Answer:
The boiling point is 308.27 K (35.27°C)
Explanation:
The chemical reaction for the boiling of titanium tetrachloride is shown below:
Ti[tex]Cl_{4(l)}[/tex] ⇒ Ti[tex]Cl_{4(g)}[/tex]
ΔH°[tex]_{f}[/tex] (Ti[tex]Cl_{4(l)}[/tex]) = -804.2 kJ/mol
ΔH°[tex]_{f}[/tex] (Ti[tex]Cl_{4(g)}[/tex]) = -763.2 kJ/mol
Therefore,
ΔH°[tex]_{f}[/tex] = ΔH°[tex]_{f}[/tex] (Ti[tex]Cl_{4(g)}[/tex]) - ΔH°[tex]_{f}[/tex] (Ti[tex]Cl_{4(l)}[/tex]) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol
Similarly,
s°(Ti[tex]Cl_{4(l)}[/tex]) = 221.9 J/(mol*K)
s°(Ti[tex]Cl_{4(g)}[/tex]) = 354.9 J/(mol*K)
Therefore,
s° = s° (Ti[tex]Cl_{4(g)}[/tex]) - s°(Ti[tex]Cl_{4(l)}[/tex]) = 354.9 - 221.9 = 133 J/(mol*K)
Thus, T = ΔH°[tex]_{f}[/tex] /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C
Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.
The boiling point of titanium tetrachloride was calculated from the thermodynamic values as 308.27 K.
The boiling point of the liquid can be given as:
[tex]\rm Boiling\;point=\dfrac{\Delta H}{\Delta S}[/tex]
The values of enthalpy and entropy for liquid and gaseous titanium chloride are given in the table attached.
The value of change in enthalpy ([tex]\Delta H[/tex]) for titanium chloride boiling is given as:
[tex]\Delta H=\Delta H_{gas}-\Delta H_{liquid}\\\Delta H=-763.2-(-804.2)\;\text {kJ}\\\Delta H=41\;\rm kJ/mol[/tex]
The value of change in entropy for the reaction is given as:
[tex]\Delta S=\Delta S_{gas}-\Delta S_{liquid}\\\Delta S=354.9-221.9\;\text {J/mol.K}\\\Delta S=133\;\rm J/mol.K[/tex]
The value of boiling point of titanium tetrachloride can be given as:
[tex]\rm Boiling \;point=\dfrac{41000\;J/mol}{133\;J/mol.K}\\ Boiling \;point=308.27\;K[/tex]
The boiling point of titanium tetrachloride was calculated from the thermodynamic values as 308.27 K.
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