The normal boiling temperature of benzene is 353.24 K, and thevapor pressure of

Question

The normal boiling temperature of benzene is 353.24 K, and thevapor pressure of liquid benzene is 1.19x10^4 Pa at 20 degreesCelsius. The enthalpy of f.95 fusion is 9.95 kJ/mol, and thevapor pressure of solid benzene is 137 Pa at -44.3 degreesCelsius. Calculate the following:

a. Hm vaporization

b. Sm vaporization

c. Triple point temperature and pressure

Explanation / Answer

ln [ P2/P1] = (H/R)*[1/T1 - 1/T2] P1 = 1 atm = 1.01325 e5 Pa; T1 = 353.24 K P2 = 1.19e4 Pa,   T2 = 20 C = 293 K ln[ 1.19e4 / 1.01325 e5] = (H/R)*[ 1/353.24 - 1/ 293] -2.14179477 = (H/R)*(-0.000582032809) H = R*3679.8523 K = (8.314 J/mol/K)*(3679.8523 K) = 30594.292J/mol = 30.59 kJ/mol b) S = H/T = 30.59 kJ/mol/ 298 K = 103 J/mol/K c) ln[P2/P1] = (H1/R)*[1/T1 - 1/T2] ln[P4/P3] = (H2/R)*[1/T3 - 1/T4] P4 = P2      T2 = T4 P1 = 101325 Pa      T1 = 353.24 K P3 = 137 Pa    T3 = -44.3 C = 228.7 K => ln[P4/P1] = (H1/R)*[1/T1 - 1/T4] = ln[P4] -ln[P1]       ln[P4/P3] = (H2/R)*[1/T3 -1/T4] = ln[P4] - ln[P3] ln[P3] - ln[P1] = (H1/R)*[1/T1 - 1/T4] -(H2/R)*[1/T3 - 1/T4] ln[P3/P1] = [H1/(R*T1) - H2/(R*T3)] +[-H1/R + H2/R]/T4 -6.60610753 = 5.18301291 -4876.11258/T4 T4 = 413.605904 K => P4 = 463320.761 = 4.63 e5 Pa

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