The enthalpy of vaporization for Benzene is 30.8 kJ/mol. 39.42 kJ is the energy change when 100g of Benzene boils at 80.1 degrees Celsius.
What is Enthalpy of Vaporization ?
The amount of enthalpy or energy that must be added to a liquid substance into gas substance is called Enthalpy of Vaporization. It is also known as Latent heat of vaporization.
How to find the energy change from enthalpy of vaporization ?
To calculate the energy use this expression:
[tex]Q = n \Delta H_{\text{vapo.}[/tex]
where,
Q = Energy change
n = number of moles
[tex]\Delta H_{\text{Vapo.}}[/tex] = Molar enthalpy of vaporization
Now find the number of moles
Number of moles (n) = [tex]\frac{\text{Given Mass}}{\text{Molar mass}}[/tex]
= [tex]\frac{100\ g }{78\ g/mol}[/tex]
= 1.28 mol
Now put the values in above formula we get
[tex]Q = n \Delta H_{\text{vapo.}[/tex]
= 1.28 mol × 30.8 kJ/mol
= 39.42 kJ
Thus from the above conclusion we can say that The enthalpy of vaporization for Benzene is 30.8 kJ/mol. 39.42 kJ is the energy change when 100g of Benzene boils at 80.1 degrees Celsius.
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Question: The enthalpy of vaporization for Benzene is 30.8 kJ/mol. What is the energy change when 100g of Benzene boils at 80.1 degrees Celsius?
