Chemistry - Thermodynamics - II Question with Solution | TestHub

Explain the question:

Calculate the entropy of fusion ( $\Delta {\mathrm{S}}_{\text{fusion }}$ ) for 9.0 g of ice at $0^{\circ }\mathrm{C}$ using the provided thermodynamic data.

Concept:

For a phase transition occurring at a constant temperature and pressure, the change in entropy is defined as: $\Delta \mathrm{S}=\frac{{\mathrm{Q}}_{\text{rev }}}{\mathrm{T}}=\frac{\Delta \mathrm{H}}{\mathrm{T}}$. For the fusion of a specific mass of ice, the formula becomes $\Delta {\mathrm{S}}_{\text{fusion }}=\frac{\mathrm{m}\cdot \Delta {\mathrm{H}}_{\text{fusion }}}{{\mathrm{T}}_{\mathrm{m}}}$, where ${\mathrm{T}}_{\mathrm{m}}$ is the absolute melting temperature.

Solution:

1. Identify given values:

Mass of ice ($m$) = 9.0 g

Enthalpy of fusion ($\Delta H_{fusion}$) = 335 J g⁻¹

Melting temperature ($T_m$) = 0°C = 273.15 K

2. Calculate the total enthalpy change ($Q$) for melting:

$Q=m\cdot \Delta H_{fusion}=9.0g\times 335J{g}^{-1}=3015J$

3. Calculate the entropy of fusion ($\Delta S_{fusion}$):

$\Delta S_{fusion}=\frac{Q}{T_m}=\frac{3015J}{273.15K}\approx 11.0379J{K}^{-1}$

Rounding to two decimal places gives 11.04 J K⁻¹. Final Answer: Option (A)