Question
please answer the question number 2
Spring 2017 1. The metal Eborium (Eb) has three solid phases: a, B, Y. (See the phase diagram). At a pressure of 0.450 atm, the o and B phase co at 77.0 K. The molar volume of Eb. is 1.23 liter/mole and that of Eb, is 1.47 liter/mole. The heats of transformation are as follows: Eb AH... 375.8 J/mole Eb AH,, 24.3 J/mole At a pressure of 0.590 a three phases may be present. Determine the temperature at which the system must be brought to insure that three phases are present. (Assume this is close to 77.0 Phase Diagram for Eborium 0.95 0.85 0.75 S 0.65 0.60 0.55 0.50 0.45 0.40 0.35 0.30 77.0 Temperature (K) Consider the attached phase diagram for carbon. a) AHfo for diamond (d) is 1.895 kJ/mole. What does this tell you about whether the diamond (d) graphite (g) transition is spontaneous? (20 b) words or less.) 2.377 JyoK-mole, while AS S(2980K)- S(00K). For diamond AS0 for graphite AS 5.740 JMOK-mole. What does this tell you about whether the d- g transition is spontaneous? (20 words or less). About which state is more ordered? (20 words or less c) Determine AGf for diamond. Are diamonds forever? diamond is 3.53 d) The density of graphite is 2.25 gms/cm while that for Sd Sg in the gms/cm What does this tell you about the sign of AS at all vicinity of the triple point? [Assume the densities are constant temperatures and pressures of interest in this problem]
Explanation / Answer
(2)
(a) Since deltaH = + 1.895 kJ, (+ve) with respect to enthalpy the reaction is not spontaneous as it needs energy to get converted.
(b)
deltaS = deltaS0g - deltaS0d = 5.740 - 2.377 = + 3.36 J/K.mol = + 0.00336 kJ/K.mol
Since deltaS0 = + ve , the reaction is spontaneous with respect to entropy.
(c)
DeltaG0 = deltaH0 - TdeltaS0
deltaG0g - deltaG0d = 1.895 - 298(0.00336)
0.000 - deltaG0d = 0.894
deltaG0d = - 0.894 kJ/mol
Diamond is not for ever, at higher temerature, TdeltS term overcomes deltaH term giveng negative sign for dletaG0, which indicates the spontaneity of reaction.
