Respuesta :
Answer: The change in entropy change when 207 g of toluene boils at [tex]110.6^{o}C[/tex] is 223 [tex]J/K[/tex].
Explanation:
It is given that mass of toulene is 207 g and its molar mass is 92.14 g/mol. So, its moles will be calculated as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{207 g}{92.14 g/mol}[/tex]
= 2.25 moles
As we are given that heat of vaporization for 1 mol is 38.1 kJ/mol. So, heat of vaporization for 2.25 moles will be calculated as follows.
[tex]2.25 moles \times 38.1 KJ/mol[/tex]
= 85.725 KJ
Now, we know that the relation between enthalpy change and entropy change is as follows.
[tex]\Delta S = \frac{\Delta H}{\Delta T}[/tex]
= [tex]\frac{85.725 kJ}{(110.6 + 273) K}[/tex]
= 0.223 [tex]kJ/K[/tex]
or, = 223 [tex]J/K[/tex] (as 1 kJ = 1000 J)
Thus, we can conclude that the change in entropy change when 207 g of toluene boils at [tex]110.6^{o}C[/tex] is 223 [tex]J/K[/tex].
