Question

The density of rhombic sulfur is 2.070 g cm3 with a standard molar entropy of 31.80J K mol-1. The density of monoclinic sulfur is 1.957 g cm-3 with a standard molar entropy of 32.60J Kmol a Can an increase in temperature be expected to make monoclinic sulfur more stable than rhombic sulfur? If so, at what temperature will the transition occur at 1 bar? Can an increase in pressure be expected to make monoclinic sulfur more stable than rhombic sulfur? b) c) d) If so, at what pressure will that occur at T 273.15 K?

Show all work. Just part b and d please. Thank you in advance.

Answer 1

Answer:

a) Gm depends on temperature via: dGm = –SmdT . Hence as T increases, Gm decreases in proportion to Sm. Since Sm(mono) > Sm(rhom), Gm(mono) will decrease faster than that of the rhombic form, so eventually, a suitable increase in temperature will make monoclinic sulfur more stable than the rhombic form.

b) Assuming the molar entropies are independent of temperature for the range of interest, ∆Gm = –Sm∆T

At the transition, Gm(rhom) = Gm(mono)

For Gm relative to the ∆Gf of the rhombic form (i.e., the monoclinic form lies 0.33 kJ/mol above the rhombic form): 0.0 + ∆Gm(rhom) = 0.33 + ∆Gm(mono)(∆Gm is the change in Gm due to T)

0.0 – Sm(rhom)(Tf–298) = 0.33 – Sm(mono)(Tf-298)

Solving for the final temperature: Tf = 710 K

c) ∆ Gf (Srho)=0.0 kJ / mol ,  Gf (Smono) = 0.33 kJ / mol

At constant T, G depends on p via:

dGm = VmdP,

hence as p increases, so does G in proportion to Vm. The molar volume is related to the density by:

V m = MW/ρ, therefore,

Vm ∝ 1/ρ

ρ(rhom) > ρ (mono), hence Vm(rhom) < Vm(mono) so increasing the pressure will make the monoclinic form less stable relative to the rhombic form rather than more stable. (i.e., G of the monoclinic form will increase faster than G of the rhombic form)

d)

At constant T, G depends on p via:

dGm = VmdP,

0.0 + VmdP1 = 0.3 +  VmdP2

0.0 + dP1/ρ1 = 0.3 + dP2/ρ2  

dP1/ρ1 =0.3 dP2/ρ2 => (P-1bar) /1.95= 0.3 + (p-1bar) /2.07

(P- 1bar) 1.06 = 0.621 + (P-1bar)

0.06 P = 0.621

Solving for the final Pressure = 10.35 bar

I hope this helps...

plesae rate this answer