Mercury(II) sulfide, HgS, occurs in two crystalline forms, called “red” and “bla
ID: 1050392 • Letter: M
Question
Mercury(II) sulfide, HgS, occurs in two crystalline forms, called “red” and “black.” For the conversion of the red form to the black form at 525C, G = 0.157 kJ mol1 and
H = 4.184 kJ mol1.
a. Assuming that H is independent of temperature, find the temperature at which the two forms can coexist at equilibrium at 1.000 bar. Which is more stable above this temperature? Which is more stable below this temperature?
b. The densities are 8.1 g cm1 for the red form and 7.7 g cm1 for the black form. Find the pressure at which the two forms can coexist at equilibrium at 525C. Which form is more stable above this pressure? Which is more stable below this pressure?
Explanation / Answer
a) we can calcualte DeltaS by using Delta G and Delta H as
DeltaG0 = DeltaH0 - T Delta S0
T = 298.15 K
Delta S0 = [4.184 -(-0.157) ] / 298.15
Delta S0 = 5.504 Joules / K mole = 0.005504 KJ / mole K
Now when the two phase exist together, the Delta G0 = 0
DeltaG0 = DeltaH0 - T Delta S0= 0
DeltaH0 = T Delta S0
T = DeltaH0 /Delta S0
T = (4.184 KJ / mole)/ (0.005504 KJ / mole K)
T = 760.17 Kelvin
b) the molar volume can be calculated as
Vm = Molecular weight / density
Vm(red) = 232.66 g / mole / 8.1 = 28.72
Vm(black) = 232.66 g / mole / 7.7 = 30.22
Vm = 30.22-28.72 = 1.5
We will use following equation
P2-P1 = H /Vm (lnT2 - LnT1)
P2 = /
P1 = 1 atm
H = 4.184 kJ mol1.
Vm = 30.22-28.72 = 1.5
T1 = 298.15 K
LnT1 = 5.69
T2 = 525C = 525 + 273.15 K = 798.15 K
Ln T2 = 6.68
P2-1 =4.184 / 1.5 (6.68 - 5.69) = 2.76
P2 = 3.76 atm