Using the data in the tables and Coulomb's law, the energy change for the given reaction is; ΔE = 95 KJ/mol
The question is a bit scattered but from online sources, the right order is observed and the question solved as below;
We want to calculate the energy change (ΔE) for the given reaction. We are given their respective ionization energies and electron affinities.
Now, the formula for change in Energy here is;
ΔE = Ionization Energy + Electron Affinity
Our given reaction is;
K(g) + Br(g) → K⁺(g) + Br⁻(g)
Looking at the reaction, we can see that K transformed to positive from neutral while Br transformed to negative from neutral.
Thus, we can say that;
- K lost Electrons
- Br gained Electrons
Energy required to lose electrons is called ionization energy and as such, we are given;
Ionization Energy of K = 0.696 aJ
From online conversion using the given ionic radius of 138 pm, we can convert 0.696 aJ to KJ/mol to give 419 KJ/mol
Thus; Ionization Energy of K = 419 KJ/mol
Since Br gained electrons, the energy that is involved is Electron affinity.
We are given;
Electron affinity of Br = -0.54 aJ
Again, from online conversion using the given radius of 196 pm, we have -0.54 aJ = -324 KJ/mol
Thus, Electron affinity of Br = -324 KJ/mol
Plugging in the relevant values into the change in energy equation earlier stated gives us;
ΔE = 419 + (-324)
ΔE = 95 KJ/mol
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