The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
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{80.2\ g }{159.8\ g/mol}[/tex]
= 0.5 mol
Now put the values in above formula we get
[tex]Q = - n \Delta H_{\text{vapo.}[/tex] [Negative sign is used because Br₂ condensed here]
= - (0.5 mol × 15.4 kJ/mol)
= - 7.7 kJ
Thus from the above conclusion we can say that The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
Learn more about the Enthalpy of Vaporization here: https://brainly.com/question/13776849
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