Use the Born–Haber cycle and data from Appendix IIB and Table 9.3 to calculate the lattice energy of CaO. (ΔH_sub for calcium is 178 kJ/mol; IE₁ and IE₂ for calcium are 590 kJ/mol and 1145 kJ/mol, respectively; EA₁ and EA₂ for O are -141 kJ/mol and 744 kJ/mol, respectively.)

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Welcome back everyone to another video use the Born Haber cycle and data from appendix to B and Sable 9.3 to calculate the lattice energy of strontium oxide. The entropy of sublimation for strontium is 164 kilojoules per mole. The first and the second ionization energies for strong TM are 550 kilojoules per mole and 1064 kilojoules per mole respectively. Electron affinity one and electron affinity. Two for oxygen are negative 141 kilojoules per mole and 744 kilojoules per mole respect respectively. We're given four answer choices. A 2038 kilojoules per mole B negative 3222 kilojoules per mole C 2287 kilojoules per mole and the negative 3471 kilojoules per mole. So now we can begin this problem by using the entropy of formation of strontium oxide. Let's recall that the equation for that is strontium solid. It's the standard state of strontium. It's a metal plus one half oxygen. So we're taking oxygen in a standard state. It's a gas to form strontium oxide in its solid state, which is the standard state. And what we find out is that the entropy of formation would be negative 592.0 kilojoules per mole. We get this from our data tables. And now we can essentially think about the target equation. Well, essentially we want to form the strontium cion from strontium. How do we do that? Well, essentially, first of all, we have to form its gaseous form. So we have to go into a gas form from its gas form. We want to ionize it into strontium plus cion in its gaseous form. And finally, we want to get strontium two plus, we want to remove the second electron. What about oxygen? Well, essentially we want to get oxide and ion in in its gaseous form. And to do that, first of all, we want to get oxygen atom in its gaseous form, then we want to add one electron to get oxygen negative in its gaseous form. And eventually, we want to get our oxide, an ion in its gaseous form. Now, if we add the resultant cion and anion together, we will actually get chromium oxide in its solid form. And this will correspond to our U which will be the lattice energy. And now we can simply use the cycle to understand that the entropy of the formation is simply the addition of all of these processes to arrive at strontium oxide solid. Now, what are those processes? Well, essentially starting with strontium solid. We're first of all using the entropy of sublimation to turn our solid to a gas. Then we are using the first ionization energy to remove the first electron and the second ionization energy to remove the second electron for oxygen. First of all, we're using the bond energy, right, we're breaking our oxygen oxygen bond to form an oxygen atom. So that would be one half of the entropy of the association. The reason why we're using one half is because we're only releasing one oxygen atom, we don't need to use two of them. Now, for the next part, we're using electron affinity one because we're adding one electron and then we're adding our second electron meaning we need to use the second electron affinity. OK. So now we can actually set up our equation. We know that the entropy of formation can be achieved using all those processes. We are going to start with the entropy of formations. We're going to say that delta H F or the entropy of formation will be equal to, if we go backwards, that will be the entropy of sublimation. So we're going to say delta H sub, then we need to add ionization energy one ionization energy to, we can also see that we need to add one half of the entropy of the association of oxygen. So when I have delta HD also, we are adding electron affinity, one electron affinity two, right? And finally, we're adding our lattice energy you. OK. So now all that we have to do is just use the data and solve for you. So let's see what we can get. We can, first of all rearrange our expression and we can say that you is nothing else but the entropy formation minus D are based on the sum of all of these terms except of you. So delta H sublimation plus ie one plus ie two plus one half, the entropy of the association plus E A one plus E A two. We have our final expression. Let's go ahead and substitute the givens. So first of all, what do we know? Well, we know that the entropy of formation is given to us. It's negative 592 kilo spro mole, we are going to subtract our sum. First of all, the entropy of sublimation. So that would be 164 kilojoules per mile in parentheses, we are adding the first ionization energy this be 550. Now the second ionization energy 1064 then we are adding one half multiplied by the bond association energy of oxygen which is 496. Now, the first affinity or the first electron affinity and finally, the second electron affinity, 744 killer jules. Now, if we do the calculation, we essentially end up with negative 3222 kilojoules per month. Now, if we look at the answer choices, we can clearly see that the correct answer to this problem corresponds to choice B negative 3222 kilojoules per mole. That would be the lattice energy of strontium oxide. Thank you for watching.

Use the Born–Haber cycle and data from Appendix IIB and Table 9.3 to calculate the lattice energy of CaO. (ΔH_sub for calcium is 178 kJ/mol; IE₁ and IE₂ for calcium are 590 kJ/mol and 1145 kJ/mol, respectively; EA₁ and EA₂ for O are -141 kJ/mol and 744 kJ/mol, respectively.)

Verified Solution

Key Concepts

Born-Haber Cycle

Lattice Energy

Ionization Energy and Electron Affinity

Use the Born–Haber cycle and data from Appendix IIB and Table 9.3 to calculate the lattice energy of CaO. (ΔHsub for calcium is 178 kJ/mol; IE1 and IE2 for calcium are 590 kJ/mol and 1145 kJ/mol, respectively; EA1 and EA2 for O are -141 kJ/mol and 744 kJ/mol, respectively.)

Verified Solution

Key Concepts

Born-Haber Cycle

Lattice Energy

Ionization Energy and Electron Affinity

Use the Born–Haber cycle and data from Appendix IIB and Table 9.3 to calculate the lattice energy of CaO. (ΔH_sub for calcium is 178 kJ/mol; IE₁ and IE₂ for calcium are 590 kJ/mol and 1145 kJ/mol, respectively; EA₁ and EA₂ for O are -141 kJ/mol and 744 kJ/mol, respectively.)