Born Haber Cycle, Basic Introduction, Lattice Energy, Hess Law & Enthalpy of Formation - Chemistry

This chemistry provides a basic introduction into the born haber cycle. It explains how to calculate the enthalpy of formation of a compound using the lattice energy, first ionization energy and second ionization energy values, electron affinity, and bonding energy for the substances that make up the compound. It explains how to use hess law to calculate the enthalpy of combustion as well as the enthalpy of formation of certain substances.
Ionization Energy:
• Ionization Energy - Ba...
Electron Affinity:
• Electron Affinity Tren...
Atomic Radius:
• Atomic Radius - Basic ...
Bond Energy & Bond Length:
• Bond Energy & Bond Len...
Electronegativity:
• Electronegativity, Bas...
Periodic Trends:
• Ionization Energy Elec...
__________________________________
Polar & Nonpolar Covalent Bonding:
• Polar Covalent Bonds a...
Bond Polarity & Dipole Moment:
• Bond Polarity, Electro...
Ionic Radius:
• Ionic Radius Trends, B...
Lattice Energy:
• Lattice Energy of Ioni...
Born Haber Cycle:
• Born Haber Cycle, Basi...
Bond Energy Calculations:
• Bond Energy Calculatio...
___________________________________
Lewis Structures - Mega Review:
• Lewis Structures, Intr...
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Great video but I am a bit confused. I thought we are meant to divide the bond energy by 2 since we are using an atom of fluorine?

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You have to use the fluorine in its naturally occuring state and it occurs as a diatomic gaseous molecule ...so 2 fluorine atoms.

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Hey in my coursebook's solution, the ∆Hatm of F2 is multiplied by 2 while you are not multiplying it by 2. So, are both solutions correct in their own ways or the solution in my coursebook is incorrect? Help needed ASAP

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What about for the electronegative atom with two electron affinities, such as oxygen? I am trying to solve a problem with two electron affinities, thank you

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How would you draw the diagram?

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When he writes the Mg+2 +2F-=MgF2 I am confused about where the charges so because I am assuming that F2 comes from 2F, does the 2+ charge on Mg disappear.

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It's because there is bond breaking in 2F which is di atomic
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When we are finding bond formation for Nacl why we are writing 1/2 cl2?

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Can we cut out substances in the equations if they're on the same side (i.e. both on products side but they're different equations)

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