Given the following information, construct a Born–Haber cycle to calculate the lattice energy of CaCl₂(s). (LO 6.13)

Net energy change for the formation of CaCl₂(s) form Ca(s) and Cl₂(g) = -795.4 kJ/mol

Heat of sublimation for Ca(s) = +178 kJ/mol

E_i1 for Ca(s) = +590 kJ/mol

E_i2 for Ca(g) = +1145 kJ/mol

Bond dissociation energy for Cl₂(g) = +243 kJ/mol

E_ea1 for Cl(g) = -348.6 kJ/mol

(a) 2603 kJ/mol (b) 2254 kJ/mol (c) 2481 kJ/mo (d) 1663 kJ/mol

Video transcript

everyone in this video. We're using the born Haber cycle to determine the lightest energy of lithium oxide. So the last energy equation it's going to be when our product L. I 202 lithium oxide is going to go ahead and associate into two moles of lithium plus Karen's and its gaseous form as well as one mole of +02 minus an ion in its gaseous form. So we can see here from the right side that our delta H of F, which is delta age of formation, signifies that this reaction here is going to be our next goal reaction. So we want everything to cancel to give us a total of this and this will not be manipulated. So we can see here from our first region two moles of our lithium solid, our only source of lithium solids from this equation here. So you have to multiply this whole reaction by two. So you can go ahead and give us a net of two moles of lithium solid. Do the same thing. I multiply the delta age of sublimation by two as well. Now we can see that from our first reaction, since we multiply this by two. The product will give us two lithium to moles of lithium gas. And because this needs to cancel out with this second reaction here will again go ahead to multiply the second reaction by two as well as the ionization energy. Now, lastly See here that the second starting material, we have half a mole of 02 gas and that comes from this year, which is the bond dissociation energy to go ahead and multiply this third reaction by half or 0.5 will do the same thing for our bond dissociation energy, multiplied by one half. All right, from our last energy equation here, we see that we have our source of our product and this is on the right side. So we need to go ahead and make a note to flip this equation. Alright, so now doing the math for these three equations, since we multiply it by half or by two, let's go ahead and write this all out and see if everything does cancel to our net reaction. So now for our first reaction, we have two moles of our lithium solid To sublimate into two moles of our lithium gas. Second equation here, we have two moles of our lithium gas and that goes on to form two moles of lithium plus Karen's as well as two electrons. For a third reaction. Now we have half a mold of our 02 gas and that goes to half. We'll have times two here because we have two moles of our auction that equals to one. So just write it as just one mole of auction gas scrolling down a little bit to give us more space here than the last three or other two reactions that are going to be just the same. So we have auction gas reacting with one mole of auction to give us oh minus Now, taking the Oh minus in his gasses formal at another electron. And that gives us 02 -. And lastly our lattice energy equation we said that to be flipped. So now we have two moles of our lithium plus caden reacting with one mole of +02 minus in its gaseous form. And that gives us our product of L. I. +20. lithium oxide in its solid state. So now for the numerical values then we have delta age of sublimation equaling to 59 times two. And all these values the unit is going to be killed, jules per mole. So for simplicity sake, I won't add in now but we can go ahead and add it later when we're running out the answer. Now for our ionization energy, we have 5 20 times two for our bonds association energy. We have 4 98 multiplied by one half for our first electron affinity value we have negative 1 41 For a second electron affinity value. We have 744. And then lastly for a lot of energy we have no value. That's what we're selling for that. So that's equal to X. Maybe this should actually align with the last value. All right. And again we're just scrolling down. We have to go ahead and input our net reaction or goal reaction which is from our delta age information. So that's two moles of lithium solid reacting with half a mole of R. 02 gas And that gives us L. I. 20. Which is lithium oxide in solid state the delta H of formation for this is very solid for you to be negative 5 98. All right. So, we can see here for our products. We need to go ahead and make sure that does not cancel for the first region. Again, we have two moles of lithium solid. So you want this to stay as is And then we also have half a mole of 02 gas. So that's this right here. And for our product this comes from the lattice energy. So we're gonna try to make sure that these three do not cancel out because we want internet reaction here. So, I see here for our two lithium gas formed neutral. We want those two to cancel out. And it does because the first one is on the right side of the arrow and the other one is on the left side of the arrow. And then we also see that our two electrons will cancel out. So we have to here and then we have one from our electron affinity one as well as electron affinity to. So these two are on the left side and this one is on the right side. So that does cancel. Next will be our two moles of our lithium Plus we see that this will go ahead and cancel with this. Go ahead and mark that next there are one mole of auction gas. This will go ahead and cancel out with this here. And Lastly you can see that the O - will go ahead and cancel as well as the 02 -. Alright, so everything does cancel out except for what we want internet reaction. This is all correct. And our calculations also correct solving for X now. So go ahead and put it into one big equation. And that's just when our delta H of F is equal to two times the delta H of sublimation adding with the two moles or two times our organization energy plus half of our bond dissociation energy. And then our electron affinity values are just as is as well as are a lot of energy. So not playing in those numerical values, we have the delta age of formation being negative 5 98 and that equals 22 times 1 59 plus two times 5, 20 plus half times for plus, negative 1 41 plus 44. And our last energy. So our unknown is just X. So here we're simply trying to solve for X taxes again equal to our lattice energy. Once we solve for S. X, we get the value of negative 2808. And again units just being killed, jules per mold. So this right here is going to be my final answer for this problem