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Charpter 4 equilibrium

By Steven Alexander,2014-05-24 12:16
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Charpter 4 equilibrium

    Equilibrium

1. At 400C, liquid zinc has a vapor pressure of 10-4 atm. Estimate the boiling

    temperature zinc, knowing that its heat of evaporation is approximately 28

    kcal/mol. (4.2, P116)

    18g/mol363V19.57cm/mol19.5710m/mol Solution: (a) ice30.92g/cm

    18g/mol363V18cm/mol1810m/mol water31g/cm

    63 V1.5710m/molfus

    According to the Clapeyron equation:

    HdP1fus(VdTTfus

    H1fusdP(dTVTfus

    take definite integration of the above:

    5H501.01310T1fusdPdT 5~~1.01310273VTfus

    VTfus5ln491.01310H273fus

    651.5710491.01310 6009

    0.013

     T272.8K

    1503236 (b) P5010lb/in.50106897Pa34510Pa 0.013

    6 P34510Pa

     (c )

    VTfus6ln34510H273fus

    661.5710345106009

    0.09

     T249.46K

    -42. At 400C, liquid zinc has a vapor pressure of 10 atm. Estimate the boiling temperature of zinc, knowing that its heat of vaporation is approximately

    28kcal/mol. (4.3,117)

    3 Solution: H28kcal/mol4.182810J/mol117.04J/molvap

     According to Claperon equation in vapor equilibrium:

    H1vapd(lnP)d() RT

    HP112vap dP(ln)()~P1RTT21

    HP11vap1 ln()RTTP212

    31117.041011 ln()4T8.314673102

    T1202K 2

    The boiling point of zinc is 1202K.

    H90T3. Troutons rule is expressed as follows: in joules per mole, where vapb

    T is the boiling point (K). The boiling temperature of mercury is 630K. Estimate b

    the partial pressure of liquid Hg at 298K. Use Troutons rule to estimate the heat of vaporization of mercury.

    H90TSolution: vapb

    HP11vapdP(ln)() ~1R298630

    6823 lnP;10.8322.90;10.8312.07 298-6 P=5.7310atm

    4. Liquid water under an air pressure of 1 atm at 25C has a large vapor pressure that it would have in the absence of air pressure. Calculate the increase in vapor

    pressure produced by the pressure of the atmosphere on the water. Water has a

    density of 1g/cm3; the vapor pressure ( in the absence of the air pressure) is

    3167.2Pa. (4.5, p116)

    18g/mol363V18cm/mol1810m/mol Solution: l31g/cm

    GGvapor pressure changes with the total external pressure, vl

    Pe,2 RTlnV(PP)lTe,1Pe,1

    Pe,26 RTln1810(101303167.2)Pe,1

    Pe,2 1.000051Pe,1

    P3167.36Pa e,2

    P = 0.16Pa

    the vapor pressure increase is 0.16Pa.

    5. The boiling point of silver (P=1 atm) is 2450K. The enthalpy of evaporation of

    liquid silver is 255,000 J/mol at its boiling point. Assume, for the purpose of this

    problem, that the heat capacities of liquid and vapor are the same. (a) Write an

    equation for the vapor pressure of silver, in atmospheres, as a function of kelvin

    temperature. (b). The equation should be suitable for use in a tabulation, NOT in

    differential form. Put numerical values in the equation based on the data given.

    (4.7, p117)

    HP11vapd(lnP)()Solution: ~1RT2450

    25500011lnP() 8.314T2450

    30685lnP;104.08 T

    6. Zinc may exist as a solid, a liquid, or a vapor. The equilibrium pressure-temperature relationship between solid zinc and zinc vapor is giben by the vapor pressure equation for the solid. A similar relation exists for liquid zinc. At the triple point all three phases, solid, liquid, and vapor exist in equilibrium. That means that the vapor pressure of the liquid and the solid are the same. The vapor pressure of solid Zn varies

    15755with T as: and the vapor pressure of liquid lnP(atm)0.755ln(T);19.25T

    15246Zn varies with T as:. Calculate: (a) The lnP(atm)1.255ln(T);21.79T

    boiling point of Zn under 1 atm; (b) The triple-point temperature; (c) the heat of evaporation of Zn at the normal (1 atm) boiling point; (d) The heat of fusion of Zn at the triple-point temperature; (e)The differences between the heat capacities of solid and liquid Zn. (4.8, p118)

    Solution :(a) At boiling point, P=1 atm, that is lnP=0

    15246 1.255ln(T);21.790T

     152461.255Tln(T);21.79T0

    To solve the equation, let y1 = 21.79T-15246, y2 = 1.255TlnT. Plot y-T of the functions, and the intersection is the answer.

    3000028000 y126000 y224000220002000018000160001400012000y1000080006000400020000-2000-4000-6000

    4006008001000120014001600180020002200

    T, K

    From the plot, the intersection is 1180 K. So at 1180K, zinc boils.

     (b) At triple point, vapor pressure of solid Zn equals that of liquid Zn;

    15246157551.255ln(T);21.790.755ln(T);19.25 TT

    5090.5ln(T);2.540 T

    5090.5Tln(T);2.54T0

    To solve the equation, assume two functions, y1=2.54T+509; y2 = 0.5TlnT. Plot y1-T and y2-T. The intersection is the answer.

     y11 y228000

    7500

    7000

    6500

    6000

    5500

    5000

    4500y4000

    3500

    3000

    2500

    2000

    1500

    10004006008001000120014001600180020002200

    T,K

    The intersection is 695.3K, and this is the triple point temperature.

    (c )

    152461.255d(lnP);21.792TT

    1(152461.255)();TdT 2T

    1(152461.255)();TdT

    H1vapd(lnP)d() RT

    Hvap15246;1.255T R

    HR(152461.255T)12675510.43T vap

    3 At T =1180K, H12675510.43T114.410J/mol114.4kJ/molbvap

     (d) For solids,

    15755lnP0.755ln(T);19.25T

    157550.755d(lnP)()dT2TT 1(15755;0.755)()dT2T

    1(157550.755)();dT

    H1fusd(lnP)d() RT

    Hfus15755;0.755T R

    HR(157550.755T) fus

    At triple point, T = 695.4K tr

    3 H8.314(157550.755T)126.610J/mol126.6kJ/molfus

    d(H)CdT (e) fusP

    d(H)fusC0.755 PdT

    7. A particular material has a latent heat of vaporization of 5000J/mol. This heat of vaporization does not change with temperature or pressure. One mole of the material exists in a two-phase equilibrium (liquid-vapor) in a container of volume V=1L, a temperature of 300K, and pressure of 1 atm. The container (constant volume ) is heated until the pressure reaches 2 atm. (Note that this is not a small P.) The vapor

    phase can be treated as an ideal monatomic gas and the molar volume of the liquid can be neglected relative to that of the gas. Find the fraction of material in the vapor phase in the initial and final states. (4.9, P118)

    5T300K,P1atm1.01310Pa,V1LSolution: In the initial state, 111

    3PV1013010311 PVnRT,n4.0610mol1111RT8.3143001

    mol(vapor)%4.06%

    In the final state, P=2 atm, V= 1L 22

    According to Clayperon equation:

    H?P11vap2?RTln?PRTT?122

    2500011 ln()18.314T3002

    T458.6K2

    3PV21013010322PVnRT,n5.310mol22222RT8.314458.62

    mol(vapor)%5.3%

    8. The melting point of gold is 1336K, and vapor pressure of liquid gold is given by:

    43522. (a) Calculate the heat of vaporization of lnP(atm)23.7161.222lnT(K)T

    gold at its melting point; Answer parts b, c, and d only if the data given in this problem statement are sufficient to support the calculation. If there are not enough data, write solution not possible. (b) What is the vapor pressure of solid gold at its melting point? (c) What is the vapor pressure of solid gold at 1200K ? (d) What is the v

    Solution: (a)

    435221.222 dPdT(ln)()2TT

    1;dT(435221.222)()2T

    1(43522;1.222)d()T

    HR(43522;1.222T)36184110.19T vap

    3(a) At 1336K, H36184110.19133634810J348kJvap

    (b) Solution not possible;

    (c) Solution not possible.

    9. (a) At 298K, what is the Gibbs free energy change for the following reaction?

    CC graphitediamond

     (b) Is the diamond thermodynamically stable relative to graphite at 298K?

     (c ) What is the change of Gibbs free energy of diamond when it is compressed isothermally from 1atm to 1000 atm?

    (d) Assuming that graphite and diamond are incompressible, calculate the pressure

    at which the two exist at equilibrium at 298K.

    (e) What is the Gibbs free energy of diamond relative to graphite at 900K? to

    simplify the calculation, assume that the heat capacities of the two materials are

    equivalent.

    3 DATA Density of graphite is 2.25g/cm

    3 Density of diamond is 3.51g/cm

    oo H(kJ/mol)H(J/mol.K)(298)(298)ff

    Diamond 1.897 2.38

    Graphite 0 5.74

    CCSolution: (a) graphitediamond

    oo HHH1897kJ/mol,,fdiamondfgraphite

    oo SSS5.74;2.383.36J/(mol.K),,fdiamondfgraphite

     GHTS1897298(3.36)2898.28J/mol(b) No, diamond is not thermodynamically stable relative to graphite at 298K.

    61210GVP991013034.29J/mol(c ) diamand3.51

    (c ) Assuming N atm , G = 0, reversible processes as following can be designed to

    realize this,

     ;4 graphite, 298K, N atm diamond, 298K, N atm

    (3) ;1

     ;2 graphite, 298K,1atm diamond, 298K,1atm

    GGGG(4)(1)(2)(3)

    VP;2898.28;V(P)graphitediamond

    (VV)(1N)10130;2898.28graphitediamond

    66 ?12101210?N10130;2898.28?2.253.51?

    0.194N;2898.280

    N14939(atm)

    CT'T'pC0,CdT0,dT0 pp~~TTT

    HH1897J/mol 900298

    SS3.36J/mol.K 900298

     GHTS1897;9003.364921J/mol

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