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 525◦C, ΔG◦ = −0.157 kJ mol−1 and
ΔH◦ = 4.184 kJ mol−1.

a. Assuming that ΔH◦ 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 cm−1 for the red form and 7.7 g cm−1 for the black form. Find the pressure at which the two forms can coexist at equilibrium at 525◦C. Which form is more stable above this pressure? Which is more stable below this pressure?

Answer 1

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 mol−1.

∆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