Question

Calculate the entropy of fusion of a compound at 25°C given
that

its enthalpy of fusion is 32 kJ mol−1 at its melting point of
146°C and

the molar heat capacities (at constant pressure) of the liquid
and solid

forms are 28 J K−1 mol−1 and 19 J K−1 mol−1,
respectively.

Answer #1

Given that the entropy of fusion at 25.0°C can be calculated by using the enthalpy of fusion at 25.0°C, which is determined by the following path.

1) Heating in the solid phase from 25°C to 146°C.

Will use ΔH = q_{p}= C_{p}ΔT C_{p}= 19 J
K^{-1}mol^{-1}( (19*121)/1000) Converting it in
Kj/mol

2) Phase transition from solid to liquid at 146°C

Will use Δ_{fus}H at melting temp = 32 kJ/mol.

3) Cooling in the liquid phase to 146°C →25°C

Will use ΔH = q_{p}= C_{p}ΔT Cp= 28
J K^{-1}mol^{-1}((28*121)/1000) Converting it in
Kj/mol

**Δ _{fus}H = (C_{p},_{m}(solid))
ΔT + Δ_{fus}H+ (C_{p},_{m}(liquid))
ΔT**

**Δ _{fus}H =**19*121+32+28*121

**Δ _{fus}H** =2.299+32+3.388

**Δ _{fus}H =37.687 Kj/mol**

Calculate the entropy of fusion of a compound at 25°C given
that
its enthalpy of fusion is 32 kJ mol−1 at its melting point of
146°C and
the molar heat capacities (at constant pressure) of the liquid
and solid
forms are 28 J K−1 mol−1 and 19 J K−1 mol−1,
respectively.

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