The heat capacity, CP, of liquid carbon disulfide is a relatively constant 78 J/

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 99 K to its melting point at 161 K, the heat capacity of solid carbon disulfide increases linearly from 44 J/(mol·K) to 57 J/(mol·K). The enthalpy of fusion of carbon disulfide is Hfus = 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 99 K.

Explanation / Answer

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 that temperature.
For CS:

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

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 + 13•[(T – 99)/62] = 23.24 + (13/62)T

(check that increases linearly from 44 to 57)

Then, S(99 K 161 K, s) = (CpdT)/T = 23.24 • ln(161/99) + (13/62)(57) = 23.25 J/mol•K

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

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

Adding up all the terms, S(0 94 K, s) = 151 – 23.25- 27.27 - 48.02 = 52.46 J/mol·K

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.