Question

The heat capacity, CP, of liquid carbon disulfide is a relatively constant 78 J/(mol·K). However, the heat capacity of solid carbon disulfide varies greatly with temperature. From 75 K to its melting point at 161 K, the heat capacity of solid carbon disulfide increases linearly from 39 J/(mol·K) to 57 J/(mol·K). The enthalpy of fusion of carbon disulfude is ΔHfus = 4390 J/mol. The absolute entropy of liquid carbon disulfide at 298 K is S = 151 J/(mol·K).

Estimate the absolute entropy of carbon disulfide at 75 K.

Answer 1

In this question we use Third Law of Thermodynamics it 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(75 K) = ∆S(0 → 75 K, s)
and
S°(298, ℓ) = ∆S(0 → 75 K, s) + ∆S(75 K → 161 K, s) + ∆Sfus(s → ℓ) + ∆S(161 K → 298 K, ℓ)
or, solving for ∆S(0 → 75 K, s),
∆S(0 → 75 K, s) = S°(298, ℓ) – ∆S(75 K → 161 K, s) – ∆Sfus(s → ℓ) – ∆S(161 K → 298 K, ℓ)

∆S(75 K → 161 K, s) = ∫(CpdT)/T (75 K → 161 K)
The information given about Cp in this temperature range can be expressed (in J/mol•K) as
Cp = 39 + 18•[(T – 75)/86] = 23.302 + (18/86)T (check that increases linearly from 39 to 57)
Then, ∆S(75 K → 161 K, s) = ∫(CpdT)/T = 23.302 • ln(161/75) + (18/86)(86) = 35.8 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 → 75 K, s) = 151 – 111.09 = 39.91 J/mol·K