Answer:
[tex]\Delta _{fus}H=3255.3J/mol[/tex]
[tex]\Delta _{fus}S=7.62\frac{J}{mol*K}[/tex]
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:
[tex]\frac{dp}{dT}=\frac{\Delta _{fus}H}{T\Delta _{fus}V}[/tex]
As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:
[tex]p_2-p_1=\frac{\Delta _{fus}H}{\Delta _{fus}V}ln(\frac{T_2}{T_1} )[/tex]
Solving for the enthalpy of fusion we obtain:
[tex]\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[/tex]
Finally the entropy of fusion is given by:
[tex]\Delta _{fus}S=\frac{\Delta _{fus}H}{T_1} =\frac{3255.3J/mol}{427.15K}\\ \\\Delta _{fus}S=7.62\frac{J}{mol*K}[/tex]
Best regards.
We have that for the Question " calculate the enthalpy and entropy of fusion of the solid" it can be said that the enthalpy and entropy of fusion of the solid are
From the question we are told
Generally the equation for the Clausius-Claepyron equation: is mathematically given as
[tex]\frac{dP}{dT}=\frac{dSfus}{dVfus}[/tex]
Where
Average temperature = 0.5(427.15+429.26) K
Average temperature = 428.205K
Therefore
[tex]dT = 2.11 K\\\\dVfus = 1.06*10-5 m^3/mol\\\\dSfus = 5.526 Pa.m^3/K.mol \\\\Where\\\\dSfus = \frac{dHfus}{average T}\\\\dHfus = 5.526 J/K.mol * 428.205 K\\\\[/tex]
For more information on this visit
https://brainly.com/question/17756498
