∆Hrxn = ∆Hvap + ∆Hbond Question
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∆Hrxn = ∆Hvap + ∆Hbond Question
Hi! What did it mean in class when Dr. Lavelle explained using bond enthalpy for breaking bonds for molecules in the gas phase and using the equation ∆Hrxn = ∆Hvap + ∆Hbond. Where does this equation come from?
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Re: ∆Hrxn = ∆Hvap + ∆Hbond Question
Hi Maham!
I think we just get this equation intuitively; if we change phases *and* change chemical compounds, then we need to take into account the enthalpy of the phase change in addition to the bond enthalpy.
So, for example, if you have Br2 (g) --> 2Br (g), you don't need to have the deltaH of vaporization because it's not changing phases—just the bond enthalpy is good. However, if it's like the in-class example where Br2 (l) --> 2Br (g), we also need to add the phase change enthalpy.
I think we just get this equation intuitively; if we change phases *and* change chemical compounds, then we need to take into account the enthalpy of the phase change in addition to the bond enthalpy.
So, for example, if you have Br2 (g) --> 2Br (g), you don't need to have the deltaH of vaporization because it's not changing phases—just the bond enthalpy is good. However, if it's like the in-class example where Br2 (l) --> 2Br (g), we also need to add the phase change enthalpy.
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Re: ∆Hrxn = ∆Hvap + ∆Hbond Question
Hello, as previously mentioned with the Br2 (l) --> 2Br (g) example, we include the enthalpy of vaporization because there is heat that is required to change bromine from a liquid phase to a gas phase. In addition, we also need to add the bond enthalpy of the Br-Br bond, since we are breaking apart a Br-Br bond to get two individual bromine molecules.
Hope this helps!
Hope this helps!
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