pressure and considering an increase in the

11 11 10 Consider the phase diagram of sulfur given below (rhombic and monoclinic are two different solid forms of sulfur) Starting at 298 K and 1 atm pressure and considering an increase in the temperature (at constant pressure), comment on the entropy change as the sulfur goes from the rhombic solid phase to the monoclinic sold phase Is the entropy change expected to be positive or negative? On the basis of the Second Law of hermodynamics, is the phase transition expected to be spontaneous in an isolated system? For the Solid-Rhombic to Solid-Monoclinic phase transition, the phase diagram shows that the slope of the coexistence curve dp/d is positive, dp > 0 d From the Clapeyron equation, we have dp ΔSm d ΔVm he slope of the coexistence curve is positive, ΔSm > 0 ΔVm We can assume that ΔVm is positive because the molar volume of the monoclinic phase would be predicted to be larger than the molar volume of the rhombic phase at 25 C his is because at a given temperature, the phase with the smaller molar volume (and therefore higher density) will be stable at higher pressure his means that ΔSm is also positive, ΔSm > 0 from the Second Law, a positive entropy change corresponds to a spontaneous process in an isolated And system

Exam 3 Solutions. ClO g. At 200 K and a total pressure of 1.0 bar, the partial pressure ratio for the chlorine-containing compounds is p ClO2

Exam 3 Solutions. ClO g. At 200 K and a total pressure of 1.0 bar, the partial pressure ratio for the chlorine-containing compounds is p ClO2 Chemistry 360 Dr. Jean M. Standard Fall 2016 Name KEY Exam 3 Solutions 1.) (14 points) Consider the gas phase decomposition of chlorine dioxide, ClO 2, ClO 2 ( g) ClO ( g) + O ( g). At 200 K and a total