1) Use the Born-Haber cycle and given data ΔH∘fKCl = -436.5 kJ/mol, IE1(K) = 419 kJ/mol, ΔHsub(K)=89.0 kJ/mol, Cl2(g)bondenergy = 243 kJ/mol, EA(Cl) = -349 kJ/mol to calculate the lattice energy of KCl. 2) Use the Born-Haber cycle and data from Appendix IIB and Table 9.3 in the textbook to calculate the lattice energy of CaO. You may need the following data: electron affinity of oxygen EA1=−141kJ/mol, EA2=744kJ/mol; ionization energy of calcium IE1=590kJ/mol, IE2=1145kJ/mol, the enthalpy of sublimation for calcium ΔHsub=178kJ/mol.

Explanation:

1 ) Let the lattice energy per mole be x

2 K + Cl₂ =  2KCl

Applying Born- Haber cycle

2 x 89 + 2 x 419 + 243 - 2 x 349  + Δx = - 2 x 436.5

178 + 838+ 243 - 698 + Δx = - 873

Δx = 1434 kJ

Lattice energy per mole = 1434 / 2 = 717 kJ / mol

2 )

2Ca + O₂ = 2 CaO

Heat of formation of CaO = - 635 kJ

Let lattice energy be x / mol

Bond energy of oxygen = 498 kJ / mol

178 x 2 + 2 ( 590 + 1145 ) + 498 + 2 ( 744 - 141 ) + x = - 2 x 635

356 + 3470 + 1206 + 498 + 2 x = - 1270

5530 + 2 x = - 1270

x = - 3400  kJ

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

- decrease the number of plants and animals living in the water.