Use the following information plus the data given in Tables 6.2 and 6.3 to calculate the second electron affinity, Eea2, of oxygen. Is the O2-ion stable in the gas phase? Why is it stable in solid MgO?
Heat of sublimation for Mg1s2 = +147.7 kJ/mol
Bond dissociation energy for O21g2 = +498.4 kJ/mol
Eea1 for O1g2 = -141.0 kJ/mol
Net energy change for formation of MgO(s) from its elements = -601.7 kJ/mol

Question

Use the following information plus the data given in Tables 6.2 and 6.3 to calculate the second electron affinity, Eea2, of oxygen. Is the O2-ion stable in the gas phase? Why is it stable in solid MgO?
Heat of sublimation for Mg1s2 = +147.7 kJ/mol 
Bond dissociation energy for O21g2 = +498.4 kJ/mol 
Eea1 for O1g2 = -141.0 kJ/mol 
Net energy change for formation of MgO(s) from its elements = -601.7 kJ/mol

Accepted Answer

Hello everyone today. We have the following problem given the following day. To calculate the second electron affinity of sulfur Is the sulfide to -1 more stable as a free ion or in calcium sulfide. So the first thing we have to do when we see this sort of data is figure out what it means. So this is from the born Haber cycle which just essentially allows us to quantifying and qualify the data given the processes in a chemical reaction. And so from this born haber cycle we have a formula that we can use. And that formula is we have the entropy change is equal to the entropy of sublimation for our potassium plus the first two ionization energies. So we have ionization energy one and ionization energy too. And we're also going to add that to the entropy change of sublimation for our sulfur. We then add these two values to the electron affinity For the first one as well as the 2nd electron affinity. And then we add that lastly to the lattice energy latticed entropy. Until then go and plug in our values. So we have our we have that Our entropy change. Our total entropy is going to be -482.4. And since these are all in units of killing joules per mole, we're gonna leave units out to save on space. And so we're going to equal this to the entropy change for our potassium which is 77.8 kg joules per mole. And then we're gonna add the first two annotation energies which is the 5 89. and the 1145.4 kg joules per mole. We are then going to add these To the entropy change for sublimation for our sulfur which is 277.2 kg per mole, We're going to add that to the two values for our electron affinity one and 2. So for our first one that's going to be our negative, 200.4 kg per mole. And then we're gonna add that to our 2nd 1. However, we do not have the second electron affinity which is what we are trying to find. So we're gonna leave that as is and then we're gonna add that to our lot of energy Which is negative. 3,043 kg joules per mole. Once we solve all of this out we end up with a value for our second electron affinity being 570. kg joules per mole. And so we have to evaluate if this sulfide ion is more stable as a free ion or in our calcium sulfide. So we have to note that this electron affinity for sulfur is positive. And so since it is positive, this is going to implicate that this So for 2 - is going to be more stable. So it is more stable when it is in a compound such as calcium sulfide and not by itself when it is negative, that is going to be when it is going to be stable on its own or in as an as an ion. And so we have our electron affinity value here and that we have that our sulfide molecule is more stable in our calcium sulfide. And with that we've answered our question overall, I hope this helped, and until next time.

Verified Solution

Key Concepts

Electron Affinity

Stability of Ions

Lattice Energy