Question

The heat capacity, Cp, of liquid carbon disulfide is a relatively constant 78J(molK). However, the heat capacity of solid carbon disulfide varies greatly with temperature. From 99K to its melting point at 161K, the heat capacity of solid carbon disulfide increases linearly from 44 J/(molK) to 57 J/(molK).

The enthalpy of fusion of carbon disulfide is 4390 J/mol.

The absolute entropy of liquid carbon disulfide at 298K is 151 K/(molK).

Estimate the absolute entropy of carbon disulfide at 99K.

Answer 1

The Third Law of Thermodynamics tells us that as T → 0 K, S → 0. Therefore, for any substance, the absolute entropy at any temperature can be obtained by summing up all the entropy changes from 0 K to to that temperature. For CS₂,

S(99 K) = ∆S(0 → 99 K, s)
and
S°(298, ℓ) = ∆S(0 → 99 K, s) + ∆S(99 K → 161 K, s) + ∆Sfus(s → ℓ) + ∆S(161 K → 298 K, ℓ)
or, solving for ∆S(0 → 99 K, s),
∆S(0 → 99 K, s) = S°(298, ℓ) – ∆S(99 K → 161 K, s) – ∆Sfus(s → ℓ) – ∆S(161 K → 298 K, ℓ)

∆S(99 K → 161 K, s) = ∫(CpdT)/T (99 K → 161 K)
The information given about Cp in this temperature range can be expressed (in J/mol•K) as
Cp = 44 + 14•[(T – 99)/67] = 20.68 + (14/67)T (check that increases linearly from 44 to 57)
Then, ∆S(99 K → 161 K, s) = ∫(CpdT)/T = 20.68 • ln(161/99) + (14/67)(67) = 24.05 J/mol•K

∆Sfus = ∆Hfus/Tm,
∆Sfus(s → ℓ) =(4390 J/mol)/(161 K) = 27.27 J/mol•K

∆S(161 K → 298 K, ℓ) = ∫(CpdT)/T, but Cp is constant so
∆S(161 K → 298 K, ℓ) = (78 J/mol·K) • ln(298/161) = 48.02 J/mol·K

Adding up all the terms, ∆S(0 → 99 K, s) = 151 – 101.86 = 51.66 J/mol·K ANSWER