Can someone please help with this chem question

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 89 K to its melting point at 161 K, the heat capacity of solid carbon disulfide increases linearly from 42 J/(mol·K) to 57 J/(mol·K). For the purposes of this question, the heat capacity for solid carbon disulfide can be estimated as the average of the given values at 89 K and 161 K.

The enthalpy of fusion of carbon disulfide 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 89 K."

Answer 1

To estimate the absolute entropy of carbon disulfide at 89 K, we can use the following equation:

ΔS = ∫(C/T) dT + ΔSfus/R

where ΔS is the change in entropy, C is the heat capacity, T is the temperature, ΔSfus is the entropy of fusion, and R is the gas constant.

Since the heat capacity of liquid carbon disulfide is constant at 78 J/(mol·K), we can write:

ΔS_l = ∫(C_l/T) dT = C_l ln(T/T_ref)

where T_ref is a reference temperature, which we can take to be 298 K. Therefore:

ΔS_l = C_l ln(T/T_ref) = 78 J/(mol·K) ln(89 K/298 K) = -154.5 J/(mol·K)

To estimate the entropy of the solid at 89 K, we can use the average heat capacity value given as 49.5 J/(mol·K):

ΔS_s = ∫(C_s/T) dT = C_s ln(T/T_ref) + ΔSfus/R

where we can take the heat capacity of the solid to be constant over the temperature range of interest. Therefore:

ΔS_s = C_s ln(T/T_ref) + ΔSfus/R = 49.5 J/(mol·K) ln(89 K/298 K) + (4390 J/mol)/(8.314 J/(mol·K)) = -62.7 J/(mol·K)

Finally, we can estimate the absolute entropy of carbon disulfide at 89 K by adding the entropies of the liquid and solid phases:

ΔS = ΔS_l + ΔS_s = -154.5 J/(mol·K) - 62.7 J/(mol·K) = -217.2 J/(mol·K)

Therefore, the estimated absolute entropy of carbon disulfide at 89 K is -217.2 J/(mol·K).