Question

The molar volume of a certain solid is 142.0 cm3 mol−1 at 1.00 atm and 427.15 k, its melting temperature. the molar volume of the liquid at this temperature and pressure is 152.6 cm3 mol−1. at 1.2 mpa the melting temperature changes to 429.26 k. calculate the enthalpy and entropy of fusion of the solid

Answer 1

Answer:

$\Delta _{fus}H=3255.3J/mol$

$\Delta _{fus}S=7.62\frac{J}{mol*K}$

Explanation:

Hello,

Clausius Clapeyron equation is suitable in this case, since it allows us to relate the P,T,V behavior along the described melting process and the associated energy change. Such equation is:

$\frac{dp}{dT}=\frac{\Delta _{fus}H}{T\Delta _{fus}V}$

As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:

$p_2-p_1=\frac{\Delta _{fus}H}{\Delta _{fus}V}ln(\frac{T_2}{T_1} )$

Solving for the enthalpy of fusion we obtain:

$\Delta _{fus}H=\frac{(p_2-p_1)(V_2-V1)}{ln(\frac{T_2}{T_1})} =\frac{(11.84atm-1.00 atm)(156.6cm^3/mol-142.0cm^3/mol)}{ln(\frac{429.26K}{427.15K} )} \\\\\Delta _{fus}H=32127.3atm*cm^3/mol*\frac{101325Pa}{1atm}*(\frac{1m}{100cm} )^3\\\Delta _{fus}H=3255.3J/mol$

Finally the entropy of fusion is given by:

$\Delta _{fus}S=\frac{\Delta _{fus}H}{T_1} =\frac{3255.3J/mol}{427.15K}\\ \\\Delta _{fus}S=7.62\frac{J}{mol*K}$

Best regards.