Answer:
[tex]\Delta _VS=3.48x10^{-3}kJ=3.48J[/tex]
Explanation:
Hello,
In this case, the change in the entropy, by knowing the change in the enthalpy and the temperature for a vaporization process is given by:
[tex]\Delta _VS=\frac{\Delta _VH}{T_V}[/tex]
Thus, we start by computing the change in the enthalpy for the 3.3g of dichloromethane as shown below:
[tex]\Delta _VH=3.3gCH_2Cl_2 *\frac{1molCH_2Cl_2}{84.9gCH_2Cl_2}*28\frac{kJ}{molCH_2Cl_2} =1.09kJ[/tex]
Finally, the change in the entropy, considering the temperature in kelvins:
[tex]\Delta _VS=\frac{1.09kJ}{(39.8+273.15)K}\\\\\Delta _VS=3.48x10^{-3}kJ=3.48J[/tex]
Best regards.
