The first ionization energy of the oxygen molecule is the energy
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Question

The first ionization energy of the oxygen molecule is the energy
required for the following process:
O21g2¡O2 +1g2 + e-
The energy needed for this process is 1175 kJ>mol, very similar
to the first ionization energy of Xe. Would you expect O2 to react
with F2? If so, suggest a product or products of this reaction.

Accepted Answer

Hey everyone in this example, we're told that Xenon and oxygen have the similar first ionization energy values. And if Xenon reacts with bromine gas, will oxygen gas reacts with bromine gas? If yes, what is their product? So when we look up the first ionization energies of Xenon in our textbooks, we would see that it's equal to a value of about 12 point 13 electron volts. Whereas when we look at the first ionization energy of oxygen, or rather for oxygen gas, we would see that it has a first ionization energy equal to a value of point oh eight electro volts in our textbooks. And we can say that these numbers are about the same value for first ionization energy between xenon and oxygen. Or rather oxygen gas. So based on the similarity and based on the fact that Xenon does react with roman gas, we would say that therefore oxygen will likely react With Bromine Gas Br two. And as far as our product that would form, we would say that we have 02 plus B. R two, Which would likely form 02, B R two as a product. Or we can say that we form anna locus to our product, Xenon BR two. Because that is what would form if Xenon reacts with romain gas, We would also form therefore O. b. R. two as a product. And so for our final answers for this example, we would say that yes, oxygen is likely going to react because it has a similar ionization energy to Xenon and Xenon does react with bromine gas. And we can also say that the product that can form is either going to be Oh to be R two or OBR two, which is analogous to what would form from xenon reacting with bromine gas. So everything boxed in represents our final answers. If you have any questions, please leave them down below, and I will see everyone in the next practice video.

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